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.

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[Shrike]

Um, I thought I said after the fourth, folks.

Thanks for the confidence. I'll write something up soon, but not immediately. If you want to suggest a topic, please PM me.

 

[gujuDoc]

Shrike,

I hope you have a great vacation. I have a quick request for when you get back.....

And I would prefer to get an answer from you specifically, because I know that I can trust your explanations better then some of these other people......

The first is, whether you could tell me whether that other person's answer to my question about work and its relation to kinetic energy and potential energy is true, or explain it more clearly in conceptual terms.

I understand the calculus and algebra of how they derive the kinetic energy formula from work and what not, but......

my friend started asking me for the conceptual reasons as to why they are related, and I didn't know how to answer her. She's taking the test in August.

Secondly,

I know you said that when you get a chance you would explain how torque works. Again, this was a section that confused me beyond all means and I didn't know how to explain it to her, so was wondering if you could explain it when you get a chance.

Last but not least, I would like to thank you in advance for your responses because they have been helpful in the past and I tend to think will be helpful in the future.

Have a great fourth of July weekend and vacation.

 

[Nutmeg]

Okay sorry to stay stuck on this one but how does a dielectric increase the charge without affecting the electric field? I am still not clear on this one... am I missing something?

The dielectric is composed of an insulating material that will shield the capacitor plates from one another. You end up with increased charge, which should raise the electric field, yes; but the dielectric will polarize in the opposite direction, making the net electric field reduced. Hence, the charges needed to sustain the same voltage, because they are sheilded from one another, must be greater.

Imagine needing twice as many trumpters to blow their horns to get the same volume of sound in an adjacent room when the door is shut versus when it is open.

 

[freesolo]

i suck at optics i know it, im always confused between mirrors and lenses, concave mirrors are converging , and convex mirrors are diverging, but converging lenses are convex, and diverging lenses are concave? Do i have that right? I get all the signs stuff it should be plug and chug but i always manage to mess it up.
oh and kaplan doesnt have the 1/f=Power on our formula sheet but we still need to know it right

 

[Shrike]

i suck at optics i know it...

These questions are answered, I think, in my optics writeup in the FAQ thread. Please check it, and if there's a subject I didn't cover adequately, come back and ask here.

 

[Shrike]

The first is, whether you could tell me whether that other person's answer to my question about work and its relation to kinetic energy and potential energy is true, or explain it more clearly in conceptual terms.

I understand the calculus and algebra of how they derive the kinetic energy formula from work and what not, but......

The calculus is irrelevant, of course, but the answer of which you speak is pretty much right on -- work is change in KE, as long as you count all the forces (including gravity). PE is there to make it look like nerergy is conserved -- it's just a mathematical convenience, which is why it's difficult to understand conceptually.

 

[gujuDoc]

The calculus is irrelevant, of course, but the answer of which you speak is pretty much right on -- work is change in KE, as long as you count all the forces (including gravity). PE is there to make it look like nerergy is conserved -- it's just a mathematical conveninece, which is why it's difficult to understand conceptually.

Thanks a bunch for clearing that up. That's what I thought and what someone else had said, but one of the other poster's kept confusing me.

I'll give the response to the friend that asked me. Oh and are we allowed to post actual TPR sci workbook questions here if we have a question on a specific problem???

 

[Shrike]

... are we allowed to post actual TPR sci workbook questions here if we have a question on a specific problem???

Yes.

 

[gujuDoc]

Yes.

Cool thanks.

 

[doctorjoy]

This is not an actual MCAT question from a book. I t is a practice problem from a Physics book that I am using.
A solid wooden cube 30.0cm on each edge, can be totally immersed in water if it is pushed doenward witha force of 54.0N What is the density of the wood? They give an answer of 800kg/m^3. I have tried working this problem backwards and forwards and do not come up with the same answer. Can someone help me with this please

 

[freesolo]

so you dont recommend the ray diagram technique? My kaplan teacher said all i need to do is ray diagrams to answer the questions. It seems like it would be faster, but the diagrams come out skewed when i draw them.

 

[gujuDoc]

Shrike,

I was wondering if you knew if there were any mistakes in the TPR sci workbook Solutions for Physics??

On page 337 there is a question that says

A 2 meter long organ pipe, closed at one end is resonating at its 5th harmonic. How many times greater is the resonant frequency then the fundamental resonant frequency. (Speed of sound through air is 340 m/s)

A. 1.25
B. 2.5
C. 5.0
D. 10.0

It says the answer is C, but this is where the confusion presides..........

Shouldn't n=9, since the resonant harmonic numbers go up as 1,3,5,7, etc. etc.

Your help is much appreciated. Thanks for clarifying. BTW, this is for a friend. She shall create a username soon. Thanks,
guju

 

[freesolo]

im having trouble with pascals principle

In the hydraulic lift shown below, piston A has a
cross-sectional diameter of 2 m and piston B a
cross-sectional diameter of 4 m. If the force
exerted by piston A on the liquid is doubled, the
force lifting up piston B is:


answer is doubled, im plugging 2F/2=XF/4 and getting the 2nd force as being 4 times less.

 

[frany584]

this is probably a silly question but I was doing a problem in my physics book, and i was wondering in which cases should I used the formulas 1/2at^2=x and v(t)=x to solve for the distance? Why can't I use either one and get the correct answer? Thanks!

 

[Shrike]

the hydraulic lift shown below, piston A has a
cross-sectional diameter of 2 m and piston B a
cross-sectional diameter of 4 m. If the force
exerted by piston A on the liquid is doubled, the
force lifting up piston B is:

answer is doubled, im plugging 2F/2=XF/4 and getting the 2nd force as being 4 times less.

Your problem is you're answering the wrong question. You are right that the force on piston B is four times that on A, but that isn't the question; the question is what happens to B when A is changed. Pressure is the same everywhere, so doubling it at A doubles it at B. Force, for a given area, is proportional to pressure.

 

[Shrike]

this is probably a silly question but I was doing a problem in my physics book, and i was wondering in which cases should I used the formulas 1/2at^2=x and v(t)=x to solve for the distance? Why can't I use either one and get the correct answer? Thanks!

You can use either one. Whichever you have, a or v (as long as it's constant), use it.

 

[Shrike]

so you dont recommend the ray diagram technique? My kaplan teacher said all i need to do is ray diagrams to answer the questions. It seems like it would be faster, but the diagrams come out skewed when i draw them.

I don't recommend it, but I suppose if it worked it could be helpful for some questions. The problem is that it won't give you numerical answers, which are what's required a lot of the time.

 

[Shrike]

A solid wooden cube 30.0cm on each edge, can be totally immersed in water if it is pushed doenward witha force of 54.0N What is the density of the wood? They give an answer of 800kg/m^3.

The easiest way to think about a cube of this size is in liters -- a liter is 10x10x10 cm. Hence our cube is 3x3x3 = 27 liters. 27 liters of water would have a mass of 27 kg, hence would weigh 270N; that's therefore the magnitude of the buoyant force when it's submerged.

270N is balanced against the weight of the cube + 54N, so the cube weighs 216N, and has a mass of 21.6kg. 21.6kg/27L = 0.8kg/L, or 0.8 times that of water. It's then easy to convert to whatever units you want: 0.8 x 1000kg/m^3 = 800kg/m^3.

For more on the use of liters in doing this sort of problem, see the FAQ thread.

 

[Lindyhopper]

Hi,
Not sure if this is right forum but, here goes.
I a first time bio teacher for TPR. This week we will cover the nervous system. The bio review pretty much skips any discussion of how or why myelination speeds the velocity of conductance. I do not want to go beyond the already detailed curriculum.
My remembrance is that myelin increases the resistance of the membrane &, therefore, reduces its capacitance. If we compare this with an unmyelinated axon; in the unmyelinated (larger capacitance) more charge must be deposited on the membrane to change the potential across the membrane, so the current must flow for a longer time to produce a given depolarization.
I'm trying to come up with a quick, simple analogy that might possible be helpful in both their physics & bio review.
I was thinking, perhaps, of comparing the unmyelinated axon to a soaker hose & the myelinated to a normal garden hose. But this has problems. The ions charging the capacitor do not normally flow through the membrane. I'm also not sure if the water flowing through the soaker hose flows significantly more slowly.
Any ideas are appreciated.

 

[Shrike]

... a first time bio teacher for TPR. This week we will cover the nervous system. The bio review pretty much skips any discussion of how or why myelination speeds the velocity of conductance. I do not want to go beyond the already detailed curriculum....
Any ideas are appreciated.

This is really a bio issue, though as you observe it has physics ramifications.

The answer is clear to me, as far as the class goes -- this is beyond the scope of the MCAT. (In general, most rate stuff is BSM.) Unfortunately, no analogy I can think of is both helpful and even remotely accurate. I suggest just saying that the signal jumps along, and that this makes it go faster -- they'll accept it.

For an actual answer to the question (which I don't know off the top of my head), try posting in the BSM thread.

 

[gujuDoc]

Your problem is you're answering the wrong question. You are right that the force on piston B is four times that on A, but that isn't the question; the question is what happens to B when A is changed. Pressure is the same everywhere, so doubling it at A doubles it at B. Force, for a given area, is proportional to pressure.

Hey did you get a chance to answer my question about harmonics, that my friend asked me to ask you???

She's confused because she thinks that when it says n=odd numbers, that means that if it says 5th harmonic, this should mean.......

that it will be the fifth odd number up starting from 1, such that you get 1,3,5,7,9 making n = 9. Is she misunderstanding something about harmonics??

Also, are there wrong solutions in the science workbook???

 

[Kussemek]

I have a question that asks: If T denotes the total time of flight of the projectile, which of the following expressions correctly gives the initial vertical velocity?

a) gT/2
b)gtsin45
c)gt
d)2gt

I was able to eliminate b and d. I selected c because I know the projectile equation is v-y=v-initial-gt.

The book answers iot as gt/2 and explains that I need to think of it as the point at the top of the trajectory and thus t is divided by two. I dont get why I need to look at it from the top of the trajectory. Where did I go wrong?

 

[smiley98]

Hi I have a question regarding momentum... if two guns pointed towards each other and at the same angle they shoot bullets simultaneously and the bullets collide with each other and fall to the ground... is momentum conserved? The answer says no because of the gravitional force working on the system... I thought in ineleastic or elastic collisions momentum is always conserved... why is this answer right? Thanks in adavance for the help!

 

[Nutmeg]

Hi,
Not sure if this is right forum but, here goes.
I a first time bio teacher for TPR. This week we will cover the nervous system. The bio review pretty much skips any discussion of how or why myelination speeds the velocity of conductance. I do not want to go beyond the already detailed curriculum.
My remembrance is that myelin increases the resistance of the membrane &, therefore, reduces its capacitance. If we compare this with an unmyelinated axon; in the unmyelinated (larger capacitance) more charge must be deposited on the membrane to change the potential across the membrane, so the current must flow for a longer time to produce a given depolarization.
I'm trying to come up with a quick, simple analogy that might possible be helpful in both their physics & bio review.
I was thinking, perhaps, of comparing the unmyelinated axon to a soaker hose & the myelinated to a normal garden hose. But this has problems. The ions charging the capacitor do not normally flow through the membrane. I'm also not sure if the water flowing through the soaker hose flows significantly more slowly.
Any ideas are appreciated.

QofQuimica: could you move this answer and the question to the bio thread or the BSM thread? Thanks.

If you're looking for an anology to understand this by, i'll give you the one I aalways use for wave propagation. Imagine you have a huge, heavy, chain and a long, unassembled fiberglass tent pole (the kind you use for a dome tent, where you have lengths about a foot long, connected by a single elastic cord that runs through the whole length) of equal length. The tent pole will sag a bit, yes, but the chain sags completely. Now imagine trying to hold one end and getting the other end to respond to a quick snap--basically, you're trying to whip it. With the chain, it will take enormous effort to get any degree of propagation because there are so many places where the movement gets transferred. With the tent pole, each segment is relatively stiff, and movement of one section sends propagates the energy much further than does a single chain link.

I usually think of these sorts of matters in terms of stiffness when trying to propagate a wave. You end up with a continuum, from a heavy steel chain, to a cotton jump rope, to a stiff nylon rope, to a bamboo pole, to a steel rod. The things that matter are the number of joints (more jointed things will be slower and take more nergy), the weight (the heavier it is, the more energy it needs), and the stiffness. In the case of the axon, the myelin makes something have fewer "joints" in that instead of a constant in-out flux down the entire length of the membrane, the local depolarization at the nodes of Ranvier can cause brief ionic diffusion along the length of the axon. The "weight" in this situation is analogous to the amount of charge required to propagate the signal. The decreased capacitance means that less charge is needed, as you state above. Likewise, the heavy chain needs more work to get the same pulse size as the light chain. Finally, there is the stiffness. In the world of ropes and rods, a region of something stiff will be very like neighboring regions. Something floppy can easily bend, and it doesn't transfer energy as quickly or efficiently. The internodal regions, in this sense, are very "stiff". They do not have local variations because they can't communicate with the extracellular space very well, so a when one area changes in polarity, since the membrane cannot pass ions to or from the extracellular space, the adjacent intracellular regions will pass inons instead. You've reduced the degrees of freedom--change will go up or down the length of the axon, but not in and out of the cell.

Or, if all that was too complicated, imagine a chain of people playing telephone. If you place people at two meters apart and have them relay a message, one to the next, it will take significantly longer to send a message one kilometer than it would if you had people passing a message in the same manner spaced 25 meters apart.

Hope that helps (and that I didn't just make everything worse).

 

[Nutmeg]

I have a question that asks: If T denotes the total time of flight of the projectile, which of the following expressions correctly gives the initial vertical velocity?

a) gT/2
b)gtsin45
c)gt
d)2gt

I was able to eliminate b and d. I selected c because I know the projectile equation is v-y=v-initial-gt.

The book answers iot as gt/2 and explains that I need to think of it as the point at the top of the trajectory and thus t is divided by two. I dont get why I need to look at it from the top of the trajectory. Where did I go wrong?

I don't have the workbook, but I'm assuming that the projectile starts at ground level and ends at the height where it started.
Since V(t) = Vinitial - g*t, we try to find a scenario to solve for Vinitial. When the projectile is at its highest point, there is an instant when the vertical velocity is exactly zero. So if we call the time needed to get to the apex Tapex, then that means that, setting V(Tapex) = 0 gives 0 = Vinitial - g*Tapex, or Vinitial = g*Tapex.

But Tapex is not the same as T (total time of flight). It is exactly one half, since it should take as much time to go up as it takes to go down.

If you think in terms of conservation of energy, KE = (1/2)*m*v^2, and the PE at the start and finish is zero. The path of the projectile upward converts kinetic energy from velocuty in the vertical direction into potential energy, and then back into kinetic energy. So at the end of the flight, assuming that velocity in the x direction is constant, than the velcity in the vertical direction at the end should be equal and opposite to the velocvity at the beginning. From this perspective, V(T) = -Vinitial, so

-Vinitial = Vinitial -g*T
-2*Vinitial = -g*T
Vinitial = (1/2)g*T

 

[Shrike]

Hey did you get a chance to answer my question about harmonics, that my friend asked me to ask you???

No, I don't know what's wrong with that one. May be a misprint.

 

[Shrike]

Hi I have a question regarding momentum... if two guns pointed towards each other and at the same angle they shoot bullets simultaneously and the bullets collide with each other and fall to the ground... is momentum conserved? The answer says no because of the gravitional force working on the system... I thought in ineleastic or elastic collisions momentum is always conserved... why is this answer right? Thanks in adavance for the help!

Please see the discussion of conservation laws in the FAQ. According to AAMC, momentum is conserved in the collision but not when the bullets fall. The reason, they say, is the external force from the Earth; in fact, it is conserved, but they're refusing to recognize the Earth's movement.

 

[RoxyKaur]

Hi pplz......well i have two questions

1) i can do most of the physics....but i have problem calculating the math part...i know of only two methods...one is just doing the streight math and the other one is rounding the numbers and then when it comes to the answer round down

2) when do u use kinamatics equations or d=st

i was working thorugh my kaplan book and this was the questions i was originaly using v= Xo + VoT + (1/2)at^2

sorry ill get to the point the question is....A submarine sends out a sonar signal in a direction directly downward. it takes 2.3s for the sound wave to travel from the submarine to the ocean bottom, and back to the submarine. How high up from the ocean floor is the submarine? (The speed of sound in water is 1,490m/s)

i know this is a simple questions but i get suck in which one to choose D=st or v= Xo + VoT + (1/2)at^2

but any how i just wanted to say u guys are the best so much love from me always

 

[smiley98]

Okay I don't know if this question is appropriate on this site but can anyone go over the concept of question 9 on the kaplan topical on work, energy and momentum... I don't think I would ever have thought of doing it the way they did it in their explanation. I know sometimes there is more than one way to look at a problem and I am hoping to have it explained in a different light, maybe more concept based. Thanks!

 

[Nutmeg]

Hi pplz......well i have two questions

1) i can do most of the physics....but i have problem calculating the math part...i know of only two methods...one is just doing the streight math and the other one is rounding the numbers and then when it comes to the answer round down

I'm afraid all I can recommend is the way to Carnegie Hall: practice, practice, practice.

2) when do u use kinamatics equations or d=st

i was working thorugh my kaplan book and this was the questions i was originaly using v= Xo + VoT + (1/2)at^2

sorry ill get to the point the question is....A submarine sends out a sonar signal in a direction directly downward. it takes 2.3s for the sound wave to travel from the submarine to the ocean bottom, and back to the submarine. How high up from the ocean floor is the submarine? (The speed of sound in water is 1,490m/s)

i know this is a simple questions but i get suck in which one to choose D=st or v= Xo + VoT + (1/2)at^2

but any how i just wanted to say u guys are the best so much love from me always

The key to the conundrum is the "a" for acceleration. Sound does not accelerate, so you don't use that formula.

 

[Nitya2284]

k i need some major help..So so far i have been reading the princeton review chapters and Im doing EK 1001 physics problems which correspond to the chapter i read. I couldn't do nearly half the problems. They go into so much detail, it's ridiculous. Do you think I should go to the library and check out the EK Physics book and then do the problems.. I don't know what to do and it's quite frustrating to know that you understand the material but can't answer the questions in the book..please let me know

 

[RoxyKaur]

I'm afraid all I can recommend is the way to Carnegie Hall: practice, practice, practice.

2) when do u use kinamatics equations or d=st

i was working thorugh my kaplan book and this was the questions i was originaly using v= Xo + VoT + (1/2)at^2


The key to the conundrum is the "a" for acceleration. Sound does not accelerate, so you don't use that formula.

but the problem i guess im having is what about the acceleration due to gravity?....that is why i was thinking about using a kinamatics equation....thanks

 

[Nutmeg]

but the problem i guess im having is what about the acceleration due to gravity?....that is why i was thinking about using a kinamatics equation....thanks

The sound wave does not "fall" and does not accelerate. Sound travels through a physical medium, and if that medium is static (as we would assume the water with the submarine should be) then there is no gravitational acceleration.

You should expect sound to travel the same speed both up and down.

 

[faluri]

In the case of myopia, the image falls too far in front of the retina, right? So you would need a diverging lens to adjust the image onto the retina. The opposite is the case for hyperopia. However, how does this variance effect the focal length. I read in an answer key that for individuals with myopia, their focal length is too short. What is the relationship there? I can't seem to make the connection.


August MCATer testing sans Physics II

 

[freesolo]

some question aamc was like why does sound we hear get weaker as we go through the wall, i put down it slows down as it goes through the wall b/c of greater index of refraction but the answer was something about wavelength decreasing.
Frequency dosnt change for sound right, so c is constant too and wavelength only changes? so thats why sound is weaker?

 

[Righty123]

In the case of myopia, the image falls too far in front of the retina, right? So you would need a diverging lens to adjust the image onto the retina. The opposite is the case for hyperopia. However, how does this variance effect the focal length. I read in an answer key that for individuals with myopia, their focal length is too short. What is the relationship there? I can't seem to make the connection.

August MCATer testing sans Physics II

The lenses in our eyes are converging lenses (fat in the middle). A lens, essentially, has two surfaces through which light travels, therefore, there are two focal points (one on each side of the lens, which are usually the same distance in a symmetric lenses.) The rule is that if an object is located before the focal point on the side through which light rays enter the lens, so that the setup is: 1. object, 2. focal point, 3. lens, then the rays all meet (converge) at the focal point on the other side (real side) of the lens, and that is where the image will be formed. In the case of myopia, the focal length is too small (where the image will be produced), meaning that the image does not fall directly onto the retina.

Hope that sorta helps!

 

[Shrike]

some question aamc was like why does sound we hear get weaker as we go through the wall, i put down it slows down as it goes through the wall b/c of greater index of refraction but the answer was something about wavelength decreasing.
Frequency dosnt change for sound right, so c is constant too and wavelength only changes? so thats why sound is weaker?

Neither frequency or wavelength is going to change a sound's volume. The volume of a sound changes when it gets to a wall because (1) some of the sound reflects off the wall; and (2) some of the remaining sound energy is absorbed by the wall (and the energy converted to thermal energy).

You are correct about speed not changing. Frequency changes only when there is a doppler effect, from a moving source or bouncing off a moving object, or perceived by a a moving detector. None of this has anything to do with the stregth of sound waves.

 

[Shrike]

In the case of myopia, the image falls too far in front of the retina, right? So you would need a diverging lens to adjust the image onto the retina. The opposite is the case for hyperopia. However, how does this variance effect the focal length. I read in an answer key that for individuals with myopia, their focal length is too short. What is the relationship there? I can't seem to make the connection.

See the FAQ on lenses. You are correct about myopia. In the lens equation, 1/o + 1/i = 1/f, i is too small so 1/i is too big; this is because 1/f is too big, or f too small.

The corrective lens makes 1/f be what it should be to get the right i, in other words, to focus the image on the retina. Adding a diverging lens (f < 0) will do this. Use the additivity of lens power (power = 1/f) to figure out how much of an effect you get.

As I said, see the FAQ.

 

[MarzMD]

I thought I understood how to do these problems, until I found this:

A cube, composed of substance X and having a mass of 50g and side length of 5cm, hangs from a string while fully submerged in saltwater(p=1.1g.cm^3). The tension in the string is 11N. What is the specific gravity of substance X?

The way the book explains the problem assumes that the value of m in your free body diagram of the forces is the mass of X. This makes no sense to me. I thought the value of m would just be the mass of the block.

 

[snakeplissken6]

there was this crazy pully question in the EK books, with two pulleys, same wt, one with twice the tension of the other. what i dont understand is the answer. the question was to find the tension in the rope.


ther answer was T + m2a= mg because mass will accelerate at twice the acceleration of the other mass. why?

for the second pully it was 2T=mg+ ma. and then you substitute in.

why is it m2a? i cant reason it out.

 

[snakeplissken6]

I thought I understood how to do these problems, until I found this:

A cube, composed of substance X and having a mass of 50g and side length of 5cm, hangs from a string while fully submerged in saltwater(p=1.1g.cm^3). The tension in the string is 11N. What is the specific gravity of substance X?

The way the book explains the problem assumes that the value of m in your free body diagram of the forces is the mass of X. This makes no sense to me. I thought the value of m would just be the mass of the block.

doesnt it have to do with bouyant force= mass of fluid displaced? so the value of m would be fluid displaced by substance x? and if they want the sg of x, its sgx/sgfluid ....i think. was this in the ek books ..or..?

 

[Shrike]

I thought I understood how to do these problems, until I found this:

A cube, composed of substance X and having a mass of 50g and side length of 5cm, hangs from a string while fully submerged in saltwater(p=1.1g.cm^3). The tension in the string is 11N. What is the specific gravity of substance X?

The way the book explains the problem assumes that the value of m in your free body diagram of the forces is the mass of X.

This problem is screwed up. As written, the specific gravity is both 0.4 and 10:

5x5x5cm = 125cm^3. 50g/125cm^3 = 0.4g/cm^3, or 0.4 times as much as water.

On the other hand, the forces supporting the block are 11N from teh string, and the buoyant force, which is rho x g x V, or 1.1g/cm^3 x 9.8 x 125, or 1.35N (correcting for grams vs. kilograms). Total force = 12.4, approx; weight = 12.4/125cm^3, or about 0.1N/cm^3. Thus mass = 10g/cm^3, and specific gravity = 10.

 

[Shrike]

there was this crazy pully question in the EK books, with two pulleys, same wt, one with twice the tension of the other. what i dont understand is the answer. the question was to find the tension in the rope.

I think it's safe to say we're going to need a little more by way of problem description. I have no idea, from what you've said, what the setup in the problem is.

 

[MarzMD]

This problem is screwed up. As written, the specific gravity is both 0.4 and 10:

5x5x5cm = 125cm^3. 50g/125cm^3 = 0.4g/cm^3, or 0.4 times as much as water.

On the other hand, the forces supporting the block are 11N from teh string, and the buoyant force, which is rho x g x V, or 1.1g/cm^3 x 9.8 x 125, or 1.35N (correcting for grams vs. kilograms). Total force = 12.4, approx; weight = 12.4/125cm^3, or about 0.1N/cm^3. Thus mass = 10g/cm^3, and specific gravity = 10.

From the way they word it, I solved it using your first method. I understand using Fb=Weight of liquid displaced, but I still fail to see why you cant just use Newtons second law to solve for Fb. If the acceleration=0, all you need to do is say netF= 11+Fb-mg=0. I guess the main problem is that I fail to see why mass in the above equation is not the value that they give you in the problem. The fact that it is composed of substance X should not make a difference, the cube still has a mass of 50g, so the density of the cube should be the same as the density of X(whatever that is).

 

[Righty123]

An electric dipole consists of two charges, +Q and -Q, where Q = 4microC, separated by a distance of d=20cm. Fine the electric field at the point midway between the charges.

The answer says that you just add up the respective E-files due to the principle of superposition. Therefore it says E=2kQ/((0.05d)^2)

My question is that since the the charge is midway between two opposing charges shouldn't the E-fields cancel out since when plugging into the equation for E-field: kQ/r, a + and - value of Q needs to be plugged in and since everything is the same besides the same, when you add it would equal zero.


Also in a pendulum, why is the tension in the string during an oscillation larger than mg? The explanation says that at the bottom of the oscillation, tension equals mg plus centripetal force. Centripetal acceleration is toward the center of the circle, meaning force of tension would have to be in direction of mg and then to counteract mg, F sub t would be in the direction of F sub c. I guess I just don't understand the explanation given and was hoping I could get some clarification.

Thanks!

 

[SensesFail]

An unknown solid weighs 31.6N. When submerged in water, its apparent weight is 19.8 N. What is the specific gravity of the unknown sample?

I never know where to start on these random Fluid discretes (and there always seems to be at least 1 of these on every Full Lenght I've taken!) I draw out a diagram, usually end up doing W-Fb = ma but then get lost from there.

 

[frankrizzo18]

From AAMC 3R: This is the provided solution to the question (Based on information in the passage how many centuries will be required for Mercury's perihelion to precess 360 degrees?) We know only 500 arcsec/century. I want to know how an arcminute is 1/60 of a degree? Thanks

The perihelion will have moved through an angle &#952; = &#937;·t after a time t has elapsed. The conversion of angles is given by the following. An arcsecond is 1/60 of an arcminute. An arcminute is 1/60 of a degree. The time for 500 arcseconds per century to accumulate to &#952; = 360 degrees is 360/(500/(60 x 60)) = 360 x 60 x (60/500) centuries. Thus, answer choice C is the correct answer.

 

[Nutmeg]

An electric dipole consists of two charges, +Q and -Q, where Q = 4microC, separated by a distance of d=20cm. Fine the electric field at the point midway between the charges.

The answer says that you just add up the respective E-files due to the principle of superposition. Therefore it says E=2kQ/((0.05d)^2)

My question is that since the the charge is midway between two opposing charges shouldn't the E-fields cancel out since when plugging into the equation for E-field: kQ/r, a + and - value of Q needs to be plugged in and since everything is the same besides the same, when you add it would equal zero.

It seems like it sould be zero to me. Don't know what to say about that.

Also in a pendulum, why is the tension in the string during an oscillation larger than mg? The explanation says that at the bottom of the oscillation, tension equals mg plus centripetal force. Centripetal acceleration is toward the center of the circle, meaning force of tension would have to be in direction of mg and then to counteract mg, F sub t would be in the direction of F sub c. I guess I just don't understand the explanation given and was hoping I could get some clarification.

Thanks!

I'm going to take a stab at this, but feel free to ignore me if I only make you more confused.

Imagine for a moment that you are looking at a weight on a string that's bouncing, and imagine that you are also looking at a pendulum swinging, but that you're looking at it from the side, so that it's swinging toward and away from you. Now let's imagine that the string length, the masses of the weights, and the spring constant, etc., are all chosen so that the period of the oscillation is the same for both, and that the maximum and minimum heights of each weight is the same.

From your perspective, they should look the same. Each weight should rise and fall in lock-step.

Now, when the spring-ball is at the bottom, it is easy to think that the tension on the spring must be greater than mg. If the tension were only mg, then the ball would sit motionless at the bottom. The added potentional energy exists in the stretching of the spring, and this potential energy will convert to kinetic energy in lifting the ball up again. When it gets to the top, the potential energy comes from it's height. This is how an oscilation works: conservation of energy, which converts from one form, to another, then back. In the case of the spring, it's potential (height) into kinetic (falling down) to potential (of the spring tension) to kinetic (the ball springing back up) and back to the start. This is a bit easier to understand intuitively than the pendulum, but it's the same principle.

With the pendulum, let's imagine that the center of rotaion is the center of a clock, where the pendulum swings from eight o'clock, down to six, up to four, and then reverses back to six, and up to eight. At eight o'clock and four o'clock, the ball has no kinetic energy--it pauses motionless for an infinitessimally short moment. All of it's energy is potential, from the height. At six o'clock, there is no potential energy. All of it's energy is kinetic, as it reaches it's maximum speed. At the time of it being at the bottom, the direction of the velocity is perpendicular to its arc. The string has the job of redirecting that kinetic energy to make the ball go up to either 8 or 4 o'clock. To do this, the direction of acceleration is perpendicular to the direction of the motion. This is centripetal force.

Now, if you imagine a ball hanging motionless from a string, it has no kinetic energy. The total tension on the string in this situation is just -mg, as it is at equilibrium. The swinging pendulum at the bottom of it's arc is not at equilibrium at all. Specifically, it has the same force of gravity as the ball at rest, but since the ball in the pendulum is swinging, it also has the force created by the acceleration--that is, the change in direction of the velocity--that comes at the bottom.

I think the key here is to remember that acceleration is not just the change in the speed, but the change in velocity. When you first hear about speed and velocity, it seems like a nit-pick to say that they are not the same, and that velocity is speed + direction. But a point on a wheel spinning at constant speed still accelerates. When you see that F = m*a, remember that acceleration doesn't care if it's a change in speed or a change in direction--it creates a force all the same. Since the pendulum ball changes direction of velocity, it must necessarily be accelerating, and since it has mass, there must be a resultant force. If the only tension on the string when the ball was at the bottom was -mg, then the system would be at equilibium, and the ball would be completely motionless.

 

[Nutmeg]

An unknown solid weighs 31.6N. When submerged in water, its apparent weight is 19.8 N. What is the specific gravity of the unknown sample?

I never know where to start on these random Fluid discretes (and there always seems to be at least 1 of these on every Full Lenght I've taken!) I draw out a diagram, usually end up doing W-Fb = ma but then get lost from there.

Specific gravity = (weight of the object)/(weight of an equal volume of water).

The Fb that you calculate is the weight of that equal volume of water. Hence, even without knowing the volume of the solid--or even the density of water--you know that the weight of the same volume of water must be the weight "lost" by submerging the object. In this instance, the water displaced weights 31.6N - 19.8N. So
Specific gravity = (weight of the object)/(weight of the object minus the apparent weight of the object when submerged in water)

Specific gravity = (31.6N)/(31.6N - 19.8N)

 

[Nutmeg]

From AAMC 3R: This is the provided solution to the question (Based on information in the passage how many centuries will be required for Mercury's perihelion to precess 360 degrees?) We know only 500 arcsec/century. I want to know how an arcminute is 1/60 of a degree? Thanks

That's just a definition. As stated, An arcsecond is 1/60 of an arcminute, an arcminute is 1/60 of a degree. It's just something you have to know, and if you didn't know, now you do.