Physics question thread

[Shrike]

All users may post questions about MCAT and OAT physics here. We will answer the questions as soon as we reasonably can. If you would like to know what physics topics appear on the MCAT, you should check the MCAT Student Manual (http://www.aamc.org/students/mcat/studentmanual/start.htm), though be warned, there are subjects listed there that are rarely tested, or that appear in passages only and need not be learned.

Be sure to check the Physics FAQs and Topic Writeups thread if you have a general question; eventually, many of your answers will be located there. Also, a request: to keep this thread at least somewhat neat, when replying to someone else's post please refrain from quoting anything more than what's necessary for clarity.

Acceptable topics:

• general, MCAT-level physics
• particular MCAT-level physics problems, whether your own or from study material
• what you need to know about physics for the MCAT
• how best to approach to MCAT physics passages
• how best to study MCAT physics
• how best to tackle the MCAT physical sciences section

Unacceptable topics:

• actual MCAT questions or passages, or close paraphrasings thereof
• anything you know to be beyond the scope of the MCAT

Side note: anyone who knows how to post subscripts and superscripts in this system, please PM me the method. I also wouldn't mind knowing how to post some obvious symbols, such as Greek letters and an infinty sign. Should be a matter of changing fonts, but I don't know what's available; again, a PM would be appreciated.

If you really know your physics, I can use your help. If you are willing to help answer questions on this thread, please let me know. Here are the current official contributors to the this thread -- a team to which I hope to add more people:

Thread moderated by: Shrike. Shrike is a full-time instructor for The Princeton Review; he has taken the MCAT twice for no good reason, scoring 14 on the physical sciences section each time. He majored in mathematics, minored in physics, and spent several years accumulating unused school experience (in economics and law).

Also answering questions: Xanthines, a Kaplan instructor. He scored 13 on the PS section of the MCAT and 34 overall.

---

Members don't see this ad.

 

[PalmettoGuy]

From TPR's Practice MCATs Book:

An object is placed 6 cm from a convex lens whose radius of curvature is 4 cm. How far is the image from the lens?

A. 1.5 cm
B. 2.4 cm
C. 3 cm
D. 12 cm

Now I would have thought to solve this, use the lensmakers equation:

1/f=(1-n)[1/R1-1/R2]

Which isn't possible because we are not provided with the index of refraction of the lens or the radius of curvature R2.

According to the book just determine the focal length using f=1/2R and then use the thin lens equation 1/f=1/o+1/i.

If we were talking about a mirror, I would think that the above approach would work, however, since we are talking about a lens, the index of refraction has to be taken into account?

Who is wrong?

 

[NontradICUdoc]

From TPR's Practice MCATs Book:

An object is placed 6 cm from a convex lens whose radius of curvature is 4 cm. How far is the image from the lens?

A. 1.5 cm
B. 2.4 cm
C. 3 cm
D. 12 cm

Now I would have thought to solve this, use the lensmakers equation:

1/f=(1-n)[1/R1-1/R2]

Which isn't possible because we are not provided with the index of refraction of the lens or the radius of curvature R2.

According to the book just determine the focal length using f=1/2R and then use the thin lens equation 1/f=1/o+1/i.

If we were talking about a mirror, I would think that the above approach would work, however, since we are talking about a lens, the index of refraction has to be taken into account?

Who is wrong?

I believe that the lense maker's equation is used when you have an object infront of 2 lenses not one. In this example, you only have 1 lense. So you are incorrect.

 

[gridiron]

I believe that the lense maker's equation is used when you have an object infront of 2 lenses not one. In this example, you only have 1 lense. So you are incorrect.

Yep. You should use the mirror equation.

 

[BrokenGlass]

From TPR's Practice MCATs Book:

An object is placed 6 cm from a convex lens whose radius of curvature is 4 cm. How far is the image from the lens?

A. 1.5 cm
B. 2.4 cm
C. 3 cm
D. 12 cm

Now I would have thought to solve this, use the lensmakers equation:

1/f=(1-n)[1/R1-1/R2]

Which isn't possible because we are not provided with the index of refraction of the lens or the radius of curvature R2.

According to the book just determine the focal length using f=1/2R and then use the thin lens equation 1/f=1/o+1/i.

If we were talking about a mirror, I would think that the above approach would work, however, since we are talking about a lens, the index of refraction has to be taken into account?

Who is wrong?

f = r/2 = 2cm

i = (o*f)/(o-f) (we just solved 1/f=1/o+1/i for i since "it's all about the image")

Since we are dealing with a converging lens, focal lengh is positive. In single lens systems, object distance is always positive.

i = (6*2)/(6-2) = 3 cm

Focal distance of lenses is found using the lens maker's equation. However, f = r/2 is usually a good approximation for lenses, even though strictly speaking f = r/2 is for mirrors.

 

[wes431]

This has to do with air resistance. Okay so I understand resistance acts like friction. If an object is thrown, and air resistance is present, air resistance will slow it down shortening it's path.

Well what happens with mass of an object when air resistance is present. Lets say two balls with the same size and shape are thrown with the same initial velocity, what will happen to the ball with greater mass? Will it have a longer/shorter flight time and greater/less maximum height?

 

[BrokenGlass]

This has to do with air resistance. Okay so I understand resistance acts like friction. If an object is thrown, and air resistance is present, air resistance will slow it down shortening it's path.

Well what happens with mass of an object when air resistance is present. Lets say two balls with the same size and shape are thrown with the same initial velocity, what will happen to the ball with greater mass? Will it have a longer/shorter flight time and greater/less maximum height?

Mass doesn't affect the force of air resistance. It affects acceleration. a = F/m. The heavier ball has more inertia and air resistance will have less of an effect on the change in its velocity.

Also the force of air resistance is the same for these 2 balls only if they are travelling at the same velocity. When their velocities start to differ (which they will due to different accelerations), they will experience different force of air resistance.

 

[wes431]

So the more massive ball will have which one? longer/shorter flight time and greater/less maximum height?

 

[Dave_D]

So the more massive ball will have which one? longer/shorter flight time and greater/less maximum height?

Assuming they both start at the same velocity the more massive ball will fly longer and higher.(Another way to look at is that more massive ball has more kinetic energy. So air resistance has to do more work to affect it. So in order to do more work either the force has the increase or the distance the force is applied has to increase. Since the force is the same for two balls of the same size distance must increase.)

 

[wes431]

Well what would happen on the downward flight of the balls if velocity is the same? Wouldn't you say that the more massive ball will actually experience a shorter flight time and a greater max height? Taking this to extremes. A feather would have a lot of air resistance causing it to fall extremely slow while a metal ball would not take that long fall. But you just said it would have a longer flight time and greater max height.

 

[Dave_D]

Well what would happen on the downward flight of the balls if velocity is the same? Wouldn't you say that the more massive ball will actually experience a shorter flight time and a greater max height? Taking this to extremes. A feather would have a lot of air resistance causing it to fall extremely slow while a metal ball would not take that long fall. But you just said it would have a longer flight time and greater max height.

Well it is a little more complex like you've said. I mean if you shoot a bowling ball and balloon the same size as bowling ball the out of a balloon at say 15000' the balloon would take longer to hit bottom. Still the bowling ball would have a greater height and travel a further distance.(Well unless you shot them horizontal then the height would be the same. If you shot both straight down the distance would be the same though.) Actually without more info we can't say totally. (I mean you could always theorize about a situation where we give enough energy to put the bowling ball in orbit yet bleed enough energy off the lighter object so it doesn't go into orbit.)

 

[wes431]

so your saying if we threw a larger mass object in the air, it would have a greater maximum height and would be in the air longer. but on the way down if we were to compare with a smaller mass object, it would be faster and would be in the air for a less amount of time.

 

[gridiron]

so your saying if we threw a larger mass object in the air, it would have a greater maximum height and would be in the air longer. but on the way down if we were to compare with a smaller mass object, it would be faster and would be in the air for a less amount of time.

Hey! What you have to understand is the distinction between mass, time and height. It is important to understand these concepts because the MCAT can trick you with questions on them. First, the maximum height a object reaches in the air is independent on the mass of the object but dependent on the initial velocity it is given. To understand this, recollect the scalar quantities of potential and kinetic energy. From this:

mg(h) = 0.5m(v^2)

where h is the height attained by the object given an initial velocity v. The mass, m, can be eliminated from both sides. This shows the height attained is independent of the mass. This can also be shown using the kinematic equations like: y = vt + 0.5at^2. Use this for an example to understand this visually: take a tennis ball and throw it underhand into the air. Do this three times but each time throw it underhand with a greater velocity. What happens? The ball will go higher each time.

There is also a distinction between mass and time. If you were to throw two balls of different mass off a building, with the same intial velocity, they both will be in the air for the same amount of time. If you use the kinematic equation: v = vinitial + at, you can see that mass and time are indepenent. You can also see it visually if you were to take a tennis ball in one hand and ping pong ball in the other and drop them from the same height. What will happen? They both will drop at the same time. I hope this helps and good .

 

[Dave_D]

Hey! What you have to understand is the distinction between mass, time and height. It is important to understand these concepts because the MCAT can trick you with questions on them. First, the maximum height a object reaches in the air is independent on the mass of the object but dependent on the initial velocity it is given. To understand this, recollect the scalar quantities of potential and kinetic energy. From this:

mg(h) = 0.5m(v^2)

where h is the height attained by the object given an initial velocity v. The mass, m, can be eliminated from both sides. This shows the height attained is independent of the mass. This can also be shown using the kinematic equations like: y = vt + 0.5at^2. Use this for an example to understand this visually: take a tennis ball and throw it underhand into the air. Do this three times but each time throw it underhand with a greater velocity. What happens? The ball will go higher each time.

There is also a distinction between mass and time. If you were to throw two balls of different mass off a building, with the same intial velocity, they both will be in the air for the same amount of time. If you use the kinematic equation: v = vinitial + at, you can see that mass and time are indepenent. You can also see it visually if you were to take a tennis ball in one hand and ping pong ball in the other and drop them from the same height. What will happen? They both will drop at the same time. I hope this helps and good .

Well he was asking about this with air resistance though. (You're right that without air resistance they fall the same.) However with resistance things change. So for example if you threw a basketball and and balloon the same size (so the resistance is the same) at the same speed the basketball would go higher and farther.(Because it has more kinetic energy.)

Wes, as for time I'm going to say that depends. If the initial velocity is parallel or downward then the heaver ball will hit first.(Because air resistance on both is the same. Basically they both accelerate downward until air resistance cancels wt and that requires the heavier ball to go faster. IE higher terminal velocity.)

If you launch it upwards it really depends. It depends how high from the ground you are and how much resistance. So for example if you threw a balloon and a basketball at ground level I wouldn't be surprised if the balloon hit first.(Because it's not going to go up anywhere near as high as the basketball because of air resistance.) However if you were in say a hot air balloon at 10K feet and chucked both then I'd expect the basketball to hit first.(Because its terminal velocity is so much higher than a balloon's)

Actually gridiron come to think of it do they even ask about air resistance on the MCAT?(I know when I took it they didn't.)

 

[gridiron]

Well he was asking about this with air resistance though. (You're right that without air resistance they fall the same.) However with resistance things change. So for example if you threw a basketball and and balloon the same size (so the resistance is the same) at the same speed the basketball would go higher and farther.(Because it has more kinetic energy.)

Oops....I didn't read the initial posts !

Actually gridiron come to think of it do they even ask about air resistance on the MCAT?(I know when I took it they didn't.)

It isn't officially listed on the AAMC physical sciences topics for the MCAT. However, anything can be fair game. When I took the exam there was a passage on transformers--not the movie. That also isn't listed on topics for the MCAT, but it was on there. The best thing to know is the basics on what air resistance does---it is a force which opposes motion. It isn't necessary to know the full details because it can get tricky. For a list of AAMC topics for physical sciences:

http://www.aamc.org/students/mcat/topics.pdf

 

[jochi1543]

Wow, this has been killing me, and it's only a 2-star problem in my physics textbook.

A compass needle points 23 degrees E of N outdoors. However, when it is placed 12.0 cm to the east of a vertical wire inside a building, it points 55 degrees E of N. What are the magnitude and direction of current in the wire? The earth's field is 0.5 x 10(-4) T and is horizontal.


The answer is 19.4 A, down, but I can't seem to get there in any way.

 

[gridiron]

Wow, this has been killing me, and it's only a 2-star problem in my physics textbook.

A compass needle points 23 degrees E of N outdoors. However, when it is placed 12.0 cm to the east of a vertical wire inside a building, it points 55 degrees E of N. What are the magnitude and direction of current in the wire? The earth's field is 0.5 x 10(-4) T and is horizontal.


The answer is 19.4 A, down, but I can't seem to get there in any way.

Hey! The fact that you are told that the earth's field is horizontal is very important. That means once the compass is placed inside the building, there must be some field pointing downward, in the vertical direction, to force the compass needle further E of N. Thus, according to the right hand rule, the current must point downward in the vertical wire. Why? If the current is pointing downward, the field points in on west side of the wire and outward on the east side of the wire. So, the current moves down. Knowing this, you need to solve for the magnetic field produced by a downward carrying current wire that is produced 0.12 m away. You need to use the following formula:

B = ui/2piR

Where u is a constant, i is the current and R is the perpendicular distance from the wire. So, what magnetic field produced by a downward carrying wire will change the position of a compass needle by approximately 30 degrees? (32 exactly). You need to use vectors. The earth's field is horizontal but the field produced by the current carrying wire is vertical--points down. Thus, tan 30 = field produced by wire/field earth. Knowing this, you can solve for the magnetic field produced by the wire 0.12 m away. From this value, you know B, R, u so you should be able to solve for i. Try this method. If it doesn't work, there might be an alternative way. Tell me if it works. Also, in general, it is important to know the sine, cosine and tangent of 30, 45 and 60 degrees. I hope this helps and good .

 

[mrmilad]

Hmmm I am not sure if I am screwing up the calculations but I cant get your method to work Grid. For some reason I keep getting 52 A .... Wow this tricky sucker is a two star problem?

 

[jochi1543]

Hmmm I am not sure if I am screwing up the calculations but I cant get your method to work Grid. For some reason I keep getting 52 A .... Wow this tricky sucker is a two star problem?

I get 18.7...close to 19.4, but far enough not to make any sense.

 

[mrmilad]

Can you put up your calculations step by step, would appreciate it greatly.

 

[jochi1543]

Can you put up your calculations step by step, would appreciate it greatly.

tan 32 = 0.6248693519
This is equal to B wire : B earth, so B wire = B earth x tan 32... 0.5 x 10 (-4) x tan 32 = 3.12434676 x 10 (-5). So, now I set this equal to ui I/ 2 pi R. ui over 2 pi is equal to 2 x 10 (-7). So then I = B wire x R / 2 x 10(-7). Substitute 0.12 for R and use the B wire value from above, and it comes out to 18.746 A.


I e-mailed my prof....it's too far off from 19.4 for this to be a matter of significant figures lost somewhere.

 

[mrmilad]

Dope I was using tan 60.... Wow I need a major break.

 

[gridiron]

Sorry guys......let me think of an alternative approach. If anyone else has an idea please share.

 

[jochi1543]

Hey, I caved in and e-mailed the prof, and he said you can't use tan 32. I tried copy/pasting his solution, but it's not working since it's a PDF file - it gets all jumbled together. I'll go through this solution myself later tonight, and if anyone is still interested, I will rewrite it in Word so I can paste it here.

 

[BrokenGlass]

This has to do with air resistance. Okay so I understand resistance acts like friction. If an object is thrown, and air resistance is present, air resistance will slow it down shortening it's path.

Well what happens with mass of an object when air resistance is present. Lets say two balls with the same size and shape are thrown with the same initial velocity, what will happen to the ball with greater mass? Will it have a longer/shorter flight time and greater/less maximum height?

Air resistance does not depend on mass. They will initially experience the same resistance. Acceleration however will differ. The heavier one will decelerate at a slower rate. As the velocities change so will the forces of air resistance. The heavier object will go higher.

 

[jochi1543]

AAARGGGHHH! I need home lab help!

I got a bunch of parallel resistors on the breadboard, and I need to measure the overall resistance with the multimeter....but I can't figure out where in the hell to put the probes to get this value! To get resistance for just one resistor, it's straightforward - just attach the clips to the resistor. But how do I do this for 6 resistors?


GODDDDDD, I thought orgo labs were bad. *dies*

 

[gridiron]

AAARGGGHHH! I need home lab help!

I got a bunch of parallel resistors on the breadboard, and I need to measure the overall resistance with the multimeter....but I can't figure out where in the hell to put the probes to get this value! To get resistance for just one resistor, it's straightforward - just attach the clips to the resistor. But how do I do this for 6 resistors?


GODDDDDD, I thought orgo labs were bad. *dies*

Hey! Each resistor is in parallel, so each will be at the same potential difference. If all resistors are in parallel, the way I used to do it was connect the multimeter to each individually and measure the current going through each. The potential difference across each is the same so the resistance can be found. Make sure to connect the red lead to the battery and the black lead to the resistor. The process is a bit lengthy, but that is what I used in lab. Good .

Also, sorry I couldn't be of more help on the magnets question.

 

[jochi1543]

Hey! Each resistor is in parallel, so each will be at the same potential difference. If all resistors are in parallel, the way I used to do it was connect the multimeter to each individually and measure the current going through each. The potential difference across each is the same so the resistance can be found. Make sure to connect the red lead to the battery and the black lead to the resistor. The process is a bit lengthy, but that is what I used in lab. Good .

Also, sorry I couldn't be of more help on the magnets question.

Hmmmm....I'm supposed to make those resistance measurements WITHOUT there being a battery attached there. Is there a way to do this?

 

[dill_punjabi]

Is coefficient of friction is independent of area of contact between surfaces?
if so, then why friction increases with increase in area of contact.
f (friction) = u n n = normal force (which i think only depend on weight), u cofficient but it is constant!!!!
it's better if you can explain it in terms of kinetic friction

Thanks,

 

[BrokenGlass]

Is coefficient of friction is independent of area of contact between surfaces?
if so, then why friction increases with increase in area of contact.
f (friction) = u n n = normal force (which i think only depend on weight), u cofficient but it is constant!!!!
it's better if you can explain it in terms of kinetic friction

Thanks,

http://en.wikipedia.org/wiki/Friction

 

[pezzang]

Is it right to say that buoyant force is equal to the weight of the object as long as the density of the object is less or equal to the density of the water? I realize that when the densities of water and object are equal, w = F(b). How can we measure the weight of the object in the case where the density of the water is less or more than that of the object? Would it be possible to find w of the object without knowing the density of the object when the density of the water is less or more than that of the object? Thank you for any help!

 

[jochi1543]

Quick induction question....In the situation below, the current is decreasing. In which direction will the induced current in the loop flow? The way I see it, the applied magnetic field due to the straight wire is out of the page. Since I is decreasing, so is the applied magnetic field, so the induced current will flow in the direction that would cause the induced magnetic field to also point out of the page. The answer is clockwise, which makes total sense. HOWEVER - isn't it going to be out of the page if the current is counterclockwise, as well? Am I not using my hand correctly for the right-hand rule?



Clockwise current....fingers point at me, i.e. out of the page.



Counterclockwise....same deal? Or is it?

 

[BrokenGlass]

Is it right to say that buoyant force is equal to the weight of the object as long as the density of the object is less or equal to the density of the water? I realize that when the densities of water and object are equal, w = F(b). How can we measure the weight of the object in the case where the density of the water is less or more than that of the object? Would it be possible to find w of the object without knowing the density of the object when the density of the water is less or more than that of the object? Thank you for any help!

First, you may want to read this: http://forums.studentdoctor.net/showpost.php?p=2887364&postcount=16

Here are my 2 cents. There are 3 cases to consider. In what follows W = weight, p = density, Fb = buoyant force, N = normal force.
Fb is weight of the fluid displaced in all cases.

===========================================================================
*CASE 1* If p object < p fluid, then the object floats:

p object / p fluid = fraction of the object submerged.

If the fluid is water, p object / p fluid = specific gravity.

W = Fb, which is achieved by displacing less fluid BY VOLUME than the object's volume.

===========================================================================

*CASE 2* If p object = p fluid, then the object floats:

W = Fb, which is achieved by displacing the amount of fluid BY VOLUME equal to the object's volume.

===========================================================================

*CASE 3* If p object > p fluid, the object sinks:

p object / p fluid = W / Fb

W > Fb

When the object sinks all the way to the bottom, W = N + Fb
===========================================================================

 

[BrokenGlass]

Quick induction question....In the situation below, the current is decreasing. In which direction will the induced current in the loop flow? The way I see it, the applied magnetic field due to the straight wire is out of the page. Since I is decreasing, so is the applied magnetic field, so the induced current will flow in the direction that would cause the induced magnetic field to also point out of the page. The answer is clockwise, which makes total sense. HOWEVER - isn't it going to be out of the page if the current is counterclockwise, as well? Am I not using my hand correctly for the right-hand rule?



Clockwise current....fingers point at me, i.e. out of the page.


Counterclockwise....same deal? Or is it?




. . . . . . . . . . . . . . .
. . . . . . . . . . . . . . .

-----------------------------------current-------->

X X X X X X X X X X X X X X X
X X X X X X X X X X X X X X X

Using your right hand, point your thumb in the direction of the current. Your fingers curl in the direction of the magnetic field.
Think of dots as arrowtips, and think of the X's as arrow tails. So dots represent magnetic field that's out of the page and X's represent magnetic field that's into the page.


The strength of the magnetic field decreases with distance. So the loop of wire is affected more by the X's then by the dots,
so lets ignore the dots.

When current in the straight wire decreases, number of x's will also decrease. In other words, the number of magnetic field lines
running through the wire loop (i.e. flux) will also decrease. The wire will induce current in order to resist changes in flux (this is
sort of similar to La Chatelier's principle).

So which way is the induced current in the wire? Well, we want the current in the direction that will create X's inside the wire
loop. In other words, the magnetic filed inside the loop should be into the page. So the current in the wire loop should be clockwise (this follows from the right hand rule).

HTH

 

[jochi1543]

Thanks, BrokenGlass, it makes sense now. I'm gonna remove my pix now, since they take up so much screen space.

 

[pezzang]

Thank you, BrokenGlass. Another question about circuits. So if the total resistnace of a network increases as more resisters are added in series, do both I and V decrease? I am not sure how changing R would affect the values of I and V. I can see how I decreases because there is more reistance but should V (potential differnece) decrease as well or does it remain the same because changing R affects potential difference of both ends? Also the answer to one problem in which we try to find which circuit uses the least amount of power says that it the one with highest Resistance because I is the smallest by V = IR. So here why is V constant? I am just not sure in which case we can assume something is constant. Is there any easier way to understand this concept?
Any insights would be really appreciated! Thanks!

First, you may want to read this: http://forums.studentdoctor.net/showpost.php?p=2887364&postcount=16

Here are my 2 cents. There are 3 cases to consider. In what follows W = weight, p = density, Fb = buoyant force, N = normal force.
Fb is weight of the fluid displaced in all cases.

===========================================================================
*CASE 1* If p object < p fluid, then the object floats:

p object / p fluid = fraction of the object submerged.

If the fluid is water, p object / p fluid = specific gravity.

W = Fb, which is achieved by displacing less fluid BY VOLUME than the object's volume.

===========================================================================

*CASE 2* If p object = p fluid, then the object floats:

W = Fb, which is achieved by displacing the amount of fluid BY VOLUME equal to the object's volume.

===========================================================================

*CASE 3* If p object > p fluid, the object sinks:

p object / p fluid = W / Fb

W > Fb

When the object sinks all the way to the bottom, W = N + Fb
===========================================================================

 

[jochi1543]

Thank you, BrokenGlass. Another question about circuits. So if the total resistnace of a network increases as more resisters are added in series, do both I and V decrease? I am not sure how changing R would affect the values of I and V. I can see how I decreases because there is more reistance but should V (potential differnece) decrease as well or does it remain the same because changing R affects potential difference of both ends? Any insights would be really appreciated! Thanks!

V=IR. So if R increases, the current will decrease so as to keep V constant (V is usually constant by virtue of a battery that provides a constant V). If R doubles, the current will be halved.

 

[jochi1543]

OK, is this a trick question?

2 rails are x meters apart and carry the same current I, but in opposite directions. One section of the rail is y meters long. What is the magnitude and direction of the total force acting between the rails along one complete section?


It's zero, isn't it? The rails repel each other because the currents are the same magnitude, but in opposite directions...but they repel each other with the same force, and while Rail 1 will exert F12 to the right on Rail 2, Rail 2 will exert F21 of the same magnitude to the left on Rail 1 - so the net force is 0. Correct?

 

[BrokenGlass]

Thank you, BrokenGlass. Another question about circuits. So if the total resistnace of a network increases as more resisters are added in series, do both I and V decrease? I am not sure how changing R would affect the values of I and V. I can see how I decreases because there is more reistance but should V (potential differnece) decrease as well or does it remain the same because changing R affects potential difference of both ends? Also the answer to one problem in which we try to find which circuit uses the least amount of power says that it the one with highest Resistance because I is the smallest by V = IR. So here why is V constant? I am just not sure in which case we can assume something is constant. Is there any easier way to understand this concept?
Any insights would be really appreciated! Thanks!

By definition, P = I*V. You can remember this as VIP spelled backwards.

If we use Ohm's law, i.e. V = I*R, we get P = V^2/R and also P = I*R^2. For resistors connected in parallel, V is the same (this follows from Kirchoff's loop rule), and for resistors connected in series, I is the same.

So for 2 unequal resistors connected in parallel, the one with greater resistance uses less power. For 2 unequal resistors connected in series, the one with greater resistance uses more power.

Now, there could be a question which asks "what happens to the power if we replace this resistor with a resistor of greater resistance?" Voltage will remain the same because the battery is the same. V = I*R and so if R goes up, I goes down.

From P = I*V, we can see that P goes down.

 

[BrokenGlass]

OK, is this a trick question?

2 rails are x meters apart and carry the same current I, but in opposite directions. One section of the rail is y meters long. What is the magnitude and direction of the total force acting between the rails along one complete section?


It's zero, isn't it? The rails repel each other because the currents are the same magnitude, but in opposite directions...but they repel each other with the same force, and while Rail 1 will exert F12 to the right on Rail 2, Rail 2 will exert F21 of the same magnitude to the left on Rail 1 - so the net force is 0. Correct?

No, it's not zero. Newton's third law forces never act on the same object, so they can never cancel each other out. Only forces which act on the same object could cancel out each other.

I believe the answer is F = B perpendicular * I * L, where

where

• B is the external, perpendicular magnetic field measured in Tesla,
• I is the current measured in amps, and
• L is the length of the current segment (in meters) that lies in the external magnetic field, B.

What does the answer key say? Maybe I am misunderstanding what's being asked....

 

[jochi1543]

No, it's not zero. Newton's third law forces never act on the same object, so they can never cancel each other out.

It says "total force," though, so it looks like it's F12 AND F21 together, and since they are the same magnitude and in opposite directions, I think it should be zero for Fnet.

 

[BrokenGlass]

It says "total force," though, so it looks like it's F12 AND F21 together, and since they are the same magnitude and in opposite directions, I think it should be zero for Fnet.

It cannot be zero. The only forces that can possibly cancel each other are (external) forces which act on the SAME object (or system of objects).

Can you post the entire question and the answer key? Otherwise, it's hard to know what the question's asking....

 

[jochi1543]

It cannot be zero. The only forces that can possibly cancel each other are (external) forces which act on the SAME object (or system of objects).

Can you post the entire question and the answer key? Otherwise, it's hard to know what the question's asking....

I posted the entire question (just took out the #s for redundancy purposes). There is no answer, it's just a practice problem I decided to do and wish I didn't...I e-mailed my tutor, but the ***** decided to take an unannounced 1-week vacation. And I don't wanna wait that long to find out the solution.


So then you're saying I should just calculate F12 and F21 (which are the same in magnitude) and just add the magnitudes? So basically, the total force will be double the value of F12 (or F21)?

 

[BrokenGlass]

I posted the entire question (just took out the #s for redundancy purposes). There is no answer, it's just a practice problem I decided to do and wish I didn't...I e-mailed my tutor, but the ***** decided to take an unannounced 1-week vacation. And I don't wanna wait that long to find out the solution.


So then you're saying I should just calculate F12 and F21 (which are the same in magnitude) and just add the magnitudes? So basically, the total force will be double the value of F12 (or F21)?

It only makes sense to add force vectors if they are acting on the same object. If you mean that there could be some charged particle between the wires and both wires are exerting a force on this particle, then the forces could cancel out because they are both acting on this charged particle and they are both external to this charged particle. But the force ON each wire will not be zero..

 

[146233]

It only makes sense to add force vectors if they are acting on the same object. If you mean that there could be some charged particle between the wires and both wires are exerting a force on this particle, then the forces could cancel out because they are both acting on this charged particle and they are both external to this charged particle. But the force ON each wire will not be zero..

The question asks for the 'total force'. Given that the currents are equal in magnitude and opposite in direction, the force acting on each wire will be equal in magnitude to the force it exerts on the other. As such, the "total force", if that's what the question truly asks, between the two wires would be the sum of the magnitudes of each individual force. The two wires would repel each other.

-z

 

[BrokenGlass]

The question asks for the 'total force'. Given that the currents are equal in magnitude and opposite in direction, the force acting on each wire will be equal in magnitude to the force it exerts on the other. As such, the "total force", if that's what the question truly asks, between the two wires would be the sum of the magnitudes of each individual force. The two wires would repel each other.

-z

That would be like asking "what's the total force if the harness is pulling on the horse with the force of 100 N?" There is no way in hell any physicist worth his/her salt would ask a question like that. When a question asks about "total force" it has to state explicitly or implicitly what it means, i.e. total force on what? And yes, wires will repel each other (that's how the rail gun works), but that's beside the point.

 

[jmart]

not sure if this is in the right place anyways,

In the kaplan review notes online it talks about the buoyant force.

Here is what it says in a few paragraphs

The magnitude of the buoyant force is equal to the weight of the fluid displaced; Fb = &#961;Vg, where &#961; is the density of the fluid and V is the submerged volume of the object.

ok, I get that part but the next they throw in another equation that is equal to the one above and that is where they lose me.

Any fluid applies a buoyant force to an object that is partially or completely immersed in it; the magnitude of the buoyant force equals the weight of the fluid that the object displaces.

now for the odd equation

F(buyoncy) = (Wfluid) / (Wfluid) = pvg

Is that a mistake? I don't understand the weight of fluid part. I'm guessing one of those is supposed to be from something else but from the wording I can't tell what that is.

Any ideas?

 

[146233]

That would be like asking "what's the total force if the harness is pulling on the horse with the force of 100 N?" There is no way in hell any physicist worth his/her salt would ask a question like that. When a question asks about "total force" it has to state explicitly or implicitly what it means, i.e. total force on what? And yes, wires will repel each other (that's how the rail gun works), but that's beside the point.

Agreed, it's ridiculous.

 

[pezzang]

TPR Q:
Assume e- in deuterium can be viewed as orbitting nucleus in uniform circular motion. If k is coulomb's const, e is the charge mag. of the electron and r is the radius of the orbit then what's the KE of the electron.

The answer says Fc = Fe so ke^2 / (2r) is equal to KE. How about Fm? Shouldn't magnetic force be accounted for because the change e is moving and thus creates current and magnetic force?

 

[BrokenGlass]

not sure if this is in the right place anyways,

In the kaplan review notes online it talks about the buoyant force.

Here is what it says in a few paragraphs

The magnitude of the buoyant force is equal to the weight of the fluid displaced; Fb = &#961;Vg, where &#961; is the density of the fluid and V is the submerged volume of the object.

ok, I get that part but the next they throw in another equation that is equal to the one above and that is where they lose me.

Any fluid applies a buoyant force to an object that is partially or completely immersed in it; the magnitude of the buoyant force equals the weight of the fluid that the object displaces.

now for the odd equation

F(buyoncy) = (Wfluid) / (Wfluid) = pvg

Is that a mistake? I don't understand the weight of fluid part. I'm guessing one of those is supposed to be from something else but from the wording I can't tell what that is.

Any ideas?

F buoyancy = W fluid displaced = mg = pVg (since m = pV)

I think that's what they meant...

 

[BrokenGlass]

TPR Q:
Assume e- in deuterium can be viewed as orbitting nucleus in uniform circular motion. If k is coulomb's const, e is the charge mag. of the electron and r is the radius of the orbit then what's the KE of the electron.

The answer says Fc = Fe so ke^2 / (2r) is equal to KE. How about Fm? Shouldn't magnetic force be accounted for because the change e is moving and thus creates current and magnetic force?

k*q1*q2/r^2 = m*V^2/r (because electrostatic force is the centripetal force in this case).

Remember that centripetal force is just the NET force that makes an object undergo uniform circular motion.

In hydrogen, the charge on the nucleus is equal in magnitude to the charge of the electron since in hydrogen the
nucleus has only 1 proton. Deuterium is a hydrogen isotope, which means that it too has 1 proton in its nucleus
(but it has a different # of neutrons).

k*e*e/r^2 = m*V^2/r

k*e^2/r = m*V^2

k*e^2/2*r = (m*V^2)/2

so KE = k*e^2/2*r

Now onto your question. We are interested in forces acting ON the electron. Whether this electron creates a magnetic field
or not is irrelevant because the force due to this field is not an external force as far as this electron is concerned. We are only interested in EXTERNAL forces acting on an object when we try to figure out a NET force acting on this object.