Return Home Members Area Experts Area The best AskMe alternative!Answerway.com - You Have Questions? We have Answers! Answerway Information Contact Us Online Help
 Wednesday 23rd April 2014 04:07:58 AM


 

Username:

Password:

or
Join Now!

 
These are answers that bluto has provided in Physics

Question/Answer
denberg asked on 04/12/08 - Does light have mass?




Does light have mass?

Thanks.

bluto answered on 02/18/09:

Yes, but it has zero rest-mass.

denberg rated this answer Excellent or Above Average Answer

Question/Answer
AhmadBalkhi asked on 12/25/07 - light waves & sound waves

Our eyes can only detect the visible wavelengths of light. So, if somehow we could see inside of the sun, we would basically see it dark since it is filled only by gamma rays which are not in the visible range. Here is my question:
Can this also be applied to the sound waves? In other words, can a sound wave with higher or lower frequencies pass though our ears and yet not be detectable by our ears?

bluto answered on 12/25/07:

Absolutely!

Here's a good and somewhat surprising example:

Radio waves cannot be heard!

Their frequencies are so high that our hears cannot detect them. This is why we must have the help of electronics that *can* hear them, and transform them a little so we can hear their 'signals'.

Hope this helps you.

:)

AhmadBalkhi rated this answer Excellent or Above Average Answer

Question/Answer
cheriskae asked on 09/08/07 - Can you please help me

A ballplayer catches a ball 3 seconds after throwing it vertically upward.With what speed did he throw it,and what height did it reach?

bluto answered on 09/09/07:

Hi cheriskae...

The ball will take the same amount of time to go to its peak in the air as it does coming back down again.

So it will take (3/2) s to reach the top.

With this info you can find its height.

y = ½gt²

Plug the numbers in:

You know t....g = 9.8 m/s²

==> y is about 11 meters in height.

The speed with which it was thrown is given by: v = gt

To reach the top it took 3/2 seconds, so plug the numbers in and you will have v.

I get v to be 14.7 m/s

Hope this helps.

cheriskae rated this answer Excellent or Above Average Answer

Question/Answer
AhmadBalkhi asked on 09/30/06 - Learning math by experiments from science

I am teaching system of linear equations this week. I am thinking of taking my students out of the class for one day to do a ground experiment in which they use the topic they are learning (systems of linear equations). I cannot think of one but I am sure there should be something they can use from physics? Can you help me on this?

bluto answered on 01/11/07:

It's good of you to do this. Sounds like something I'd try and do!

The only that that pops into my mind at the moment is maybe an application of Linear Programming. Maybe visit a small manufacturing company and use that as your experiment. Try and get some data from the company, mock up a reasonable model of what you have and then apply the linear programming methods to it. it involves syustems of linear equations and might be practical enough for you and your class.

I do this on the side and it seems to help, but it has its limits and pitfalls.

Hope that helps....good luck.

:)

AhmadBalkhi rated this answer Excellent or Above Average Answer

Question/Answer
pgray23607 asked on 09/14/06 - Electromagnetic Radiation

It took 11 minutes tor a command from the controllers on Earth to reach the Pathfinder space vehicle on the surface of Mars July 4, 1997. How many miles did the message have to travel?

bluto answered on 09/14/06:

It takes about 8 minutes for light to reach earth from the sun, so it's further than that.

The signal travels at light speed (c) = 3x10^8 m/s

So the distance is given (roughly) by: d = ct

And d = 1.98 x 10^11 m

-or-

1.23 x 10^8 miles

For reference; the sun is about 9.3 x 10^7 miles from earth.

:)

pgray23607 rated this answer Excellent or Above Average Answer

Question/Answer
raezzoul asked on 08/07/06 - sound waves

Hello bluto!
I was reading some of the questions you answered and wondered if you could help me too.
I know that sound waves reflect but could you please give me a detailed explanation on how sound waves travel inside a dome?

bluto answered on 08/07/06:

Hi raezzoul......

Well, generally speaking they ricochet and bounce around not unlike a rubber ball when you throw it into a corner. It bounces up, down, left, right and exits and some angle.

Sound, however is a complicated wave because each is composed of numerous other frequencies that are all mixed together to give it its particular sound. When these are bouncing around each frequency reflects a little differently and eventually they separate and dissipate until they essentially vanish.

Upon each reflection there is a little energy lost due to absorption of the wall material. This decreases the energy the wave has when it encounters it's next bounce.

Does this help you?

Hopefully it's a little more clear, but let me know otherwise. :)

raezzoul rated this answer Excellent or Above Average Answer

Question/Answer
cifs1 asked on 07/27/06 - Coin toss and bounce simulation - contact point to floor

I am simulating a coin flipping in the air. Is there a relativly simple formula for determining the contact point of the flipping coin with a flat surface, given the coins current x,y and z rotation angle?
Is there a second formula which once the contact point is found, determine the velocity vectors given the rotations and the current velocities?

bluto answered on 07/27/06:

If you already know the coordinates then all you need to find is the angle. Depends on what coordinate system you are using tho.

You have to have some kind of reference. A coin spinning and moving thru space could have numerous orientations. You have to be able to orient the coin to some reference plane, and then to some origin on the surface upon which it will land.

Like I said, this is not trivial. This is a mechanics problem, and they can get quite messy to deal with.

The tilt of the coin has to be relative to a reference, like I said......a reference plane or coordinate system at the coin. This way you can set a constraint. For example, taking the Sine of the angle between the radius of the coin relative to a horizontal you will get a small deflection pointing downward. The largest such deflection is probably the coordinate you need, but again....to actually calculate this requires a bit of mechanics and math to take place.

Question/Answer
cifs1 asked on 07/27/06 - Coin toss and bounce simulation - contact point to floor

I am simulating a coin flipping in the air. Is there a relativly simple formula for determining the contact point of the flipping coin with a flat surface, given the coins current x,y and z rotation angle?
Is there a second formula which once the contact point is found, determine the velocity vectors given the rotations and the current velocities?

bluto answered on 07/27/06:

It would be a daunting and tedious problem to work out. Not if you want it to actually represent reality I guess.

You would have to use the coin's moment of inertia, weight, and initial conditions; how much force was used to start it off, how high was it flipped, how hard is the surface, what kind of surface is it landing on, coefficient of elasticity and so on.

So no there really isn't a simple formula for this unless you scour the web and find something where someone has already worked this out.

cifs1 rated this answer Average Answer

Question/Answer
5incere asked on 05/12/06 - energy, work, power

i dont know how to find the input and out put and wasteed energy please help!

can you show it to me in steps because i have other questions like is andd i want to apply the sname procedure as well thnx


Source Electric kettle

power (w) 1500

∆t (s) 173


Output energy (J)
Input energy (J)
Waste energy (j)
efficiency %

bluto answered on 05/14/06:

Hello.....

I don't think you've given me all the info for the problem.

You're asking about efficiency and input/output power and all that, and you've given me a delta 't' of 173 seconds, and an input power of 1500 Watts, but you've not given any other information.

Can you review the question again and make sure you have included everything. To know that I am looking at an electric kettle with power rating of 1500 W and know it's on for 173 secs doesn't really tell me a whole lot.

Will look for your update.

:)

5incere rated this answer Excellent or Above Average Answer

Question/Answer
asraiblu asked on 02/01/06 - Hi bluto...2 questions...


If light can behave like a particle, can particles behave like light? If they can, why do we not measure wavelengths of things like toasters and trucks.?


If I am wearing a teal blue (cyan) shirt under normal light, I see cyan because it is reflecting green and blue light. If I go into a room that has blue light, what color does my shirt appear and why? What if the room is lit with red light and why? Green light and why?
Thanks in advance for your help. :)

bluto answered on 02/01/06:

Hello there asraiblu.....

1) The difference in wavelength between large (macroscopic) objects and smaller (microscopic) objects is vastly different. The macroscopic objects have wavelengths so small they are insignificant compared to the mass. And that mass could be something as small as a baseball.



2) The color of your shirt changes because it has to reflect the light coming from the source in the room. A red object appears red to us because white light falling upon it contains light of a wavelength that is associated with red. That light strikes our eyes and causes the sensation of what we perceive as 're'. Same goes for blue, green etc.

If the light contains only a small amount of red light then the object will appear less red to us. The shirt will reflect red most strongly and the other colors less, if at all. This is why a pure red surface, if illuminated with pure blue light will appear dark and sort of colorless.

However there is a difference between color adding and color subtracting. Against my usual judgment to throw websites at people who ask me questions, I think this one may help you a little bit. I think it will convey the pertinent information to you quicker than having you suffer thru my long, blathering explanation. This is in addition to what I've mentioned already tho. k?

:)

http://home.att.net/~B-P.TRUSCIO/SUBCOLOR.htm



Hope this help you a little, but as always....if something isn't clear, ask a follow-up question.

asraiblu rated this answer Excellent or Above Average Answer

Question/Answer
TRIGNOMETRY asked on 12/08/05 - trignometry

(1)given that Y= ASin(2pi/lambda(ct-x)), are measured in meters , which statement is true (please choose and write the solution)

bluto answered on 12/09/05:

Hi TRIG.....

Once again, I don't understand what your actual question is. If you clarify what you want/need and re-submit I'll try and help, but I cannot do so unless your question is clear.




Question/Answer
TRIGNOMETRY asked on 12/04/05 - VECTORS AND ONE DIEMENSION MOTION

(1)the component of A =2i +3j along the vector B = 3i +4j is
(2)a particle undergoes three successive displacements given by S_1=2^1/2m north east s_2=2m due to south and s_3 = 4m 30degree northwest then magnitude of net displacement
(3)

bluto answered on 12/05/05:

T-

Here's a stab in the dark.

1) Take a dot product of A & B.....thusly:

A.B [meaning the dot product, not multiplication]

A.B = (2i + 3j).(3i + 4j)

==> 2*3i^2 + 3*4j^2
==> 6 + 12 = 18.

This is the length of the component of vector A onto vector B.

That's as close as I can get with your question, and I have nothing to offer you on the others unless you can clarify them.

bluto

TRIGNOMETRY rated this answer Excellent or Above Average Answer

Question/Answer
sexybodyjulie asked on 11/26/05 - A capacitor...

Hi Bluto
I really appreciated the answers you gave to me.They were really helpful. Ok, i was wondering if you might be so kind as to help me with another one.

A capacitor is connected across the terminals of an a.c generator that has a frequency of 580 hz and supplies a voltage of 25V. When the second capacitor is connected in a parralel with the first one, the current from the generator increases by 0.23 A. Find the the capacitance of the second capacitor.
note: the AC circuit and the voltage are rms values and power is the average value unless indicated otherwise.

bluto answered on 11/30/05:

Hi ya Julian......

Too bad I can't embed graphics into this window, but anyway.......this is what's going on.

When you add the second cap and the current increases by 0.23 A, this is the current thru this cap. With this info you can then go ahead and solve the problem.

A cap offers resistance to current, but it's not technically a resistance, it's more of a reaction to current. So, this is termed "Reactance"....sometimes called "Capacitive Reactance", and it's given the symbol X_c (Captical "x" with a subscript c).

X_c = 1/[2(pi)fC]

You need to find "C" but you don't know what X_c is.

You can, however use the fact that you really only need the peak (or rms) voltage given in the problem, which is 25V, and the fact that X_c can be used essentially the same way resistance can.....as follows:

Usually you see: V = IR right?

Well, you can also say: V = IX_c [since X_C is in units of Ohms]

I = 0.23 A and V = 25V, so X_c = V/I

==> 25V/0.23A= 108.7 Ohms

Now that you know X_c you can find C.

X_c = 1/[2(pi)fC]

==> C = 1/[2(pi)fX_c]

Plug in your numbers for X_c and f (f = 580 Hz) and you get:

X_c = 2.52 x 10^-6 Farads

-or-

X_c = 2.52 uF (microfarads)

(done)

Hope that helps, but let me know if something's not clear.

bluto

Question/Answer
sexybodyjulie asked on 11/06/05 - Another Magnet question

Hello Bluto
I just have another question that i could really use your help on. Hope its not a bother..

When a charged particle moves at an angle of 17 with respect to a magnetic field, it experiences a magnetic force of magnitude F. At what angle (less than 90) with respect to this field will this particle, moving at the same speed, experience a magnetic force of magnitude 2F?
Thanks Bluto
Julian

bluto answered on 11/15/05:

Hello Julian...

Ok....hip-shooting here; the force due to the mag. field is given by:

F = qvBsin(x)

Where:

F = force
q = charge
v = velocity
x = angle to the field = 17 deg

You have F1 and F2; one is given (I'm calling that one F1), the other is the one you want....which I'm calling F2 (obviously). :)

Based on the question you need to know when the force is doubled. In other words, when is F2 = 2(F1)

Well....

F1 = (q1)(v1)(B1)sin(x1)

Therefore -----

F2 = (q2)(v2)(B2)sin(x2)

Then F2 = 2(F1)

==> (q2)(v2)(B2)sin(x2) = 2(q1)(v1)(B1)sin(17)

As far as I can tell, everything else remains the same, so you have:

v1 = v2 = v (stated in the problem)
B1 = B2 = B
q1 = q2 = q

This gives:

qvBsin(x2) = 2qvBsin(17)

==> sin(x2) = 2sin(17)
==> sin(x2) = 2(0.29) = 0.58
==> x2 ~ 36 degrees.

Hope that made sense, but if not....lemme know.

Sorry for your LONG wait. :(

bluto

sexybodyjulie rated this answer Excellent or Above Average Answer

Question/Answer
sexybodyjulie asked on 11/06/05 - Current

Hello Bluto
What must be the radius of a circular loop of wire so the magnetic field at its center is 1.5 x 10-4 T when the loop carries a current of 15 A?

bluto answered on 11/15/05:

Hi sexybodyjulie.....

The magnetic field strength is given by a canned formula essentially:

B = uI/(2R)

u = a constant = 4(pi) x 10^(-7) Tm/A
I = current = 15 A
B = magnetic field = 1.5 x 10(-4) T
R = radius = unknown

Solve for R in the equation and you have:

R = uI/(2B)

Plug the numbers in and you find that:

R = 0.063 meters = 6.3 centimeters

I'm sorry it took so long for me to get to this for you. Don't worry about not giving me a rating or a low one.....I'd do the same! :)

bluto

sexybodyjulie rated this answer Excellent or Above Average Answer

Question/Answer
sexybodyjulie asked on 11/06/05 - Acce;eration

Hi Bluto
Could you please help with the following?


In a certain area, the horizontal component of the Earth's magnetic field has a magnitude of 1.1 x 10-5 T. An electron is shot vertically straight up from the ground with a speed of 2.2 x 106 m/s. What is the magnitude of the acceleration caused by the magnetic force? Ignore the gravitational force acting on the electron
I would really appreciate it thanks
Julian

bluto answered on 11/14/05:

Hi sexybodyjulie....

Sorry it took so long to get to you on this! ugh!

From Newton: F = ma

F = Force
m = mass (of the electron....9.11 x 10^(-31) kg)

a = acceleration

For a magnetic field, F = qvBsin(q)

Here, q = 90 degrees (pi radians) since the field is perpindicular to the motion, and this gives sin(90) = 1.

B = 1.1 x 10^(-5) T
v = 2.2 x 10^6 m/s
q = electron charge = 1.602 x 10^(-19) C

This will give you:

F = qvBsin(q) = 3.88 x 10^(-44)N

Now use F = ma

3.88 x 10^(-44)N = ma [solve for 'a']

==> a = F/m = (3.88 x 10^(-44) N)/(9.11 x 10^(-31) kg)

==> a = 3.161 x 10^(-14) m/s^2

I'll try to get to your others asap.

bluto

sexybodyjulie rated this answer Excellent or Above Average Answer

Question/Answer
aislinn asked on 10/30/05 - Reaction Time ∆x=vt + 1/2at

Hi Bluto,

Just been reading a few of your previous explanations and wondered if you can help me?

I have to work out the reaction time for a girl who catches a ruler after its dropped 30cm(0.3m).

I know I need to use ∆x=vt + 1/2at

However, to get t=? I am confused as to how to change the equation around. (So its kind of algebra than physics but hopefully you can still help out!)

I know vt=0 as the initial velocity=0 and a=gravity so 9.8m/s so the equation would then be 0.3=1/2(9.8)t - I think! - just not sure how to isolate t from there.

I have the answer but not the working through.

Thanks heaps in advance if you can simplify this for me :o)

bluto answered on 10/31/05:

Hi ya aislinn......

Sure, be glad to try and lend a hand.

If "t" is what you need, then you essentially have to 'solve' your equation for that quantity....namely.....t. :)

So, let's do that:

∆x=vt + 1/2at

---> ∆x=vt + 1/2gt [Note: the gravitational constant is usually represented by "g", so I made the replacement here]

==> ∆x - vt = (1/2)gt
==> 2(∆x - vt) = gt
==> (2/g)(∆x - vt) = t (almost there)
==> Sqrt[(2/g)(∆x - vt)] = t

-or-

==> t = (+ -)Sqrt[(2/g)(∆x - vt)]
(done)

Ok...now the "(+ -)" thingy I put in there means that when I took the square root ("Sqrt") the step prior, it generates two answers of equal magnitude but opposite sign. One positive, and the other negative, and I used that clunky notation to signify that. Does this make sense to you?

When you plug all your numbers into the formula(s), one will, most likely, be reasonable, and the other will not. The standard practice is to keep that one and ignore the other one.

Anyway....hope that helps, but lemme know if not....k?

bluto

aislinn rated this answer Excellent or Above Average Answer

Question/Answer
asraiblu asked on 10/22/05 - A couple of questions

Hi Bluto,
If a 10 kg ball is rolling at 5 m/s and hits head-on a 9 kg ball that is rolling at 6 m/s and the two stick together and stop, what can we say about the collision? What is conserved - momentum, energy or kinetic energy - or all 3 or a combo of two of them?

I am shooting at coconuts hanging from trees. I hit one but it doesn't fall. I know that the coconut will fall just as I shoot at it again. Why should I aim right at the coconut?
Thanks, Bluto!

bluto answered on 10/26/05:

You're welcome...glad to help!

asraiblu rated this answer Excellent or Above Average Answer

Question/Answer
asraiblu asked on 10/22/05 - A couple of questions

Hi Bluto,
If a 10 kg ball is rolling at 5 m/s and hits head-on a 9 kg ball that is rolling at 6 m/s and the two stick together and stop, what can we say about the collision? What is conserved - momentum, energy or kinetic energy - or all 3 or a combo of two of them?

I am shooting at coconuts hanging from trees. I hit one but it doesn't fall. I know that the coconut will fall just as I shoot at it again. Why should I aim right at the coconut?
Thanks, Bluto!

bluto answered on 10/26/05:

I'm stumped on that one too. I thought of both of those scenarios, but if I had to select one, based on the context of your other question, I'd say it was because of your first argument; the opossum being slower than your pellet.

Sorry I can't be more help to you on this one. It's not a very good question.


:)

asraiblu rated this answer Excellent or Above Average Answer

Question/Answer
asraiblu asked on 10/22/05 - A couple of questions

Hi Bluto,
If a 10 kg ball is rolling at 5 m/s and hits head-on a 9 kg ball that is rolling at 6 m/s and the two stick together and stop, what can we say about the collision? What is conserved - momentum, energy or kinetic energy - or all 3 or a combo of two of them?

I am shooting at coconuts hanging from trees. I hit one but it doesn't fall. I know that the coconut will fall just as I shoot at it again. Why should I aim right at the coconut?
Thanks, Bluto!

bluto answered on 10/24/05:

hi asraiblu....

For the first question:

Momentum and energy are both conserved. There are no external forces drawing energy away from the collision (none mentioned at least, even tho they are still acting); like friction for example.

As for the second one. I don't have a clue what this one is talking about. Can you clarify it at all please? Sorry. :(

I'm not sure what they're driving at with this one. Eat apples....they're easier to get. ;)

I'll be looking for your reply.

bluto --

asraiblu rated this answer Excellent or Above Average Answer

Question/Answer
asraiblu asked on 08/27/05 - collision

If a 6000 kg dump truck heads north at 10 m/s and hits head on a 2000 kg SUV that was moving south at 15 m/s, and the SUV gets stuck in the dump truck and they slide along together, in what direction and speed do they slide?

bluto answered on 08/27/05:

asraiblu......

I can't really show you diagrams here...not easily anyway, so hopefully this will make sense. If not....let me know......k?

Set it up like the following, which is actually as the problem describes.

The SUV is traveling south at 15 m/s. It hit's the truck head-on, which is going 10 m/s.

The SUV weighs in at 2000 kg, and the truck weighs 6000 kg.

What does common sense tell you about this scenario?

Think about it.

You have a much heavier truck hitting a lighter vehicle head-on. Yes the speed of the truck is less than the SUV, but......not by much. The truck weighs 3 times what the SUV does, and the speed is only 1.5 times less.....so common sense should indicate that the final velocity is going to be in the direction of the truck's original direction, and the speed will be less than the slowest vehicle speed, which is the SUV.

Let's see if that's the case.

Conservation of momentum. It's a physical fact that a body with a velocity and mass will have a property called (linear) momentum defined as the product of the body's mass and its speed. Since velocity and mass are not quantities that show up out of thin air, it stands to reason that the momentum also is something that will remain constant throughout a collision.

P1 + p1 = P2 + p2

P ---- truck momentum
p ---- SUV momentum

1 indicates before the collision
2 indicated after the collision

P1 = (6000 kg)(10 m/s) = 60000 kgm/s
p1 = (2000 kg)(15 m/s) = 30000 kgm/s

Since the SUV and the truck stick together after the collision (this is called an "inelastic" collision), then the final mass is the sum of the two masses. AND....the final velocity will be that of the 'clump'.

So: P2 + p2 = (6000 kg + 2000 kg)v

"v" is the final velocity which we need to find.

(m1 + m2)v = P1 + p1
==> v = (P1 + p1)/(m1 + m2)

P1 = 60000 kgm/s
p1 = 30000 kgm/s

NOTE: Since the SUV and the truck are traveling in exactly opposite directions, we need to decide which direction is positive and the other negative. Let's call the truck's direction positive...that means the SUV's is negative. This means the truck's velocity is 10 m/s, but the SUV's is -15 m/s.

Understand?

So this makes P1 + p1 = P1 - p1
==> 60000 kgm/s + (-30000 kgm/s) = 30000 kgm/s
==> 60000 kgm/s - 30000 kgm/s = 30000 kgm/s

v = (P1 + p1)/(m1 + m2)
==> (30000 kgm/s)/(6000 kg + 200 kg)
==> (30000 kgm/s)/(8000 kg)
==> 3.75 m/s

So......

Since the final velocity is positive that means it is in the direction of the truck. (As was suspected before we did the problem), and the final velocity is 3.75 m/s....which is less than either of the two individual velocities....also as suspected, and it's a reasonable value.

So....the final velocity is 3.75 m/s and it is in the direction of the truck's original direction.

Hope that helps.

bluto

asraiblu rated this answer Excellent or Above Average Answer

Question/Answer
asraiblu asked on 08/11/05 - Acceleration

If a 5 kg brick is dragged across the parking lot - horizontal force of 35N and a frictional force of 15N, what is the acceleration?
If this same brick is dropped from a very high building and along the way the friction because of air resistance is 15 N in strength, what is the acceleration of the brick then?

bluto answered on 08/11/05:

asr...

a) The acceleration is found using Newton's law: F = ma

"F" is the net force on the object, "m" is the object's mass and "a" is its acceleration.

You know m.....m = 5 kg. F is found easily enough as follows:

The net force is the difference between the applied force (f1 = 35 N) and the opposing force of friction given as f2 = 15 N.

F = f1 - f2 = 35 N - 15 N = 20 N

Then the acceleration is found by: a = F/m

==> a = (20 N)/(5 kg) = 4 m/s^2

(done)
-------------------------------------------------------------------------------------------------------------------------------------------------------------------------

b) Same game here.

As the brick falls the force is due to gravity and given by a form of Newton's law:

F = ma = mg

where "g" is the gravitational constant = 9.8 m/s^2

Since you know m, then you can find F, which I'll call "f1" again.

F = f1 = mg = (5 kg)(9.8 m/s^2) = 49 N

Since the opposing force is due to air resistance, I'll call that force "f2", then the NET force (F) is the difference between those two:

F = f1 - f2 = 49 N - 15 N = 34 N

Since this is the net force, and we know the mass....(m = 5 kg), the acceleration is:

a = F/m = (34 N)/(5 kg) = 6.8 m/s^2

That is your answer. It make sense because the ratio of the gravitational acceleration (9.8 m/s^2) to the net acceleration (a = 6.8 m/s^2) is the same as the Original force (F = 49 N) to the net force of 34 N.

Acceleration: 9.8/6.8 = 1.441
Force: 49/34 = 1.441

(done)

Hope this hepe, but let me know otherwise.

b-

asraiblu rated this answer Excellent or Above Average Answer

Question/Answer
darkstar asked on 05/12/05 - my camera

hi tom, i have a lap top computer, its a hp with windows 2000 professional on it and i have installed the hardware for my sony dsc-s30 camera on to the computer and talked to sony. the computer recognizes that the camera has been installed but i still can't load any of my pics onto the computer and sony said i would have to contact microsoft because microsoft does not support sony equipment but my last two computers were also microsoft, one was 98 and one was millenium and i had no problem with either, i was wondering if you knew of a place to download.....something that may let me take the pics from the camera and store them on my laptop. this has been very frustrating for me, or even if you know of an 800 number i might be able to use to contact microsoft, i have spent a couple hrs. now and can't figure it out. i never had this problem with my other pc's. thanks darlene

bluto answered on 05/12/05:

Hi.....

Hmmmmmmm........You said you installed the "hardware" onto the laptop for the camera. What hardware did you install?

I think most digital cameras come with their own CD with software (drivers and a user interface with which to interact) that you install on the computer.

Once that's done all that remains, as far as I know is to connect the camera to the computer/laptop with a cable. It's probably a USB cable, and it is probably included with the camera, but if not you'd have to get one.

So.....here are my questions to you:

1) What 'hardware' did you install?
2) Did you install the software that came with the camera?
3) Do you see a cable with which you can use to conect the camera and computer?

Check on those and then let me know what you come up with.

Hope this helps...


b-

darkstar rated this answer Excellent or Above Average Answer

Question/Answer
sexybodyjulie asked on 04/24/05 - Moment of Inertia

Hello there Bluto!
I was just wondering if you could help me to figure out the following question.

A certain merry go round is accelerated uniformly from rest and attains an angular speed of 0.4 radians/ second in the forst 10 seconds. If the net applied torque is 2000 Nxm, what is the moment of inertia of the merry go round?
CAn the answer be found based on the info above or is it that it cannot be determined since the radius is not specified?
I would greatly appreciate you imput
Thank You Much
Julian

bluto answered on 04/24/05:

Hi Julian...

you can, indeed solve it with the given info.

The figure given in the problem: 0.4 rad/sec is the angular acceleration.

Note: a = dw/dt

Where:

'a' is the angular acceleration
'dw' is suppose to be delta omega (the change in angular speed).....and....

'dt' is delta t, for (the change in) time.

'Delta' being the 'change in' those quantities....understand? If you don't lemme know k?

So, anyway: a =
You can derive the folowing formula easily enough, but I will forgo that and just say that the torque T is given by:

T = Ia

You're given 'T' and 'a'....

T = 2000 Nm
a = 0.4 rad/sec

Solve for I (the moment of inertia) and you'll have it.

I = T/a

You do the arithmetic :)

Hope that helps you.

b~

sexybodyjulie rated this answer Excellent or Above Average Answer

Question/Answer
sexybodyjulie asked on 04/12/05 - physics prob

Hell again Bluto
How have you been?! I ma asking for your help in a rather simple questio. I just want to make sure i did it right. So i would appreciate your input. Thanks

A disk ( radius = 2.95 mm) is attached to a high speed drill at a dentist's office and is turning at 8.33 x 10^4 rads/second. Dtermine the tangential speed of a point on the outer edge of this disk.

bluto answered on 04/12/05:

Hi sexybodyjulie....

hell?

I'm doing ok...thanks for asking.

Tangential speed is the speed of the outer part of the disk. Its formula is given by:

v = rw

v = tangential speed
r = radius of the disk, which is 2.95 mm
w = angular speed, which is 8.33 x 10^4 rad/sec

Plug these in and you have it:

v = (2.95 mm)(8.33 x 10^4 rad/sec) = 245,735 mm/sec -or- 245.735 m/sec

Depending on what units you want to work with.

Hope this helps, but let me know otherwise....k? :)

b-

sexybodyjulie rated this answer Excellent or Above Average Answer

Question/Answer
sexybodyjulie asked on 03/19/05 - Work

Hello Bluto!

I have a physics question that i would really appreciate you help on. I would really appreciate it.

1) A 58.8 kg box is being pushed a distance of 8.21 m across the floor by a Force P , whose magnitude of 171 N. The force P is parralel to the displacement of the box. The coefficient of the kinetic friction is .300. Determine the work done on the box by:
a) the applied force
b) the friction force
c) the normal force
d) the force of gravity
( Be sure to include all proper +/- sign for work done by each force.

Thanks Bluto
Julian

bluto answered on 03/19/05:

Hi Julian...

Work is defined *only* if there is a displacement due to the applied force.

The only work being done here is by the person pushing the box. Gravity does not displace the box, nor does the normal force. The force of friction opposes the direction the box is being pushed, so it will have a magnitude with opposite sign as that of the person pushing the box.

Friction (f) is equal to: f = -uN

u = coefficient of friction (0.300)
N = the normal force, which is equal to -mg

m = 58.8 kg
g = 9.8 m/s^2 (this is the gravitational constant)

If g is positive in the downward direction then it has the opposite sign upward.

For the applied force, the magnitude of this force is: 171 N, but since you are pushing againts friction this is less by (-uN). So the force necessary to push the box is: F = 171 N - (-uN) = 171 N + uN = 171 N + (0.300*mg) = 343.9 N (I don't see how the box can be moved if the aplied force is less than the frictional force....so check into this and make sure either the question is correct or that I am not missing something)

This is the total force applied due to friction. The work done on the box is this force times the distance it was moved:

W = Fd = (343.9 N)(8.21 m) = 2.8 kJ (kilojoules).

The force done by friction is (-uN)(8.21 m) = (-0.300*9.8 m/s^2*58.8 kg)(8.21 m) = 172.9 j

The other forces do not give rise to any displacement so they are all zero.

Hope this helps, but let me know otherwise.

:)

sexybodyjulie rated this answer Excellent or Above Average Answer

Question/Answer
sexybodyjulie asked on 03/12/05 - Dynamics

hi bluto
i was wondering if you could help me wit htis question please. it would really be appreciated.

A block is hung by a string from the inside roof of a van. When the van goes straight ahead at a speed of 25.1 m/s , the block hangs vertically down but when the van maintains the same speed around an unbanked curve (radius= 178), the block swings toward the outside of the curve. Then the string makesan angle (theta) with the vertical. Find the angle (theta)
thanks again Bluto
Julian

bluto answered on 03/13/05:

Hi sexybodyjulie....

I had to make a couple assumptions with your question. Just want to make sure you included all the information the question gives, and that the units for the radius are in meters.

I can't draw diagrams here, so it will make this a little more difficult to 'explain' things to you, but here goes.

As the car rounds the curve the mass on the string will move outward making an angle to the vertical. This is the angle theta that is asked for.

Resolving the forces on the legs of this triangle....tension and gravity you will [hopefully] be able to see how you arrive at this equation:

tan(q) = v^2/(rg)

Where: q = 'theta'
v = 25.1 m/s
r = radius = 178 [meters?]
g = gravitational constant = 9.8 m/s^2

If you plug these numbers into the above equation, you will get 19.86 for the value of theta.

Note that the centripetal acceleration is given by:

a = v^2/r

This is the acceleration of the car and the pendulum as the car goes around the circle. The car is considered to be accelerating because the direction is changing.

Hopefully my assumptions about some of the units in your question were correct, but if not let me know so I can revisit this for you. If you have any other questions let me know. Hope this helped you some.

b-

sexybodyjulie rated this answer Excellent or Above Average Answer

Question/Answer
jenny asked on 01/08/05 - i dont know

it is said that as a child Einstain asked the question
"What wuold i see in a mirror if i carried it in my hands
andran at the speed of the light?".

bluto answered on 01/11/05:

jenny.....

Light is a pretty weird thing. Without going into a lot of mumbo-jumbo, suffice it to say that light behaves the same for every person in every situation. Physicists refer to this as it being 'constant in any inertial reference frame'.

So if he himself is traveling at the speed of light then any external light, outside of himself and the mirror, would necessarily be traveling at the speed of light *relative to him and the mirror*. So, he could expect to see himself as normal in the mirror.

This concept even trips up scientists, and most often it trips up folks whom are not science-inclined. This idea brings about a lot of interesting questions and concepts, and it has also brought about many a book, which usually then go on to talk about other things like faster-than-light travel, and other kinds of things. It's a good question.

I hope this helps.

Here is a low-key website that talks about this idea (if I didn't goof it up by trying to post it as a 'live' link):

http://www.321books.co.uk/encyclopedia/physics/relativity/einsteins-mirror.htm


bluto

Question/Answer
ADCInc. asked on 10/27/04 - Can anyone tell...


Show that speed v reached by a car of mass m that is driven with constant power P is given by:

v= (3xP/m)^1/3

Thanks

bluto answered on 10/29/04:

ADC~....

I've played with this alittle bit. My findings are that you *can* relate power to velocity and mass.

Mathematically you can show anything as a function of anything, as long as it adheres to the definition of what a function is....a one-to-one mapping of indep variable to dependent......blah blah blah blah.

I agree and disagree with jm's response you have to this. I understad the point that is being made, and agree with that, but I don't think that applies to what you're looking for I don't think.

If you look at this dimensionally you can show easily enough that v^3 ~ P/m....minus the constant, which would come out of an analysis different from the one I did. Often to find out of a relationship is valid one can do a dimentional analysis to find out, which is essentially working with only the units.

Power(P) = J/s
Joules (J) = N*m [m = meters]
Newtons(N) = kg*m/s^2 [kilograms, meters, seconds]
velocity is m/s [meters & seconds]

dimensionally: P = J/s
==> N*m/s
==> [kg*m/s^2]*m^2/s^3

You say that v^3 ~ P/m [m = kilograms]

---> v^3 ~ P/kg [dimensionaly]

P/kg = [kg*m/s^2]*m^2/s^3
==> P = [kg^2*m/s^2]*m^2/s^3
==> kg^2v^2/s
==> m^2v^2/s [m is now METERS]
==> mv^3

So we have P ~ mv^3, or v^3 ~ P/m....like I said....absent the constant.

This shows that dimensionally you can end up with the relationship you have. Don't know if this helps, but hopefully so.

b-

Question/Answer
Ccl471 asked on 09/28/04 - Anti-matter

Is there such a thing as anti-matter?


Many thanks,

C.L.

bluto answered on 09/28/04:

Cd....

yep

There is antimatter. For example, when there is sufficient energy a gamma ray can interact with ordinary matter and create an electron - positron pair.


The positron is the antiparticle for the electron. It has the same mass as the electron, but opposite charge.

Gamma rays can be created in a lab, or they can occur in nature. 'Hot' radioactive materials can give off gamma rays, which is why they are so dangerous. Not because of antiparticles, but because they carry enough energy to damage and cause mutations of our body's cells (cancer).

Antiparticles don't last very long if they are produced because they will quickly encounter electrons and disappear, but in that annihilation there are other 'exotic' particles proiduced.

There are other details, but this is essentially what it's about. :)

bluto

Ccl471 rated this answer Excellent or Above Average Answer

Question/Answer
mefoley asked on 08/18/04 - heisenberg transformation

I recently ran across the phrase "heisenberg transformation" in the work of someone who I'm not sure knows what they're talking about. Googling for info hasn't turned up anything (that I can understand). My guess is that this is something to do with Heisenberg's matrix mechanics?

So -- is there such a thing as a Heisenberg Transformation? If so, what is it and what's it useful for?

Many thanks --

bluto answered on 09/09/04:

Hello mefoley....

Interesting paper (pdf). I'll have to read it over a little more carefully, but to answer your question. I think my original thought about the definition of the Heisenberg Transformation (HT) might have some merit.

It looks like they are simply operating on the Hamiltonian of the interaction (eq'n 2 on page 1 with the [time] evolution operator (basic quantum mechanics) shown on page 2, top right, and then just calling that process a HT. It seems to me like it is a definition on-the-fly, or of context, but there may be some meaning the authors of the paper are going from. Why they would choose such a thing is beyond me. Usually one defines such things so they may be referred to later, but I don't see where they do this, unless this is one paper of a series, or perhaps they are borrowing the definition from a different reference. But I do not see where there is a reference tied to this definition.

Seeing this in context helps, and I can tell you what it's referring to, but *why* they chose this particular definition, and why it is not used again is not something I have a good explanation for.......sorry!

:(


I hope this was of some help to you nonetheless!

Also, thank you for the follow-up. Good call. Normally I check back on answers I have given, but after a while I get busy, or don't see where there is activity and I just forget. Typical geek! :D

bluto

Question/Answer
mefoley asked on 08/18/04 - heisenberg transformation

I recently ran across the phrase "heisenberg transformation" in the work of someone who I'm not sure knows what they're talking about. Googling for info hasn't turned up anything (that I can understand). My guess is that this is something to do with Heisenberg's matrix mechanics?

So -- is there such a thing as a Heisenberg Transformation? If so, what is it and what's it useful for?

Many thanks --

bluto answered on 08/18/04:

Hello mefoley.....

Unless there is a document you can point me to for the usage of this term, I can only say the following:

This is not a term I am familiar with. That's not to say that it isn't a legitimate term. But.....physics is filled with colloquial terms that combine meanings of different concepts, or persons' names.

e.g. Heisenberg Uncertainty Principle, Hooke's Law, Bohr Radius, and on and on.

I did some checking too, but could not find the usage in math or physics, and any instance of it, in the resulting searches could be as valid as any of the others.

So, I would say that it is a colloquialism this person used in context and that it's probably not a formally accepted term.

Hope this helps, but check around. Perhaps another expert has some input.

bluto

mefoley rated this answer Excellent or Above Average Answer

Question/Answer
stilleb asked on 08/18/04 - guidance systems

does computer graphics programming have anything to do with guidance systems used in the military?

bluto answered on 08/18/04:

hi stilleb......

I had a hunch there would be quite a bit of uses for graphics computing for military aps, including guidance systems, but I did not have first-hand knowledge.

So, I did a quick search on the internet, e.g google; "computer graphics and military guidance systems". This and other similar searches brought up a number of hits.

This is where I am going to point you. Mostly since you can look for yourself and find resources that you may find more meaningful than anything I may throw your way. I don't like sending someone to website searches for responses, but if it's more appropriate than my 'guess', or hunch then I do it.

I hope you found this helpful, but if not, let me know.

bluto

Question/Answer
graycek asked on 08/12/04 - Is The Space (Tether) Elevator "Pie in the Sky"

I would like a simple (or any) explanation of how the men in white are going to get a payload from ground zero to 22,240 miles above earth and from an initial velocity of ~1000 MPH to a geocentric velocity of 6870 MPH by creeping up a yet-to-be-developed carbon fiber bean stalk, all while expending little of the energy of current rocket systems ? Are these guys rejects from SETI ? You may detect some sarcasm in my inquiry. I realize that air friction will be moot, and the massive weight of those vehicles and of the fuel itself will be gone, but potential and kinetic energy of the payload itself will still be there. It will not only have to be lifted up, but also pushed faster along the way up and a fuelled propulsion system will still be needed for that, won't it ? Then that, too will have to be decelerated on descent just like the shuttles. (Retired Engineer needing a physicist's viewpoint)

bluto answered on 08/13/04:

Hi graycek.....

I hope this isn't a disappointing response, but I guess I have to ask.....

...this sounds like something specific, and I can't say I know what it is you are referring to. Perhaps you can tell me what you mean when you are referring to the carbon bean stalk.

I'm aware of nanotech and C14 nanotubes, but I can't say I know what the connection is to this.

So, some additional detail would help please.

For what it's worth tho, I can say that usually launches are timed in such a way that the velocity of the earth is used in conjunction with where the rocket is to ultimately end up. In other words, if it's landing on the moon, the moons orbit has to be coincident with the path of the rocket.

For a geocentric orbit it only makes sense for the rocket to launch in a direction with the rotation of the earth. So, relative to an external reference frame, the rocket launches with its own power plus that of the earth. That doesn't help with getting it off the ground per se, but they have bloated fuel containers for that!

Ok....I'll leave it at that until I hear back from you.

bluto :)

graycek rated this answer Average Answer

Question/Answer
frank asked on 07/28/04 - Gr.11 Physics about Resistant

How come the Resistant equivalent is smaller than any orignal resistance in a parallel circuit????????

bluto answered on 07/28/04:

Hi frank...

Great question.

Mathematically it's easy to explain, but if you think about it like this, maybe it will help. The gist is the following:

In a series circuit with, say 2 resistors, the applied circuit voltage (like a battery) required to push electrons thru both resistors. Once the electrons get thru the first resistor, there is a voltage drop, and therefore a lesser voltage 'left over' to push the electrons thru the next resistor....and so on down the line for more resistors.

In a parallel circuit, the applied circuit voltage is applied to each branch in the circuit. So, the voltage that pushes the electrons (current) thru the first resistor is the same as the second resistor. So the net result is....the current 'feels' more push thru each resistor in this circuit than in the first one. If the electron feels more push, then it stands to reason it 'feels' less resistance as well. This means, the net effect is the resistance is less in a parallel circuit than in the series crcuit.

Mathematically this works out to be true.

Make sense? Does this help?

bluto

Question/Answer
Dathaeus asked on 07/12/04 - Hey Tom...

Hey there, long time no talk, been busy moving! Anyways, quick question... I found this great site http://www.acousticalsurfaces.com/ that sells all sorts of sound dampening panels, etc, which was so hard to find in any local sound advice or soundworks store, or even on ebay... I did find something on ebay but htey are scarce...
http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&category=3278&item=3735778600&rd=1
but i am worried if this is just packaging foam they are selling as "sound dampeners" or if they are really the same thing.

However, besides maybe getting one of those, if you think they are any good, would it be cheaper for me to just buy some foam or styrofoam and some loose fabric of the color I want and cover the foam in fabric and layer the wall I want? This wall wont be too visible, but i dont really wanna see white foam sticing out u know? I heard u can also spray paint the foam, so maybe I can just buy a lot of ebay and paint it.

I rather not spend the hours unless i have to, so let me know what u would do, thanks.....

bluto answered on 07/12/04:

Hi Dat....

Yes, long time no talk. I thought you were already moved? Weren't you in the progress of moving some number of months ago? Or was that just prep? Either way I wish you luck. Moving isn't easy no matter how you cut it!

Unless the picture is just that misleading (the one that shows up on the link to ebay) that stuff looks like nothing more than ordinary eggshell foam. The kind of stuff you see for pillows, bedding, and various other things. I suppose it has some sound dampening to it, but so would a pile of dirty diapers! :D I think if you are serious about getting something for your money, I'd avoid this one, unless you only need something light duty. But the price looks good. No bids, 10 $$$ for 10 sheets, but the sheets are not that large tho.

Here's a place that will make foam any shape you want:

http://www.efoamstore.com/, but you have to suffer the pain of sending them a 'contact me' email from their site.


This site may also be of use to you:

http://www.foamfabricating.net/

But I think the first site you mention is a good source as well. But you MAY be able to contact that company, the one you mentioned (http://www.acousticalsurfaces.com/) and see if they are willing to send you a sample. I would think they would. Then you can examine it yourself and see if it's similar enough to just regular ole egshell foam that you can spray paint and/or cover with cloth. This way you'll have the cost nailed down a little better. But I know what you mean about going to all the trouble of making something.

Let me know what you think, but this is my best guess. I tend to think that genuine acoustic stuff is $$$, but they normally impregnate foam with a substance that's similar to the effect you'd get to using spray paint. It's not as good, of course, but if price and time are an issue then it's worth considering.

I hope this helps, and don't be such a stranger!

See ya ----

bluto :)

Dathaeus rated this answer Excellent or Above Average Answer

Question/Answer
SCOOBY asked on 05/23/04 - glass pots stuck

Yesterday when I was washing dishes, I inserted a smaller glass pot that I had just washed into a larger glass pot. When I went to use the smaller glass pot, I couldn't pull it out. It is stuck. Is there some trick I can try to get it unstuck this morning?

bluto answered on 05/23/04:

Hi SCOOBY....

It could be from one, two, or a combination of the two following things.

1) Could be the smaller glass pot made a seal with the inside of the larger one, and a partial vacuum prevented you from pulling them apart.

2) Could be just plain ole friction.


#1 Running warm water over them would quickly alleviate this. Might help to pour some liquid soap in between where they are stuck while you do this so they don't -scrape- when separated.

#2 Most likely remedy would be to hold the handle of the larger one (probably take another person to do this, unless you can be clever and do it yourself) Grab the handle of the smaller one and then tap the outside of the larger pot....*gently*.....and wiggle the handle of the smaller pot while you do this. This should break the friction easily and free the pots.

Hope this helps!

b~

SCOOBY rated this answer Excellent or Above Average Answer

Question/Answer
dapromise asked on 05/17/04 - Bombs and Weapons

Some of this stuff requires GREAT detail and explanation, and I am unable to find a site that simply explains without pulling into complicated adult equations and statistics that I don't understand. If there is easy Internet reference for this, THANK you very much. God bless you, and have a wonderful day!

1) Why did the Chernobyl incident in 1986 occur?
2) What were the effects of the bombing of Hiroshima and Nagasaki?
3) Should scientists continue creating more efficient weapons to keep enemies from attacking them, even if the cost of human life is high?
4) What are the benefits of nuclear fission?
5) How does a Hydrogen bomb work?

ps. I SO envy people who get the hang of Physics and know it off the top of their heads!

bluto answered on 05/18/04:

Hi dapromise..

1) This is a very well documented incident that you could look up on the web easily. The gist of it is the following:

Memory told me it was 1985, but I had to check; it was in 1986, in April, that having disregarded numerous safety precautions during a test procedure, one of the four reactors managed to get out of control (meaning the chain reactions began to get out of control). It caused explosions that blew apart the heavy steel and concrete lid on the reactor. It killed 30 people, and the resulting radiation caused some 135,000 people to have to be evacuated. It has been touted as the worst nuclear disaster on record. It occurred at Chernobyl in the former USSR, which is now the Ukraine. Chernobyl is about 80 miles North of Kiev.

2) Againthis is a well documented piece of history that you can look up for yourself. As opposed as I am to answering a question like this one, and the first one, and others like itANDto answer questions by throwing out websites like I am about to do.I am doing so here partly because you normally do a good job of asking questions and this is the best way to address your question:

If you take a look at this link: http://www.uic.com.au/nip29.htm

It will give you the necessary details (I think) that will help you. I dont like to blow off questions, but for those that require only a little time searching and reading on the web I have problems.sorry.

3) This question is asking for your opinion. What do you think? This one I refuse to answer for you.sorry. I know what my opinion is. You should think about it and try and put your feelings and thoughts into words. I would essentially be doing you a disservice by answering this for you.

4) This is another question that I have trouble giving details about for obvious reasonssorry. Whomever is giving you assignments like this shouldnt be asking you to find out how a bomb works. Its a very poorly thought out question/assignment. I know that may not help you with your problem at hand, and Im sorry, but this is not an appropriate assignment, or question to be asking, and I cannot give you an answer.

Nevertheless I hope what I was able to provide to you was helpful to some degree.

bluto

dapromise rated this answer Excellent or Above Average Answer

Question/Answer
dapromise asked on 05/13/04 - Nuclide

How do you write the symbol of a Uranium nuclide with 143 neutrons? I don't understand how the numbers are formed using mass/atomic numbers, etc. Nuclide numbers are confusing.. ><" Thanks!

bluto answered on 05/13/04:

Hi dapromise....

Before I start, I'm going to try some html formatting so you can see the symbols better. The formatting I'm using should work, but I won't know if it's actually going to come out the way I intend until I send it to you. So, if it comes out wrong, or garbled, I'll follow-up, and try and clear it up for you. If it looks close, I may just rely on you to know what I'm trying to do by the context. If that happens and you're confused, I'll make it right as well....ok? I hope this works.

Uranium has the symbol U. The symbolic representation of elements is as follows:

A
X
Z N


Where X is the symbol of the element.

A = atomic mass of the nucleus
Z = atomic number (# protons in the nucleus)
N = neutron number (number of neutrons)

These are all related as follows: A = Z + N

Which makes sense, because the atomic mass would have to the how much stuff makes up the nucleus..right? So you would have to add up everything that makes up the nucleus; namely the number of protons (Z) and neutrons (N).

So, for Uranium with 143 neutrons, youd have:

A
U
Z 143


If were talking about standard Uranium, which has an atomic number of 92 (it is 92 in the periodic table), then we know that:

N = 143
Z = 92

Then we can figure out that A = Z + N = 235.

So the symbol would be:

235
U
92 143




I hope this helps

bluto






dapromise rated this answer Excellent or Above Average Answer

Question/Answer
dapromise asked on 05/10/04 - Wiring a Room?!

Hi again, bluto! (: Arg it's me again.. I'm sorry I'm asking so many questions :( Among the many experiments and physics projects I have to do, one is draw a diagram and plan the wiring of a house with 3 rooms. The house looks similar to this:

120v
[______==______][______==______]
[<> 240v.......][room2..144v <>]
[room1.........][______________]
[.......<> 240v][room3..120v <>]
[______==______][______==______]

The diagram is kind of messed up- I can't get it to work properly! :(

There is 120V coming into the house.In the large room there are two lamps, each with a resistance of 240 Ohms. A light switch controls both lamps, turning them on and off as a unit. In the second room there is a light switch and one lamp, with a resistance of 144 Ohms. In the last room there is a light switch and one lamp, with a reisistance of 120 Ohms.

I have to make the parallel or series circuit clear and include a circuit breaker (fuse box?) as well as the current flowing through each light switch when they are all turned on. I know the basic idea about series and parallel circuits, but how do I draw this? o_O Thanks and God bless. Have a great day!

bluto answered on 05/10/04:

Hi dapromise......

First of all...lemme just say that I don't mind answering your questions at all, but if/when you rate one of my answers, if you'd let me know what you think I *didn't* address for you if you felt that my answer didn't deserve 5 stars. I don't care about the ratings so much as being able to address someone's questions as fully as possible. If I'm constantly giving you answers that don't deserve 5 stars, then that tells me that I'm missing something when I give you an answer and that's not helping either of us if I don't know what's up. :)

On this question, I'm afraid you're going to have to clarify the following for me:

Your diagram doesn't help me at all. If you could describe what you're suppose to do with the information given then I can help.

For instance, I know you have an input current.....I'm assuming it's AC, or is it DC...or does it matter for this question?

I know you have lamps and switches, but which are suppose to be in series and which are suppose to be in parallel? Or does it matter?

Can you give me more an idea what you need to find and I'll try to help you as best I can.

Thanks ----

bluto

dapromise rated this answer Excellent or Above Average Answer

Question/Answer
dapromise asked on 05/06/04 - Confused on Electricity

I am so confused! I don't understand this. The more I study physics and the more the teacher explains all the stuff, the more confused I become. Physics is SO hard! I have a lot of questions, but I'm trying to find the answers myself. Here are a few I am absolutely downright confused on.

1. What is the difference in the voltage across series resistors and the voltage across parallel resistors?
2. What is the difference in the current through series resistors and the current through parallel resistors?
3. What happens if you break a magnet, i.e. where are the poles?
4. How do you make an electromagnet from an iron nail, 3m of copper wire, a battery, and a sheet of A4 paper?

I'm confused! I read the chapter and it is almost unable to be understood. Is there some really easy kid's physics web site that I can actually understand the physics?

Thank you so much and God bless. You've saved me from my misery- arg.

bluto answered on 05/06/04:

dapromise.....

1) When resistors are in series and you place a battery (which can be regarded generally as a power source (which is a misnomer), voltage source, or sometimes in physics its just called a potential) the push given by this (Im gonna call it a voltage source) voltage source to the electrons, which is the current that flows along the wire (conductor), has to flow thru each resistor. Since, in a series of resistors there is one resistor after another the individual resistances add up in the eyes of the current. Also, the resistance of a resistance at one end of a wire can be felt by the current at the other end. Imagine a 50 foot section of garden hose. Youre on one end and Im on the other. If you turn on the water at your end, the water will travel down the hose and come out the other end. All fine and good. No revelation there. Now.if I shove a piece of steel wool say, into the end of the hose, the water flow will be cut back because it now has to flow thru this resistanceright? Sowhat if I took a bunch of pieces and shoved each one into the end and pushed it way back in with a long rod, so theres some spacing between each one. There would be a series of these inside the hose. The water would get to the first one and slow down some, and then get to the next and slow down some more, and so on, until either a trickle would come out of the end of the hose, or it would be plugged up and nothing would be able to get thru. The water feels the resistance of each piece, just like current feels resistance from resistors in series. The voltage source has to have enough push to force the electrons past ALL the resistance if there is to be any current flow at all..just like with the garden hose. If you raised the water pressure at your end, then you might have enough push to force the water past all those pieces of steel wool, and if you lowered the pressure, you would not have enough pressure to get past them.understand?

Following the garden hose analogy again. A parallel system is like you at one end of a garden hose, except your end feeds into the middle of say, a long pipe.like a piece of PVC. From this section of PVC there are say 3 pieces of hose attached to it, and they are all of different size. Those each end at another long section of PVC, from which one long section of hose comes from it and leads to my end. Can you picture that?

If you turn on the water and it flows, it fills the first section of PVC pipe and it then feeds each of the three sections of hose, which then quickly fill the other section of PVC and voila.water will flow out my end of the hose. Got it?

Now, suppose Im a poop and decide to put a big clamp on one of the three pieces of hose so it blocks the water. Will that stop the water from flowing in my end? No..I hope you said no there at least! The water at your end has a certain amount of pressure in it, and blocking one of the three hoses wont stop all the water from getting thru because it has two more hoses (paths) for the water to follow.right? If I block two of the three hoses.it still wont block all the water, but of course if I block all three it will. Makes sense doesnt it? So.since the three hoses are in parallel with each other they each see the same amount of water pressure, but since they are different sizes they will only permit so much water from passing thru. It makes sense that the largest diameter hose will make it easier for the water to flow, and the skinniest will make it harder. Blocking one hose versus another will correspondingly affect the flow at the other end, but I hope you see that my goal is to illustrate the difference between series and parallel. Going into more detail is overkill I think, so Ill stop there, but if you have questions on this let me know.

2) This is similar to what I described above, except instead of the voltage which is the amount of push the current feels in a circuit (or water pressure in our garden hose example) current is the amount of electrons. Current corresponds to electrons like volume is to water.

High current = a lot of electrons
High volume = a lot of water

Low current = little current
Low volume = little water

Let me know if what I said in #1 and what I mentioned here is enough to answer this one for you, if not, let me know. Theres a LOT one can say about these, and I dont want to confuse you with to much detail.

3) If you break a magnet, the poles rearrange themselves into north and south again. No matter how small there will always be a north and south pole..always. The poles are at opposite ends of each other.

4) I dont know why you need the paper, but if you wrap the nail with the wire (leaving a loose end several inches long), and do so as carefully as you can. Making each turn not overlap a previous one, and be very neat and symmetric about it (this makes the magnetic field stronger and more uniform), and then end with a piece about as long as the first piece. Connect each end to either side of a battery.youll be able to pick up paper clips and iron filings and other similar items with the nail. I guess the paper is wither for wrapping around the nail, or maybe its to lay paper clips and iron filings on.

I will see if I can find a good website for physics, but what I like and what you like might be different. Id suggest going to Google and typing in physics for kids, and start going thru the list. Youre bound to finds one you like. If I run across one I will add it as a follow-up to this question.ok?

Hope this helps.

bluto

dapromise rated this answer Above Average Answer
raridon rated this answer Excellent or Above Average Answer

Question/Answer
dapromise asked on 05/04/04 - Electricity

I have been studying electricity lately and I'm confused about the following questions. Please help me!

1) Why does a balloon rubbed on one's head attract water?
2) After walking across carpet in a dry climate, wood a wooden or metal doorknob shock you more? Why?
3) According to Ohm's law, why should you wear rubber-soled shoes in a lightning storm?

Thank you very much and God bless.

bluto answered on 05/04/04:

Hi dapromise.....

1) rubbing a balloon on one's head transfers electrons from the rubber to your head (hair mostly). This transfer of electrons from the balloon leaves it with a net positive charged.

A water molecule is composed of one oxygen atom and two hydrogens. The hydrogen has one free elctron that it shares with oxygen's 6 free elctrons. Once this happens, the hydrogens have a net positive charge and the oxygen, which has 4 leftover electrons have a higher negative charge. This makes the water molecule highly polar. In other words, one end of the molecule is positively charged, and the otehr negative.

So, holding the balloon near a tiny stream of water from a tap causes the stream to bend inward toward the balloon because of the attraction of unlike charge.

2) Walking across the carpet when the humidity is low (water [the water that causes humidity] carries away charges that cause static electricity) causes your body to pick up excess electrons. Touching wood would do nothing. Wood is not a conductor. Touching an iron railing however, will give you a good sized shock! The reason is the railing is a terrific conductor and the excess electrons that reside on the outermost portion of your skin distributed evenly over your entire body all want to go to a place of lower energy.....which is on the railing. So as they exit your body is leaves you with that delightful snapping ~spark~ we all know and love. :)

3) Well...lemme tell ya. If the potential is great enough that lightning is going to discharge, rubber soled shoes aren't gonna help ya. A thin layer of rubber isn't going to guarantee you won't be struck, but it does help, and it's advice that should be followed. It certainly can't hurt. Reason is, the rubber is not a conductor, and therefore it will not permit charge from passing thru the ground, over your body and cause you to be a walking antenna. To lightning you appear as a tiny pointy spike on the ground. If you're the only 'tall' object in an otherwise barren landscape YOU are a prime candidate for lightning. That's why golfers get hammered so often. They walk around a somewhat barren landscape, with metal on their shoes and metal in their hands. yet, they are foolish enough to play with lightning flashing and humidity levels elevated.

Hope this helps.

bluto

dapromise rated this answer Excellent or Above Average Answer

Question/Answer
Whitestar asked on 04/19/04 - Other Theories On Teleportation

Hello again Bluto.


Your comment,


"Sorry......tunneling is a quantum mechanical effect that I assumed you were familiar with. It refers to the passage of current thru a barrier that it shouldn't be allowed to cross according to classical physics. This is analogous to our discussion of TT, so I utilized it......that's all. So, in my reference I was only referring to an object (a person) passing thru a barrier (space) that it normally isn't allowed to."


That reminds me, a few years back, the UPN network aired a tv movie entitled, "Warlord: Battle For The Galaxy", starring Rod Taylor (H.G. Wells's Time Machine from 1960). Anyway, Taylor played a soon-to-be-retired General in charge of a starship that was equipped with a teleportation device, that worked by bending and folding space around the people traveling in it. It was mentioned that an older and primitive teleporter worked by scrambling peoples molecules, but there were too many accidents. This new form of teleportation was proven to be far more reliable and safer. There is a scene where his crew teleported from their bridge and onto the surface of a planet on a specific floor in a building.

1) Is the Warlord Teleporter a form of tunneling?


2) Is it possible in theory to teleport a crew from the bridge of a ship to the inside of a building on a specific floor using the Warlord Teleporter?


3) Would the energy requirements be too great to permit this?


Your second comment,


"Violate the Uncertainly Principle? Yes and no. We discussed this a little bit. I brought it up and then remarked that you can violate it, but if the time restriction permits. Mass-mapping is my way of saying there would probably have to be some kind of clever way to scoop up all the information contained in each particle and track it, but not do it on an individual basis. This way you'd be able to manipulate it, and move it and reassemble it, etc, but not necessarily know momentum and location of each of the particles, thus avoiding violating the Uncertainty Principle."


4) Is this mass mapping the same as quantum entanglements? If not, what does mass mapping involve?


regards,

Whitestar

bluto answered on 04/20/04:

Hi whitestar.

Once again, to respond to you in order:

1) To reiteratethe notion of teleporting is like the notion of tunneling. Its just an analogy. Tunneling is real, teleportation is not. To say that the functionality of a fictitious telporter is like a real principle of science such as tunneling isnt appropriate. You seem to really want an answer to this question, so I shall repeat myself and say that, in general, tunneling is similar to the idea of teleportation

2) Once again, youre seeking, what appear to be real answers to a fictitious principle. First..youre asking: Is it possible in theory to teleport a crew from the bridge of a ship to the inside of a building on a specific floor using the Warlord Teleporter?. There is no theory for teleportation. If you want it to have this property then allow your imagination make it so. Frankly I dont know how to respond to this question. Its like asking..how many eggs can the Easter Bunny carry at one time.

3) Like the previous two questions, youre seem to be asking for real answers to a completely fictitious thing. To answer a question in any degree if seriousness would require arguments based on physics as we know it today, for a principle that exists, and as we know, it does not exist. Sorry.I dont know what to tell you here.

4) Quantum entanglement involves the instantaneous communication between two spatially isolated particles. Make changes to one, and as if by magic the other reflects the influence of that change(s). This is not what I mean by mass-mapping.

Heres what I mean by mass mapping. Say you have a picture of a birthday party. You wish to send this pic to someone. You convert the picture to a digital format, save it to a location on your hard drive in your computer and then send it. The computer can send it serially, meaning one piece at a time until the entire file has been sent, or it can send it in clusters at a timeI think they still call this multiplexing. Sending the file takes, say for argument 1 byte (8 bits) of information of the picture file and sends it, then it goes back and gets another byte, and sends it, and so one, until the entire picture has been sent. It has to keep track of each piece so it can be correctly restored on the other end. Its a mind-numbingly slow process. Sending the picture in a multiplex fashion takes maybe 8 or 16 (or whatever) bytes at a time and sends it. Repeating until the entire picture is sent. This is much faster, and more efficient. It isnt necessary to keep track of each BYTE like in the serial case, but the 8 or 16 byte pieces still have to be kept track of. Still, its a better method. This is similar to what I refer to as mass-mapping, where the teleported would have some way of working with a smaller data set, but in massive quantities as opposed to the millions and millions and millions and millions of individual pieces. Does this help clarify it?

Hope this helps.

Kind Regards,
bluto

Whitestar rated this answer Excellent or Above Average Answer

Question/Answer
Whitestar asked on 04/16/04 - More On Teleportation


Hi Bluto.


Your comment,


"I don't (and didn't) disagree with you at all regarding using deductive reasoning to draw conclusions based on today's physics for a teleportation theory (TT)."


Yes, I agree and I apologize for misunderstanding you. :)


Your second comment,


"But as I pointed out, in this case, I don't think one can gain much by basing arguments on current technology for a theory as complex as that of teleportation. Having said that tho, I don't think we will get to the point of discovery without following along standard concepts, but yet, I believe the actual discovery that leads to a TT will require a discovery, or nonstandard leap from the current state of physics."


Once again I wholeheartedly agree with you on that.


Your third comment,


"Next.....the converting of a body to energy: Would that result in the liberation of great amounts of energy........my answer is also a resounding yes. BUT.....as I said does the process of TT *require* this? I tend to think not. Indeed e = mc^2, but e^2 = c^2p^2 +(m_o)^2c^4 is the total energy in the relativistic regime, which is what I would think would be the more appropriate energy to consider. But....would this even apply to whatever technique comes to pass? This is the energy *released*........would we actually want to *release* this energy? I'd tend to think not, otherwise the poor soul being transported goes up in a puff of smoke. It could also be the supposed yield of energy, but, again......I don't think this is the appropriate direction or information for a TT."


Okay but let's say for the sake of argument that we had the technology to reverse the process.


1) What are the prospects of a person surviving such a procedure? Would it preserve the person or destroy the individual, thereby resulting in a replica?


2) Would this theorical technology be better suited for inanimate objects instead?


Your fourth comment,


"For your #1: I thought I already stated my opinion on this as well, but once again I say.....IF TT were possible and someone was tunneled across space.......in my personal opinion, I think the 'being' that exited would be an essentially seamless copy, but minus a soul. But that brings a host of questions that I touched briefly in my previous posts."


Okay, I need some clarification here. I never mentioned teleportation being accomplished through tunneling. I previously stated that in your opinion, could a person survive the teleportation process if he or she were to be converted into energy and be reconverted back into matter.


3) What does converting a person into energy have to do with tunneling?


Your fifth comment,


"Your #2: IF TT were possible I think it will involve a clever way of mass-mapping. It will have to encompass a way of keeping record of each and every state of each and every particle in the body at the time of disassembly."


4) But won't that violate the Heisenberg Uncertainty Principle?


regards,

Whitestar


bluto answered on 04/19/04:

Hello whitestar........

No need to apologize for anything, but thanks just the same. :)

The first thing to comment on is what I have in a 'box 1' just below, where you refer to my third comment:

1) To ask for the prospects of surviving such a teleportation procedure is purely fantasy. Anything goes. If you want to imagine it being survivable, then it's survivable, if you don't, then it's not. There is no basis on fact whatsoever, so to ponder the likelihood of it working successfully is pointless in my opinion......sorry. :)

2) Teleporting inanimate objects solves the problem of individuality and souls. A box of crackerjack is a box of crackerjack. BUT......would there be harm transporting anything for consumption? Would transporting molecules cause them to be in disarray upon reassembly? If so, would it be subtle to the point of being harmful, but not necessarily detectable without close scrutiny? What about other things......skin oils and creams, medicines, things to be taken into the body or applied on it. Would there be harmful rearrangements (mutations) to these kinds of items that would be harmful if used. If not short term........long term? Perhaps a chair would be safer to teleport. As long as the bonds between the atoms remain intact, what difference would it make it a bond is between atom A and C versus atom A and B? Would there be weird effects like the magnetism of some object that wasn't magnetized before. Would transporting cause some alignment in the crystal structures of objects which give rise to things like this? Your guess. So.....maybe there are problems transporting inanimate objects too.



box 1:
---------------------------------------------------------------------------------------------------------------------------------------

Your third comment,


"Next.....the converting of a body to energy: Would that result in the liberation of great amounts of energy........my answer is also a resounding yes. BUT.....as I said does the process of TT *require* this? I tend to think not. Indeed e = mc^2, but e^2 = c^2p^2 +(m_o)^2c^4 is the total energy in the relativistic regime, which is what I would think would be the more appropriate energy to consider. But....would this even apply to whatever technique comes to pass? This is the energy *released*........would we actually want to *release* this energy? I'd tend to think not, otherwise the poor soul being transported goes up in a puff of smoke. It could also be the supposed yield of energy, but, again......I don't think this is the appropriate direction or information for a TT."


Okay but let's say for the sake of argument that we had the technology to reverse the process.


1) What are the prospects of a person surviving such a procedure? Would it preserve the person or destroy the individual, thereby resulting in a replica?


2) Would this theoretical technology be better suited for inanimate objects instead?
---------------------------------------------------------------------------------------------------------------------------------------


For box 2:

Sorry......tunneling is a quantum mechanical effect that I assumed you were familiar with. It refers to the passage of current thru a barrier that it shouldn't be allowed to cross according to classical physics. This is analogous to our discussion of TT, so I utilized it......that's all. So, in my reference I was only referring to an object (a person) passing thru a barrier (space) that it normally isn't allowed to.


box 2:
---------------------------------------------------------------------------------------------------------------------------------------

Your fourth comment,


"For your #1: I thought I already stated my opinion on this as well, but once again I say.....IF TT were possible and someone was tunneled across space.......in my personal opinion, I think the 'being' that exited would be an essentially seamless copy, but minus a soul. But that brings a host of questions that I touched briefly in my previous posts."


Okay, I need some clarification here. I never mentioned teleportation being accomplished through tunneling. I previously stated that in your opinion, could a person survive the teleportation process if he or she were to be converted into energy and be reconverted back into matter.


3) What does converting a person into energy have to do with tunneling?
---------------------------------------------------------------------------------------------------------------------------------------

For box 3:

Violate the Uncertainly Principle? Yes and no. We discussed this a little bit. I brought it up and then remarked that you can violate it, but if the time restriction permits. Mass-mapping is my way of saying there would probably have to be some kind of clever way to scoop up all the information contained in each particle and track it, but not do it on an individual basis. This way you'd be able to manipulate it, and move it and reassemble it, etc, but not necessarily know momentum and location of each of the particles, thus avoiding violating the Uncertainty Principle.

box 3:
---------------------------------------------------------------------------------------------------------------------------------------

Your fifth comment,


"Your #2: IF TT were possible I think it will involve a clever way of mass-mapping. It will have to encompass a way of keeping record of each and every state of each and every particle in the body at the time of disassembly."


4) But won't that violate the Heisenberg Uncertainty Principle?
---------------------------------------------------------------------------------------------------------------------------------------

Whitestar rated this answer Excellent or Above Average Answer

Question/Answer
Whitestar asked on 04/15/04 - More On Teleportation

Hello again Bluto.


Your comment,


"In some cases I do not agree that deductive reasoning can always be relied upon to make intellectual leaps from the technology of today to the technology of tomorrow. Here's why I have this opinion *in this case*. For a technology (theory) of such magnitude as teleportation (intact) of anything, I find it difficult to believe that it will contain any (or very many) concepts we are now familiar with. Certainly it will be built (to some degree) on laws now known to us. In my opinion it will be a leap of intellect, or a guess, or wild, outside attempt that will bridge some gap between the theories of today and this theory of tomorrow."


I agree but sometimes we can use deductive reasoning as a guide for the theorical technologies.


Your second comment,


"If a person is turned into energy, why must the result be gamma rays?"


Sorry, that was just me thinking out loud. :)


Your third comment,


"Regarding destruction of the body. What do you mean by destruction? The dismantling of a human body doesn't have to imply destruction."


I mean whenever you're converting matter into energy, that is, E = MC squared. When that happens, fantastic amounts of energy is released.


1) If you converted a person into energy and you had the technology to reverse the process, would it be the same exact person as before or would it be a replica? (Note: I already stated my opinion on this one but I'd like to know yours.)


2) If teleportation were to be a possibility in the future, how do YOU think it will probably work?


regards,

Whitestar


bluto answered on 04/15/04:

Hello whitestar......

I'll respond, in order, of/to your comments. Thanks again for the thoughts too. Very interesting things to consider.

I don't (and didn't) disagree with you at all regarding using deductive reasoning to draw conclusions based on today's physics for a teleportation theory (TT). But as I pointed out, in this case, I don't think one can gain much by basing arguments on current technology for a theory as complex as that of teleportation. Having said that tho, I don't think we will get to the point of discovery without following along standard concepts, but yet, I believe the actual discovery that leads to a TT will require a discovery, or nonstandard leap from the current state of physics. Does that make sense. It may sound like I'm contradicting myself, but I'm not, if I'm describing it clearly.

Next.....the converting of a body to energy:

Would that result in the liberation of great amounts of energy........my answer is also a resounding yes. BUT.....as I said does the process of TT *require* this? I tend to think not. Indeed e = mc^2, but e^2 = c^2p^2 +(m_o)^2c^4 is the total energy in the relativistic regime, which is what I would think would be the more appropriate energy to consider. But....would this even apply to whatever technique comes to pass? This is the energy *released*........would we actually want to *release* this energy? I'd tend to think not, otherwise the poor soul being transported goes up in a puff of smoke. It could also be the supposed yield of energy, but, again......I don't think this is the appropriate direction or information for a TT.

For your #1:

I thought I already stated my opinion on this as well, but once again I say.....IF TT were possible and someone was tunneled across space.......in my personal opinion, I think the 'being' that exited would be an essentially seamless copy, but minus a soul. But that brings a host of questions that I touched briefly in my previous posts.

Your #2:

IF TT were possible I think it will involve a clever way of mass-mapping. It will have to encompass a way of keeping record of each and every state of each and every particle in the body at the time of disassembly. Not only that, a human body communicates within itself. Synapses fire, electric signals transit here and there. These occur in a finite period of time. If a signal is between points of origin and destination, these would have to be restored upon reassembly otherwise there could be compromises to the health of the subject. Think about it....what if brain signals telling the heart to contract got lost.....or to breathe. Think about the massively complex chemical reactions occurring. Any self respecting physicist knows that chemistry is kindergarten quantum mechanics (QM). That reactions occur for reasons chemists have to memorize. They lump the ways atoms and molecules bond by referring to electron clouds......orbitals....and other hand-waving, one-dimensional accounts of 'why' this compound happens versus that compound. Quantum mechanics doesn't really 'explain' that either, but at least it says that atoms are governed by QM and the location of the electrons (valence), which as you probably know are solely responsible for the chemical properties of an atom, is a probabilistic at best. This, to me is the most compelling evidence for the Uncertainty Principle, but that's a digression. Anyway......as atoms bond and unbond in the countless reactions inside the body, chemistry is a horridly lacking
So.....I think TT would have to require some massively-mapping method where all pertinent pieces of information are accounted for enmasse, and partly relying on other areas of info such as DNA.

Have a nice evening.....

bluto

Whitestar rated this answer Excellent or Above Average Answer

Question/Answer
Whitestar asked on 04/13/04 - Teleportation (Follow-up)

Hi Bluto.


Your comment,


"And your question..would the person survive the transport. It depends on what is meant by survival and death. Whats the definition?.....wheres the line? Does the resulting person look like the original? Does he/she think and act like the original? Have the same memories, same feelings, personality, same flaws, scars, mannerisms, handedness..If so.would that then be considered survival of the original person? Dont forget about religion. Does the spirit transport along with the physical body? Does the spirit re-inhabit the resulting person on the other side? So.what is actually meant by death? Is the separating of the particles of a body a cause for the spirit to vacate and not return, yet to us humans the resulting person appears, for all intents and purpose to be the exact same person, even tho that person may not have the spirit? Can a human exist without one? If so, how would we know?"


You raised some interesting points. Although we don't have the technology to teleport a person by converting an individual into energy and reversing the process, we can however make some deductive reasoning based on what we know today.


From what I understand, such conversions of particles to energy are called annihilations, that is, they are like explosions: the explosive material is completely destroyed and no memory of its original form remains. Now if you turn each person into energy, you get a cloud of gamma rays expanding outwards. There is nothing that would make them spontaneously reform the person - even if you reflected them backwards, they would not neccessarily create the original particles. It is much more likely that teleportation would involve sending the information that can be gleaned from the gamma rays, and then having the information used by a base station to construct the person, more mechanically.


In my view, when your body is destroy, you die. End of story. What comes out of the teleporter is an exact copy, with all your memories etc, and no knowledge that it isn't you, but it isn't. No one would ever notice the problem, so it only affects you when it happens. Unless, if you believe in souls, there are "conservation of souls" problems to deal with - does the same sould follow the body around?

While in an energy state, there is no consciousness, no heart to beat. Hence, once matter has converted into energy, the person who first underwent the procedure has cease to exist and once the energy is reconverted back into matter, that individual will be replaced by a replica who will literally be born into existence.


What is your opinion?

Whitestar

bluto answered on 04/15/04:

Hi whitestar....

Very thoughtfilled post, and more interesting points. Here are my thoughts:

Not to begin on a negative note, but I'm progressing thru your post chronologically. In some cases I do not agree that deductive reasoning can always be relied upon to make intellectual leaps from the technology of today to the technology of tomorrow. Here's why I have this opinion *in this case*. For a technology (theory) of such magnitude as teleportation (intact) of anything, I find it difficult to believe that it will contain any (or very many) concepts we are now familiar with. Certainly it will be built (to some degree) on laws now known to us. In my opinion it will be a leap of intellect, or a guess, or wild, outside attempt that will bridge some gap between the theories of today and this theory of tomorrow.

Classical physics did not allow for the concepts of quantum mechanics(QM), but it nevertheless was developed and is loosely founded on classical concepts.

Referring to your statement about "explosions", I'm not familiar with this term other than the standard 'creation/annihilation' operator in QM, but I don't doubt that it's been used or that you've read such a thing. If a person is turned into energy, why must the result be gamma rays? How can we assume this will be the mode of conversion in this 'new' Teleportation Theory (TT). This may be what it would be today, but does it make sense to assume it will be this way then?

Regarding destruction of the body. Again.....what do you mean by destruction? The dismantling of a human body doesn't have to imply destruction. This is the point I was making last thread and this one too......ideas of today don't necessarily apply for a theory of tomorrow. Does dismantling with the caveat of perfect reassembly imply destruction? In my opinion, not necessarily. Based on what we have available to us today I'd say yes, but in science it is often foolish to bury one's self in ruts of thought or belief. If a body is destroyed.......meaning altered in such a way as to cease normal functionality, then it is destroyed; via car accident, old age (organs aging to the point of being compromised in their functionality), gun shot, or any number of other ways someone dies. This can include the dismantling for the purpose of teleportation, if the reassembly is unsuccessful. But again, does a successful reassembly guarantee the original body (mind/spirit) will exit the teleporter?

With regard to your reference to an energy state, I'm assuming you mean the state of a body that has been disassembled and is a cloud of energy (which I also have problems with). This can be debatable too, and here's why. If a person's body were somehow able to be rendered into a ball of energy say, for the purposes of putting all the parts back together again later, wouldn't each particle of energy have to remember where it was at the time of decomposition? Wouldn't it then have to either be able to 'restart' right at the exact instant later upon reassembly? Or, would the person's body age during disassembly/transportation? Perhaps this is not worth considering anyway really. There are holes in all those ideas too. After all it assumes each particle of the body has to be handled and kept track of. It assumes each particle can continue to function on its own during this nether state.

There are a lot of questions for sure. Some worthy some not, but it's still fun to think about.

bluto

Whitestar rated this answer Excellent or Above Average Answer

Question/Answer
Whitestar asked on 04/12/04 - Theorical Physics

Hi.

Are you an expert in theorical physics?

Whitestar

bluto answered on 04/13/04:

Hi whitestar....

I will answer questions in it if I feel confident in what I'm talking about, but I am not currently signed up as an expert in this area.

bluto

Whitestar rated this answer Excellent or Above Average Answer

Question/Answer
dapromise asked on 04/08/04 - Doppler Effect

If a large dog were chasing you in the dark, how would the Doppler Effect help you?

bluto answered on 04/12/04:

Hello dapromise.....

Well, if I were being chased by a large dog in the dark, I'd rather have a big club, or at least a tree to climb!

But since that's not an option, I guess I'd opt for using sound. Since the Doppler Effect utilizes either light or sound, and the problem states there's no light, then the only other possibility is sound.

Even with this it doesn't sound very useful to me. Here's what I guess the problem is driving at. By listening to the frequency of the barks you'll be able to tell that you're either outrunning the dog, or if he(or she) is gaining on you.

If you're outrunning the dog the barks will get further apart and sound fainter, assuming he's barking at you while he's chasing you, and his (or her) barks are at regular intervals.

On the other hand if the dog is gaining on you the barks will get closer together and louder.

In my opinion it's kind of a poor analogy, but this is my take on it.

Hope it helps you, and if something isn't clear, let me know and I'll do my best to help make it more clear.

bluto

dapromise rated this answer Excellent or Above Average Answer

Question/Answer
Dathaeus asked on 04/05/04 - Biology

Hey Tom, I tried emailing you but it didnt go through.... question, are you any good at biology? I may have some questions, but I know you are great in physics, but how about high school bio? If not do you know who is the best 2-3 bio experts on Answerway who might answer within a day for most questions? Any input would be appreciated.... thanks

bluto answered on 04/06/04:

Hi Dat.....

Sorry my email didn't go thru for you. I'll send you an email, just to make sure, but I think my ISP has been doing maintenance on their web page, maybe that had something to do with it.

Thanks for the compliments....always great to get those isn't it?! Biology.....I can get away with common sense type things, but nomenclature and really knowing one's stuff in it, I'm not your guy.

I have a cousin that's a pretty good Biologist, but she's not on AW.

I'm sorry to be letting you down, but I don't know any of the experts in Biology here on AW. I don't have any good avice to offer.

I know that doesn't help you, but if there's anything else I can do ot lend a hand, all you have to do is let me know.

Good luck tho!

bluto

Dathaeus rated this answer Excellent or Above Average Answer

Question/Answer
asearls asked on 03/25/04 - Relativity and Quantum Physics

I'm attempting to learn the basics of special and general relativity and some facets of quantum physics. I would like to make contact with someone who is well-versed in both branches and could answer my questions over the course of several months, perhaps in the capacity of an online tutor at a reasonable monthly price.

bluto answered on 03/26/04:

Hello to you....

Here's what I suggest.

That you simply ask your questions and accept the answers from all the experts.

This way you can get a sample of style and content from the various expert. Perhaps then you can make your offer to the prospective expert for a more one-on-one exchange.

bluto

asearls rated this answer Excellent or Above Average Answer

Question/Answer
dapromise asked on 03/23/04 - The Nature of Sound

hello! can you please help me with the following questions? thank you!

1) Why is ultrasonic technology used? What is it used for?
2) If you strike a iron bar will it be heard more quickly if both you and the bar are under cold water, in the open air, or under hot water?
3) What is constructive and destructive interference? Desribe one way you could create each one.

bluto answered on 03/24/04:

Hello dapromise.....

1) It is used because it is an easy, inexpensive method with which to study various effects and material properties. It is used for many things. Obvious ones such as testing materials; whether it be for durability, integrity or for their properties such as heat capacities of various materials, altho that is on a very, very high frequency scale.

2) Water is a fluid, like air, like argon, like neon, etc. It is composed of molecules that are rapidly moving about. The temperature dictates what the average kinetic energy of those molecules is of that fluid, in this case water. The density of water is very interesting. It is parabolic with concavity of the curve downward for density versus temp. The density increases very slightly as the temperature rises from 0 C, to 4 C (check this to be sure, but I'm pretty sure this is correct). Then above 4 C the density begins to drop again. So, for 'normal' water temperatures, if you increase it's temp, the density decreases, and if you decrease its temp the density increases. So, to be nitpicky about this decreasing the temp of the water will increase the density of the water and thus its ability to transfer sound.

3) Constructive/destructive interference is simply how the troughs and peaks of waves coincide with each other. If, say two waves produced by two separate sources, like a firecracker for instance, go off near you, and the peaks of one reaches your ear at the same time as the other, they will "Constructively" interfere and produce a resulting peak that is twice the amplitude of either separately, assuming the two waves were equal in amplitude to begin with. Likewise, if one wave reaches your ear 180 degrees out of phase with the other, meaning that a trough of one reaches your ear at the same time the peak of the other reaches your ear, then if everything is equal, theoretically you would hear nothing. That is "Destructive interference.


Does this help?

bluto

dapromise rated this answer Excellent or Above Average Answer

Question/Answer
AliMcJ asked on 03/16/04 - Invite

Private question please.

Send me an email at same id at yahoo.com and I'll send an invite to experts group if interested (an unofficial one)

--Alison

bluto answered on 03/17/04:

Hi Ali.....

I'm sending you an email. It'll be from "hector" at infionline.net. Hector....yeah...don't ask. :)

I'm not making any promises with regard to my level of participation though. I'm lucky most of the time to just be able to read the forum posts because of my schedule. But I will try to pay you a visit there (group).

Thanks for asking!


bluto :)

AliMcJ rated this answer Excellent or Above Average Answer

Question/Answer
dayson4 asked on 02/15/04 - Vector Question

I am not exactly sure what I am supposed to do with this question?
Vector (A)is 3.00 units in length and points along the positive x axis. Vector (B) is 4.00 units in length and points along the negative y axis. Use graphical methods to find the magnitude and direction of the vectors (a) A+b and (b) A-B.

The teacher says that we do not need a compass for this class but this question seems that it needs a compass in order to answer it. Is there another way to answer this without one?

bluto answered on 02/15/04:

dayson....

Draw the (cartesian) coordinate system on a piece of paper. Give each positive and negative leg of the system equal length so you have, essentially, a giant "+" sign on your paper.

Now, scale the positive x axis up to, say 5, and do the same for the negative y axis.

Draw the x-vector 3 units long, starting from the origin, and along the x axis and put an arrow on the trailing end.

Do the same for the y-vector, making it 4 units....BUT.....begin drawing the y vector from the tip of the x-axis vector.

Let A = 3i
and B = (-4)j

Where "i" and "j" are the unit vectors along the x and y axes respectively. B is negative because the question specifically says it points along the negative y direction.

The resulting vector (A + B)....let's call it Z, is the vector from the origin to the tip of the y-vector. You can use a little trig to find its magnitude: Z = Sqrt(A^2 + B^2). I'll let you do that.
There are other ways to find this, but I'll leave it to you to let me know if what I've suggested is ok for your level or not.

For A - B, let's call that Q:

Q = A - B

-or-

Q = A + (-B)

Since we know B = (-4)j:

-B must be: -(-4)j = 4j

So (-B) must point UP, along the positive y direction. It is drawn the same as the other 'B' vector except this one is in the opposite direction. You find the magnitude of the resultant (Q) the same way as before.

Hope this help, but let me know if you are still stuck.

bluto

Question/Answer
kristi107896 asked on 02/10/04 - physics help

Hi, I just started Physics 20 and I'm not doin good in it at all and I'm really afraid of falling behind and not being able to get this. My teacher has tried to help me out but I still don't get and I was wondering if there were maybe and good links or books that I could read on my own that might help me. Right now I'm really needing help with equation manipulation, all algebra and trigonometry. Know any good links? Also links for anything else. I'd really appreciate it. Thank you so much.

bluto answered on 02/11/04:

kristi....

Here's what you have to do:

First....you are the one that has to do all the work. I'm sure you know that no one can do it for you, and a book won't tell you all you need to know.

BUT....

What you can do for yourself is, go to the library. A good, library. Find the largest one you can find (better selection). Maybe a university library, but you'd have to see of you can access books from there if you're not a student. Go to the physics section and plant yourself there. Pick up every and any book you can find that deals with the level physics class you're in...."physics 20" whatever that is. Find all the ones that look easy to read and understand. Ones that have a lot of example problems worked out. Ones that explain things clearly. Good diagrams and so on. Some might do better with one area than another, so be prepared to take home 2, 3, 4, 5... books! Sometimes math books have physics examples in them too. It might be worth while, since you're math is so poor to do the same thing with the math books...algebra.

It's a lot of work, and a royal pain no matter how you cut it, but what you need is material YOU can relate to. You can't find that any other way besides consulting various sources like this. It's slow going at first, but after a while you're get in the swing of things, and you will learn. Once you get the learning thing going, you'll see things will go much faster and smoother, but for now, this is what I recommend doing.

Also....find a tutor...as a friend to help....the more exposure you have the better, but there's absolutely no substitute for getting the information into your head other than you doing it yourself.

This website is also a good resource. Any of the math and physics experts can help you understand the concepts or show you a step in a problem or show you how to approach a problem, so don't forget about us.

Good luck...hope this helps.

bluto

kristi107896 rated this answer Excellent or Above Average Answer

Question/Answer
keon14 asked on 01/30/04 - Bernoulli's equation

Hello: I would like someone to explain Bernoulli's equation to me. In one version the equation = constant, what does constant mean? I am having trouble using it so if someone could explain it to me I would appeciate it. Thanks
Jerome

bluto answered on 01/31/04:

Hello Jerome.....

The form of Bernoulli's eq'n you're talking about, I think, is:

P + (1/2)pv^2 + pgy = constant

Where "P" is the pressure, and "p" is suppose to be the greek letter rho, which I cannot easily type here.

This form of the eq'n says that the pressure in a tube or vessel is equal everywhere. If there is a tube say, with some pressure in it, and the tube's diameter enlarges (or shrinks), and the pressure in that part of the tube is a quantity you know the numerical value of, you can then use that form of the eq'n to find velocity, density, or whatever of that fluid.

Make sense?

This is just one form of the more general eq'n:

(P1) + (1/2)p(v1)^2 + pg(y1) = (P2) + (1/2)p(v2)^2 + pg(y2)

Based on the conditions of the problem at hand, you have to tailor the eq'n to fit the problem so it's possible to find the desired quantities or see the resulting relations between the variables involved.

Hope that helps......

bluto

keon14 rated this answer Average Answer
raridon rated this answer Excellent or Above Average Answer

Question/Answer
Squall asked on 01/05/04 - Temperature Question

Hi

If something is at 0 degrees Celsius and I need to make it twice as cold, what temp do I bring it to?

Thanks.

bluto answered on 01/05/04:

squall.....

Interesting question. Since zero degrees Celsius is 32 degrees Fahrenheit. Twice as cold would be half that temp. 32/2 = 16 [degrees Fahrenheit]. This would be -8.9 degrees Celsius. Funny how units give rise to nonintuitive situations like this.

Hope it helps....


bluto

Squall rated this answer Excellent or Above Average Answer

Question/Answer
jUnOoNi asked on 10/28/03 - abt average velocity

1-If average velocity of an object is zero in some time interval, what can you say about the displacement of the object for that time interval

2-Average velocity and instant.velocity are generally different quantities. Can they ever be equal for a specfic type of motion?

bluto answered on 10/30/03:

jUnO....

1) About the only thing you can say (practically speaking) is, the initial and final positions are the same.

2) As long as the average distance per average time equal that of the single position and time value for the instantaneous velocity they can be equal.

Question/Answer
jUnOoNi asked on 10/28/03 - better than usual

Why our voice seems to sound better than usual when we sing in shower?

bluto answered on 10/28/03:

jUnO....

That's a pretty subjective question, but I'll give it my 2 cents worth.

I'd say it's mostly psychological. You *think* it sounds better. Perhaps it's the somewhat uninhibited feeling of your voice being drowned out by the running water and being enclosed in the shower in the bathroom. The acoustics are terrible actually with all the hard surfaces for the sound to bounce off of and all.

I'm guessing if it really made singing better in the shower, Pavarotti (and Carreras and Domingo) would be standing in one stage when they perform. :)

Hope that helps...

bluto

Question/Answer
jUnOoNi asked on 10/28/03 - photons having momentum?

How is it that photons that have zero rest mass, have momentum?

bluto answered on 10/28/03:

jUnO......

When a photon is at rest is has zero rest mass, and it has no momentum. It only has momentum when it is traveling with its classic velocity of 'c', which is about 3 x 10^8 m/s.

Hope that help, but let me know otherwise.

bluto

jUnOoNi rated this answer Excellent or Above Average Answer

Question/Answer
Dathaeus asked on 10/10/03 - Sound continuation...

Clarification/Follow-up by bluto on 10/10/03 9:18 am:
Hi Dathaeus.....

Sorry for the long delay....ugh!

If you assume....there goes that word again.........the impinging wave front (sound wave) strikes the object normal to its surface (at right angles), then the primary mechanisms responsible for absorbing the energy of the sound are:

* mechanical (movement of the object; deformation of its shape or movement as a whole)

* heat

* phonons (essentially crystal lattice energy waves)

* reflection

* transmission

There could be other metrics to consider I suppose, but for macroscopic (life-size) objects these are actually more than necessary to account for all the energy.

The material itself would be the main factor as to how elastic the material is. Cork would not transmit sound as readily as.....say water. There are absorption, reflection and transmission factors to consider for any material you might consider. If you have a specific material in mind I could potentially check these parameters for you, but to list them would be detrimental to the health of my typing fingers!

I'm making a huge guess that your losses would amount to about 20% or so total. Allow alot of leeway there, it may be much more than that!

I don't know of any way to create sound out of nothing. You know...the old can't-create'something-out-of-nothing business. Science can be a real stick-in-the-mud you know! Sounds needs something to create it. A disturbance of the air (gas) molecules.

Does this help?
_________________________________________________

I thought it would be better to continue like this... so u are saying that if X amount of energy was used then X would be dissipated between

* mechanical - yes, like stereo speakers, massive vibrations

* heat - heat? really? but is this due to the molecular motion of the target object that is producing heat, or is the main contact between the sound wave and the target creating the heat immediately? i.e., is heat a result of mechanical or is it a direct priduct of the sound wave inital impact... I can see another solid object hitting another creating heat through friction, but can sound actually create friction on another solid or even against another sound wave?

* phonons (essentially crystal lattice energy waves)- coooooool, I love this stuff... how does this energy compare to photons? are phonon waves somewhere bewteen photons and sound waves? and are phonons practical for use in energy for any application, theoretical or current use?

* reflection - if this is so, is there such a material that lets in more sound than it lets out? kind of like a funnel for sound, so the reflected sound off the target can be kept in more for "recycled waves" if u know what I mean

* transmission - transmission... do you mean the transmission of a new sound created by the initial sound wave hitting the target? or do u mean resultant sound from the target vibrating, etc., same concept as the heat thing above....

Thanks again... I need a thorough understanding of this as well to think of additional possibilities for a new project maybe...

bluto answered on 10/14/03:

Hi Dathaeus....

I made the mistake of just adding a follow-up to your question rather than placing it as its own answer.

Anyway.....if you look at my follow-up you will see my response to your question.

Hope it helps. :)


bluto

Dathaeus rated this answer Excellent or Above Average Answer

Question/Answer
jUnOoNi asked on 10/09/03 - why not closed loops

Why electric field lines do not form closed loops?

bluto answered on 10/09/03:

jUnO....

Actually, the reason they cannot cross is:

The electric field is a vector. Meaning it is a quantity that has direction and magnitude.

If the electric field crossed itself then it would have to have two different directions at the same time at the intersection point. Which cannot happen, and which is why they don't cross.

And.....if you rate this answer, make sure you rate ALL the answers you're given. Not just from me, but the other experts a well.....it's only fair. :)


b~

jUnOoNi rated this answer Excellent or Above Average Answer

Question/Answer
jUnOoNi asked on 10/06/03 - abt resistance

Hi All,
Will bending a wire cause its resistance to increase or decrease?
I know R is directly proportional to length, so i think it would decrease. Am i right?
Thanks

bluto answered on 10/06/03:

Hi jUnO......

Picture this:

Let's say you're riding a bike.....bicycle, or motorcycle, let's just say about 60 mph, down a fictitious road where there are rows of phone poles separated just about far enough that you can slalom down one row (zigzagging in between the poles of a row).

Picture the rows of poles on either side of you close enough that you can just comfortably weave in and out without hitting any of those poles on the sides.

Finally.....imagine now that there is a fairly sharp bend in the path that you aren't really expecting:

Question:

Do you think you could continue going the same speed without hitting a pole?

Your answer is:

"No way Jose`, I'd get creamed if I did that", and you'd be exactly right.

In order for you to make it around the curve you'd have to slow down.

Electrons aren't too much different than that simply put, however they slow down at a bend because they ARE hitting things. They are hitting each other, they are hitting atoms in the crystal lattice and all that gives rise to heat. That is what resistance would do, and that is what a bend (if sever enough) would do to propagating electrons (current) in a wire.

Hope that helps......

bluto

jUnOoNi rated this answer Excellent or Above Average Answer

Question/Answer
itz_jacki asked on 09/27/03 - Kinematics

I am having trouble with this question...
A plant growing rapidly next to a building doubles in height each week. At the end of the 25th day, the plant reaches the height of the building. At what time was the plant one fourth the height of the building?

Here is what I have, for the plant, time = 25days, height = h, but I don't know what I can do with the growth... is that the acceleration? can I say initial velocity = 0?

Please help...

bluto answered on 09/29/03:

itz jacki....


Begin by understanding what the problem is asking for, which I think it plain enough, but the one thing that is not understood is what's going on with the doubling.

Say you have something of some height 'h', and it's height doubles each day, week, month....whatever.

It's initial, at say week #0 height is h.

One week later (week #1) it's height is twice that: 2h
The next week (week #2) its height is twice what it was the week before: 2*2h = 4h
The next week (week #3) its height is twice what it was the week before: 2*2*2h = 8h
The next week (week #4) its height is twice what it was the week before: 2*2*2*2h = 16h
The next week (week #5) its height is twice what it was the week before: 2*2*2*2*2h = 32h

So....what you need to do is look for the 'mathematical' pattern. What is the little formula that will give you the 2h, the 4h, the 8h, and so on?

Well.....look at it in a more condensed way:

Week 0 (zero): Height = h
Week 1: Height = 2h
Week 2: Height = 4h
Week 3: Height = 8h
Week 4: Height = 16h
Week 5: Height = 32h
etc

If you play with this a minute or two you'll see that if you raise 2 to the power of the week # you will get the pattern you need.

Let "n" be the week #. then the doubling will be 2^n

So, to test it....after 3 days you have: 2^3 = 8

That is what we see.

After 5 days we have: 2^5 = 32

That is also what we see. You'd probably want to try more to make certain in cases like this, but for now, I will assure you that this is what is necessary.

Now that we have our formula: Height (h) after n days = 2^n -or- [h = 2^n]
We can find the height after 25 days...which will be a HUGE number:

==> 2^(25) = 33554432

What you need to know is...what is (1/4) this number. 33554432/4 = 8388608

Since you need the time at which this height is reached, you have to solve (in our formula above) for n:

h = 2^n
==> For h = 8388608
==> 8388608 = 2^n
==> ln(8388608) = nln(2)
==> n = ln(8388608)/ln(2)
==> n = 23 [weeks]

Let's check this out:

h = 2^23
==> h = 8388608 [units]


It checks out ok, so that's it. The height of this nutty plant reaches 1/4th the height of the building in 23 days. Imagine how fast the growth has to be in order to grow the remaining 3/4 the height in only 2 days!

Hope this helps, but lemme know otherwise....k?


bluto

Question/Answer
Dathaeus asked on 09/19/03 - Hey... sound query

Hey B... yea uh oh here we go again, had another light bulb in my head today.... u know when there is any contact, energy is lost through friction(i even forgot what that kind of permanently lost energy is called, used most in efficiency calcs, geez, I feel like i am back in college, with a test looming tomorrow)... so I was wondering, is the same principle true with sound? i.e., if there were some collision, and sound is produced, did part of that kinetic energy get transferred to the production of sound at all? Because I was thinking, if sound is the movement or vibration of air or the medium in which it travels, it must start with and use energy somehow, but is it DIRECTLY correlated? You must think I have this obsession with sound heh... and light I guess... but to me, it seems the key to all things related to energy can be tied to light and sound... but anyways, that is for a different time... thanks!

bluto answered on 09/20/03:

Hi Dathaeus......

Back in college? Is there a keg around close by? :D

"... so I was wondering, is the same principle true with sound?"

Exactly right. During a collision there is energy lost to a host of things, as you know. Things such as friction, heat, and mechanical to name a few.

So, the energy that might have gone into producing heat say can go into exciting the ambient air molecules, and that excitation of the air is what produces the resulting sound.

Also, depending on how much heat is liberated in the collision that too can afffect the localized region of air molecules. It may, to some degree, but probably fairly insignificantly, rarefy the air resulting in unusual pitches over the course of the collision.

Does that help??

Good question! Obsession with sound and light? Nahhhh.....just think you like to wonder and think about things.

bluto

Dathaeus rated this answer Excellent or Above Average Answer

Question/Answer
zaurus asked on 09/03/03 - Rebounce direction of a moving body

How to determine the direction of a moving body after it bounces on any irregular surface?

bluto answered on 09/03/03:

zaurus....

The short answer: trial and error.

The longer answer:

Depends on the size of the irregularities, type of surface, the size of the ball, the material of the ball may possibly also have some effect. In other words a hard ball might rebound differently than a soft 'forgiving' ball.

It's not a cut and dried answer.

Does this help you?


b~

Question/Answer
Squall asked on 08/26/03 - Energy units

What exactly is a calorie? Why are they so bad?

Thanks

bluto answered on 08/26/03:

squall....

A Calorie is a unit that measures energy. Technically it's defined as:

The amount of energy, or heat necessary to raise 1 gram of water from 14.5 degrees Celsius to 15.5 degrees Celsius.

The trouble I think you're referring to is when people don't 'watch their calories'.

People eat food to get energy. Foods with any nutritional substance has the ability to give the body a certain amount of energy. This energy is then used by the body to nourish muscles, bones and so on.

What happens when the body has more 'fuel' than it can use is it stores it, and it does so in the form of fat. In our land-o-plenty eating is more a pastime than for survival. So one of the reasons many people are obese because they are taking in more calories...energy, than they are using up.

Hope this helps....

bluto

Squall rated this answer Excellent or Above Average Answer

Question/Answer
Spider asked on 08/21/03 - Decelerating the hips and body pivot?

A user in another forum reports that in order to correct push/slices his pro has told him to slow down the pivot and use his arms more aggresively.

I disagree. Any thoughts here?

Thank you

bluto answered on 08/21/03:

Hello spider...

I'm not a golf enthusiast, so am not familiar with a "push".

Can you in general give a little more info on that and to what "pivot" you're referring to? I'm assuming you mean a hip pivot, but not knowing what you mean here I can't help you with your question.

Thanks ---

b~

Question/Answer
Spider asked on 08/02/03 - Force and motion vectors effecting the flight of a golf ball

Layman here so please consider same.

I am assuming from Newton's Second Law that a golf club striking a ball will produce acceleration to the ball in a direction (vector?) the club (force) is traveling at the point of impact. And that if the face of the golf club is aligned perpendicular to the path of the club, the ball will fly straight.

Discounting the Venturi effect, what physical elements/laws come into play affecting the vector when the clubface is angled to the direction of the force/club?

The question is prompted from discussions in the Golf Digest Instruction Forum. Some are in effect saying that the 2nd law only operates in relationship to the magnitude of the force. Example: A golf ball struck with an open face putter simply will veer to the right and is not affected by the direction of the force of the putter head.

It does not make sense to me to believe that the 2nd law has arbitrary application, that is, depending on magnitude.

Any help here will be most appreciated and please reduce any responses to my unsophisticated level.

Thank you

bluto answered on 08/02/03:

Hi Spider....

Essentially you are correct. It would have been a good idea to include the link to the golf discussion you're talking about, or include the text of that discussion if not too lengthy.

I'm not going to go into the math here. I'm sure you'll be a-ok with that (if not lemme know).

Newton's 2nd law: Acceleration is proportional to the net applied force and inversely proportional to the mass.

a = F/m

No reason to go into all kinds of nitty-gritty detail with this definition, but essentially:

If you have a golf ball sitting on a tee, and there's no unusual 'smilies' (cuts and gouges) in its surface...

...and you have a 'clean' club-face, which you can consider to be a plane, that is exactly perpendicular to the direction of travel at the moment of impact. The ball will exit the club-face in that direction.

The Venturi effect, or whatever else has nothing whatsoever to do with it. Once the ball is struck, you are at the mercy of whatever mother nature, or chance throws at the ball anyway, so those things are completely irrelevant here.

Now......put an angle on the club-face. What happens? Nothing, except:

The ball will now travel in a direction that is the trigonometric factor of the total force perpendicular to that direction, whether it be left or right, or up and down.

So, the NET force is LESS due to the angle, whether it be parallel to the ground, or at some angle, like a 5 wood has a nice loft angle that I've always liked. Another way to look at it is, the ball will (initially) travel in a direction defined by the center of the ball and the point of impact, and the club-face is angled to change this point of impact to aid in loft or directional control.

Does that help?


bluto

Jim.McGinness rated this answer Excellent or Above Average Answer

Question/Answer
Rascal asked on 07/16/03 - Acceleration in G's...

Im a drag racer having some trouble with acceleration in G force mathin drag racing we have several measured points along the track to judge performance.
One of these points is the 60 foot markwherein the only measured data offered is 60 ft ET or elapsed time it has taken to accelerate from a standing stop to a distance of 60 ft. Obviously the shorter times represent better performanceor acceleration.
Some enthusiasts have taken to calculating acceleration in G force based upon this information, most likely for no reason other than to impress themselves.

The formula used for this calculation is as follows:

Using this formula (from physics):
Xf = Xi + Vi*T + .5*A*T^2

Where:
Xi -> Initial Distance
Xf -> Final Distance
Vi -> Initial Velocity
Vf -> Final Velocity
T -> Time
A - > Acceleration

Here's the info we have
Xf -> 60
Xi -> 0
Vi -> 0
T -> Input from user (Let's say 2.0 for our example)
A - > Acceleration

This make our formula
60 = 0 + 0*T +.5*A*T^2
60 = .5*A*T^2
120 = A * T^2
A = (120 / T^2)
A = (120 / 2^2)
A = (120 / 4)
A = 30 ft/sec
G's = (30/32) => 0.9375 G's

Now that we know the acceleration we can solve for Velocity at distance.

Formula used -> Vf = Vi + A*T

Where:
Vi -> Initial Velocity
Vf -> Final Velocity
T -> Time
A - > Acceleration

Our Numbers:
Vi -> 0
Vf -> Solving for
T -> 2.0 sec
A - > 30 ft/sec squared

Vf = 0 + 2 * 30
Vf = 60 ft / sec
Convert to mph -> 60 *(3600/5280) => 40.91 mph

I feel this formula to be a misrepresentation of acceleration.at least when applied to acceleration perpendicular to the force of gravityas compared to acceleration against the force of gravity.
I am in no way a mathematician or physicist so forgive my ignorancebutwhen I stop and think of physics calculationsaccelerationand how this stuff may typically be appliedI wind up asking myself if G force us typically used to evaluate gravitys effect on moving objectssuch as bullets and suchat least thats how I remember it

I guess that would make most G force calculations relate to decelerationrather than accelerationand in that case we would probably be talking in terms of less distance traveled in a givenunchangingtimeframe?

Or maybe I simply do not understand gravitywhich is entirely possiblelolbut looking at 32.174 feet per second squared has me wondering just where the constant comes from.

I imagine someone dropping an object to the earth and measuring its speedin other words the object fallsor is pulled by gravityat a speed of 32.174 ft per second

As I recall all objects fall at this speed unless other forcessuch as aero resistanceimpede movementthe old feather vs bowling ballin air vs a vacuum chamber

As Ive seen it put an object will accelerate to 32.174 ft per second under the influence of gravitybut Im starting to see this as misleading

I do not believe acceleration is really at work under the force of gravity...and that an instant and constant speed of 32.174 ft per second is really all we get from gravity

Anything else would imply that there are variables involved.and there may be as distance from the source of the standard (earth) is increasedbut were driving on the standard and shouldnt be concerned

So to meif gravitys pull is a constant forceat least on the earths surfacethan acceleration is not truly a component of gravityand all were really trying to do is calculate whether or not we are achieving or exceeding this established force

This leads me to understand that if we were to be working with an object traveling directly away from the source of gravitythe earths surfacesay a rocketanytime we achieve or exceed 32.174 ft per secondwe have achieved or exceeded the force of gravity

But if we are traveling perpendicular to the source of gravitythings get trickyand very confusing.

We now do not actually have a force working against usbut were trying to use this force as a benchmark to judge accelerationso we kind of need to create itand since were trying to determine acceleration valuesthen acceleration we should be looking at/forand thats why I see a literal description of ft per second per second as a worthy value

We already have our benchmark speed of 32.174 ft per secand can square the time in seconds to represent a factor of resistance against accelerationwhich still gives us a one to our 32.174 ft per one second speedgiving us gravity in a horizontal direction
Its when we change factors representing acceleration in our equation things get confusingand the standard use for G force math becomes challenged
Using per second per second as a true measure of accelerationin relation to our newly laid out gravitational pullwould have us altering either distance traveled in the given time measurement of ONE second squared.or altering the squaring by entering the changes in time (60 ft ET) into our calculationright?

Please help! Thank you! Rascal

bluto answered on 07/16/03:

Hello Ras......

It looks like you have two dandy answers already. Posted while I was typing away. So, rather that throw away what I've written I will post this anyway. Normally I don't bother doing something like that, but it's possible you may find information in what I've said that will be helpful to you. If not, then you are no worse for the wear except for having to have read it! :)

Interesting question!! For a drag racer you know a lot about science. If you don't mind my trying to guide your thinking to a slightly more proper level about acceleration and g-force, perhaps you will understand what is going on a little better.....I hope.

First of all the formula you used is only going to be really meaningful if the acceleration is a constant value, which it is not, over the time period for which it is measured. You know the beginning time (zero) and you're recording the time at a distance of 60 ft.

This is like having, say two people measure your average run speed for a race by taking a speed measure just out of the box, and then, say one foot before you come to a complete stop. You'd have a very *slow* average speed wouldn't you?!

What you'd need here is many measurements along the entire run. That's what you'd need to get a measure of your true acceleration and then data-fit a best-fit curve to that data.

The formula you use assumes the acceleration does not change, in which case you would be getting a good representation of your acceleration using that formula. Here's another example: Say it's 15 miles for you to work one way. If it takes you 30 minutes to get to work, which is mighty reasonable situation around here, then that assumes you drove at 30 mph! On *average* that *is* your rate, *but* what this value takes into account is stop signs, traffic lights, traffic slow-ups, etc etc. All this goes into that value 'behind the scenes'. Now, if you took a stopwatch and recorded your speed every quarter mile or so, and plotted the data you'd see high's and low's. Where you stopped, and where you sped up. You could then use a little more math to find a better average.

Make sense?

The value of the gravitational constant is just a number. It's been verified and re-verified ad nauseam over the decades. It is a value that is *not* exactly the same over the surface of the Earth, and that is because the earth is not a perfect sphere, by and large, and it will vary slightly from mountainous areas versus flat, sea-level areas. This value is taken to be constant for most people however, without worry of significant error. So, as far as you are concerned, its value is 9.8 m/s^2, or 32 ft/s^2. It makes no difference if you are going up, down, left, right, inside, outside, this way, that way, or some other way. If the date you plug into the formula are accurate and representative of the quantities in question then the value you get for acceleration is what you get.

In your case, and in the case of your associates of yours into figuring out their g-force values, they probably would not benefit from taking wind resistance into account. If you did, you'd have to go to the trouble of knowing just what the behavior of the wind resistance is to the profile of your car. In other words does the wind resistance vary proportionally to the profile of your car, or is it exponential, logarithmic, or whatever.

Ok, I'll stop here and see how this sets with you. Hope this helps some.

bluto

Rascal rated this answer Excellent or Above Average Answer

Question/Answer
Squall asked on 06/25/03 - Watts, Volts and Amps

Could somebody explain (in laymans terms) the difference between a Watt, an Amp and a Volt. If you have time, could you then get a bit more technical. I'm a biologist and use a powerpack for stuff and it would be nice if I actually knew what the numbers on the power settings really meant!!!!!!

bluto answered on 06/25/03:

Hello squall.....

A Watt is a unit of POWER. It means AMPS multiplied by VOLTS....it's that simple.

Someone asks, "How much 'power' does that device use?"....the proper response is 'blah blah blah' WATTS, or Kilowatts.....or Megawatts.

Make sense?

An AMP is a unit that refers to some amount of charge flowing per unit time....or, a flow of current.

A VOLT is a unit that refers to how much 'push' a given amount of current has 'behind' it to *make* it flow. Voltage is not what is dangerous as most people think. It is current which can be lethal. However, when one sees a sign that says "High Voltage" its can be rightfully assumed that there is enough current present to be lethal for that advertised voltage, but technically it is the current that is has the potential for damage.
---------------------------------------------

More technically:

A Watt (W) = current*volts
==> P = IV

I = most common symbol for current (Amps)
V = most common symbol for voltage (Volts)
P = Power, the most common symbol for this
unit (Watt).
---------------------------------------------

An Amp is defined as the flow of one Coulomb of charge per second.

Where: 1 Coulomb = 6.242 x 10^18 electrons

So, 1 Amp = (6.242 x 10^18 electrons)/second

While seemingly arbitrary, this value is determined by experiment to be the number of electrons that flow in a time of one second.
---------------------------------------------

A Volt is defined as energy in Joules per Coulomb of charge:

V = J/C

J = Symbol for Joules
C = Coulomb of charge as discussed above.

Sometimes the volt and amp are defined together as:

A Volt is that potential difference that occurs when there is a current of one Amp that flows through a resistance of one Ohm.

This sounds circular, but when one defines a quantity like this, it makes sense to set forth definitions in this way. But if you think about it, if you have the ability to 'push' electrons (charge) through a wire (which has some resistance), then as long as you get 6.242 x 10^18 electrons flowing in a time of one second, that that amount of 'push' can be called the Volt, and that much current can be called a Coulomb, and that much charget flowing in that much time can be called an Amp of Current.

Do you see?

bluto

Squall rated this answer Excellent or Above Average Answer

Question/Answer
tt_turbo asked on 05/16/03 - work and potential differences



A line charge of "lambda" of length "L" lies parallel to an infinite sheet of surface charge "sigma". the line charge is seperated from the infinite sheet by a constant distance "a". find the work needed to rotate the line charge so that it is vertical with respect to the infinite sheet of charge.(clockwise rotation, please..90 degrees)

Thanks guys.
(TBT)

(I think that you find the potentials for the two configurations of the line charge(Horizontal, then vertical), subtract them, and W=qV...not too sure)

bluto answered on 05/26/03:

turbo.....

I have to apologize again. I thought I'd be able to get to your question but I won't be able to.......sorry. I'll be out of town all week and won't be available to answer your (or anyone's questions until sometime next week mid week at best. I'm very sorry for this inconvenience to you, but this happens from time to time. If you wish to wait....that's fine, if not, that's understandable.

bluto

tt_turbo rated this answer Average Answer

Question/Answer
Dbestskills asked on 05/13/03 - Physics Questions

Hi Bluto, I am using copied problems from a text book to help prepare for my physics test (and I have no solutions), but I am having problems with four of them. Can you please answer these questions as soon as possible, my test is tomorrow, and my teaacher said that these types of problems will be on the test. Can you please as detailed or step wise as possible show me how to do them. Thank you very much.

1) Find the equilibrium temperature of the system when 10 kg of boiling water is poured into a container with a 10 kg of ice at -20 degrees C. Explain your answer in detail. (Ignore the heat loss to the environment).

2)A 0.15 kg toy is undergoing simple harmonic motion on the end of a horizontal spring with a force constant k=300 N/m. The object has a speed of 0.300 m/s when it is 0.012 m away from its equilibrium position. What is a) the total energy of the object at any point of its motion b) the amplitude of the motion, and c) the maximum speed attained by the object during the motion?

3) The horizontal beam weighs 150N and its center of gravity lies at its center. Find a) the tension in the cables b) the horizontal and vertical components of the force exerted on the beam at the wall.
Ii
I_i
I__i
I___i
I____i
I_____i 5.0 m
I______i
I_3.0 m_i
I________i
I_________i
IIIIIIIIIIIi (horizontal beam)
I___4.0 m__i
I__________i
I__________i
I____(300 N barrel)
I
I
I
I
i= cables

4)A block weighs 25 N. It appears to weigh 17 N when immersed in water (and suspended from a spring scale - see diagram) and 20 N when immersed in another liquid of unknown density. (Density of water = 1000kg/m^3)
(see Figure 1 & 2)

A) Find the volume of the object


B) Find the density of the unknown liquid.


C) Before you submerge the object in water, you place the container of water on a scale. The scale reads 100 N. When you immerse the object in water, is the scale reading greater than, less than, or equal to 100 N? (Assume that the object does not touch the sides of the bottom of the container at all). Explain your reasoning. (see Figure 3)

Hint: Decide whether Newton's 3rd law is helpful here.


I___(scale 17 N)____I
I________i__________I
IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII
I_water__i__________I
I________i__________I
I________i__________I
I________i__________I
I_____(block)_______I
I___________________I
I___________________I
IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII
Figure 1

I____(scale 20 N)___I
I________i__________I
IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII
Iunknown_i__________I
I_liquid_i____________I
I________i__________I
I________i__________I
I________i__________I
I_____(block)________I
I__________________I
I__________________I
IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII
Figure 2


I_____(scale )______I
I________i__________I
IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII
I water__i__________I
I________i__________I
I________i__________I
I________i__________I
I________i__________I
I_____(block)_______I
I___________________I
I___________________I
IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII
_____HHHHHHHH_______
________HH__________
________HH__________
(scale origionally reads 100 N, before wieght is placed in water))

Figure 3

bluto answered on 05/14/03:

Hello Dbest.....

Before I do anything else with your question:

There's no way I'll be able to answer all these questions by today just because of the short notice. If you still want me to try and address them, I'd suggest putting them in descending order of importance. At least you'll maximize the situation.

b-

Question/Answer
ehsanhb asked on 05/10/03 - definitons

hey,

can you please give me the defintions for the following words:

Electron Voltage

Binding Force


Thanks a lot,

Ehsan

bluto answered on 05/11/03:

Hi ehsanhb......

An Electron Volt is a defined quantity in physics that means, formally:

The energy gained by an electron as it moves thru a potential difference (voltage) of 1 V.

So what does that mean?

As an electron moves thru, say a potential difference of 500 V, it will LOSE 500 eV of potential energy, but GAINS 500 eV of kinetic energy. Which is the connection to volts and energy.

Also, 1.602 x 10^(-19) is the amount of charge on the electron, and the change in potential energy is equal to qV (sharge times voltage)....therefore:

1 eV = 1.602 x 10^(-19) J


Binding Force:

This is usually referred to as Binding Energy, not force. I *think* you're talking about energy here, so I will answer based on that assumption.

Electrons that surround an atom have very specific orbits for which they (the electrons) need a certain amount of energy to maintain. In order to remove an electron from an atom it requires a certain amount of energy to be delivered. Electrons that are very tightly bound require a greater amount of energy than those that are more loosely bound to a nucleus.

Atomic physics is the area of study that deals with these outer atomic electrons.

Hope this helps, but lemme know if you have other questions.

bluto

ehsanhb rated this answer Excellent or Above Average Answer

Question/Answer
Navspert asked on 04/26/03 - Reaction Heat

Dear Bluto,

I just had this puzzling question about heat and since you are a physicist, maybe you can help me out:

In the following chemical reaction 2H2+O2-->2H2O, there is a lot of heat that is evolved. Where does all of this heat come from? Please be detailed in your answer.

Thank you Very much!!

bluto answered on 05/01/03:

Hi Nav.......

Sorry for the delay. Sometimes I get too busy to get to my questions as fast as I'd like, so.....sorry for that!

When atoms combine or break apart there has to be some kind of exchange of energy. Usually this energy has to do with heat. Some chemical reactions require energy to break them apart into smaller molecules. For example, you can break water apart into it's hydrogen and oxygen constituents by a process called electrolysis. The 'energy' is electrical, sort of and in this process the oxygen and hydrogen are broken apart and they form their characteristics gaseous states. It's really kinda cool how it works.

Some reactions will actually liberate energy, or heat when they occur. If you mix vinegar and baking soda together you will be able to detect a change in temperature. This is a mild example of the reaction you gave, and both are called "Exothermic" reactions because they give off heat during the reaction. The ones that require heat are called "Endothermic". Sometimes with certain reactants you can feel the container get very cold to the touch. This is because as the reaction takes place it is stealing the heat from the surrounding solute (water or whatever) to make it run. When the heat source is used up, the reaction slows and stops.

Some reactions like the one I mentioned above with the vinegar occur slowly. The one you gave occurs very fast, and that is what an explosion is.....a very, very rapid burning. The mechanisms and details are complicated, but in essence the result is the same.....reactants combining together extremely fast and then self-perpetuating until the fuel is used up.

So, in the reaction you gave, it would probably be a little more accurate to write is like this:

2H2 + O2 --> 2H2O + heat

This is essentially the kind of reaction that happened in the Hindenberg fire many years ago and the tragic Challenger explosion some year ago. Just to give you an idea of the magnitude of the power involved here.....if you took, say 1 kg of hydrogen, and 8 kg of oxygen, mixed them together so they'd 'react' you would liberate enough energy to light a 100 W light bulb for almost 3.5 weeks! This is about 2 x 10^8 joules of energy.

I hope this answers your question, and again....sorry it took this long to answer you.

bluto

Navspert rated this answer Excellent or Above Average Answer

Question/Answer
cyberbatt asked on 04/26/03 - Air resistance

In solving problems like a ball is thrown up into the air and than it falls back again onto the ground in Physics, does air resistance exist when the ball is being thrown up? If yes, how does it affect its motion?

bluto answered on 04/28/03:

Hi cyberbatt....

Kewl nickname....


Anything that moves thru the air is subject to air resistance. For some things however it may be negligible. A large object (like a bowling ball) being dropped from a ladder, or even a building probably wouldn't be influenced too much by air resistance. Tossing a baseball to a friend probably wouldn't either. Nor would tossing a ball up into the air and catching it on the way back down. The reason for this, in this case, is because the mass of the object is great enough for the relatively short fall that the force due to air resistance just isn't significant enough to impede the motion of the ball.

Dropping a baseball, bowling ball or just about anything else for that matter from say, an airplane would be affected by air resistance because of the much longer distance it would fall. The ball would travel much further and much faster than in the previous scenario.

Actually it's not the weight of the object that governs the magnitude of air resistance anyway....it's the shape. Remember Galileo's experiment with dropping a bowling ball and a feather in a vacuum? They both reach the bottom at the same time when dropped from the same height at the same time. They both fall at the same RATE, without the influence of the air. An object that has alot of open surface area will be impeded more by air resistance than by an object that is compact and streamline....like a bullet or a ball.

As the object falls it is hitting air molecules and water molecules, dust particles and who knows what all else. The higher an object is dropped the longer gravity has to 'pull' it back to Earth, so it (the object) wants to go faster and faster. The faster it goes the more 'stuff' it has to pass thru on the way down, which serves to slow the object.

Think about it like this:

If you drop a baseball from a very high ladder thru one piece of taught newspaper it will punch right thru with no trouble....right? But if you drop it from the same height and have it pass thru say.....a dozen pieces of paper spaced say...a foot apart....the ball with slow down quite a bit. It's a crude analogy, but each piece of newspaper is like a layer of air molecules. Each one slows the ball a little bit, and a little bit more, and a little bit more, eventually enough that it slows it down.

I hope this helps you understand what's going on a little better, but if you have more questions.....let me know.

bluto

cyberbatt rated this answer Excellent or Above Average Answer

Question/Answer
jenny96 asked on 04/08/03 - electricity

can you also clarify these 4 questions for me please??

1) a negative charge of 2.4 x 10^-6 C experiences an electric force of magnitude 3.2 N

acting to the left
a) what is the magnitude and direction of the electric field at that point. ans: 1.3 x

10^6 N/c right
b) what is the value of the field at that point if a charge of 4.8 x 10^-6 C replaces the

charge of 2.4 x 10^-6 C

a) the formula is E = Fe/Q right?? and u jsut sub in the numbers?
for b) what does it mean..to replaces the charge?? can you explain that?

2) at a certain point P in an electric field, the magnitude of teh electric field is

12N/C. what is the magnitude of the electric force that would be exerted on a point

charge of 2.5 x 10^-7 C, located at P.
ans: 3.0 x 10^-6N

so E = Fe/q
and 12 is the E right?? so you multiply am i correct?

3) what is the magnitude and direction of the electric field at a point 3.0 m to the

right of a positive point charge of 5.4 x 10^-4C.
ans: 5.4 x 10^5 N/C right
the formula for this is E = kq/r^2 rite?? and you plug in what you know. am i correct?

4) what is the magnitude and direction of the electric field at point Z due to the

charged spheres at points X and Y

X---------Y-----------Z

X has a charge of 50.0 uC
Y has a charge of -10.0 uC

distance between X and Y is 0.45m
distance between Y and Z = 0.30 m
distance from X to Z = 0.75m

ans: 2.0 x 10^5 N/C right

Can you explain that one for me please? Thanks!!

bluto answered on 04/08/03:

Hi jenny96.....


[1]

a) F = QE

F = Force = 3.2 N
Q = Charge = -2.4*10^(-6) C
E = Elec. field

==> E = F/Q = 3.2 N/2.4*10^(-6) C
==> E = 1.3*10^6 (N/C)

The force is positive, indicative of an 'attractive' force. It's a matter of convention really....not written in stone. therefore, this force draws the charge to it, which would be toward the right.

b) Yes....all you have to do is replace the other charge with this one, and do the arithmetic. Let me know if you have trouble....k?


[2]

a) F = QE

F = 12 N/C
C = 2.5 x 10^(-7) C

==> F = QE = (12 N/C)(2.5 x 10^(-7) C)
==> F = 3*10^(-6) N

Make sense?


[3] You're correct here, and you will get the answer given if you do as you said. Let me know if you have trouble.


[4] Elec. fields add together along lines connecting the centers of the individual charges. Here, everything lies along the same line. So the resultiong electric field E is given by:

E = e1 + e2 (sum of the two fields of the two charges)

e1 = kq1/(r1)^2

e2 = kq2/(r2)^2

q1 = 50 uC = 5.0*10^(-5) C
q2 = -10 uC = -1.0*10^(-5)
r1 = 0.75 m
r2 = 0.3 m
k = 9*10^9 (Nm^2/C^2)

Therefore:

e1 = (9*10^9)(5.0*10^(-5))/0.75^2
==> e1 = 8.0*10^5 N/C

e2 = (9*10^9)(-1.0*10^(-5))/0.3^2
==> e2 = -1.0*10^6 N/C

Finally:

E = e1 + e2
==> E = 8.0*10^6 N/C + (-1.0*10^6 N/C)
==> E = 8.0*10^6 N/C - 1.0*10^6 N/C
==> E = -2.0*10^5 N/C

So...you were right, but I wanted to show you the work on this one.

Hope this helps.


bluto

jenny96 rated this answer Excellent or Above Average Answer

Question/Answer
barhoooooom asked on 03/20/03 - charge carriers

hi,
I taking a physics II course now, and the professor is very bad, I cant understand anything from him, anyways, he asked us as a homewrok to submit a an A4 paper explaining how to calculate n (charge carriers) for an element if we new the atomic structre of that element, he wants us also to calculate n for Al and Cu, actually I dont understand the question, and I also cant fined any explanaition in my book about it. can u please explain to beriefly what he wants and what this n is? I would also be greatful if u guided me to any sites where I can read more about this.
I have to submit the homework after tommorw, so I would also be greatful if u hurry up.
thanx
bye

bluto answered on 03/21/03:

bar.....

Cool nickname!

My schedule has been so busy that I've been going thru declining almost all my questions....which I hate doing, but so you aren't sitting around waiting....you can either do just that....wait (till early/mid week next week) or you can pose your question to another expert. Normally I'm not tardy getting back to people, but now and again it happens, and my apologies for that. I can helkp you with your question, but I don't have the time right now. I understand your confusion with this question......it's tricky when you first learn how to do it.

bluto

barhoooooom rated this answer Above Average Answer

Question/Answer
cyberbatt asked on 03/18/03 - Motion

A bullet of mass 8.00*10^-3kg moving at 320ms^-1 penetrates a target to a depth of 16.0mm before coming to rest. Find the resistance offered by the target, assuming it to be uniform.

I get the answer for this question but I am rather puzzled by some parts in the answer which I get.

F=mv
=8.00*10^-3*320
=2.56N

v^2=u^2 +2as
0=320^2+2a(0.016)
a=-3200000m/s^2

F=ma
=8.00*10^-3*.3200000
=-25600N

Resistance force: 2.56N-(-25600)N
=25.6kN

How come the force gotten in the second part is negative? Does it mean that the bullet is moving backward or is it my answer to this question is wrong?

bluto answered on 03/21/03:

cyber......

My humble apologies for the long delay getting back to you with your question. Been SO busy, and in fact I've been going thru declining questions because I just can't gte to them right now. For now.....let me just say this....and maybe this will help, but if not, send me the question again. Altho I won't be able to get to it until next week...unfortunately. But...don't forget Newton's 3rd law....for every action there's an equal and OPPOSITE reaction.

Also....don't forget Hook's Law (F = -kx) where you can get positive and negative forces depending on how you define things.

If this helps...GREAT....if not, and you want to take a chance sending me another question......please do so. I promise I'm not usually tardy like this.

b-

cyberbatt rated this answer Excellent or Above Average Answer

Question/Answer
KoolPeace7 asked on 03/05/03 - Electric Circuits II

Since the first one is being too messy for me, here's the updated question:

If you can help with even one of these questions, that would be great! I tried some of these. If an expert answered one of these already, please don't answer that one unless I say that it still doesn't work out. Also, I might add clarifications in any chance that I figure out some of the questions saying that I figured it out because that happens. However, any help is very appreciated.

(1) http://www.webassign.net/hrw/28_41.gif
(a) Calculate the current through each ideal battery in Amps. (As a sign convention, assume the currents are "up" through each battery.)
I1-
I2-
I3-
Assume that R1 = 0.9 ohms, R2 = 2.2 ohms, emf1 = 2.0 V, and emf2 = emf3 = 3.5 V.
(b) Calculate Va - Vb






(5) In http://www.webassign.net/hrw/28_44.gif, emf1 = 3.00 V, emf2 = 4.00 V, R1 = 3.00 ohms, R2 = 5.00 ohms, R3 = 5.00 ohms, and both batteries are ideal.
(a) What is the rate at which energy is dissipated in R1 in W? In R2? In R3?
(b) What is the power of battery 1 in W? Of battery 2?






(6) In the circuit of http://www.webassign.net/hrw/28_46.gif, emf has a constant value but R can be varied. Find the value of R that results in the maximum heating in that resistor in ohms. The battery is ideal.

Please answer by 6PM Thursday, March 13, 2003. Thanks!

bluto answered on 03/06/03:

Hi Kool....


Here's #6:


Interesting question actually. You can argue the whole thing actually without doing any math at all, but I'll do some of the math for you anyway so you can see how things fit. I know this is long...I did that on purpose, cuz it a good 'thinking' problem. If I guessed wrong about all the detail.....then my apologies, but I took you as someone that would appreciate the info.


=====================================================================================================================
First let me argue my way to the answer:


The 2 Ω resistor is meaningless.....forget about it, it's in the way. Instead, jump to what will be some voltage applied to the parallel resistors. This voltage will push current thru each resistor. Since the one is fixed at 5 Ω, all you can do is adjust "R" to get max current (which means max heat). In order to divert most of the current away from the 5 Ω resistor, you need to make "R" very small. I'd say the smallest non-zero value permitted by the unwritten rules of your question. That is, in a nutshell the answer.
=========================================================================================================================


The problem essentially boils down to the fact that you will have some voltage at the junction of the two parallel resistors. The value is immaterial unless you want, or need values, but since there are none provided that can't be done anyway.


The total resistance of the circuit is: RT = (10 + 7R)/(5 + R) Ω


So, the total current in the circuit will be: (IT) = V/(RT), where V is the applied voltage of the battery. Which becomes:


(IT) = V[(5 + R)/(10 + 7R)]


The voltage drop (Vd) across the first resistor (in series with the battery and the parallel resistors) will be:


Vd = (IT)(R)
==> Vd = V[(5 + R)/(10 + 7R)][2 Ω]


Therefore, the available voltage (Va) to the parallel resistors will be the difference between this voltage drop (Vd) and the applied voltage (V):


Va = V - Vd
==> Va = V - V[(5 + R)/(10 + 7R)](2 Ω)
==> Va = V[1 - [(5 + R)/(10 + 7R)](2 Ω)]


This is the maximum voltage the parallel resistors will see.


So now what.....


Like I said, we didn't need to go thru all the math to make the point, but I figured you'd need it, or some of it anyway, which is why I included it.


Anyway.....now we can focus on what's really the only thing of importance....the parallel resistors.


You want to maximize the heat thru "R". What does that mean? It means that you need to maximize the current in that resistor in order to generate the heat....right?


So...if you have one resistor in parallel with another.....what happens to the current when a voltage is applied? The voltage will push the current thru each branch by an amount that is proportional to the resistance in that branch.


If you make "R" equal to 5 Ω like the other resistor the current will split equally between the two. If you make "R" greater than 5 Ω it will cause more of the current to pass thru the 5 Ω resistor. So....that tells us that "R" has to be LESS that 5 Ω .....right?


So...by how much less?


Well......think about it. What if you made it zero? That's less that 5 Ω isn't it? But there is no such thing, in this context, as a zero Ω resistor, so that idea's out the door. It can't be 5 Ω and it can't be zero Ohms. So...the answer is they're looking for I guess is...the smallest non-zero resistance you can place there. In essence....you have to short-out the 5 Ω resistor!


So help illustrate this a little better mathematically:


Let the current thru the 5 Ω resistor be "i".


-and-


Let the current thru "R" be "I".


Then, by Ohm's law:


i = Va/5 Ω and I = Va/R


Since the resistors are in parallel and a different current goes thru each, we can look at them separately like this (sort of).


Since the 5 Ω resistor is fixed, as is the voltage, its current is fixed. But the other one is not. We can tweak that one. It's current: I = Va/R


The only way to maximize "I" is to minimize "R"....which is exactly what I've been saying....even without the math.


To put it still another way:


Since you need "I" to be larger than "i", set: I > i


==> Va/R > Va/5 Ω
==> 1/R > 1/5 Ω
==> R < 5 Ω


So...again, we get the same information...not alot but it's telling us what common sense tells us, which is a good thing.


One more thing, and then I think we can declare this one dead-ass horse I've beaten on.


The current in each branch of the parallel resistors will sum to equal to the total current in the circuit. In other words:


Total Current (IT) = I1 + I2


[sorry about all the changing notation, but I'm not trying to relate any of the notation together so much as just making local points as I tell you all this]


IT = Va/5 Ω + Va/R
==> IT = Va[1/5 Ω + 1/R]


You want IT to be as large as possible...to maximize the heat in "R". That means you have to add as much as you can to ŕ/5 Ω"....right? In order to do that......you have to make R as small as possible. Since you can't make it zero....making it as small as you can will make "i/R" as LARGE as possible, which will in turn make IT as large as possible.


Ok....the horse is dead now....hope this helps and wasn't too confusing. Lemme know if you have any concerns.


bluto



KoolPeace7 rated this answer Excellent or Above Average Answer

Question/Answer
jenny96 asked on 02/28/03 - hi

When you stretch a rubber band. how do you find the velocity? How do you find the work done by hand? how do you find the force displacement graph? how do you calculate the energy when you stretch the rubber band and then energy when released?
and how do you design a catapult? please help me out.

bluto answered on 03/03/03:

Hi jenny96....

You've asked 5 questions:

1) How do you find the velocity?
2) How do you find the work done by hand?
3) How do you find the force displacement graph?

4) How do you calculate the energy when you stretch the rubber band and then energy when released?


5) how do you design a catapult?
------------------------------------------------------------------------------------------------------------------------

Most of these are too general for me to answer specifically. So I will answer as best as I can based on the information you provided.

1) Not sure what the situation is here. Generally the speed of a spring, or rubber band, for small displacements is given by:

(1/2)kA^2 = (1/2)kx^2 + (1/2)mv^2 [total energy of the spring or rubber band]

Where:

A = total displacement of the spring/band
x = intermediary displacement
k = spring const.
m = mass
v = velocity

==> v = Sqrt[(k/m)(A^2 - x^2)]
--------------------------------------------------------------------------------------------------------------------
2) Work is defined, as I said in a previous response, is force multiplied by distance:

W = Fd

Once you know what the force is, and the displacement "d", you can find the work.
--------------------------------------------------------------------------------------------------------------------
3) I'm assuming what you mean here is the force vs. displacement.

Hook's law says: F = -kx

F is the force, and "x" is the displacement.

If you simply plot the force "F" for a given displacement "x" you can easily generate the plot by connecting the resulting points by a smooth curve. Note...in physics terminology....."curve" means either a straight line or literally a curved line.
--------------------------------------------------------------------------------------------------------------------
4) Actually, I gave you the formula for the total energy of a spring, or rubber band (given certain stated limitations) above in #1. You will have to be more specific in your question if you have a specific case in point.
--------------------------------------------------------------------------------------------------------------------
5) You design a catapult by imparting some kind of force to a mass which would thereby cause it to be cast at some distance away. There are numerous ways to do this. You did not specify what materials you have to choose from, so I'm assuming you mean a spring or rubber band.

Try building some kind of holder where you can place a mass. Maybe this can ride on a some very small section of track aimed upward at some angle...say 45 degrees would be optimum. Attach a rubber band/spring to an anchor point under this holder. Pulling on the holder will stretch the spring/band to a comfortable limit, and then you'll have to decide if you want to design some kind of trigger or just 'let go' at this point. In either case the spring or rubber band will snap, pulling the holder/mass assembly along the track. Design a way to stop the holder without it crashing and causing damage, so that it will stop, but the mass will continue to upward.

This is far from well-thought out, but since I have no idea what your application is, this is what I am offering as a response for now.
------------------------------------------------------------------------------------------------------------------

If you can be more clear with some of your questions I will be able to answer you with less generalities and more accuracy.

Hope this helps tho.....


bluto

jenny96 rated this answer Excellent or Above Average Answer

Question/Answer
KoolPeace7 asked on 03/01/03 - Electric Circuits

If you can help with even one of these questions, that would be great! I tried some of these. If an expert answered one of these already, please don't answer that one unless I say that it still doesn't work out. Also, I might add clarifications in any chance that I figure out some of the questions saying that I figured it out because that happens. However, any help is very appreciated.

(1) http://www.webassign.net/hrw/28_37.gif shows a circuit containing three switches, labeled S1, S2, and S3. Find the current at a for all possible combinations of switch settings. Put emf = 140 V, R1 = 20.0 ohms, and R2 = 10.0 ohms. Assume that the battery has no resistance. Put answers in Amps for:
(S1 open, S2 and S3 closed)
(S2 open, S1 and S3 closed)
(S3 open, S1 and S2 closed)
(S1, S2, and S3 all closed)






(2) A circuit containing five resistors connected to a battery with a 12.0 V emf is shown in http://www.webassign.net/hrw/28_38alt.gif. (R1 = 8.0 ohms; R2 = 3.0 ohms; R3 = 7.0 ohms) What is the potential difference across R3 in Volts?





(3) http://www.webassign.net/hrw/28_41.gif
(a) Calculate the current through each ideal battery in Amps. (As a sign convention, assume the currents are "up" through each battery.)
I1-
I2-
I3-
Assume that R1 = 0.9 ohms, R2 = 2.2 ohms, emf1 = 2.0 V, and emf2 = emf3 = 3.5 V.
(b) Calculate Va - Vb





(4) (a) In http://www.webassign.net/hrw/28_43.gif, what is the equivalent resistance of the network shown in ohms?
(b)What is the current in each resistor in Amps? Put R1 = 105 ohms, R2 = R3 = 40 ohms, R4 = 60 ohms, emf = 6.0 V; assume the battery is ideal.
i1 -
i2 -
i3 -
i4 -





(5) In http://www.webassign.net/hrw/28_44.gif, emf1 = 3.00 V, emf2 = 4.00 V, R1 = 3.00 ohms, R2 = 5.00 ohms, R3 = 5.00 ohms, and both batteries are ideal.
(a) What is the rate at which energy is dissipated in R1 in W? In R2? In R3?
(b) What is the power of battery 1 in W? Of battery 2?






(6) In the circuit of http://www.webassign.net/hrw/28_46.gif, emf has a constant value but R can be varied. Find the value of R that results in the maximum heating in that resistor in ohms. The battery is ideal.

Please answer by 6PM Tuesday, March 4, 2003. Thanks!

bluto answered on 03/03/03:

Hi Kool......

Not to step on any toes, but here's a response for your #2 which will hopefully add to the very good information vg posted.

Given:

R1 = 8
R2 = 3
R3 = 7
V = 12 V

There are essentially two branches of resistor networks, one parallel with the other.

There is the top branch consisting of: R1 and a 12 and a 4 resistors. Then there is the lower branch consisting of: R2 and R3 which are in series.

You need to find the current in the lower branch, for it will be the same thru each resistor since they are in series, and, unless I'm missing something that is the only branch you need to consider.

Apply Ohm's Law to this branch and you have:

I = V/R

I will be the current in that branch.
V is the voltage across that branch.
R is the total resistance of that branch.

R = R2 + R3 = 3 + 7 = 10

Then the current is: I = V/R

==> 12 V/10
==> 1.2 A

Since the current in this branch must be the same thru each resistor, you can find the voltage drop across R3:

From v = ir

Where:

v = voltage drop across R3
i = 1.2 A
r = 7

r = v/i = 7 * 1.2 A = 8.4 V

Therefore....the voltage drop across R3 is 8.4 V.


As a check.....you have a total voltage of 12 V across this branch. So, the voltage drops across R2 and R3 should equal 12 V.

Voltage drop across R3 is 8.4 V (I claim).

Voltage drop across R2 is:

v = ir = 1.2 A * 3 = 3.6 V

Sum of voltage drops = 8.4 V + 3.6 V = 12 V




Hope I read your question correctly, but if not, lemme know. Hope this helps out.


bluto

KoolPeace7 rated this answer Excellent or Above Average Answer

Question/Answer
jenny96 asked on 02/28/03 - hi

how do you find the work required to stretch a spring from its natural length to a new position

given original spring = 22.7 cm
applied force 1 N = 26.8 cm
2N = 30.8 cm
3N = 34.8 cm
4N = 39 cm
5N = 42.8 cm
do you have to convert to meteres?
how do you plot a force over displacement graph? please explain.thanks

bluto answered on 03/02/03:

Hello jenny.....

Hook's law says: F = -kx

F is the force on/by the spring.
x is the distance of the spring's stretch/compression.
k is the spring constant of the spring.

Work(W) is force times distance: W = Fd

So....given that you know the force and distance in each case. All you have to do is multiply those values to get the work.

e.g The first one:

F = 1 N
d = 26.8 cm

Convert cm to m: 26.8 cm = 0.268 m

Work(W) = Fd = 1 N * 0.268 m = 0.268 J


You can do the rest.....yes?

Make sense how to do this?
------------------------------------------------------------------------------------------------------------------------

For the plot.....I don't understand what you are asking.

Plotting Force versus displacement is plotting the force in the 'y' axis and the displacement on the 'x' axis for each pair of numbers that you have.

Since you are asking for a plot of force OVER displacement....I don't know what you mean by that and you will have to clarify that for me a little more please.

Hope this helps....

bluto :)

jenny96 rated this answer Excellent or Above Average Answer

Question/Answer
jurplesman asked on 02/19/03 - Solar heating of water

I have always been wondering about the correct answer to a question that some my fellow campers in Australia have been debating.
We used plastic water bags to heat water in the sun. The bag is black on one side, and transparent on the other side. It was always my understanding when heating water in the sun, that the transparent side should face the sun, with the black side on the ground. I used to heat water in a black plastic bucket quite efficiently on the same principle.
The theory is that the sun' light was attracted to the black water, because of the black side of the bag. A debate arose among my fellow campers that the black side of the bag should face the sun and not the transparent side. I disputed that but as I am not very educated in physics I could not convince the others of my point of view. The opinion was divide 50/50, so who is right?

The question: What is more efficient in heating water, with the transparent side up or on the ground?

Jurriaan Plesman

bluto answered on 02/25/03:

Hello jurian....

I just ran across this....so...sorry if my follow-up is not timely.

I'm sorry you are disappointed with the response I gave you, but I think you may have missed something I said, or I was just yakking so much that it blew past you...lol.

I sidestepped talking about UV specifically by lumping it in with simply saying that the water is 'warmed' by the sun's rays.

There is more than just UVthat can account for the heating of the water directly. I'm not certain, but the hamster on the wheel in my brain is telling me that about half (give or take) of the energy of the sun is visible light, most of the remaining half is infrared, and then some/all of the rest of that is UV.

I just didn't bother to go into all of those details because it really doesn't exactly *which* metric of the sun's electromagnetic outpouring of energy is responsible for heating the water in the bag....just that it gets heated, and I was worried my response would become still longer and just too wordy.
------------------------------------------------------------------------------------------------------------------------


Ok...so here it is again:

The bag will heat faster with the clear side UP...toward the sun and the black part DOWN. That's what I said before, and here's a more succint list of why this is so.

* You have the sun warming the water by the 'greenhouse effect'

* You have the sun warming the water by sheer impingement of the energy (kinetic)water molecules.

* You have the sun warming the black part of the bag since the sun's light will pass thru the water and strike the black part, and having that heat re-radiate UP into the water.

======================================================================================================================

With the black side UP. The black portion of the bag will screen much of the sun's energy eliminating, or reducing some of the above channels of heating. Therefore, this method, or position of the bag (black side up), will not allow the water to heat as quickly as with the transparent side up.

Was this more helpful? Lemme know otherwise....

bluto

jurplesman rated this answer Excellent or Above Average Answer

Question/Answer
jurplesman asked on 02/19/03 - Solar heating of water

I have always been wondering about the correct answer to a question that some my fellow campers in Australia have been debating.
We used plastic water bags to heat water in the sun. The bag is black on one side, and transparent on the other side. It was always my understanding when heating water in the sun, that the transparent side should face the sun, with the black side on the ground. I used to heat water in a black plastic bucket quite efficiently on the same principle.
The theory is that the sun' light was attracted to the black water, because of the black side of the bag. A debate arose among my fellow campers that the black side of the bag should face the sun and not the transparent side. I disputed that but as I am not very educated in physics I could not convince the others of my point of view. The opinion was divide 50/50, so who is right?

The question: What is more efficient in heating water, with the transparent side up or on the ground?

Jurriaan Plesman

bluto answered on 02/22/03:

Hello jurplesman.....

Very interesting scenario and question!

Let's spare the math-argument and allow me to carry you thru some science
to answer your question.

Let's say the black part of the bag occupies the same amount of area as the
clear part of the bag. Let's further say that the bag is aimed at the most
direct sunlight possible.....and.....that there is little or no 'lensing'
by the water onto the black surface. By that I mean.....that the water
doesn't act like a magnifying glass.

Ok.....now, given these perfectly fair conditions, which change nothing
about the situation you're asking about we can go ahead with things.

A black object will absorb heat more readily, than a white, or clear
object. That is a fact and it's discussed, in pain-inducing detail in an
area of physics called thermodynamics. But let's not even consider the
black part of the bag at all! Let's agree on this first......that the water
itself will indeed heat up if placed in the sun.....correct? (Say
yes...[hahaha]) It's true......a lake in Spring is mighty chilly, but as
hot weather goes on the lake will heat up to decent temperatures.

However...at the bottom of the lake....it's still kinda chilly....right?
Same thing happens to any amount of water placed in the sun, or exposed to
some heat source on one side. This deals with something in physics
(thermodynamics) called "heat capacity" and "specific heat". Same thing
will happen to the bag of water....black on one side or not.

So....we know that the sun will, and does, heat up the water the side
exposed to the sun....fine.....so what? Well, not let's say you put the bag
on a hotplate, in total *darkness*......and turn it on low (so it doesn't melt
the bag obviously)....
.....what will happen?

Exactly!....it will heat up....no mystery there.....and the side exposed to
the hotplate will get warm first and then eventually the other side of the
bag (water) will achieve the same temperature. Makes perfect sense right?

Why a hotplate? Because it's a heat source we all (I hope) are familiar with, and
it will help make the connection to the case of the black part of the plastic bag.

You now actually have the answer to your question......do you see?! :)

Ok...lemme tie things together for you.
If you place the bag with the black side to the sun and the clear side
down.....the black will absorb the heat and transfer it to the water....and
the water will heat up....blah blah blah right?

If you place the bag with the clear side to the sun......the sun will heat
the water AND pass thru the bag and heat up the black part of the bag. So
the water will be receiving heat from two places for the price of one! And
they say you can't get something for nothing! ha ha

So....the water will heat faster if the black side is away from the sun
such that the sunlight can strike the bag on one side, pass thru and strike
the black part on the other side...that's your answer!

Also....since heat tends to also warm the air around a source.....the heat
absorbed by the black part of the bag when that part faces the sun will
tend to have its heat radiate AWAY from the bag/water. When the black part
is away from the sun opposite to the clear part facing the sun, the heat
absorbed by the black part will radiate INTO the bag/water, so that will
give you a radiation AND conduction source of heat, which is another reason
why the water will heat faster with the clear side toward the sun and the
black part opposite.

Make sense? Kewl question......hope this helps! :)


bluto

jurplesman rated this answer Average Answer

Question/Answer
Dathaeus asked on 02/10/03 - Hello my friend....

Hi there, its me... I have not had the time to go thru with that LED flashlight yet, but maybe soon, life has been hectic, and you didnt seem as active on the forum as I thought u would be, is everything ok?

Anyways, to the main q here. Now, I need to be refreshed, that, is it true most earthly energy cannot travel through a vacuum? I know they use vacuum seals for thermoses to preserve temperature most effectively. However, is the exact same true for sound? I.e., if we made a incredibly loud noise inside an almost perfectly sealed vacuum, like a thermos, would the sound be 99% contained?

If not, what is the most effective way of containing sound? The contraints are

Flexibility
(in manageability, i.e., I dont want a solution where you state " 10 foot thick steel and concrete walls" know what I mean? smaller/simpler the better)

Effectiveness
I need the finished solution to be not perfect, but enough so that at least a 120-150 db sound is contained within, say, a 10 foot radius to the level of a whisper, give or take a few db's.

Practicality
I dont mind hearing, but, telling me to use NASA materials that may cost hundreds of thousands or millions is not a viable solution either...

I hope you get my drift here... i.e., my perfect solution would be to somehow contain vacuum in small "strips" or "modules" and thus, use these to form anything I want; however, the downside is, how do I soundproof the seams of these modules? See my dilemma? Heh, I know I am being a little vague with my final purpose, but there is a reason, and I hope you can work with me like this, well, I know you can... thanks! (ask me for any clari of course!)

Dathaeus

bluto answered on 02/19/03:

Good afternoon Dat......

Sorry, but there isn't a 'magical' solution. You are trying to contain the sound-level of a jet engine in a space that is almost laughably small....in a *flexible* container with dollar-store materials....isn't gonna happen Dat! Sorry. :(

Like I said...I'm not even sure how my suggestions would work out for you realistically speaking. I think they will help, but the likelihood of containing that much sound energy given the constraints/supplies is......well....unlikely.....lol. The sand idea, or even better....the use of some kind of liquid surrounding the sound source will be your best.....cheap....common 'solution'. Part of what further complicates things is the fact that this needs to be rolled up........'deformed' if you will. Not only is the material itself important when soundproofing, but the *shape*. Soundproof rooms are composed of foam-like tiles that are made up of tiny little cones, or pyramids. You need a soundproof material made into a cylinder which radiates spherically! I'm guessing about the shape of your source. I'm assuming that since it's small it can be assumed essentially spherical for all intents and purposes. Acoustic tiles work on flat surfaces, but if you were to bend them into various shapes (if they were able to be bent) they would not perform the same.......that's my point.

I've been looking around the web for some materials to suggest to you, but most of what I've seen thus far are materials for rooms, and buildings. If not that, the others don't give pricing. I will continue to look for you tho!

Also.....I use to have a decibel meter, but it did not have the functionality to emit. If I still had the thing I'd just send it to you! I've not heard of meters that emit as well as measure, and have not found any on the web either. I have some emails out there requesting that information for you......so in the meantime, I will continue to look and ask. When I find out something I will let you know. Just as a reference....Radio Crap....errr....I mean Radio Shack sells decibel meters. They are probably fine for the accuracy level you cited. (www.radioshack.com), and they'd be economical..... ~ $40.00.

As for my previous suggestions....to answer you on that:

Yes it is better to have as little air in between the layers as possible. Remember........air = sound propagation. Thermal insulation....like fiberglass is made to catch air and hold it, so that would probably be working against you. One of the other things that's killing you on this thing is the fact that the sound needs to die in such a short distance. Sound intensity falls off as the inverse square of the distance away: I o< 1/d^2 ("o<" means proportional too.....clever huh!? lol). You will receive no benefit from this in such a small space.

Oh....you asked about the layers. There's no hard fast rule on the number of layers or thickness. Depends on the loudness of the source, distance away from it, etc. Come to think of it.....well...re-think I should say.......the thin cotton weave suggestion is probably not going to do a whole lot for you. Something more dense would be better......something that will reflect/deflect/absorb is what you will have to choose from. But you're right.....multiple layers would be better than one thick one if you can get away with it.

Aside from poking around on the web....which I will continue to do, I'm running out of ideas for you. :(

Sorry I don't have the silver bullet, but I'll keep working on this. Talk to ya later!

Maybe I'm thinking TOO frugal!?

bluot

Dathaeus rated this answer Excellent or Above Average Answer

Question/Answer
jenny96 asked on 02/18/03 - hi

a linear elastic spring is 15.0 cm long. when the upper end is held in the hand and a 500 g mass is suspended from the lower end, its length becomes 22.0 c. if the hand is not jerked quickly upward, the spring first extends to a length of 28.5 cm, the the mass starts to move up. the hand is then held still.
a) what is the acceleration with which the mass first begins to move ans: -9.1 m/s^2
b) what is the speed of the mass when the length of the spring becomes 22.0 cm again?
ans: 0.77?
c) what is the length of the spring when the mass comes to rest at its highest point?
can you help me with the above question please? thnaks! please show me all the steps and formulas used for this question. i would really like to see how u do it thanks!

bluto answered on 02/18/03:

Hello jenny.....

These questions all deal with Hook's Law....which isn't even a 'law' per se!

Hook's law is: F = -kx

F = force on the spring.
x = distance the spring is stretched/compressed
k = the spring constant individual to each spring, kind of like a fingerprint.
The negative sign just means the force is opposite to that of the distance of spring stretch.

a) The mass of the weight is: 500 g = 0.5 kg

The spring stretches: 0.15 m - 0.22 m = -0.07 m

At 0.022 m spring stretch, the spring constant can be found:

F = -kx
==> k = -F/x = -mg/x = -(0.5 kg)(9.8 m/s^2)/(0.07 m) = 70 N/m

Then by Newton's law: F = ma -and- Hook's law: F = -kx

You can find the acceleration (a):

==> -kx = ma
==> a = -kx/m
==> a = -(70 N/m)(-0.065 m)/(0.5 kg)
==> a = 9.1 m/s^2

Does that make sense to you?


b) For this one, you can make use of this kinematics eq'n: v^2 = (vo)^2 - 2ax

vo = initial velocity, which is zero.

v = Sqrt(-2ax)

a = 9.1 m/s^2
x = -0.065 m (opposite direction now, so you add the negative sign)

v = 1.1 m/s

[Not getting your answer here]



c) The velocity in part "b" is the initial velocity for depressing the spring, which will serve to oppose the upward force of the moving mass.

The mass will have total kinetic energy: KE = (1/2)mv^2

==> KE = (1/2)(0.5 kg)(1.1 m/s)^2 = 0.3 J (Joules of energy)

The spring will stop compressing when this kinetic energy is completely converted to potential energy (PE).

The PE of a spring is: PE = (1/2)kx^2

Therefore, when PE = KE.....the spring will stop.

KE we know, we just found it above. So we can now set PE = 0.3 J

==> (1/2)kx^2 = 0.3 J [Now solve for x]
==> x = Sqrt(2*0.3 J/ 70 N/m) = 0.1 m, or about 10 cm.




Hope this helps, but let me know if something's not clear....k?

bluto :)

jenny96 rated this answer Excellent or Above Average Answer

Question/Answer
Dathaeus asked on 02/10/03 - Hello my friend....

Hi there, its me... I have not had the time to go thru with that LED flashlight yet, but maybe soon, life has been hectic, and you didnt seem as active on the forum as I thought u would be, is everything ok?

Anyways, to the main q here. Now, I need to be refreshed, that, is it true most earthly energy cannot travel through a vacuum? I know they use vacuum seals for thermoses to preserve temperature most effectively. However, is the exact same true for sound? I.e., if we made a incredibly loud noise inside an almost perfectly sealed vacuum, like a thermos, would the sound be 99% contained?

If not, what is the most effective way of containing sound? The contraints are

Flexibility
(in manageability, i.e., I dont want a solution where you state " 10 foot thick steel and concrete walls" know what I mean? smaller/simpler the better)

Effectiveness
I need the finished solution to be not perfect, but enough so that at least a 120-150 db sound is contained within, say, a 10 foot radius to the level of a whisper, give or take a few db's.

Practicality
I dont mind hearing, but, telling me to use NASA materials that may cost hundreds of thousands or millions is not a viable solution either...

I hope you get my drift here... i.e., my perfect solution would be to somehow contain vacuum in small "strips" or "modules" and thus, use these to form anything I want; however, the downside is, how do I soundproof the seams of these modules? See my dilemma? Heh, I know I am being a little vague with my final purpose, but there is a reason, and I hope you can work with me like this, well, I know you can... thanks! (ask me for any clari of course!)

Dathaeus

bluto answered on 02/18/03:

Hey Dat.....

Ok ok ok...no more cell-talk! lol You want to stay away from thin, rigid materials if you want to sound proof something. Given the new info you provided, I'd say....go to a fabric shop and look around and see what you can get away with as far as thickness. Roll some of it up and see how easy/hard it is. While you do this.....keep this main idea in mind. You want as many layers as possible.

-AND-

You will want to interleave layers of different materials in the roll. For this you will want some kind of plastic material with small (~BB-size), preferably randomly placed holes in it, kinda like those plastic bags you get rice in these days, except those aren't random, but see what you can find. I'm sure a fabric shop would have something like that.

Try to find as many layers as you can fit. The more the better and make each one as different from the next as possible. Now....this isn't gospel by any means, but try the layering as follows:

Thick - soft
thin - soft
thin - semi-hard, yet flexible, plastic w/holes

-- repeat --

For the soft material, try felt, or a semi-dense cotton weave. If send you to buy soundproofing material, you're gonna yell at me for being too exotic or too expensive! lol I'm just teasin' you a little Dat. To be honest....the suggestions I gave you follow closely with the concepts that sound engineers try to follow themselves when they design something to kill sound.

What I don't know off the top of my head is by how much one material deadens sound over another. I could look that up, but for this I think a little experimentation on your part along with some of the above suggestions will get you there.

You're looking at a sound gradient of 40 to 50 db per inch of material. I DO know that a whisper is *about* 40 db, but to drop to that level or less from 120-150 db will be a feat for such a small area.

Look for material in that store that has alot of little peaks and valleys to it......and something with holes in it......like FOAM. Like packing material. I'm sorry that I don't have a particular material to say, "Go buy THIS....part number blah blah blah"....but I don't....unfortunately. If I were doing this myself I'd probably go about it the same way I'm describing to you......experimentally.

I'm still gonna recommend to you to do what you can to evacuate as much air out of this thing as you can. That's like a sound-leak!

One other suggestion.....for the next to the last, outside layer. Use aluminum foil....no holes, and nothing thick. Just standard, everyday stuff......k? Oh...and use the shiny side inward to the sound source. Then when you get it all zipped up.......go throw the thing in the ocean! lol.......kidding, kidding!

I'm intrigued....as usual! I hope these ideas are a little more appropriate for you, however the other idea would have probably done ok....messy and a pain, but could have worked. I gotta say tho.....120-150 db.....that's mighty loud to try and squelch with a little bit of felt, and cotton and what-not.

At this point I will look around and see what kinds of sound-proofing materials are available.....just in case, and we'll go from there. Take care and will talk to ya later. :)

bluto

Dathaeus rated this answer Excellent or Above Average Answer

Question/Answer
Dathaeus asked on 02/10/03 - Hello my friend....

Hi there, its me... I have not had the time to go thru with that LED flashlight yet, but maybe soon, life has been hectic, and you didnt seem as active on the forum as I thought u would be, is everything ok?

Anyways, to the main q here. Now, I need to be refreshed, that, is it true most earthly energy cannot travel through a vacuum? I know they use vacuum seals for thermoses to preserve temperature most effectively. However, is the exact same true for sound? I.e., if we made a incredibly loud noise inside an almost perfectly sealed vacuum, like a thermos, would the sound be 99% contained?

If not, what is the most effective way of containing sound? The contraints are

Flexibility
(in manageability, i.e., I dont want a solution where you state " 10 foot thick steel and concrete walls" know what I mean? smaller/simpler the better)

Effectiveness
I need the finished solution to be not perfect, but enough so that at least a 120-150 db sound is contained within, say, a 10 foot radius to the level of a whisper, give or take a few db's.

Practicality
I dont mind hearing, but, telling me to use NASA materials that may cost hundreds of thousands or millions is not a viable solution either...

I hope you get my drift here... i.e., my perfect solution would be to somehow contain vacuum in small "strips" or "modules" and thus, use these to form anything I want; however, the downside is, how do I soundproof the seams of these modules? See my dilemma? Heh, I know I am being a little vague with my final purpose, but there is a reason, and I hope you can work with me like this, well, I know you can... thanks! (ask me for any clari of course!)

Dathaeus

bluto answered on 02/13/03:

Top-o the morning to you Dat....

So, I take it you're all settled in and adjusted to things now?

So you finally spilled a few more beans eh? I'm intrigued for sure, but just outta curiosity, and even tho I hate to admit being a mensa dude, and I rarely do admit it... (it's like being a member of a lookie-at-how-smart-I-am, or a snob society) I *gotta* ask......so, why *don't* you turn the sound down, or off on the cell? lol

I dunno how you would be able to evacuate/re-evacuate a sort of portable pouch like you described and still have the thing be small and easy to handle. But I have an idea you can try. I've never tried this, so I don't know how well this will work, but it's cheap! lol and the only way NASA would want it is if you charged'em 40 million for it.

There is one gizmo you might have to hunt for tho, but I'll tell you in a minute what that is. Ok, here's the scoop with the vacuum first:

Pulling even a small vacuum would help your cause in theory. In practice, it would probably be cumbersome, expensive and a big ole hassle for you. The problem with it is, since the vacuum *isn't* perfect, more than likely far from it, there's still gas in the 'chamber' which will vibrate with the sound of the cell, which will in turn, resonate the walls of whatever the chamber is made of, and then 'escape' to the outside. This is your single biggest nemesis with this project.

To kill the sound, you have to either, prevent/reduce the pressure waves from striking the walls of the container, or make multiple barriers to reduce the sound in stages. Really....you can't prevent the sound from reaching the inside walls of the chamber, so that leaves you with only two those options.

My suggestion will be a combination of these two I guess. I will leave the actual final product idea to your very clever mind...and that's not sarcasm, but will tell you what I have in mind and you can build it as you think best.

** If you can find a large thermos type thing that would be great, but it would have to be large. If not, then maybe you can come up with something else. A flexible pouch probably. Get some heavy plastic bags. I doubt that you'll want to trust ziplock bags but they may be useful to work with. Take two bags, and place one inside the other. Ok....here's the gizmo thingy I mentioned earlier. You need to find something that will be able to seal these two bags together around the top. One of those electrical resistance things that uses a wire to heat/melt the plastic together. Know what I mean? They do make those things, and I think you could find one, altho I do not think they are ubiquitous.

Ok, I'm breaking the paragraph there so you can sit and think....or conjure up colorful adjectives about me for that!

If that's not a show-stopper, then proceed to seal those two , so far empty bags together.....but....leave a space open. Say.....3"-4" or so.....whatever you think's best once you see what I'm getting at.

** Now......don't laugh, but *you* asked for it.......fill the bag with some fine sand and flour. Fill it so it's even all over but not over filled, and so it's not easy to make bare spots either. Once you have this done, get as narrow a tube as you can find. Super small tygon tubing, or a micropipette (plastic....get some, and other stuff like that at 1-800-225-3739 --- http://www.freyscientific.com/index.jsp). Here's why you need something tiny. You need to insert this part way into the bag and suck as much air out of it as you can, and then....at the same time...yank it out and seal it up real fast before the air goes back in! See....and you thought this would be hard! ;)

** You will need to make about 3, or possibly 4 of these. You could make them all the same, but here's a thought that I'd recommend going with. Maybe make the first one with mostly flour and some sand, and the next about 50/50, and the next more sand than flour and the last one mostly sand. You don't have to use sand...you *could* use sugar, but that's probably too expensive and harder to clean up in case of an accident. But....if you used ziplock bags, leave the *inside bag's ziplock seal intact so you can open and close it like a normal ziplock bag, but the outer bag would be heat-sealed to that bag......does that make sense?

** Furthermore, try and use progressively larger bags. So that you will have a small'ish bag to put the phone into. Then 'zip' it shut and place it inside a larger (sand/flour-filled) bag....zip that shut, and place *that* into a third, or however many you want to do. Insert the first one 'open' end down tho, so the 'open' ends of the bags aren't all lined up. Place a lining of thick, soft cloth.....like felt, heavy cotton, or the like, as a lining in each bag, so the plastic walls of one bag don't contact the other. You can even try to suck the air out of these as you go too....use a standard straw for that if you think it will help.

When you're done....place the whole thing into a soft gym bag of some kind, and pack stuff around that if you want. It will all help. If you do all this I would wager that the cell would be sufficiently silenced.

I would envision the whole thing.....gym bag and all would be only about 18" inches long, and maybe 12" in diameter...give or take.

As an added measure.....just in case. Set the ringer to as high a frequency as you can. Not loudness....frequency. The lower wavelengths (long wavelengths) penetrate more than will the shorter ones (high freq.).

The easiest approach would probably be to just fill the ziplock bags without all the screwing around with the heat-sealing, and then pack these around the cell.....and that might work, but the other approach would be *much* more sound proof for you since there are no holes thru which the sound can escape.

So Dat....there ya have it. Cheap, not *too* difficult, and I bet it will work, but lemme know what you think.

Looking forward to hearing back from you whenever you get a chance. Have fun and thanks for the cool stuff to think about!

bluto :)

Dathaeus rated this answer Excellent or Above Average Answer

Question/Answer
Dathaeus asked on 02/10/03 - Hello my friend....

Hi there, its me... I have not had the time to go thru with that LED flashlight yet, but maybe soon, life has been hectic, and you didnt seem as active on the forum as I thought u would be, is everything ok?

Anyways, to the main q here. Now, I need to be refreshed, that, is it true most earthly energy cannot travel through a vacuum? I know they use vacuum seals for thermoses to preserve temperature most effectively. However, is the exact same true for sound? I.e., if we made a incredibly loud noise inside an almost perfectly sealed vacuum, like a thermos, would the sound be 99% contained?

If not, what is the most effective way of containing sound? The contraints are

Flexibility
(in manageability, i.e., I dont want a solution where you state " 10 foot thick steel and concrete walls" know what I mean? smaller/simpler the better)

Effectiveness
I need the finished solution to be not perfect, but enough so that at least a 120-150 db sound is contained within, say, a 10 foot radius to the level of a whisper, give or take a few db's.

Practicality
I dont mind hearing, but, telling me to use NASA materials that may cost hundreds of thousands or millions is not a viable solution either...

I hope you get my drift here... i.e., my perfect solution would be to somehow contain vacuum in small "strips" or "modules" and thus, use these to form anything I want; however, the downside is, how do I soundproof the seams of these modules? See my dilemma? Heh, I know I am being a little vague with my final purpose, but there is a reason, and I hope you can work with me like this, well, I know you can... thanks! (ask me for any clari of course!)

Dathaeus

bluto answered on 02/11/03:

Hi Dat!!

Glad to receive your note and thanks for the correspondence!

Not a problem regarding the flashlight project. Whenever it's convenient for you, drop a line and I'll do my best to help out.

I don't participate in the forums anymore. I read thru them as best I can to 'keep up', but there is/was too much juvenile behavior, ego-tripping, and bickering going on amongst the experts that I just didn't see that I was spending my time in a constructive way. But I am answering questions as they come in and doing my best to go thru the site and 'test' things, even tho Vijay is always ahead of the game on that. I've been doing some advertising as well with posters and word-of-mouth, but don't know what real impact it's had.

As for your question. Sound certainly will *not* travel thru a vacuum, but any electromagnetic 'radiation' will; light, radio waves, microwaves, TV signals, etc. In a perfect vacuum, the sound would be 100% contained!

Sound is a longitudinal disturbance of the air. That means that if you take, say a paper plate in both hands and hold it up in front of you so the plate's edge is in your hands and the flat part is aiming outward, and then move the plate toward and away from you, you'd be simulating how a longitudinal wave would be made. How was THAT for a run-on sentence! lol

That motion is what happens in a stereo speaker. If you look closely at one working you can actually see it moving in and out.

To contain sound is pretty tough. You need a soft material with LOTS of holes in it....lots and lots. Like a piece of pumice say.

There are acoustic materials made just for this kind of thing. Depending on how effecient this project has to be, you might be able to fashion sometihng out of 'good' acoustic ceiling tile from say Lowe's or Home Depot. Just an idea at this point.

The shape of the container may help, but that depends on where the source is. Sound will definitely leak thru the seams, but depending on the shape and size of the cavity within which you want to contain the sound it may not be too difficult to 'insulate' them.

I don't have a feel for how much suppression you could expect, but it wouldn't be perfect. If you can pull some kind of vacuum, that would help too.

......so I'll leave it at that for now and will try and do a bit of digging to see what's out there, and whenever you are ready you can send some more info!

You always come up with cool stuff Dat! Looking forward to the next installment.

bluto :)


Dathaeus rated this answer Excellent or Above Average Answer

Question/Answer
kammy asked on 01/21/03 - hello

hi bluto thanks so much for your answers. I will rate you hehe. Can you please answer these for me please?

1) in a amusement park ride, a person stands in a cage and is turned in a vertical circle with a constant speed of 5 m/s. if the person's feet are 25 m from the centre of the ride, how do you calculate the apparent weight when the cage is at its lowest position. The person has a mass of 65 kg. ans: 7.0 x 10^2 N

2) a record album has a radius of 32 cm and a playing "speed" of 33.3 rpm. A 10g insect is standing at rest, relative to the record, at the extreme edge.
a) what is the magnitude of the frictional force needed to keep the insect at this point? ans: 3.8 x 10^-2 N
b) what is the coefficient of friction? ans: 0.39

3) a 1.0 kg mass is sitting at rest at the edge of a circular table. A 2.0 kg mass moving 5.0 m/s along a diameter, collides head on with the 1.0 kg mas and the two objects stick together. if the table is 1.2 m high, how far from the edge are the objects when they hit the floor? ans: 1.6 m

4) a 5.0 kg mass is dropped from rest from a height of 2.8m
a) what is the momenturm of the object just before it strikes the earth ans: 37 kg m/s down
b) if the object comes to rest in a time of 0.12 s while striking the earth, what is the impulse and the average net force acting on the object? ans: 3.1 x 10^2 N up

5) a 0.50 kg puck is moving at 4.0 m/s north when it collides with a second 0.50 kg puck, initially at rest. after teh collision the first puck is observed to be travelling at 1.0 m/s north 30 degrees west. what is the velocity of the second puck.
ans: 3.3 m/s E 15 degrees N

6) a 3.3 kg object experiences a force that varies as follows; for the first 3.0 s, the force rises linearly from 0 to 10 N. for the next 3.0 s the force is constan t. the force then falls linearly to 0 N in the next 6.0 s.
a) what is the force time graph assuming force is positive?
b) what is the impulse exerted on teh object for the entire 12 s? ans: 75 Ns
c) if the object was initially at rest, what must be the magnitude of its velocity at the end of the 12s? ans: 23m/s

bluto answered on 02/01/03:

kammy.....

Thanks for being patient. Here's the first one, but would you please select half the remaining questions as ones you particularly need help with and submit those instead.

I have other folks that need some time and it will be difficult to answer all these for you at once.

Thanks :)
-------------------------------------------------------------------------------------------------------------------

1) For this one, you need to calculate the centripetal acceleration "a":

a = v^2/r

v = 5 m/s
r = 25 m

Plug these into the above eq'n and you will see that: a = 1 m/s^2

Given that the person's mass "m" = 65 kg, this gives a force of F = ma = 65 N

Since at the bottom of the ride the weight of the person will ad with this centripetal force to give:

F(total) = F(centripetal) + w(weight)
==> F = 65 N + mg
==> F = 65 N + 637 N = 702 N
=============================================

Sorry this is short on what you asked, but hope that what I've provided has helped just the same. :)

bluto

Question/Answer
kammy asked on 01/21/03 - hi

hello this is for review once again. I'm really sorry for the amount. These were the ones..that i had trouble figurign out. PLease show me the steps in how to do each one of these. Thank you thank you!!

1) a runner is competing in a 1600 m race that involves 4 laps around a 400 m track. he runs the first 3 laps at a constant speed of 6.86 m/s. at the end of the third lap, he accelerates "insanteneously" to some new constant speed with which he finishes the race. if his overall average speed for the entire race was 6.95 m/s, what is the constant speed during the fourth lap?
ans: 7.23m/s

2) a hiker walks 2.0 m/s south for 10 min, then at 2.3 m/s west for20 minutes, and finally at 1.8 m/s west 30 degrees north for 15 minute. what is the
a) total time for the walk? ans: 2.7 x 10^3s
b) the total distance of the walk? ans: 5.6 x 10^3 m
c) the average speed? ans : 2.1 m/s
d) the displacement? ans: 4.2 x 10^3 m west 5 degrees South
e) the average velocity? ans: 1.5 m/s west 5 degrees south

3) a student slides a dime across a cafeteria table. the coin travels with constant velocity of 0.752 m/s while on the table, after sliding off the edge, the coin lands on the floor, 0.283 m horizontally from table's edge. how tall is the table?
ans: 0.694m

4) a football is punted by a kicker, leaving his foot at a height of 0.80 m above the ground and travelling at 20m/s up 40 degrees right. a receiver is standing 50 m away from teh kicker and runs to catch the ball. he does so at a height of 0.80 m above ground.
how can you answer these questions without the range/time flight equations?
a) for how long was the balli nthe air? ans: 3.1s
b) what horizontal dispalacement did ball traveL? ans: 40m
c) if the receiver began to run when the ball reaced its heighest point,at what constant velocity must he have been travelling? what direction?ans: 6.4 m/s left

5) how do you convert the following a) 3.98 Ms = ?ns
ans: 3.98 x 10^15 ns
b) 4.0 mm^2 = ? m^2 ans: 4.0 x 10^-6 m^2

6) momenturm (p) is defined as follows p = mv, where m is mass and v is velocity. what are the dimensions of momentum?
ans: M L T ^ -1

7) when a current (I in units of amperes , A) is flowing in a wire, it produces a magnetic field ( B in units of teslas, T). the size of the magnetic field at any point around the conductor varies directly with the size of teh current and inversely as the distance ( d, in units of metres , m) of teh point from the centre of teh conducting wiring.
a) what is the proportionality equation relating B to I and d? ans: B proportional I/d or B = kI/d
b) a certain wire with a current of 25 A has a magnetic field of 5.0 x 10^-5 T at a point that is 0.10 m from teh wire. what would be the magnetic field at a distance of 0.20 m assuming the same current? ans: 2.5 x 10^-5 T
c) the theorectically equation for the magnetic field near a wire is B = U oI /(2*pi*d), where Uo is a constant called the permeability of free space". base on given data in (b) what is the value of Uo, along with proper units? ans: 1.3 x 10^-6 T * m/A

8) a 50.0 kg pilot is flying an airplane in a vertical circle. at the top of teh loop, her apparent weight is 300 N . waht is her apparent weight at the bottom of the loop, assumign she maintains a constant speed around the circle? ans: 1.3 x 10^3 N

9) a student is standing at rest relative to a moving van. the van is travelling east. the student drops a 15 g marble from ceiling . what is the magnitude and direction of the fictitious force (if any) that would be observed under each of the following conditions?
a) the van accelerates forward at 1.5 m/s^2 east ans: 2.3 x 10^-2 N west
b) the van travels at constant velocity of 8.0 m/s east ans: 0 N
c) the van slows down at 1.5 m/s^2 west ans: 2.3 x 10^-2 N east

bluto answered on 02/01/03:

Kammy....

Here's the remainder of your questions answered:

7a) The equation governing this relation is: B = kI/d

The question given is stated incorrectly. It is not a "proportionality equation". It's a proportionality.........period.

To answer the question: B o< I/d. In other words, it varies directly with current "I" and inversely with separation "d". The symbol I mocked up "o<" is suppose to be the proportionality symbol. :)

So, the second 'asnwer' you've given is NOT correct. That is an equation because OF the proportionality CONSTANT "k". Subtle perhaps, but worthy of making the distinction.

7b) Since velocity and the radius of the circular path of the plane are not given. You will have to proceed as follows:

The Total force "F" is a combination of the centripetal (center-seeking) force "f" and the weight of the object in question "w".

==> F = f + w

w = mg

m = 50 kg and we all know what "g" is. This gives w = 490 N.

At the top of the curve, the apparent weight.....'total force' is given as 300 N. This says:

300 N = f - mg (taking gravitation as negative downward this time)

==> f = 300 + 490 N = 790 N

Note: "f" and "mg" have differnt signs here. This is why the apparent weight is LESS than the actual weight "w", which is 490 N.

Likewise, at the bottom of the curve, the apparent weight and the weight add to give an increased effective 'g'-force.

==> F = f + w
==> F = 790 N + 490 N = 1280 N


---------------------------------------------

9) Could you offer a bit of clarification about this please?
=============================================

Ok....this finishes this batch of problems. Hope it helps, however I will not be able to accept large lists of problems like this again. Perhaps you can send one or two representative problems and we can go from there. :)

bluto

Question/Answer
KoolPeace7 asked on 01/31/03 - Electric Charge

I am completely stumped on this question. I've tried every possible way I am capable to thinking, and stressed with the due date, I request that you answer this by 6PM on Saturday, February 1, 2003. Thank you.

In the diagram shown here: http://www.webassign.net/hrw/22_23.gif

What are the horizontal and vertical components of the resultant electrostatic force on the charge in the lower left corner of the square if q = 0.9 x 10^-7 C and a = 6.8 cm? (Assume the positive directions are upward and to the right.)
Please answer in Newtons for each of the horizontal and vertical components.

Thank you!

bluto answered on 02/01/03:

Hi kool......

These are tricky for folks it seems, but if you take things one at a time, it's not so bad.

Before we calculate ANYTHING, let's see if we can GUESS where the resulting force should end up....k?

The charge in question has a value of q". Above it is a positive charge "q", which means they are repulsed by each other.....right?

At the opposite corner we have a neqative charge "q", which is attractive, but it's attraction is weaker than the positive charge because it is further away, even tho they are of the same magnitide.

The last charge is to the right and has a value of q". That is also atractive. More so than the one above it. So....based on this I'd say the resulting force should be to direct our charge "Q" to the right and slightly below the charge to its right......make sense? Let's see if it works out.
=============================================

You need to use Coulomb's Law for this:

F = kQq/r^2

Where k is the constant 4(pi)eo = 9 x 10^9 Nm^2/C^2

Q is a charge
q is a charge
r is the distance separating the charges.

For the charge in the lower left corner of that diagram, I will call it "q". the other three charges I will call: a, b and c respectively from the top/left & right.

The Total force on Q due to the other charges is F, and given by:

F = Fa + Fb + Fc

Meaning, the total force "F" is the sum of the forces from charge a "Fa", b "Fb" and c "Fc".

Let the origin of a cartesian coordinate system be at our charge. Then charge a will be directly overhead in the vertical direction, Charge b will be in the opposite corner and c will be directly to the right horizontally. Make sense?

You need to find the vertical and horizontal components of each of the three charges. Let's do this starting with charge a, the upper left one:

It has no horizontal components since it lies directly along a vertical from 'our' charge Q. It is a distance 'a' away. NOT to be confused with our label 'a' for that same charge!

The force on Q due to this charge will be:

Fa = kQq/r^2y = k(-2q)(q)/a^2y = -2k q^2/a^2y

Note that I used "y" to signify a unit vector in the vertical direction. This will help a great deal in showing where everything will end up.
=============================================
Next is the force on Q due to charge b:

These charges are separated by a distance of Sqrt(2).

This interaction will have vert. and horiz. components.

Fb = (kQq/r^2)(Cos45)x + (kQq/r^2)(Cos45)y = [k(-2q)(-q)Sqrt(2)/[Sqrt(2)a]^2]x + [k(-2q)(-q)Sqrt(2)/[Sqrt(2)a]^2]y

==> Fb = Sqrt(2)kq^2/(2a^2)x + Sqrt(2)kq^2/(2a^2)y

Note my use of "x" and "y" to denote unit vectors in the horizontal "x" and vertical "y" directions.

Also note the result is positive. Meaning....it's a erpulsive interaction.
=============================================
Last charge:

Fc = kQq/r^2x = [k(-2q)(-2q)/a^2]x = [4kq^2/a^2]x

=============================================

Now, let's put these results togetehr and see what they tell us:

The total force "F" is:

F = Fa + Fb + Fc

==> F = [-2k q^2/a^2]y + [Sqrt(2)kq^2/(2a^2)]x + [Sqrt(2)kq^2/(2a^2)]y + [4kq^2/a^2]x

==> F = (kq^2/a^2)[-2y + [Sqrt(2)/2]x + [Sqrt(2)/2]y + 4x]

==> F = (kq^2/a^2)[(Sqrt(2)/2 + 4)x + (Sqrt(2)/2 - 2)y]

==> F = (kq^2/a^2)[4.7x - 0.59y]

So, the resultig force appears to be directed to the right and down, which is what we 'guessed' it to be. So at least it makes sense!

I will leave you to put the numbers into the variables and do the arithmetic, which I'm quite sure you're capable of. I think it's clear what's what, so I hope this helps. Look it over and lemme now if you have any concerns or questions. :)

bluto

KoolPeace7 rated this answer Excellent or Above Average Answer
sscfirst rated this answer Excellent or Above Average Answer

Question/Answer
kammy asked on 01/21/03 - hi

hello this is for review once again. I'm really sorry for the amount. These were the ones..that i had trouble figurign out. PLease show me the steps in how to do each one of these. Thank you thank you!!

1) a runner is competing in a 1600 m race that involves 4 laps around a 400 m track. he runs the first 3 laps at a constant speed of 6.86 m/s. at the end of the third lap, he accelerates "insanteneously" to some new constant speed with which he finishes the race. if his overall average speed for the entire race was 6.95 m/s, what is the constant speed during the fourth lap?
ans: 7.23m/s

2) a hiker walks 2.0 m/s south for 10 min, then at 2.3 m/s west for20 minutes, and finally at 1.8 m/s west 30 degrees north for 15 minute. what is the
a) total time for the walk? ans: 2.7 x 10^3s
b) the total distance of the walk? ans: 5.6 x 10^3 m
c) the average speed? ans : 2.1 m/s
d) the displacement? ans: 4.2 x 10^3 m west 5 degrees South
e) the average velocity? ans: 1.5 m/s west 5 degrees south

3) a student slides a dime across a cafeteria table. the coin travels with constant velocity of 0.752 m/s while on the table, after sliding off the edge, the coin lands on the floor, 0.283 m horizontally from table's edge. how tall is the table?
ans: 0.694m

4) a football is punted by a kicker, leaving his foot at a height of 0.80 m above the ground and travelling at 20m/s up 40 degrees right. a receiver is standing 50 m away from teh kicker and runs to catch the ball. he does so at a height of 0.80 m above ground.
how can you answer these questions without the range/time flight equations?
a) for how long was the balli nthe air? ans: 3.1s
b) what horizontal dispalacement did ball traveL? ans: 40m
c) if the receiver began to run when the ball reaced its heighest point,at what constant velocity must he have been travelling? what direction?ans: 6.4 m/s left

5) how do you convert the following a) 3.98 Ms = ?ns
ans: 3.98 x 10^15 ns
b) 4.0 mm^2 = ? m^2 ans: 4.0 x 10^-6 m^2

6) momenturm (p) is defined as follows p = mv, where m is mass and v is velocity. what are the dimensions of momentum?
ans: M L T ^ -1

7) when a current (I in units of amperes , A) is flowing in a wire, it produces a magnetic field ( B in units of teslas, T). the size of the magnetic field at any point around the conductor varies directly with the size of teh current and inversely as the distance ( d, in units of metres , m) of teh point from the centre of teh conducting wiring.
a) what is the proportionality equation relating B to I and d? ans: B proportional I/d or B = kI/d
b) a certain wire with a current of 25 A has a magnetic field of 5.0 x 10^-5 T at a point that is 0.10 m from teh wire. what would be the magnetic field at a distance of 0.20 m assuming the same current? ans: 2.5 x 10^-5 T
c) the theorectically equation for the magnetic field near a wire is B = U oI /(2*pi*d), where Uo is a constant called the permeability of free space". base on given data in (b) what is the value of Uo, along with proper units? ans: 1.3 x 10^-6 T * m/A

8) a 50.0 kg pilot is flying an airplane in a vertical circle. at the top of teh loop, her apparent weight is 300 N . waht is her apparent weight at the bottom of the loop, assumign she maintains a constant speed around the circle? ans: 1.3 x 10^3 N

9) a student is standing at rest relative to a moving van. the van is travelling east. the student drops a 15 g marble from ceiling . what is the magnitude and direction of the fictitious force (if any) that would be observed under each of the following conditions?
a) the van accelerates forward at 1.5 m/s^2 east ans: 2.3 x 10^-2 N west
b) the van travels at constant velocity of 8.0 m/s east ans: 0 N
c) the van slows down at 1.5 m/s^2 west ans: 2.3 x 10^-2 N east

bluto answered on 01/29/03:

Hi kammy.....

Thanks for being patient, and hope I can continue to help you. Fortunately things are slow right now, so here's a batch of answers for you:

1) This problem isn't difficult, but it's subtle, and that often times makes things tricky.

On the surface it appears as tho you can simply take the average of the two velocities given, but that does not work. Why? Well, for the sake of simplicity it's due to the abrupt change in the car's position from one given speed to the next. Essentially, the car has to have to different velocities with no elapsed time in between. That's why the problem used the term 'instantaneous' acceleration, and that's why one can't simply add up a bunch of velocities for a given trip and figure out what the average velocity (speed) was. Calculus is used for that.

So, here's how to 'solve' this one:

Using the dist/time/rate equation for each segment of the race, you have: d = rt in general.

You have three velocities to consider: v1, v2, and V (which I'm calling the average)

Let v1 = the first velocity: v1 = 6.86 m/s

v2 = unknown
V = 6.95 m/s

Using the dist/time/rate eq'n for each we have:

d1 = (v1)(t1)
d2 = (v2)(t2)
D = VT (Which I'm denoting as the average)

We need to find the times for each leg of the race, so let's do that:

t1 = (d1)/(v1) = 1200 m/6.86 m/s = 174.93 s
T = D/V = 1600 m/6.95 m/s = 230.22 s

==> T - t1 = 230.22 - 174.93 = 55.29 s

[Note: I often leave off units in my calculations for clarity. Make sure you include them all in your work :)]

So we have 55.29 s as the 'missing' chuck of time that will allow us to account for the abrupt change in velocity.

Since we know there is only 400 m during this 'missing' portion of the race, and we know that part of the race took place in 55.29 s, then we can find a velocity that corresponds to that information.....like this:

d2 = (r2)(t2)
==> r2 = (d2)/(t2) = 400/55.29 = 7.23 m/s (remember....units units units)



Make sense?

-------------------------------------------------------------------------------------------------------------------------------

2) Ok.....at this level I think you can start to think for yourself. These first two require almost no thought or math skills at all, so I will *tell* you what needs to be done....*you* can do the work for yourself. *but*, if you have a question about it, all you have to do is ask.....k?

For the first two, if you can ADD, you can do these. The first one requires that you convert minutes to seconds.

c) You already have the information to answer this one in parts "a" and "b". Dive the total distance by the total time and you got it. Lemme know if you get stuck.

d) This requires that you find the distance from the starting point to where you end up.

These are hard because a diagram would really help here! Find the total horizontal distance "X" from the starting point. This would be:

X = 2760 m + 1402.96 m = 4162.96 m

The second term is found using some trig on that last leg of the trip. You have a 30 degree right triangle for which you find the horizontal component: x = 1620*Cos(30) = 1402.96 m

(The ��" is the distance of that leg of the trip found using the info in the problem)

You'll eventually need the corresponding vertical distance "y" of this triangle:

y = 16520*Sin(30) = 810 m

You need to find the horizontal and vertical components of the distance from start to end. We have the horiz part...."X". We can use our 'y' value to find the vertical distance "Y".

Y = 1200 - y = 1200 - 810 = 390 m [Remember.....UNITS!]

The first term (1200) is the total distance given in the problem as the first leg of the trip.

Now we can use the Pyth. Thrm. to find the total displacement distance "D":

D = Sqrt( 390^2 + 4162.96^2) = 4181.18 m....or...about 4.2 x 10^3 m as the 'given' answer.

e) The avg velocity "V" is the total distance divided by the total time. Both of which we know.

V = D/T = 4200/2700 = 1.5 m/s



------------------------------------------------------------------------------------------------------------------------------

3) These are classic problems, and kinda fun to play with, but they cause students alot of grief.

The trick here, and with alot of these kinds of problems is this:

**There is NO acceleration in the horizontal direction**

What does that mean? It means you can use the good ole standard time/distance/rate formula to get some missing, or needed information. This problem is no different.

You know the initial velocity, which, by the way, is in the *horizontal* direction. And, you know how far....horizontally......it went.

So, odd as it may seem, use it to find how long the dime was in flight:

d = rt

==> t = d/r = 0.283/0.752 = 0.376 s

Now, since there's no vertical component of velocity, this means the motion is strictly free-fall! Great....now you can use the simplistic free-fall equation now that you have the time.

y = (1/2)gt^2 (positive direction downward)

Plug in your numbers and voila`...you find that y = 0.694 m


-------------------------------------------------------------------------------------------------------------------------------

4a) Here you have an object with vertical (vy) and horizontal (vx) components of velocity. They are found by resolving the initial velocity into its vert/horiz components in the usual way:

vx = vCosq = 15.32 m/s
vy = vSinq = 12.86 m/s

Where:
v = initial velocity = 20 m/s
q = angle = 40 deg.

For such an object there is no acceleration along the horizontal, but there is in the vertical. And since you know the vertical comp. of velocity you can use the kinematics eq'n: v = vo + gt

Here, "v" is the velocity after some time "t" and "vo" is the initial velocity, which we know now as vy = 12.86 m/s.

Set v = 0 since this will be the velocity at the peak of the flight of the ball and solve to find "t".

==> 0 = vy + gt
==> t = vy/a = 12.86/9.8 = 1.31 s

Since the flight is symmetric, the time it takes to come back down is equal to the time it too to reach the peak. So if you double this quantity you will get the total time-of-flight:

2t = 2.62 s

The value you give as the answer is not correct. That value is derived from using the horizontal component of the velocity, which is not correct.

4b) To find how far (horizontally) the ball goes, you can now use the standard dist/time/rate formula:
d = rt

The velocity is the horizontal component we found earlier vx:

d = (vx)t = 15.32*2.62 = 40.21 m [remember the units!]

4c) I'd say...if the catcher started running as soon as the ball reached the highest point, then he'd better be the highest paid sports figure in history! Since he/she started running from 50 m away, and the ball's total distance traveled was 40.21 m, then he/she'd have 50m - 40.21 m = 9.79 m. Given the time to fall would be 1.31 s (from the peak height), he/she'd have to run at a rate of:

r = d/t = 9.79/1.31 = 7.47 m/s

I think the answer you've given is flawed like the previous one I mentioned. Check things out carefully and make sure you have the correct info for the problem. I can only give you answers based on the data you provide. This goes with all of them. :)


-----------------------------------------------------------------------------------------------------------------------------

5) "M" means 'Mega, or million. "n" means 'nano', or billionth.

1 x 10^6 is one 'Mega" of something. Likewise, 1 x 10^-9 is one 'billionth' of something.

To go from Mega to nano, you must simply move the decimal point to the left by 15 decimal places. In numbers it means: 1 mega = 10^15*(1 nano)

More formally: 1 x 10^6 = (1 x 10^15)(1 x 10^-9) -or- 10^6 = 10^15*10^-9

So, for 3.98 Ms, you have:

3.98 Ms = 3.98 x 10^6 s

==> (3.98 x 10^6 s)/(10^15) = (3.98 x 10^6)*(10^-15)
==> 3.98 x 10^-9 s
==> 3.98 ns

------------------------------------------------------------------------------------------------------------------------------

6) Since momentum is given by: P = MV, replace the quantities "M" and "V" by their respective units:

V ~ m/s
M ~ kg

==> MV = (m/s)(kg) = m*kg/s -or- kg*m*s^-1


I have no idea what your answer means. You may want to check that one over as well.
--------------------------------------------------------------------------------------------------------------------------------------------------------

Ok kammy.........this is enough for now. I wanted to do the first batch all at once but there is ALOT of questions here!

So you know....I will not make a habit out of answering this many questions at once time in the future. There are questions in here which, based on the level of the questions themselves, I think you could answer yourself, or at least attempt. Please limit the questions to only a few at a time. You will get faster responses due to my availability, but by all means, if you have a question about a response.....ask. Sorry this took so long, but I've been sick with the flu and was out of town for a couple days. :)

Will start looking at the rest of the first batch and then the second after that. I'll add the rest of these as a follow-up to this page, so keep watch for those.

bluto :)



Question/Answer
kammy asked on 01/17/03 - help!

3) a tank travelling at 6 m/s east relative to the earth, fires a shell at 15m/s east 60 degrees south relative to the tank. with components, how do you find teh velocity of the shell with respect to the earth?
ans: 19 m/s (east 44 degrees south)

4) a ferris wheel is moving in a vertical circle with a constatn speed of 3.0m/s. people on the ride are a distance of 20m from the centre of rotation. the coaster turns in a clockwise manner.
a) what is the instantaneous velocity of a person when she is at a position located on the far left side of the ride?
ans: 3.0m/s up
b) what is the change in velocity as teh person moves from the far right position to the lowest position on the ride?

5) a helicopter hovering at 100m above the earth's surface (hovering is at rest relative to earth) there is a wind blowing at 5 m/s west. a small object is dropped from teh helicopter and falls to earth. due to the object's cone shape, it falls vertically with simple free fall mothion, but is pushed sideways at the velocity of the wind. how long does it take the object to reach the earth?
ans: 4.52 s



11) a 100 kg crate is acted upon by a 250 N horizontal force adn is observed to accelerate at 1.90 m/s^2. what is the magnitude of the frictional force acting on the crate?
ans: 60.0 N

12) how long would it take to lift a 12 kg bucket of water through a vertical distance of 28m starting from rest, if the tension of the rope doing the lifting was 125 N? (the bucket is not at rest when it reaches the 28m point)
ans: 9.5 s

13) what is the tension in the single cable supporting an empty 1300 kg elevator moving upwards at a constant velocity of 1.2m/s?
ans: 1.3 x 10^4N down

14) a 30 kg child standing in a moving elevator stands on a scale and has an apparent weight of 280 N. what is the acceleration (magnitude and direction) of the elevator?
ans: 0.47 m/s^2 down

15) a 20 kg wooden block is pushed across a floor by an applied force of 150 N acting at 30 degrees. the coefficient of kinetic friction between block and the floor is 0.40 calculate the magnitudes of blocks' acceleration?
ans: 1.1 m/s^2

bluto answered on 01/21/03:

Hi kammy....

Here are the remainder of your questions:


12)) Let the total force applied to the bucket be "F", the tension "T" and the weight of the bucket be "W".
Then the total force will be: F = T - W
Substitute your known values into this and you will find:
F = T - mg
==> F = 125N - 12 kg * 9.8 m/s^2
==> F = 7.4 N
Now, Since F = ma
==> a = F/m = 7.4 N/12 kg = 0.62 m/s^2
With this you can now plug this in to the standard 'freefall' equation....yes it works with objects are moving UP!
y = (1/2)gt^2 (I'm letting the positive direction be upward)
==> t = Sqrt(2y/g) plug in your numbers and you will find that:
==> t = 9.5 s
--------------------------------------------------------------------------------------------------------------------------

13)) This one is a little tricky because it's simple. You're told that there's a velocity of blah blah m/s. Find and dandy, but you don't need that! All you need to do is look at the weight of the elevator and the tension in the cable. These two quantities are opposite in direction, and must add up to zero. Otherwise, either there will be slack in the cable, or the cable will break!
So, you can write this as: T + W = 0
==> T = -W
Since W = mg (taking the positive direction downward this time)
==> T = -mg = -1300 kg * 9.8 m/s^2 = 1.27 x 10^4 N
-or-
About 1.3 x 10^4 N
Make sense?

[I'm not sure where you're getting the "down" direction for the tension]
--------------------------------------------------------------------------------------------------------------------------

14)) A child that has a mass of 30 kg has a weight of: W = mg = 30*9.8 = 294 N
So this tells us that the elevator must be moving AWAY from him/her in order to make then lighter than that. Make sense? So we know the direction from this fact.......downward.
Take the difference between these two forces: 294 N - 280 N = 14 N
Since F = ma, and you need to know "a" (acceleration) solve this for "a" and plug in your numbers.
==> a = F/m = 14 N/30 kg = 0.47 m/s^2
Do you see? Hope this is helping you.
--------------------------------------------------------------------------------------------------------------------------

15)) There's an important piece of information left out of this problem. You may want to check this over again and make sure tho. The piece of info I'm talking about is the direction of the applied force on the box. Is it 30 degrees upward, downward, left, right....

You need to know this so you can account for it, otherwise you will not get the correct answer for the particular situation.

In this problem I surmise the force must be acting downward at 30 degrees to the horizontal on the box.

Now we can proceed:

Find the horiz and vert. components of the applied force:

x = 150Cos(30) = 130 N
y = 150Sin(30) = 75 N

The resulting, total force "Ft" acting on the box will be the difference between the horizontal component of the applied force "Fh" and the frictional force "f".

Ft = Fh - f

Where: f - uN

u = 0.4
N = The normal force.

This is where you need to know which direction the original, applied force acts.

Usually, N is equal in magnitude, and opposite in direction to the object's weight. But when a force is applied at some angle, this can have an affect on this quantity.

Since, I'm guessing the applied force is acting downward, this will *increase* the value of the normal force. The amount by which this will increase is eqial to the vertical component of the applied force:

y = 75N

Therefore: N = mg + 75 N = 196 N + 75 N = 271 N

Then, taking the coefficient of friction (u) into account: f = uN = 0.4*271 N = 108.4 N

Since the only component of the applied force acting to move the box along the horiz direction is the value we found above:

x = 130 N

We can now plug our information in and find the resulting force acting on the box:

Ft = 130 N - 108.4 N = 21.6 N

Since we need to find the acceleration we can use Newton's 2nd law: F = ma:

Ft = ma
==> a = (Ft)/m = 21.6 N/20 kg = 1.08 m/s^2

-or-

About 1.1 m/s^2
-------------------------------------------------------------------------------------------------------------------------------

Ok....hope these helped you, but if not let me know....k?


bluto

kammy rated this answer Excellent or Above Average Answer

Question/Answer
kammy asked on 01/17/03 - help!

3) a tank travelling at 6 m/s east relative to the earth, fires a shell at 15m/s east 60 degrees south relative to the tank. with components, how do you find teh velocity of the shell with respect to the earth?
ans: 19 m/s (east 44 degrees south)

4) a ferris wheel is moving in a vertical circle with a constatn speed of 3.0m/s. people on the ride are a distance of 20m from the centre of rotation. the coaster turns in a clockwise manner.
a) what is the instantaneous velocity of a person when she is at a position located on the far left side of the ride?
ans: 3.0m/s up
b) what is the change in velocity as teh person moves from the far right position to the lowest position on the ride?

5) a helicopter hovering at 100m above the earth's surface (hovering is at rest relative to earth) there is a wind blowing at 5 m/s west. a small object is dropped from teh helicopter and falls to earth. due to the object's cone shape, it falls vertically with simple free fall mothion, but is pushed sideways at the velocity of the wind. how long does it take the object to reach the earth?
ans: 4.52 s



11) a 100 kg crate is acted upon by a 250 N horizontal force adn is observed to accelerate at 1.90 m/s^2. what is the magnitude of the frictional force acting on the crate?
ans: 60.0 N

12) how long would it take to lift a 12 kg bucket of water through a vertical distance of 28m starting from rest, if the tension of the rope doing the lifting was 125 N? (the bucket is not at rest when it reaches the 28m point)
ans: 9.5 s

13) what is the tension in the single cable supporting an empty 1300 kg elevator moving upwards at a constant velocity of 1.2m/s?
ans: 1.3 x 10^4N down

14) a 30 kg child standing in a moving elevator stands on a scale and has an apparent weight of 280 N. what is the acceleration (magnitude and direction) of the elevator?
ans: 0.47 m/s^2 down

15) a 20 kg wooden block is pushed across a floor by an applied force of 150 N acting at 30 degrees. the coefficient of kinetic friction between block and the floor is 0.40 calculate the magnitudes of blocks' acceleration?
ans: 1.1 m/s^2

bluto answered on 01/20/03:

Hi kammy....

It's usually hard for me to answer this many questions posed all at once. Maybe next time send half this many, or less, and then send them in lots like that until you have asked all you wish to ask. That way I can send you responses faster without having to add the remaining answers as follow-ups, which is what I'm going to do now.

I'm sending you what I have so you have *something* to go on. I'll add the rest as follow-ups, so keep checking back. If something doesn't make sense......let me know.
======================================================

3)) With this one you really need to find the horizontal(x) and vertical(y) components of the shell's velocity.

If you draw a right triangle on your paper such that the right angle is in the upper right corner. You will have the horizontal portionat the top as the horizontal component (x), and the hypotenuse as the 15 m/s shell velocity component. The 60 degree angle will be the angle in the upper left of that triangle.

So, here's how you find the horiz/vert. components:

x = 15Cos(60) = 15*0.5 = 7.5 m/s
y = 15Sin(60) = 15*Sqrt(3)/2 = 13 m/s

Note: [I normally leave off units while I do calculations so it's more clear, but YOU need to include them with all your work. So make sure you don't forget to include them when you're doing these. :)

Since the tank had a velocity of 6 m/s to begin with, you will have to add this to the components of the shells velocity components we just found. So the 'new' x and y values are:

x = 7.5 m/s + 6 m/s = 13.5 m/s
y = 13 m/s + 6 m/s = 19 m/s

---------------------------------------------------------------------------------------------------------------

4)) I'm not sure what level question this is, so let me know if I miss the mark with this:
An object rotating in a 20 m circle at a constant speed of 3 m/s will always possess an instantaneous velocity of 3 m/s around the circle. If not, the path would no longer be a circle! Here's why:
Stats: Circle diameter(r) = 20 m
Speed around circle(v) = 3 m/s
You can relate the angular velocity "w" to linear velocity "v" by: v = rw
Since w = 2(pi)fr Where: (pi) = 3.141592...blah blah blah, and f = the frequency of the rotation. Don't panic, that's just how many revolutions per unit time that is made.
==> v = 2(pi)r/T Where "T" is the time for one complete revolution: T = (distance)/(velocity)
==> T = 2(pi)r/v
==> v = 2(pi)r/[2(pi)r/v]
==> v = v
So, all this means is that the instantaneous velocity is the velocity given in the problem. That's because you're dealing with a circle and the velocity is constant.
The CHANGE in velocity is the following:
Let the radius from the center of the circle to any point on the outside be given as "r"....like this:
r = xi + yj Where "i" and "j" are unit vectors.
At the rightmost point y = 0, so let this be "r1", r1 = xi
Similarly, let the lowest point be "r2". Here x = 0 so you have: r2 = yj
The change in magnitude of velocity would be: v2 - v2.
==> v2 = 3 m/s, and v1 = 3 m/s
Since the magnitude of the velocity = constant 3 m/s, that means the change is zero, even tho the direction changes.
---------------------------------------------------------------------------------------------------------------

5)) This one is a little tricky, and subtle, but if you think.....it's not too bad:
Since the object will fall at the same rate regardless if it falls with or without wind is a subtle concept you can use to your advantage. As such, you can use the freefall equation directly to compute the required time:
y = (1/2)gt^2
==> t = Sqrt(2y/g) Plug in your numbers and you will get:
==> t = 4.52 s
---------------------------------------------------------------------------------------------------------------

11)) The total force on the object is the difference between the applied force and the frictional force:
Let Ft be the total force, F be the applied force and f be the frictional force. Then we have:
Ft = F - f
Since F = ma Both quantities of which are given....and....
f = uN Where u = the coefficient of friction, and N is the normal force....and...
Ft = 250 N
Cobine these and you get: Ft = ma - uN
m = 100 kg
N = -100 kg (this is the force equal to, but opposite to that of the object's weight)
a = 1.90 m/s^2
Therefore: u = (Ft - ma)/N = 60
---------------------------------------------------------------------------------------------------------------

More to follow, so check back!

Hope this helps, but lemme know otherwise.


bluto

kammy rated this answer Excellent or Above Average Answer

Question/Answer
s1n1st3r_ur9e asked on 12/15/02 - Decreasing current...

Would a 2.3 V bulb decrease more current than a 6.3Vbulb????? which bulb will decrease cureent more?

cheers,
Igor

bluto answered on 12/16/02:


Igor...

This is a bit ambiguous, so one can't conclusively say for certain which can stand the most current. However, if we assume a couple things we can come to an assumption, which is the following:

Let's say that: v1 = i1*r1 for the 2.3 V bulb....and....

v2 = i2*r2 for the 6.3 V bulb.

In other words: v1 = 2.3 V, and v2 = 6.3 V

If v2 > v1, then, i2*r2 > i1*r1

So...the possibilities are:

If r1 = r2, then i2 > i1

-or-

i2 and r2 are both greater than i1 and r1

The problem comes in with the resistances of each of the bulbs. We don't know what r1 and r2 are. If we assume they are equal, which is probably *not* true, then i2 would indeed be greater than i1, which means that the 6.3 V bulb could handle more current. But most of the time, the reason a bulb *can* handle more current is because

The reason one bulb could handle more current than another is because of the resistance in the bulb. One would have to know what the power rating is of the bulb.......Watts. Then, for say a 25 W bulb, with a maximum voltage rating of 6.3 volts, it would be able to handle a current of:

Power(in Watts) = iv

==> 25 W = i*6.3 V
==> i = 25/6.3 = 3.9 (Amps) [This would be our i2]

On the other hand, for a 2.3 V bulb we'd get:

==> i = 25/2.3 = 10.9 (Amps) [This would be our i1]

So, as you can see, it's hard to know for sure which is which unless there is more information. You could 'prove' it one way or the other, depending on which approach and what assumptions are being made.

Hope this helps....lemme know otherwise please.



b~

s1n1st3r_ur9e rated this answer Excellent or Above Average Answer

Question/Answer
stchitnis asked on 12/10/02 - Fluid Resistance

Hi!


I am doing a experiment to find the resistance provided water to a moving object.

The experiment is as follows:
I drop a sperical metal object(O) known mass(M) and dimentions(radius R) in the water and then I let it cover a certain distance(D) in a given time(T).
I can find the acceleration(A) of the object and the force(F) it experienced over the distance(D). Acceleration will be given by the fomulla:
A=2D/T2 (this is a square).
Force(F)=mass(M) X accleration(A).
The object will have acclerated by 9.80m/s2 (this is a square) if it were in vaccume. Therefore, the change in accleletation of the object will help us find the force exerted by the water.
But my question is that is this a good way to find 'drag' or fluid resistance.??




SSCFIRST

bluto answered on 12/10/02:

st....

I guess it depends on what you mean by, 'a good way to find resistance of the fluid'.

This is certainly an easy way to do it. Your most critical variable is the accuracy of the time. If you're not worried about superlative accuracy, then there isn't anything wrong with this method. It's easy, uncomplicated and it will give you a good 'feel' for the resistance offered by whatever fluid you choose.

Also, the object you drop should be as free from flaws as possible. But, again, for a 'back of the envelope' type experiment, there isn't anything wrong with this and a flaw or two won't hurt you much.

There are of course other methods for figuring out resistance that have been contrived over the years, but unless you have a need for something more sophisticated, what you have is fine.

Does this help? Lemme know otherwise.....

b~

stchitnis rated this answer Excellent or Above Average Answer

Question/Answer
Sepehr asked on 12/02/02 - laws of thermodynamics

Would the 1st and 2nd laws of thermodynamics disallow two spherical objects to orbit each other indefinitely? Would the system eventually "break down" and the two objects collide? Thanks.

bluto answered on 12/03/02:

Would the 1st and 2nd laws of thermodynamics disallow two spherical objects to orbit each other indefinitely? Would the system eventually "break down" and the two objects collide? Thanks.

Sephr...



The simple reasoning is....there is no such thing as a perpetual motion machine, period.



There is always friction, temperature fluctuations, minute changes in charge, position, mass, internal energy and so on for real objects. As such, mechanisms like friction, for example, rob the system of energy, causing changes which, over
time, would cause the two objects to spiral into each other and collide.



In theory, if one conjures up all kinds of fictitious scenarios, anything is possible. Even in deep space there is energy, and there is friction, and while some binary systems, which is what astronomical 'planetary' systems such as you described
are called, are so massive they can continue orbiting each other for millions of years. Even their orbits will erode and the bodies collide some day.



The first law alludes to the idea of efficiency, which is the ratio of the work out, to the heat put in to the system. A perfectly efficient system will be able to convert all the work out, from all the heat put in. If the system is closed,
containing only the two bodies then the heat must be taken from the system itself and put back in and converted completely as work. Cyclic systems the total internal energy....say...U is zero (because the system must always return to
the same state). As such, the heat in must equal the work: Q = W. While this may work fine and dandy for part of a cycle, it cannot hold for the entire process, otherwise this will violate the first law.



The second law incorporates the idea of entropy. It says, essentially that the entropy of the universe must increase. You may be able to take an isolated system and show that it's entropy is constant, or decreases, but the entropy of the rest
of the universe must account for those situations. For the two bodies to continuously rotate about each other without colliding, the entropy of an initial state, to a final state has to remain constant. For that to be true energy must flow
to/from the rest of the universe to maintain that equilibrium. If that's true, then the system cannot be considered closed. If the system isn't considered closed, then the entropy must be considered for the entire universe. Entropy must increase
with time for the universe. Otherwise, odd things would be possible. Clearly, this makes one have to decide whether to allow fictitious possibilities or not. To be considered a real system, one must allow for the fact that entropy increases.
To allow the two bodies to be in a closed system by themselves forces one to sat that entropy for the system is constant, but then the first law says that in order to do so, all the work generated by the system must be converted completely to work, which violates that law.



Hope this made sense. If not, lemme know and I will try and help.




b~

Derrex rated this answer Excellent or Above Average Answer
Sepehr rated this answer Excellent or Above Average Answer

Question/Answer
itz_jacki asked on 11/25/02 - Kinetic Energy in collision

When I collide two objects together, how come the kinetic energy after the collision is less than the kinetic energy before the collision? Does that mean the conservation of energy does not apply in my experiment? Or are there other things that I have to take into account?

bluto answered on 11/25/02:

jacki....


Most of the energy is lost due to friction/heat, some may be absorbed in your hands

(if you're holding the objects) or supporting apparatus.



Some of the energy may be absorbed into the objects themselves without necessarily

being significantly measureable.



Conservation of energy holds, it's just not all being accounted for is all.





bluto

itz_jacki rated this answer Excellent or Above Average Answer

Question/Answer
Dathaeus asked on 11/23/02 - Lumens, etc..

Hey guys, first let me thank you all for doing a great job on these science boards, hope all the kids appreciate your help.

I need some info for an experiment I am doing, and I just cant remember from my old engineering days, what is the unit to measure PRACTICAL brightness of light, because I know flashlights use candlepower and then now the white car lights use some unit to compare to sunlight (which is supposed to be around 4500), etc... I am dealing with white LED lights and I need a measurement for what will light up the brightest in the real world. You can disregard the angle of effectiveness because I already know about that part, just plain brightness, thanks in advance.

Dathaeus

bluto answered on 11/23/02:

Hi Dathaeus...... Thanks for the kudos! One of the units you're looking for is a unit of measure called the Candela...."cd" for short. It represents the total number of "Lumens" from the surface of some light source. Depending on the manufacturer, the output of a LED is somewhere in the neighborhood of ~700 mcd, which is 700 millicandelas. This will depend on mnfgr and color of the LED. This is a good seguay to the next unit you're asking about....Color Temperature. I'll spare you the gory details. Essentially the color of an object corresponds to the color of a hot ingot at a given temperature. Here's a rough guide that shows the relation of color to temperature: Some typical color temperatures are: 1500 k Candlelight 2680 k 40 W incandescent lamp 3000 k 200 W incandescent lamp 3200 k Sunrise/sunset 3400 k Tungsten lamp 3400 k 1 hour from dusk/dawn 5000-4500 k Xenon lamp/light arc 5500 k Sunny daylight around noon 5500-5600 k Electronic photo flash 6500-7500 k Overcast sky 9000-12000 k Blue sky The "k" in the temperatures refers to a scale we scientists like to refer to as the 'absolute temperature' scale and it's called "Kelvin". 0 k = -273.15 C = -459.67 F You may want to refer to the following websites for a bit more info: You have to keep in mind what the application will be used for. I say this because the huma eye is most sensative to the green end of the visible spectrum, and while it may appear that one light is 'brighter' than another, it's all quite relative, and dependent on the units you use to measure, and with what you use to take your readings with. There are radiometric units and photonic units. You cannot convert from one to the other. Units measured in one, cannot be correlated to those in the other. believe me.....I played that game. I've spent a good bit of time with LEDs and units and lumens oh my! Perhaps if you have more specific questions down the road I could be a little less nebulous, and more specific, but I wanted to make you aware of some things that are not usaully apparent to folks delving into light and color. It's quite a comlpicated field, but also a fascinating one. :) Hope this helps....lemme know otherwise..... bluto

Dathaeus rated this answer Excellent or Above Average Answer

Question/Answer
_JacquelineA asked on 11/20/02 - Pigmentation of skin

I read a true story of this man who was white and had changed the pigmentation of his skin to black during the 1950's to do an observation in the south so he could make history of what he's done. I would like to do the same so I can make an observation for myself in the present day and I'd like to then write a book on it. What I want to know is where can I find more information on doing this, what will the costs be, and exactly how do I do this. Thanks.

bluto answered on 11/21/02:

Here's another site to try: http://www.emedicine.com/derm/topic528.htm Looks pretty high-brow, but it may be useful also. b-

_JacquelineA rated this answer Excellent or Above Average Answer

Question/Answer
KoolPeace7 asked on 11/06/02 - Work and Energy

A 1.5 kg block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of an x axis is aplied to the block. The force is given by F(x) = (2.5 - x)i N, where x is in meters and the initial position of the block is x = 0. (a) What is the kinetic energy of the block as it passes through x = 2.0 m in Joules? (b) what is the maximum kinetic energy of the block between x = 0 and x = 2 m in Joules?

bluto answered on 11/11/02:

kool..... So you're saying the applied force is imaginary?? F(x) = (2.5 - x)i Newtons Or are you trying to convey a vector?? bluto

KoolPeace7 rated this answer Bad/Wrong Answer

exper   © Copyright 2002-2008 Answerway.org. All rights reserved. User Guidelines. Expert Guidelines.
Privacy Policy. Terms of Use.   Make Us Your Homepage
. Bookmark Answerway.