copy physics posts

[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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[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]

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 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. 🙂

 

[liverotcod]

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

Note that the charges are not balancing. The fields generated by +Q and -Q act in the same direction.

 

[Nutmeg]

Note that the charges are not balancing. The fields generated by +Q and -Q act in the same direction.

You add this field to the field of the same equation with negative Q. Directionality is only applicable to the slope of the field or the voltage. The electric field should be zero, though, like it is at the halfway point in this graph:

Am I missing something?? 😕

 

[liverotcod]

You add this field to the field of the same equation with negative Q. Directionality is only applicable to the slope of the field or the voltage. The electric field should be zero, though, like it is at the halfway point in this graph:

Am I missing something?? 😕

Is that not a graph of potential, rather than e-field? And in your equation, wouldn't r-hat resolve to 1 for one of the charges, and -1 for the other? That seems like common sense to me, but electromagnetism is sometimes pretty counterintuitive to my mind.

Here's my take: imagine that we're calculating electrical force, rather than field. +Q would push on a positive test charge, and -Q would pull on it with the same amplitude as +Q, and the test charge would move toward -Q. The force is just the interaction between the charge and the field, right? So there has to be a net field between the charges.

 

[frankrizzo18]

Questions like that give me gas! I had a question this april on a tarzan passage where the answers were A) Sin(theta) B) Cos(theta) C) Sinsquared(theta) D) Cossquared(theta)

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. 🙂

 

[stoleyerscrubz]

I finished studying all of EK and I have the basics down for the most part but there are a number of things mentioned in the AAMC MCAT Topic List that were not covered in EK materials. Could you comment on what we need to know about some of these. Thanks.

WORK-Work and Energy:mechanical advantage
Waves and Periodic motion-sound:attenuation

 

[Caboose]

O.k. so there's this guy - he's just nuts; he wants to get from point A to point B on a river flowing 8m/s east. He's a little off and just assumed he could cross the 1800m wide river directly going at 15m/s. Well, his efforts were in vain as he ended up 2040m downstream, (totally far off from his drinkin' buddies). He forgot his watch and was really into timing things. He turns to you in a panic, his wall-eyed gaze frantically looking to you for help.
Being the brilliant mind that you are, you chuckle and say, "Never you worry. We simply use the equation t=d/v. 1800north / 15m/s yielding the solution 120 seconds, or 2 minutes."

A temporary wave of relieve washes the anxiety from his face, but suddenly is replaced by doubt, "But what if ye jist take 2040east / 8m/s that the river's goin'... 'cause there's only one eastward force, so shouldn't that give ya the same answer... but thien it's like... 255 seconds!"

Suddenly the birds cease to chirp, all is silent and your staggered breath betrays your confident charm... you... you don't know.

Caboose.

 

[Nutmeg]

Is that not a graph of potential, rather than e-field? And in your equation, wouldn't r-hat resolve to 1 for one of the charges, and -1 for the other? That seems like common sense to me, but electromagnetism is sometimes pretty counterintuitive to my mind.

Here's my take: imagine that we're calculating electrical force, rather than field. +Q would push on a positive test charge, and -Q would pull on it with the same amplitude as +Q, and the test charge would move toward -Q. The force is just the interaction between the charge and the field, right? So there has to be a net field between the charges.

Ah ha! Gotcha. The problem with the equation i put up is that I forgot abot the r vector quantity. By superposition, Q is the same for both with the r vector being the same, or Q is opposite with the r vector being the opposite.

Nice work. I was completely wrong on that one, question answered. 🙂

 

[yalla22]

I'm reviewing the periodic motion, waves, and sound portion of my kaplan book and I'm wondering how many of those equations one has to have memorized..There are so many of them and they ALL seem exactly alike!!!

Like for standing waves, wavelength = 2L/n and f = nv/2L
and then for waves we have y=Ysin(kx-wt) where k = 2pie/wavelength, etc etc🙁!!

Any suggestions about how to tease these apart?

 

[PneoDr]

I have a question about physics. It's hard for me to put my finger on what exactly has me upset. But it has to do with the idea of voltage or potential. My first problem: In a simple circuit, say one with just a single resistor and a battery with no internal resistance. If the battery is 6V then there must be a "potential drop" across the single resistor equal to 6V. That seems like such a simple idea, but it is hard for me to see. What is losing potential across the resistor? I have not found a book that makes this idea too clear. I arrived at this conclusion: I just think of it as the charge that is flowing through the resistor collides with the atoms of the resistor they lose some of thier "potential," so at the entry point of the resistor there is a higher potential then the exit point. I usually get circuit problems right, but I just wonder if there is a better way to understand this.

The other idea that has me confused is voltage at a point. The way I am thinking of this is that at a given point there is a given potential and the work required to move a point charge to that point is given by the (voltage times the charge). Is this the correct way to think of this concept?

Thank you for your help.
Bryan

 

[yalla22]

I have a question about physics. It's hard for me to put my finger on what exactly has me upset. But it has to do with the idea of voltage or potential. My first problem: In a simple circuit, say one with just a single resistor and a battery with no internal resistance. If the battery is 6V then there must be a "potential drop" across the single resistor equal to 6V. That seems like such a simple idea, but it is hard for me to see. What is losing potential across the resistor? I have not found a book that makes this idea too clear. I arrived at this conclusion: I just think of it as the charge that is flowing through the resistor collides with the atoms of the resistor they lose some of thier "potential," so at the entry point of the resistor there is a higher potential then the exit point. I usually get circuit problems right, but I just wonder if there is a better way to understand this.

The other idea that has me confused is voltage at a point. The way I am thinking of this is that at a given point there is a given potential and the work required to move a point charge to that point is given by the (voltage times the charge). Is this the correct way to think of this concept?

Thank you for your help.
Bryan

For the first paragraph-I think you are actually thinking too hard about it! As long as you can solve the problems you should be okay...