you are very much correct.. this is from kaplan's Qbank, and I just happened to cross this Q last night. The answer IS C (not B).

Do you mean the answer is not 3.8N?

Two objects A and B are placed on a spring, mass A has twice the mass of B. If the spring is depressed and released, object A will:

ExamKrackers states that they will both rise to the same height, but I'm having trouble making this connection. They state that mass is not proportional to height, but won't the elastic potential energy from the spring be converted into GPE (mgh), which will compensate heights due to the masses for each?

Can you post the problem statement and the answer key word for word?

A question.

When light travels into a denser medium, its velocity decreases (opposite to sound which velocity increases). The textbook says that the wavelength is the one that is changing, not the frequency. Why is that? I could not think of any intuitive answer and the text book does not really explain it.

Also, sound vs. light. Why sound is opposite to light? Sound is based in vibrations so the denser media should have a greater resistance to vibrations. Isually solids are made of heavier molecules than say air and more dense. I was thinking of Young's modulus. The only explanation is that the sound may travel faster in denser media, but it would not travel as far as in a less denser medium -- that's kind of a payback for the increased velocity.

Thank you.

A question.

When light travels into a denser medium, its velocity decreases (opposite to sound which velocity increases). The textbook says that the wavelength is the one that is changing, not the frequency. Why is that? I could not think of any intuitive answer and the text book does not really explain it.

Also, sound vs. light. Why sound is opposite to light? Sound is based in vibrations so the denser media should have a greater resistance to vibrations. Isually solids are made of heavier molecules than say air and more dense. I was thinking of Young's modulus. The only explanation is that the sound may travel faster in denser media, but it would not travel as far as in a less denser medium -- that's kind of a payback for the increased velocity.

Thank you.

Velocity=Frequency x wavelength

Frequency= velocity/wavelength

Frequency does not change in such a case- its just a matter of how nature behaves.

Using the formula, if velocity increases, wavelength should decrease accordingly to keep frequency constant.
-------------------------
Part B

Again, part of just how nature behaves. Speed of sound is directly proportional to the density of the material (ie speed of sound in steel>speed of sound in air). However, within a phase of a given material (ie solid/liquid/or gas), speed of sound is inversely proportional to density of that material.

Good luck on the MCAT. I took it when it was given in the paper format and eight hours long. Glad to hear theyve changed things.

A question.

When light travels into a denser medium, its velocity decreases (opposite to sound which velocity increases). The textbook says that the wavelength is the one that is changing, not the frequency. Why is that? I could not think of any intuitive answer and the text book does not really explain it.

Also, sound vs. light. Why sound is opposite to light? Sound is based in vibrations so the denser media should have a greater resistance to vibrations. Isually solids are made of heavier molecules than say air and more dense. I was thinking of Young's modulus. The only explanation is that the sound may travel faster in denser media, but it would not travel as far as in a less denser medium -- that's kind of a payback for the increased velocity.

Thank you.

Wavelength of any wave is determined by the medium. Frequency and amplitude of any wave are determined by the wave source.

Speed of any wave is proportional to the square root of the ratio of the elastic component to the inertial component. Sound travels faster in denser media because elastic component is greater in denser media. Here elastic component and inertial component are properties of the medium.

Note: electromagnetic waves don't require a medium in order to travel; in fact, a medium slows EM waves down. EM waves travel the fastest in a vacuum.

Its #67 in the EK physics book if you have that:

Objects A and B are placed on the spring as shown. Object A has twice as much mass as object B. If the spring is depressed and released, propelling the objects into the air, object A will:

A) rise 1/4 as high as object B
B) rise half as high as object B
C) rise to the same height as object B
D) rise twice as high as object B

Answer is C, reasoning:
Take the example to extremes. Here the examples are reasonably close in mass. If object A were 1 million times as massive as objective B, object B will not be propelled 1 million times more into the air than object A, therefore mass is not proportional to height. Since all answers are such, only C can be correct.

I have a problem with that reasoning. Since GPE is involved in the question, which relates mass and height, I try to picture the energy in 2 states: initial and at when both objects are some point in the air. Therefore you'll have GPE (of both masses) + KE (possibly) = Elastic Potential of Spring. From this I think the height one will rise in relation to the other is dependent on mass?

Its #67 in the EK physics book if you have that:

Objects A and B are placed on the spring as shown. Object A has twice as much mass as object B. If the spring is depressed and released, propelling the objects into the air, object A will:

A) rise 1/4 as high as object B
B) rise half as high as object B
C) rise to the same height as object B
D) rise twice as high as object B

Answer is C, reasoning:
Take the example to extremes. Here the examples are reasonably close in mass. If object A were 1 million times as massive as objective B, object B will not be propelled 1 million times more into the air than object A, therefore mass is not proportional to height. Since all answers are such, only C can be correct.

I have a problem with that reasoning. Since GPE is involved in the question, which relates mass and height, I try to picture the energy in 2 states: initial and at when both objects are some point in the air. Therefore you'll have GPE (of both masses) + KE (possibly) = Elastic Potential of Spring. From this I think the height one will rise in relation to the other is dependent on mass?

I don't have the 1001 questions books. What exactly does the picture show?
Is the spring depressed by exactly the same amount in both cases? U = 1/2 * k * x^2, so if x is different, the elastic potential energy will be different for A and B and therefore gravitational potential energy will be different for A and B (since elastic potential energy is converted to gravitational potential energy; I am assuming that objects are launched straight up). So they can rise to the same height. That's certainly possible. What exactly does the picture show?

Its #67 in the EK physics book if you have that:

Objects A and B are placed on the spring as shown. Object A has twice as much mass as object B. If the spring is depressed and released, propelling the objects into the air, object A will:

A) rise 1/4 as high as object B
B) rise half as high as object B
C) rise to the same height as object B
D) rise twice as high as object B

Answer is C, reasoning:
Take the example to extremes. Here the examples are reasonably close in mass. If object A were 1 million times as massive as objective B, object B will not be propelled 1 million times more into the air than object A, therefore mass is not proportional to height. Since all answers are such, only C can be correct.

I have a problem with that reasoning. Since GPE is involved in the question, which relates mass and height, I try to picture the energy in 2 states: initial and at when both objects are some point in the air. Therefore you'll have GPE (of both masses) + KE (possibly) = Elastic Potential of Spring. From this I think the height one will rise in relation to the other is dependent on mass?
Base your reasoning on the equations.

If you call the equilibrium point of the spring y=0, initial velocity=0, assume each mass is propelled from the same spring (read: same spring constant k), and x=the compression of the spring from equilibrium and assume x is different for A and B then:

(1/2)kx^2 + mg(-x) = (1/2)mv^2 + mgh

if A is twice B then:
A = 2m
B = m

Solve the equilibrium equation for h. Plug in 2m, Plug in m. Now you should have two slightly different looking equations solved for h. Set them equal to each other - since h reaches the same height for both - and you'll see that in order for them to reach the same height, mass B will compress the spring to approximately 0.707 the compression of mass A.

Therefore when the masses launch from a spring with the same constant k, they will rise in relation to the compression of the spring.


IF you assume that they are both placed on the spring at the same time and launch, then you plug in 3m instead. You'll see that for the masses to reach the same height, they need to be traveling the same velocity. No way.

Someone feel free to correct me if I'm wrong.:thumbup:

Base your reasoning on the equations.

If you call the equilibrium point of the spring y=0, initial velocity=0, assume each mass is propelled from the same spring (read: same spring constant k), and x=the compression of the spring from equilibrium and assume x is different for A and B then:

(1/2)kx^2 + mg(-x) = (1/2)mv^2 + mgh

if A is twice B then:
A = 2m
B = m

Solve the equilibrium equation for h. Plug in 2m, Plug in m. Now you should have two slightly different looking equations solved for h. Set them equal to each other - since h reaches the same height for both - and you'll see that in order for them to reach the same height, mass B will compress the spring to approximately 0.707 the compression of mass A.

Therefore when the masses launch from a spring with the same constant k, they will rise in relation to the compression of the spring.


IF you assume that they are both placed on the spring at the same time and launch, then you plug in 3m instead. You'll see that for the masses to reach the same height, they need to be traveling the same velocity. No way.

Someone feel free to correct me if I'm wrong.:thumbup:

Well, since energy is a scalar (not a vector), it cannot be negative. So I don't think -x is correct. But your overall approach looks OK. I think you just need to choose a different point as a zero reference point for gravitational potential energy...so the left side of the equation will have only elastic potential energy. Also, my understanding is that we want to know if the final heights are the same for objects A and B, so when they reach their maximum heights, they will no longer have any KE...so the right side should probalby have only graviational potential energy...

but then again, I have been known to be wrong on occasion

Well, since energy is a scalar (not a vector), it cannot be negative. So I don't think -x is correct. But your overall approach looks OK. I think you just need to choose a different point as a zero reference point for gravitational potential energy...so the left side of the equation will have only elastic potential energy. Also, my understanding is that we want to know if the final heights are the same for objects A and B, so when they reach their maximum heights, they will no longer have any KE...so the right side should probalby have only graviational potential energy...

but then again, I have been known to be wrong on occasion
Hmm...
I'm sticking by the -x, because the equation is basically saying that the energy used between the compression and equilibrium of the spring ((1/2)kx^2 + mg(-x)) is converted into KE plus the potential energy of whatever height it obtains.

I also can't think of another reasonable zero reference point, because at equilibrium you can seperate the initial and final mechanical energies to what happens while the mass is on the spring versus what happens when the mass is off the spring.

Also if you where looking at maximum heights, then you'd just get rid of the KE equation on the right side of the equation. In that case, masses will still rise in relation to the compression of the spring. The KE equation only factors in if the masses happen to reach the same height while they're still traveling upwards.

Both masses are on the same spring with a platform (neglect the platform)

|B| | A |
\ \ \ \ \ \ \
/ / / / / / /
\ \ \ \ \ \ \


IF you assume that they are both placed on the spring at the same time and launch, then you plug in 3m instead. You'll see that for the masses to reach the same height, they need to be traveling the same velocity. No way.

That's what I'm saying too, I don't understand why it's B.

Both masses are on the same spring with a platform (neglect the platform)

|B| | A |
\ \ \ \ \ \ \
/ / / / / / /
\ \ \ \ \ \ \



That's what I'm saying too, I don't understand why it's B.
perhaps I spoke too soon about the "no way" part. They start off with the same initial velocity and both are subject to gravity with this equation:

vf^2 = vi^2 -2gy, so basically if height y were the same, the vf is the same and vice versa


blah.

Hmm...
I'm sticking by the -x, because the equation is basically saying that the energy used between the compression and equilibrium of the spring ((1/2)kx^2 + mg(-x)) is converted into KE plus the potential energy of whatever height it obtains.

I also can't think of another reasonable zero reference point, because at equilibrium you can seperate the initial and final mechanical energies to what happens while the mass is on the spring versus what happens when the mass is off the spring.

Also if you where looking at maximum heights, then you'd just get rid of the KE equation on the right side of the equation. In that case, masses will still rise in relation to the compression of the spring. The KE equation only factors in if the masses happen to reach the same height while they're still traveling upwards.

I made a mistake. Potential energy can be negative. One obvious example is U = -GMm/r. Kinetic energy is always >= 0. E = mc^2 is always >=0, but potential energy can be negative...

Regarding this spring question, now that I understand the problem statement I'll post more comments later...

Hi everybody! I had a Capacitance/voltage question from Nova Physics - Chapter 15, Passage2, Question#5:

In experiment 1 we place 2 copper circular disks of area A1 a distance d1 apart, thus creating a capacitor with capacitance C1. We connect the two plates to oppostie terminals of a battery which produces a potential Vbat. This produces a positive charge Q1 & an electric field E1 between the plates.

In experiment 3 we reproduce the setup in Experiment 1. THen the wires are removed from the copper plates. We place cellulose nitrate (a dielectric with dielectric constant K=9) between the plates.

What is the magnitude of the potential difference between the plates @ the end of Experiment 3.
A. V1/9
B. 3V1
C. 9V1
D. 81V1

My take: I understand that because the wires are removed in experiment 3 before dielectric is put in, Q3 = Q1. The two equations are:
C1 = Q1*V1

C3 = Q3*V3; but C3 = 9C1 & Q1 = Q3
therefore,: 9C1 = Q1*V3

Divide: 9C1 = Q1*V3 by C1 = Q1*V1

(9C1 / C1) = (Q1 * V3) / (Q1 * V1)
9 = V3 / V1
9V1 = V3

So I said the answer was C but the answer is actually A. WHY?!?!?!?

^^ right... so it turns out I had a complete brain fart ... I looked @ the question again & it finally dawned on me that Q = VC NOT C=QV...

this is a bit frightening ... my exam's on May31 :'(:eek:

^^ right... so it turns out I had a complete brain fart ... I looked @ the question again & it finally dawned on me that Q = VC NOT C=QV...

this is a bit frightening ... my exam's on May31 :'(:eek:

:) I'm making stupid mistakes too, like thinking C[eq] {and not 1/C[eq]}= 1/C[1] = 1/C[2].

Spent maybe 5 minutes trying to figure how I got a problem testing this wrong :rolleyes:.

perhaps I spoke too soon about the "no way" part. They start off with the same initial velocity and both are subject to gravity with this equation:

vf^2 = vi^2 -2gy, so basically if height y were the same, the vf is the same and vice versa


blah.

The kinematics interpretation makes sense, but I still can't see it from the energy perspective. Is this something that should be looked at in terms of kinematics rather than work at first?

The kinematics interpretation makes sense, but I still can't see it from the energy perspective. Is this something that should be looked at in terms of kinematics rather than work at first?

for vertical speed v, an object's highest height is independent of its mass.

PE = mgy
KE = .5mv^2

begin with just KE, @highest point with just PE. so conserve energy:

.5mv^2 = mgy

m cancels out.
y = .5 v^2/g.

Since they both start with the same v, and g is the same for everybody on my planet's surface, they both have the same highest height y.

begin with just KE, @highest point with just PE. so conserve energy:

.5mv^2 = mgy


ahhh there you go, since it Uel goes all into KE, it is velocity dependent and mass independent. Thanks to everyone that helped.

Hello
Is that topic not covered in MCAT? I'll gladly forego that topic if we are not responsible.
Thanks!

Hey guys
So really easy question:
A ball thrown straight up in the air and returns to inital position. You are given the v, t, a (gravity). So in figuring out the total time the ball travels the equation is given as t= 2v/g. What I want to know is where this equation comes from?? How does one deriive it?
Thanx

Would 3H+ and 3He+ (the 3's are both superscript aka mass number) have the same mass?? If so what is the breakdown of p, n, e??



Thanx

Hey guys
So really easy question:
A ball thrown straight up in the air and returns to inital position. You are given the v, t, a (gravity). So in figuring out the total time the ball travels the equation is given as t= 2v/g. What I want to know is where this equation comes from?? How does one deriive it?
Thanx
yf=yi + vi*t - 0.5gt^2
since the ball returns to its initial position
yf=yi

thus we have 0=vi*t - 0.5gt^2
now calculate
vi*t = 0.5gt^2
vi=0.5gt
2vi/g = t

i think this is correct...

yf=yi + vi*t - 0.5gt^2
since the ball returns to its initial position
yf=yi

thus we have 0=vi*t - 0.5gt^2
now calculate
vi*t = 0.5gt^2
vi=0.5gt
2vi/g = t

i think this is correct...
cool. you rock. i was doing this :
height = v^2/2g (from KE = PE at max heigth)
then
h=vt (from x=vt)
vt=v^2/2g
t=v/2g
times by 2 for the time required to fall down
t=v/g which is incorrect. I think my initial mistake was assuming h=vt which is only true for horizontal motion.

Would 3H+ and 3He+ (the 3's are both superscript aka mass number) have the same mass?? If so what is the breakdown of p, n, e??



Thanx

H is always protons = 1
He is always protons = 2

In neutral species, protons = electrons.
For H+, you have 1 less electron than neutral.
So, H is electrons = 0.
Same thing with He
He electrons = 1.

Neutrons equal mass number - proton numbers.

3H+ = 2
3He = 1

Yeah, their masses would be just about equal, since neutrons have just about the same mass as protons (you also should consider mass defect, but there's not much difference there either).

For the optics equation, 1/f = 1/d[o] + 1/d[i], how do you determine if f is positive or negative?

For the optics equation, 1/f = 1/d[o] + 1/d[i], how do you determine if f is positive or negative?

converging (convex) lens +f

diverging (concave) lens -f

converging (convex) lens +f

diverging (concave) lens -f

Thanks, what about mirrors? Always positive?

Thanks, what about mirrors? Always positive?


converging (concave) mirror +f
diverging (convex) mirror -f

If a clown is catapulted straight up in air (which is 1.5 m off ground) at 4 m/s. How would you find the velocity at which he hits the ground at (0 meters)??? I was able to figure out his max height as 1.5 + 0.8 =2.3. I also know that it will be more than 4 m/s.
Thanks

If a clown is catapulted straight up in air (which is 1.5 m off ground) at 4 m/s. How would you find the velocity at which he hits the ground at (0 meters)??? I was able to figure out his max height as 1.5 + 0.8 =2.3. I also know that it will be more than 4 m/s.
ThanksUse (v final)^2 = v^2 + 2 a y

v = velocity at max height = 0

y = -2.3 (negative since using Cartesian Coordinate w/max ht as ref. pt.)

a = -10 (negative since direction is down) = gravitational acceleration

v final = velocity just before hitting the ground.

If a clown is catapulted straight up in air (which is 1.5 m off ground) at 4 m/s. How would you find the velocity at which he hits the ground at (0 meters)??? I was able to figure out his max height as 1.5 + 0.8 =2.3. I also know that it will be more than 4 m/s.
Thanks

Conserve energy. If he was shot up at 4m/s, then when he returns to where he was shot he will be going down at 4m/s, and use gy = .5v^2 to figure out what additional speed he gets in 1.5m.

initial e = final e
PE + KE = PE + KE
mgy + .5mv^2 = mgy + .5mv^2
m cancels out, fill in everything you can
10(1.5) + .5(4)^2 = 0 + .5v^2
15 + 8 = .5v^2
46 = v^2
v < 7m/s.

Hey guys
So really easy question:
A ball thrown straight up in the air and returns to inital position. You are given the v, t, a (gravity). So in figuring out the total time the ball travels the equation is given as t= 2v/g. What I want to know is where this equation comes from?? How does one deriive it?
Thanx


v = v0 + at
v = v0 - gt

0 = v0 - gt (since velocity at max height is zero)
t = v0/g

So now double this time because you want the time for the round trip.

If a clown is catapulted straight up in air (which is 1.5 m off ground) at 4 m/s. How would you find the velocity at which he hits the ground at (0 meters)??? I was able to figure out his max height as 1.5 + 0.8 =2.3. I also know that it will be more than 4 m/s.
Thanks

Chose positive y direction to be down. The magnitude of velocity at 1.5m
is the same whether the clown is going up or down.

v^2 = v0^2 + 2ay (where a is g)
v^2 = 4^2 + 2*10*1.5
V^2 = 46
v is a little less than 7

ahhh there you go, since it Uel goes all into KE, it is velocity dependent and mass independent. Thanks to everyone that helped.

The velocity of the masses at separation is the same because they are launched from the same platform.
Mv^2/2=Mgh, so you can see that M cancels out. The height is independent of mass.

Waves:

One end of a string is shaken each second sending a wave with an amplitude of 10 cm toward the other end. The string is 5 meters long, and the wavelength of each wave is 50 cm. How many waves reach the other end of the string in each 10 second interval?

A. 2
B. 5
C. 10
D. 50


Examkrackers said that the frequency that each wave is sent will be equal at the beginning and end of the string (1 Hz), which I agree with, therefore 10 waves are sent in 10 seconds.

But don't you have to take into account how long it takes for each wave sent to travel the 5 meters, meaning that you would have to first find the velocity of each wave, see how long each takes to go down the string, and then determine how many will reach the other side in 10 seconds?

Waves:

One end of a string is shaken each second sending a wave with an amplitude of 10 cm toward the other end. The string is 5 meters long, and the wavelength of each wave is 50 cm. How many waves reach the other end of the string in each 10 second interval?

A. 2
B. 5
C. 10
D. 50


Examkrackers said that the frequency that each wave is sent will be equal at the beginning and end of the string (1 Hz), which I agree with, therefore 10 waves are sent in 10 seconds.

But don't you have to take into account how long it takes for each wave sent to travel the 5 meters, meaning that you would have to first find the velocity of each wave, see how long each takes to go down the string, and then determine how many will reach the other side in 10 seconds?

There are often many ways to solve a given problem. I don't have the complete solution that EK provided for this problem, but here are my 2 cents:

If you are sending one wave per second, then in 10 seconds you are sending 10 waves.

v = f * lambda = 1 * 0.5 = 0.5 m/s (Unit check: m/s = 1/s * m).

velocity = distance/time

time = distance/velocity = (5 meters)/(0.5 meters/second ) = 10 seconds

So it takes one second for a wave pulse to travel the entire length of the string. Therefore, in 10 seconds, 10 wave pulses reach the other end of the string.

At time = 0 sec, wave pulse 1 is generated.
At time = 1 sec, wave pulse 2 is generated.

At time = 1 sec wave pulse 1 reaches the end of the string
At time = 2 sec wave pulse 2 reaches the end of the string
.
.
.
At time = 10 sec wave pulse 10 reaches the end of the string

There are often many ways to solve a given problem. I don't have the complete solution that EK provided for this problem, but here are my 2 cents:

If you are sending one wave per second, then in 10 seconds you are sending 10 waves.

v = f * lambda = 1 * 0.5 = 0.5 m/s (Unit check: m/s = 1/s * m).

velocity = distance/time

time = distance/velocity = (5 meters)/(0.5 meters/second ) = 10 seconds

So it takes one second for a wave pulse to travel the entire length of the string. Therefore, in 10 seconds, 10 wave pulses reach the other end of the string.

At time = 0 sec, wave pulse 1 is generated.
At time = 1 sec, wave pulse 2 is generated.

At time = 1 sec wave pulse 1 reaches the end of the string
At time = 2 sec wave pulse 2 reaches the end of the string
.
.
.
At time = 10 sec wave pulse 10 reaches the end of the string

The velocity calculated is the velocity for each wave, therefore wouldn't that mean that the 10 seconds is the time required for each wave to traverse the length of the string?

if you have the EK physics book ed.6, page 144 question 173.
It says light travels through a lens and hits point B. Point A is to the left of point B. They then ask what could be done to the lens to make the light hit point A. The correct answer choice is making the lens thicker. The wrong answer choices are bringing the lens closer to the light source, bringing the lens closer to the object (A/B), and making the lens thinner. Im having a really hard time understanding why making the lens thicker is correct and why the other answer choices are wrong. I understand that light wants to travel the shortest possible amount of time, but I dont see how that applies here. Thank you.

if you have the EK physics book ed.6, page 144 question 173.
It says light travels through a lens and hits point B. Point A is to the left of point B. They then ask what could be done to the lens to make the light hit point A. The correct answer choice is making the lens thicker. The wrong answer choices are bringing the lens closer to the light source, bringing the lens closer to the object (A/B), and making the lens thinner. Im having a really hard time understanding why making the lens thicker is correct and why the other answer choices are wrong. I understand that light wants to travel the shortest possible amount of time, but I dont see how that applies here. Thank you.

Hey! I don't have the EK book but I can offer an explanation behind the concept at hand. The purpose of a lense is to diffract light by changing its direction upon entering a lense. This is done by the geometric shape of the lense. Suppose you have light incident on a convex lense (converging lense) from the air. What is happening? The light is passing into a more dense medium of glass. This means the velocity of light will slow down because it is passing from a less dense medium (air) into a more dense medium (glass). By the principle of refraction, this means that light will bend toward the normal line of the lense. The normal line of the lense is the "thick" center of the lense by convention. Conversely, as the light exits the lense it is moving from a more dense medium to a less dense medium which means the light will bend away from the normal line of the lense. This means that the thicker the lense, the more light will bend toward the normal--the more it will be refracted. I hope this helps and good :luck:.

Hey! I don't have the EK book but I can offer an explanation behind the concept at hand. The purpose of a lense is to diffract light by changing its direction upon entering a lense. This is done by the geometric shape of the lense. Suppose you have light incident on a convex lense (converging lense) from the air. What is happening? The light is passing into a more dense medium of glass. This means the velocity of light will slow down because it is passing from a less dense medium (air) into a more dense medium (glass). By the principle of refraction, this means that light will bend toward the normal line of the lense. The normal line of the lense is the "thick" center of the lense by convention. Conversely, as the light exits the lense it is moving from a more dense medium to a less dense medium which means the light will bend away from the normal line of the lense. This means that the thicker the lense, the more light will bend toward the normal--the more it will be refracted. I hope this helps and good :luck:.
i kinda think i got it now, thanks alot.

The velocity calculated is the velocity for each wave, therefore wouldn't that mean that the 10 seconds is the time required for each wave to traverse the length of the string?

When you think of how fast a wave is moving, think of amount of time it takes for each wave pulse to go from point A to point B. If you shake the rope only once, you send a single wave pulse down the rope. If you keep shaking it, you send multiple wavepulses.

http://www.kettering.edu/~drussell/Demos/reflect/reflect.html

http://www.cord.edu/dept/physics/p128/lecture99_33.html

http://electron9.phys.utk.edu/phys135d/modules/m10/waves.htm

When you think of how fast a wave is moving, think of amount of time it takes for each wave pulse to go from point A to point B. If you shake the rope only once, you send a single wave pulse down the rope. If you keep shaking it, you send multiple wavepulses.

http://www.kettering.edu/~drussell/Demos/reflect/reflect.html

http://www.cord.edu/dept/physics/p128/lecture99_33.html

http://electron9.phys.utk.edu/phys135d/modules/m10/waves.htm

The way im thinking of it is like this:

A pulse is generated. The velocity that we calculated from f*lamba is a measure of the speed of the pulse traveling from where it originated to the end of the string. From that we see that it takes 10 seconds from origination to ending, so the first pulse will reach the end at 10 sec, the next at 11 sec, etc.

The way im thinking of it is like this:

A pulse is generated. The velocity that we calculated from f*lamba is a measure of the speed of the pulse traveling from where it originated to the end of the string. From that we see that it takes 10 seconds from origination to ending, so the first pulse will reach the end at 10 sec, the next at 11 sec, etc.

What's the answer key provided by EK?

C. 10

In the first interval, won't the answer be 1? Since it takes 10 s for the pulse to arrive?

C. 10

In the first interval, won't the answer be 1? Since it takes 10 s for the pulse to arrive?

Lets call the left end of the rope point A and the right end of the rope point B. It takes 10 seconds for each pulse to go from point A to point B

@ t = 0 pulse 1 is generated at point A
@ t = 1 pulse 2 is generated at point A
@ t = 2 pulse 3 is generated at point A
@ t = 3 pulse 4 is generated at point A
@ t = 4 pulse 5 is generated at point A
@ t = 5 pulse 6 is generated at point A
@ t = 6 pulse 7 is generated at point A
@ t = 7 pulse 8 is generated at point A
@ t = 8 pulse 9 is generated at point A
@ t = 9 pulse 10 is generated at point A
@ t = 10 pulse 11 is generated at point A, pulse 1 reaches point B
@ t = 11 pulse 12 is generated at point A, pulse 2 reaches point B
@ t = 12 pulse 13 is generated at point A, pulse 3 reaches point B
@ t = 13 pulse 14 is generated at point A, pulse 4 reaches point B
@ t = 14 pulse 15 is generated at point A, pulse 5 reaches point B
@ t = 15 pulse 16 is generated at point A, pulse 6 reaches point B
@ t = 16 pulse 17 is generated at point A, pulse 7 reaches point B
@ t = 17 pulse 18 is generated at point A, pulse 8 reaches point B
@ t = 18 pulse 19 is generated at point A, pulse 9 reaches point B
@ t = 19 pulse 20 is generated at point A, pulse 10 reaches point B
@ t = 20 pulse 21 is generated at point A, pulse 11 reaches point B
.
.
.
.
.
.
So I guess the answer to this question depends on interpretation. In the first 10 second interval only one wave pulse goes from point A to point B. In all subsequent 10 second intervals 10 wave pulses go from point A to point B.

There is a term in Computer Science called "full pipeline" and it has to do with parallelism (my undergrad and grad degrees are in Computer Science)...I guess EK wants an answer when the pipeline is full (I am just drawing an analogy here).

Maybe other people can comment on this question in case I am misunderstanding what's going on here...

EK says "in each 10 second interval"; im guessing that means time-independent, in other words, once waves are reaching the end already. in which case, math is unnecessary; if every second a wave is sent out, every second a wave has to arrive.

EK says "in each 10 second interval"; im guessing that means time-independent, in other words, once waves are reaching the end already. in which case, math is unnecessary; if every second a wave is sent out, every second a wave has to arrive.


That's certainly one way to look at it, but xlr8 (http://forums.studentdoctor.net/member.php?u=101603)'s point was that during the inital 10 seconds, it's not the case that 10 waves arrive at the end of the rope. I think EK should/could be more clear with their wording...

you are right- but in such a case on an exam, would we be right to assume that "each 10 second interval" only makes sense if you have multiple 10 second intervals, in which case, the only time you can have only one correct answer is if those multiple 10 second intervals all occur after the initial 10 seconds?

you are right- but in such a case on an exam, would we be right to assume that "each 10 second interval" only makes sense if you have multiple 10 second intervals, in which case, the only time you can have only one correct answer is if those multiple 10 second intervals all occur after the initial 10 seconds?

I would. But I hope AAMC would be clear enough as to what the question is asking so that we can keep the number of assumptions to a minimum...

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?

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.

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.

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.

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?

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.

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

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.)

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 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.)

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.

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 :luck:.

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 :luck:.

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.)

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 :o!

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

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.

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 :luck:.

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?

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.

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

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.

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

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

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.

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.

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*

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 :luck:.

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

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 :luck:.

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?

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,

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

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! :)

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?

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

[quote=jochi1543;5277898]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

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

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? :mad: 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
================================================== =========================

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? :mad: 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.

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?

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? :mad: 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.

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....

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.

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....

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 moron decided to take an unannounced 1-week vacation.:rolleyes: 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)?

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 moron decided to take an unannounced 1-week vacation.:rolleyes: 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..

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

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.

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 = ρVg, where ρ 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?

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.

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?

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 = ρVg, where ρ 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...

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.

Q 86: Which of the follwoing represents the maximum hieght h reached by the projectile?
A v2/2g
B 1/8 g t^2
C vt
D. vtsin(x)

Answer B, I m confused that why are they ignoring initial velocity in B. If I use Distance formula then answer should be = vsin(x)t+1/8gt^2

Q 86: Which of the follwoing represents the maximum hieght h reached by the projectile?
A v2/2g
B 1/8 g t^2
C vt
D. vtsin(x)

Answer B, I m confused that why are they ignoring initial velocity in B. If I use Distance formula then answer should be = vsin(x)t+1/8gt^2

Post the complete problem statement. Not everybody has EK books.

I am studying for EK Osmosis at the moment, and here's my question.

Here's the problem: It is about torque.

There is a 2 kg mass tied to the left end of a 1 m stick, and a 6 kg mass tied to the right end of the meter stick.

Solution: Jordan and Jon pick the left end of the stick or 2 kg mass as their lever arm.

Now here's my question: How do you find the direction of torque?

Jordan and Jon say that to find the direction of the torque, imagine that the rotation point on the left end of the stick is held motionless while we push the stick upwards--the stick would rotate counterclockwise, so torque is counterclockwise?

Confusion: Why would it rotate counterclockwise? I would think the the heavier block would pull it in its direction, and make it go clockwise.

Help, I am scared :scared:!

I am studying for EK Osmosis at the moment, and here's my question.

Here's the problem: It is about torque.

There is a 2 kg mass tied to the left end of a 1 m stick, and a 6 kg mass tied to the right end of the meter stick.

Solution: Jordan and Jon pick the left end of the stick or 2 kg mass as their lever arm.

Now here's my question: How do you find the direction of torque?

Jordan and Jon say that to find the direction of the torque, imagine that the rotation point on the left end of the stick is held motionless while we push the stick upwards--the stick would rotate counterclockwise, so torque is counterclockwise?

Confusion: Why would it rotate counterclockwise? I would think the the heavier block would pull it in its direction, and make it go clockwise.

Help, I am scared :scared:!

They pick the left end of the stick as the rotation point, not as lever arm.

What they mean is IF the only force acting were a force due to the person pushing the stick upward, the torque would be
counterclockwise.

The NET torque is zero, so the stick would not rotate.

BTW, there is a much easier way to do this problem. Since the ratio of weights is 6/2 = 3, the ratio of distances also has to
be 3, or 75cm/25cm (it's a meter stick). So the stick needs to be supported 75 cm away from the 2kg mass.

HTH

Time to close this thread up and start a new one. Please visit the Physics Question Thread 2 to ask your physics questions. :)