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 (http://forums.studentdoctor.net/showthread.php?t=222481) 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.

So I'm reading my kaplan books (and others) and for example snell's law ( or any thing with a trig function) will come up and Ill derive it down (because after all I do have all equations memorized ) and I'll get something like Sin@=.3/1.8 (@=theta) and then the next thing I know the answer reads @=9.5 degrees. How do I do that? I believe its inverse sin, but in my head, come on?? Any help on how to figure problems like this out in a hurry would be great! Thanks!

So I'm reading my kaplan books (and others) and for example snell's law ( or any thing with a trig function) will come up and Ill derive it down (because after all I do have all equations memorized ) and I'll get something like Sin@=.3/1.8 (@=theta) and then the next thing I know the answer reads @=9.5 degrees. How do I do that? I believe its inverse sin, but in my head, come on??
Good question, but the answer may be unsatisfying: you don't have to be able to do this on actual MCAT questions. All you would need to be able to do, at the very most, on a problem like the one you quoted is to see that sin(theta) is (a lot) less than 0.5, the sin of 30 degrees, so the angle is also (a lot) less than 30 degrees. That'll be enough to choose the answer using process of elimination.

So, here's the general aproach to MCAT questions that ask for an angle as an answer. First, the only angles you need to know anything about are 30, 45, and 60 degrees (plus things like 0 and 90, plus equivalent angles like 135 [which is the same as 45]). Moreover, you don't need to know them with any precision: on MCAT physics questions, one significant figure is plenty. This means that for 30 degrees, sine is 0.5 and cosine is 0.9; for 60 degrees these two are reversed, and for 45 degrees the sine and cosine are each 0.7. (Warning: MCAT chemistry questions sometimes do require more than one significant figure, but there's no trig needed for them.)

How, then, do you answer questions whose answers are other than these? Glad you asked. The rule is this: angular answers to MCAT problems can only be angles you know about, plus angles that are actually discussed in the question or the passage, unless only one of the four answer choices is possible. In other words, any answer choice that is an angle that you don't know anything about (i.e., haven't memorized the sine and cosine of) and that isn't listed in the passage, cannot be the answer unless all other choices are obviously wrong.

For the problem listed, either they gave you an abbreviated trig table listing, among other angles, 9.5 degrees, or that was an angle that was discussed in the passage, say an initial angle or an experimental value, or all the other choices were 30 degrees or more (or 0). No, that probably wasn't the case in your practice material, but on the MCAT, it will be so.

Thanks Shrike!!! I appreciate you setting this up!

Next question if you have time. What equations are a must for the MCAT, as my review material contains seriously A LOT of equations, and if I need to commit them to the small memory I have left I should start now ;) .

What equations are a must for the MCAT, as my review material contains seriously A LOT of equations, and if I need to commit them to the small memory I have left I should start now ;) .
That one's going to take some time to put together; give it a while, please, but eventually I'll have it in the FAQs.

Re my editing of your second post: I'm just trying to keep the thread fairly uncluttered, so I took out the big quotation of my answer.

Thanks for moving your question in here, and congratulations on being the first to take advantage of the new forum. I appreciate the opportunity to help.

To others: I'll be looking for help. If you know your physics and want to lend a hand, please PM me.

Hey Shrike, I can put together some equations if you want. I scored a 13 on the PS section a few years back. Just tell me if you are busy

Hey Shrike,
I guess my question to you would be about how best to tackle Physics.....it's been a loooong time since I took physics, and I'm having a hard time trying to learn all this stuff again. I guess the stuff that's giving me the most trouble is Newtonian Mechanics, but I really do need to go over everything. So what would you suggest is the best way to tackle it? Thanks.

I guess my question to you would be about how best to tackle Physics.....it's been a loooong time since I took physics, and I'm having a hard time trying to learn all this stuff again. I guess the stuff that's giving me the most trouble is Newtonian Mechanics, but I really do need to go over everything. So what would you suggest is the best way to tackle it? Thanks.
You will need to two things by the time you're done: (1) the basic equations, a list of which I'll be compiling (there are somwhere between thirty and fifty that I believe you absolutely must know) and some related definitions; and (2) an understanding of the major concepts. For example, with Newtonian mechanics, you need F = ma, F(G) = GMm/R^2, the equations for uniformly accelerated motion (e.g., d = v(0)t + 0.5at^2), and an understanding of three concepts: velocity and acceleration can be in the same or different directions; all forces and motion can be broken into x- and y-components, which you then deal with separately; and with projectiles, once you've broken the motion down into components the x-component is trivial, because v(x) remains constant always.

That's the general idea, though of course it doesn't really answer your question. Give me a while; I'll be writing it up. Essentially, I need to put nine two-and-a-half-hour lectures into print, and I don't type fast. Meanwhile, the more specific questions are easier to address, but keep asking whatever you want answers to.

Thanks for the help......Just a suggestion, but could you do a thread that's "General Physics Concepts" and put it in there......this thread could be more for specific questions and the other one could be a thread exclusively for mods where you add general info that you feel would be beneficial for the average test taker to know.

alright torque problem:
a massless plank with a length of 4 m has two weights placed on it. One with a mass of 10 kg on the left edge of the plank, while the other mass of 20 kg is placed 1.5 m inward from the opposite right edge. Where should the fulcrum be located if the plank with the weights on it were to remain horizontal?
Also do i have to use the CCW and CW to determine whats negative and positive torque?

a massless plank with a length of 4 m has two weights placed on it. One with a mass of 10 kg on the left edge of the plank, while the other mass of 20 kg is placed 1.5 m inward from the opposite right edge. Where should the fulcrum be located if the plank with the weights on it were to remain horizontal?
The plank is massless, so only the two weights matter. Never mind the equation for this problem; one's twice as heavy, so the fulcrum needs to be only half as far from it. With the numbers given, one third of the distance between the two weights is 0.83m from the larger mass, 1.67 from the smaller mass (and therefore the left edge).

When using the center of mass equation/center of gravity equation, I recommend always putting the zero at the left-most mass or force (because you're used to zero being on the left). If you do that, you don't have to mess with the signs so much when handling torques: up is one sign, down is the other; no clockwise or counterclockwise.

The center of mass (which is where you want the fulcrum) is at

CoM = (m1X1 + m2X2)/(m1 + m2) = (10kg x 0m + 20kg x 2.5m)/(20kg + 10kg) = (5/3)m = 1.67m

Also do i have to use the CCW and CW to determine whats negative and positive torque?
Don't worry about using a different sign for different directions. In most problems, you care only about equilibrium, so everything zeros out anyway. If you have to calculate a torque, figure out its magnitude and then ask yourself what direction it's going.

Just a suggestion, but could you do a thread that's "General Physics Concepts" and put it in there......this thread could be more for specific questions and the other one could be a thread exclusively for mods where you add general info that you feel would be beneficial for the average test taker to know.
We may eventually arrange it that way; we'll see what we come up with by way of general concepts (and how quickly).

Thanks for the suggestion.

I could really use some help in understanding the buoyancy force in relation to figuring out the density of an submerged object. The archimedes principle.

Why is it that no current runs through a capacitor in a circuit if the circuit has been on for a while?
Because that's the nature of a capacitor. The way a capacitor stores charge, and therefore energy, is by keeping charges crossing from one part (one plate, in the usual parallel plate capacitor) to the other, even though they want to. In the parallel plate capacitor, the plates are separated by a non-conductor, or by vacuum (which doesn't conduct current either). As long as the capacitor doesn't fail, charges can't cross through it.

The reason there is current in the short run is that charges are filling up the capacitor -- none of them is going through it.

When a capacitor fails -- it's often called dielectric breakdown -- the voltage has become high enough that those charges can get across the non-conductor (or vacuum) between the plates. Then, current flows, and all bets are off. This doesn't happen on the MCAT, as far as I know.

In the subject of machines(ramp, pulley, lever). Isn't the ramp inefficient in the way that the work(which leads to force) is being wasted on the horizontal component as opposed to lifting an object straight up in which everything goes to the vertical component?
Not really. Pushing and pulling is conservative on the MCAT: no energy is lost. On the other hand, if energy is used to accelerate the block (up the ramp, or otherwise), then that energy isn't available to lift the block while the speed is increased (relative to its starting speed). On a complex problem this could matter, but in a simple problem we would assume either constant or negligible speed. In either of these simpler cases, all of the energy that goes into pushing or pulling is translated into potential energy by lifting the block, rather than kinetic energy by speeding it up.

im confused with the topic if of weightlessness
a 5kg object is considered weightless when:
I chose when its accelarting up at 9.8 m/s^2
the answer is down, im just thinking weightless means mg is 0 so Forces acting on it should cancel out right? I dont know what im trying to say the answer and explanation they gave is confusing.

Weight is the force due to gravity, which on earth is mg. An object has no weight when there is no acceleration due to gravity.

Alternatively, I guess an object could be considered weightless if something cancels out mg, like objects with densities identical to water that are submerged in water. This is why NASA conducts some of its training underwater and why senior citizens excercise in water. They don't have to deal with the forces being exerted by that pesky gravity.

Tell us what question/answer/explanation is giving you trouble and we can probably give a better response.

-X

im confused with the topic if of weightlessness...

thats what im saying if mg is pointing down and the answer i chose is the opposite of g than why isnt it weightless. Instead the answer is if it accelerates downward with g???

im confused with the topic if of weightlessness
a 5kg object is considered weightless when:
Weight is defined as the force of gravity on an object, and always equals mg. In popular parlance, an object is "weightless" when its apparent weight is zero. Apparent weight is just what a scale would read if the object were sitting on it; i.e., it's the force exterted on it, in an upward direction, by what it's resting on (or, occasionally, hanging from). An object is therefore weightless when the floor et al is not pushing it up; in other words, when it's in free fall.

Another example of apperent weight being zero is a neutrally buouyant object submerged in a fluid, or any object floating in equilibrium on the surface of the fluid. Because the buoyant force balances the force of gravity, no more force is needed to support it and a scale under the object would read zero.

Abe: if the object were accelerating upward at 10m/s^2, then that would be because there was a force pushing it up. That force would be sufficient to overcome gravity, and still push it upward just as fast as gravity wanted to push it down. Hence, its apparent weight would be doubled from what it was at rest -- the scale under it would read 2mg, because that's how strong the force would have to be.

Thanks for the help......Just a suggestion, but could you do a thread that's "General Physics Concepts" and put it in there......this thread could be more for specific questions and the other one could be a thread exclusively for mods where you add general info that you feel would be beneficial for the average test taker to know.
Check out the new Physics FAQs and Topic Writeups thread; I'm pretty much following your suggestion to the letter. Thanks for the help.

I could really use some help in understanding the buoyancy force in relation to figuring out the density of an submerged object. The archimedes principle.
Sorry; I've posted a fairly comprehensive answer to this somewhere, but now I can't find it. Be patient and I will get to it.

for some reason, i cant figure out why a car is pushed away from the center of a circle when it goes around in a circle. the centripital force and i want to say that the static friction force are both facing into the middle (preventing the car from skidding away from the center)... so, i guess this should be some type of equilibrium situation. what force am i missing that pushes away from the center?

Something happened when I first tried to post this, so I apologize if I left anything out.

The car is not being pushed away from the center of the circle. Without a centripetal force, objects such as cars and yo-yo's on strings will move in a direction that is tangential to the circular path.

In the case of your car making donuts in a level parking lot, the engine is providing the power for forward motion (by forward, I mean the front of the car). When you let go of the steering wheel, the car will move in a straight line that is tangential to the donuts you were making before.

You are correct in thinking the friction force points towards the center. That is because on level surfaces, the centripetal force is the friction force...

Fc = uFn = mv^2/r

Without friction, there can be no circular motion for cars. This is what happens when you try to make turns on icy patches of the road. There is no friction to provide a centripetal force, and thus you skid off the road into a ditch.

Note: This is only for level surfaces. On banked highways and race tracks, gravity comes into play.

Also Note: There is no such thing as a "centrifugal force." This is the name commonly given to another physics term known as inertia.

-X

for some reason, i cant figure out why a car is pushed away from the center of a circle when it goes around in a circle. the centripital force and i want to say that the static friction force are both facing into the middle (preventing the car from skidding away from the center)... so, i guess this should be some type of equilibrium situation. what force am i missing that pushes away from the center?

thanks for the quick reply. so when what force is greater than the inward static frictional force does the car start skidding away?

There is no such thing as a "centrifugal force." This is the name commonly given to another physics term known as inertia.
This part is not exactly true. You're right that by "centrifugal force," non-physicists usually mean something else: either inertia, or centripetal force. But there is such a thing as centrifugal force:

Recall that, by Newton's Third Law, for every force there's an equal and opposite force, and that the two constitute an action-reaction pair. Centrifugal force is the force that forms a pair with the centripetal force -- it's the force exerted by the body that's moving in a circle, on whatever's making it turn.

For example, in the case of a yo-yo whirled on a string, the force of the string on the yoyo is centripetal, and the force of the yo-yo on the strong, and thus your finger, is centrifugal. In the case of the car driving in a circle, the ground exerts a centripetal foce on the tires, while the tires exert a centrifugal force on the ground.

You might guess from these examples that it's centripetal force that we usually worry about, not centrifugal. You'd be right. Centrifigal force tends to be a little odd, and mostly irrelevant. I've never seen it matter on an MCAT problem.

Sorry, I meant to say something like the Centripetal force isn't really a new type of force, but just the name used to indicate the force pointing inwards. In this case, it is the static frction force.

Since the static coefficient of friction is normally given as the maximum, you know that any more force applied to that object will overcome friction and will begin to slide or in the case of the car, skid off the road.

The centripetal force equation Fc=(m)(v^2)/r describes the force required to keep the object in a circular path. When the car moves fast enough, the force due to static friction will not be high enough to maintain circular motion. Incidentally, I've heard the term "centripetal force requirement" to describe the situation.

Likewise when a satellite or any other thing moves fast enough it will escape the earth's gravity:

G(Mearth)(Msatellite)/(r^2) is less than (msatellite)(v^2)/r

Again, I apologize for the incompleteness of my answer. Hope that helps!

-X

thanks for the quick reply. so when what force is greater than the inward static frictional force does the car start skidding away?

Shrike is right. I meant to say "There is no such thing as centrifugal force acting on you in the car...."

That is still inertia "pushing" you up against the side of the car. Centrifugal force is indeed the force the car exerts on the road.

Mea culpa.

-X

thats why this is confusing to me. i dont know if this is beyond the scope of the MCAT or what. so if you draw a free body diagram (ignore the centrifugal force of the car on the ground because the car is the object, not the ground) with a car moving in a circle at constant velocity, then the only one force vector would be inward (=friction=centripital force), right? im sure this has to do with the velocity because intuitively i know that at a higher velocity, it would be harder for a car to stay in its circle. this is more out of curiousity now more than worrying that something like this will be on the mcat. thanks again for all your help.

Pretty much!

The freebody diagram should show mg pointing downwards, the Normal Force (Fn) going up, and the Friction force (Ff) going towards the center of the circle. The centripetal force (Fc) is the force required to keep the car in a circular path at that velocity. Your intuition is correct. The fast you go, the higher the Fc. When the required Fc surpasses the Frictional force, the car begins to skid and circular motion is no longer maintained. When this happens, you'd better be wearing your seatbelt!

-X

Check out the new Physics FAQs and Topic Writeups thread; I'm pretty much following your suggestion to the letter. Thanks for the help.


Thanks Shrike....I've been looking over it and it's really good stuff. Keep up the good work.

... with a car moving in a circle at constant velocity...
A minor point, but just in case: the car may be moving at constant speed, but its velocity is changing because the direction is changing.

All these threads will be up at least until August, right?
I know it is a dumb question, but I am debating about whether or not I should retake the exam.
This study area is great!

A minor point, but just in case: the car may be moving at constant speed, but its velocity is changing because the direction is changing.


Hard to picture that because in real life we'd have to slow down to change directions(ie turn) while driving the car. An object that is being swung in a ciscular motion at constant speed would be easier to picture. Speed constand but the direction is changing.

Hard to picture that because in real life we'd have to slow down to change directions(ie turn) while driving the car.
You apparently haven't driven with me.

I could really use some help in understanding the buoyancy force in relation to figuring out the density of an submerged object. The archimedes principle.
Look here (http://forums.studentdoctor.net/showthread.php?p=2721216&posted=1#post2721216), and tell me whether it answers your question.

All these threads will be up at least until August, right?
I know it is a dumb question, but I am debating about whether or not I should retake the exam.
This study area is great!

They will be up permanently, as far as we know. I'm glad that you are finding them helpful. Best of luck for those taking the test in Aug. :luck:

TPR asks in one of its hw problems the following:

In a series of experimental trials, a projectile is launched with a fixed speed, but with various angles of elevation. As the angle is increased from 0 to 90, the vertical component of the velocity:

Now, I picked the option that said: increases, while the horizontal component remains constant.

The answer, accodring to TPR is: increases, while the horizontal component decreases.

I thought horizontal velocity was constant during the duration of the flight.

For some reason, I have a lot of trouble with some of the more basic physics subjects i.e. kinematics, Newtonian mechanics, conservation of energy, etc. I don't have an intuitive understaning of these topics and on MCAT problems that are more conceptual, I often waste a lot of time trying to overanalyze them and often end up picking the wrong answer. I'm not sure what do do about it, because I understand the basic concepts and know the formulas but I just can't seem to do well on the more conceptual problems -- an example would be how changing the angle of inclination would affect velocity, acceleration, etc. Basically problems that are really easy if you have an intuitive understanding of the concepts and how they are interrelated but extremely difficult if you do not.

I'd appreciate any advice on how I can improve on this. Thanks.

Horizontal velocity IS contant during the duration of a flight (assuming no air resistance), but it is only constant with regard to your starting angle of elevation. Put another way, going from 0 to 90 degrees in one degree increments, will give you 90 different horizontal velocities, but they will be constant, ie they won't slow down or speed up.

As for your specific problem. Descriptively, at zero degrees all of the energy from the cannon will be directed in the horizontal component, making the projectile travel in a straight line parallel to the ground. Whereas at 90 degrees all of the energy will go into the vertical component, making the projectile travel straight up. The horizontal velocity would HAVE to decrease since you know that firing anything straight up into the air will come straight back down again. It won't move left or right. At 45 degrees both components will have equal values. This is why 45 degrees give you the best distance in the presence of gravity and the absence of aire resistance.

Remember that sin and cos of theta (ie angle of elevation) will tell you how much of the total starting velocity is composed of its horizontal and verical components. Just give an arbitrary starting velocity of 10 m/s and compare the sin and cos values as go from 0 to 90 degrees in 15 degree increments.

Example:
Vertical Component --> 10 m/s * sin(0) = 0 m/s
Horizontal Component --> 10 m/s * cos(0) = 10 m/s

Shrike is a TPR teacher. Perhaps he can comment better since he will probably know the HW Question you are talking about.

-X

TPR asks in one of its hw problems the following:

In a series of experimental trials, a projectile is launched with a fixed speed, but with various angles of elevation. As the angle is increased from 0 to 90, the vertical component of the velocity:

Now, I picked the option that said: increases, while the horizontal component remains constant.

The answer, accodring to TPR is: increases, while the horizontal component decreases.

I thought horizontal velocity was constant during the duration of the flight.

Which of the follwing gives the percent change to the Young's Modulus for a substance when its cross sectional are is increased by a factor of 3?

0
33%
300%
900%

I put 33% since the equation is F/A/ (Delta L/L)

But of course it was wrong and the answer was 0.

Also for another stress q:
The sole of a certain tennis shoe has a shear modulus of 4x 10 to the 7th, I f the height of the sole is double the strain will.

Decrease by factor of 2
Same
Increase by f of 2
Increase by f of 4

I put increase by f of 2 since, which is wrong, but is it 0 because young's modulus is height, and shear is length? Thanks
__________________

Shrike is a TPR teacher. Perhaps he can comment better since he will probably know the HW Question you are talking about.
Naah. We don't look at homework problems until someone asks about them, and then we usually promptly forget the problem -- it's just not a big part of our class.

Xanthine's explanation is fine. For a given set of launch conditions, vx remains constant, but you're changing launch conditions.

Which of the follwing gives the percent change to the Young's Modulus for a substance when its cross sectional are is increased by a factor of 3?
Young's modulus is a property of the material, like coefficient of friction or index of refraction. It doesn't change just because there's more stuff.

The equation is then used, with this fixed value, to figure the strain for a given stress.

...I don't have an intuitive understaning of these topics and on MCAT problems that are more conceptual, I often waste a lot of time trying to overanalyze them and often end up picking the wrong answer. I'm not sure what do do about it.
Tough problem. Some solutions ideas:


Move to Dallas, and take my class.

Arrange to get better instruction at your school.

Keep an eye on the FAQs and Topic Writeups thread, where I'll be trying to give ways to deal with conceptual problems.

Find a self-study source that promotes conceptual understanding. I'm told Examkracker's physics text is good for this, but I can't tell you from firsthand knowledge. As an extreme solution, you could get a book of the Feinman lectures.

Well, a less ideal way would be to just memorize what happens if you do X. Since you already know the euquations and concepts, you could memorize a few key "happenings" despite the lack of intuition. Again, it would be better if you could "just get it", but in the absence of that (or Shrike's suggestions) this may help you through the physics Q's.

-X

For some reason, I have a lot of trouble with some of the more basic physics subjects i.e. kinematics, Newtonian mechanics, conservation of energy, etc. I don't have an intuitive understaning of these topics and on MCAT problems that are more conceptual, I often waste a lot of time trying to overanalyze them and often end up picking the wrong answer. I'm not sure what do do about it, because I understand the basic concepts and know the formulas but I just can't seem to do well on the more conceptual problems -- an example would be how changing the angle of inclination would affect velocity, acceleration, etc. Basically problems that are really easy if you have an intuitive understanding of the concepts and how they are interrelated but extremely difficult if you do not.

I'd appreciate any advice on how I can improve on this. Thanks.

is torque covered on the mcat? it's not listed in the aamc list, but it's still discussed in most mcat review books.

Yes, torque is covered. And it is listed on the MCAT topic list, as is rotational equilibrium, which requires an understanding of torque.

Does anyone one know tricks to figuring out tangent, sine, and cosines in your head?

Does anyone one know tricks to figuring out tangent, sine, and cosines in your head?
I don't know when you'd need tangent on the MCAT. Sine and cosine, you need only for a few angles:


sin(30) = 0.5; cos(30) = 0.9
sin(45) = 0.7; cos(45) = 0.7
sin(60) = 0.9; cos(60) = 0.5

Any other angle, you will be given the trig functions for, or you will only have to see that the function is bigger/smaller than one of the angles listed above.

The mnemonic for trig functions is SOH CAH TOA: sine = opposite over hypotenuse, cosine = adjacent over hypotenuse, tangent = opposite over adjacent.

In the future, please ask your questions in this thread, rather than starting a new thread. Thanks.

Hi shrike, can you please go over torque and discuss rotational equilibrium? and how do u approach torque problems? ty

i dread seeing sound passages, i dont get this stuff about open end tube closed end tube string tied on both ends! I dont get any of the sound stuff at all other than v=fwavelength =/. What equations do we use for each situation, and what is n! its not nodes is all i remember

Questions pending about torque and rotational equilibrium, and sound. I'm sorry, I haven't had time to get to these, but I will. Unfortunately, neither is a simple subject; they take some work to write up. May not get to them before the 4th.

I have a pulley question. I understand that pulley is 2 ropes of a pulley supporting the weight mg of an object as opposed to having a single rope with tension T. The pulley is 2 ropes with 1/2 the tension of the single rope supporting weight mg, but I don't understand how to get the force needing to tug the rope in the pulley on the other end of the other end of the pulley. :confused:

I have a pulley question... :confused:
I reluctantly admit that I have no idea what you're asking. Could you rephrase, please?

I reluctantly admit that I have no idea what you're asking. Could you rephrase, please?

What is the relation between the force applied on the rope to tug it so that a weigh gets lifted on a pulley?

I reluctantly admit that I have no idea what you're asking. Could you rephrase, please?
:laugh:

I don't know when you'd need tangent on the MCAT. Sine and cosine, you need only for a few angles:


sin(30) = 0.5; cos(30) = 0.9
sin(45) = 0.7; cos(45) = 0.7
sin(60) = 0.9; cos(60) = 0.5



There is something I wanted to add to this that TPR teaches:

Sin 0 = sq rt.(0)/2 = 0
Sin 30 = sq rt. (1)/2 = .5
Sin 45 = sq rt (2)/2 = .7
Sin 60 = sq rt (3)/2 = .85 or roughly .9 as Shrike pointed out above
Sin 90 = sq rt. (4)/2 = 1

For cosine, it is the exact reverse of the values for Sine. The TPR books listed it this way in the appendix with all the formulas for the Physics section as well as in the math appendix at the end of the Physical sciences book. Thinking of it this way helped me a bit to remember it, so I thought it would be useful to point that out.

The very opposite of that is the trend for Cosine.

I was looking at the examkrackers book and it doesn't do go into much detail about the center of mass and center of gravity. what do we have to knwo about the center of mass and the center of gravity?

This might be a stupid question, but can you explain the relationship between work and kinetic enery/potential energy?????? A friend of mine studying for the MCAT wanted me to ask you here.

This might be a stupid question, but can you explain the relationship between work and kinetic enery/potential energy?????? A friend of mine studying for the MCAT wanted me to ask you here.

Work is defined as force * distance and in other words it is related to potential energy and kinetic energy by a theory called work/energy theory where work = mgh or (1/2 mv^2). Let me know if that helps..

Okay this question deals with circuits... Let E be the strength of the electric field between the plates at the end of the set-up procedure, and let E1 be teh strength of the field after the insertion of the dielectric in Experiment #1. How does E1 compare to E? The set up procedure just charges the capacitor using a battery then exp. 1 just inserts a insulator while the battery is still connected (dielectric of value 4).
THe answer is that they equal each other E=E1 but what I don't understand is when an dielectric is inserted doesn't that decrease the electric field thus decreasing the voltage and increasing the capacitance? How could the electric fields be equal to each other?

Work is defined as force * distance and in other words it is related to potential energy and kinetic energy by a theory called work/energy theory where work = mgh or (1/2 mv^2). Let me know if that helps..

WRONG
work is NOT always equal to the change in potential energy but it is always equal to the change in the kinetic energy

here is the proof for W = delta KE

I assume V1 = 0 to make calculations simpler but the calculations for V1 != 0 is the same:

w = f * d = (m * a) d (eq. #1)
d = [(V1 + V2)/2] * t (eq. #2)
Assuming V1 = 0 & given eq #1 & eq. #2
W = (m * V2 - 0 / t) * (V2 + 0)/2 * t = (m * V2 / t) * (V2) / 2 * t
W = 1/2 m * V2^2

sound travels faster in what? less dense or denser mediums?

Faster in denser media. The more closely packed the atoms, the more quickly the compression will propagate.

Okay this question deals with circuits... Let E be the strength of the electric field between the plates at the end of the set-up procedure, and let E1 be teh strength of the field after the insertion of the dielectric in Experiment #1. How does E1 compare to E? The set up procedure just charges the capacitor using a battery then exp. 1 just inserts a insulator while the battery is still connected (dielectric of value 4).
THe answer is that they equal each other E=E1 but what I don't understand is when an dielectric is inserted doesn't that decrease the electric field thus decreasing the voltage and increasing the capacitance? How could the electric fields be equal to each other?
The function of the dielectric is to create an opposing electric field by assuming a polarity in the opposite direction. For a given charge, the electric field would end up lower with the dielectric than without. But in the set up you describe, the voltage drop is controlled by the battery--the voltage drop of the capacitor must remain equal to the voltage drop of the battery (remember Kirchhoff's voltage law: The directed sum of the electrical potential differences around a circuit must sum to zero). Hence the electric field is fixed because the voltage is fixed. The dielectric merely acts to increase the charge on the plates.

Work is defined as force * distance and in other words it is related to potential energy and kinetic energy by a theory called work/energy theory where work = mgh or (1/2 mv^2). Let me know if that helps..


No offense, but I was hoping shrike could give a more detailed response which is why I posted it here. I know what the work-energy theorem is but I was asking a theoretical/conceptual question about why it is what it is. Something he or psuying might have better insight on. But thanks.

WRONG
work is NOT always equal to the change in potential energy but it is always equal to the change in the kinetic energy

here is the proof for W = delta KE

I assume V1 = 0 to make calculations simpler but the calculations for V1 != 0 is the same:

w = f * d = (m * a) d (eq. #1)
d = [(V1 + V2)/2] * t (eq. #2)
Assuming V1 = 0 & given eq #1 & eq. #2
W = (m * V2 - 0 / t) * (V2 + 0)/2 * t = (m * V2 / t) * (V2) / 2 * t
W = 1/2 m * V2^2

THANKS A BUNCH. That makes more sense.

Hi, I've been reading about Standing waves in my Kaplan book and I'm pretty confused! It discusses the equations for finding the wavelength or length of standing waves, which for a wave fixed on both ends the wavelength =2L, 2L/3, 2L/4....2L/n (n=1,2,3....) I know n is the designated harmonic, meaning n=1 is the first harmonic, but if you are given a question with only a diagram of a string, how do u know what n equals? Can the same apply with open/closed pipes?

Faster in denser media. The more closely packed the atoms, the more quickly the compression will propagate.

merging this thread with the physics question thread.

WRONG
work is NOT always equal to the change in potential energy but it is always equal to the change in the kinetic energy

Actually it depends on what type of work you're talking about. The work done by a force (for MCAT purposes, a linear force) over a given displacement is equal to the change in kinetic energy. But the work done by the gravitational force is indeed equal to the change in potential energy.

Actually it depends on what type of work you're talking about. The work done by a force (for MCAT purposes, a linear force) over a given displacement is equal to the change in kinetic energy. But the work done by the gravitational force is indeed equal to the change in potential energy.

Well, like I said, you can "ALWAYS" say work is equal to change in kinetic energy but you can "SOMETIMES" say work is equal to the change in potential energy. In other words, work is equal to the change in potential energy only when delta KE = delta PE.

WRONG
work is NOT always equal to the change in potential energy but it is always equal to the change in the kinetic energy

here is the proof for W = delta KE

I assume V1 = 0 to make calculations simpler but the calculations for V1 != 0 is the same:

w = f * d = (m * a) d (eq. #1)
d = [(V1 + V2)/2] * t (eq. #2)
Assuming V1 = 0 & given eq #1 & eq. #2
W = (m * V2 - 0 / t) * (V2 + 0)/2 * t = (m * V2 / t) * (V2) / 2 * t
W = 1/2 m * V2^2

I'm pretty sure to the work energy theory, its equal to both potential and kinetic energy. I don't quite understand you're so called proof??

I'm pretty sure to the work energy theory, its equal to both potential and kinetic energy. I don't quite understand you're so called proof??

k yeah i get it..thanks! that helps..i just didn't look it at right

I'm pretty sure to the work energy theory, its equal to both potential and kinetic energy. I don't quite understand you're so called proof??

tell you what...directly derive "work = f * d = mg* delta h" without setting a to g and I'll agree with you.

sound travels faster in what? less dense or denser mediums?


so how come exankrackers said that...

when its humid, water vapor goes into the air. air is made up of co2 and n2 that has a greater mass than water. so if water vapor goes into the air, the mass decreases (18g/mol).. which decreases the density. ok, that makes sense. but then they said that sound travels faster in that.. it increases speed in *less* dense material... why?

thats why i was confused, cuz i thought the same as you.

1 m^3 = 1m x 1m x 1m = 100 cm x 100 cm = 100cm = 1000mm x 1000mm x 1000 mm[/b]

That should be 100 cm x 100 cm x 100cm. Just thought I'd help. I'm pretty stoked, it's 4th of July weekend and Shrike's supposed to be posting some new writeups this weekend :D

Is light a wave, particle or both? When I first took physics 100 years ago, it was a wave only. EK says it's a wave AND particle. Which is right? Thanks!

Is light a wave, particle or both? When I first took physics 100 years ago, it was a wave only. EK says it's a wave AND particle. Which is right? Thanks!

its neither. :idea:

Actually, EK is right. Light exhibits properties of both waves and particles. For more information, check here:

http://en.wikipedia.org/wiki/Wave-particle_duality

Actually, EK is right. Light exhibits properties of both waves and particles. For more information, check here:

http://en.wikipedia.org/wiki/Wave-particle_duality


oh i thot he was being sarcastic. :confused:

oh i thot he was being sarcastic. :confused:
I SAID I took physics many,many years ago (14 to be exacrt) so dude, gimmie a break? :rolleyes: Now if you ask me about the dysregulation of apoptosis in renal cell carcinoma and my thoughts on whether or not this is a funtion of the extrinsic, intrinsic, or both pathways, I could help you. Unfortunately, this ain't on the MCAT!!

So is this the thread of physics MCAT questions or physics MCAT questions YOU guys think are "good/smart" questions? :confused:

And by the way, light being ONLY a wave is from the physics book I used at the time: Physics: Concepts and Connections by Art Hobson.

Actually, EK is right. Light exhibits properties of both waves and particles. For more information, check here:

http://en.wikipedia.org/wiki/Wave-particle_duality

Thanks!!

Would someone explain what the normal force is supposed to be? Where does it come from, and what does it represent? I never got a clear definition in Physics class, and still don't really understand what it is. I did a search, and didn't find anything defining it. Thanks!

I SAID I took physics many,many years ago (14 to be exacrt) so dude, gimmie a break? :rolleyes: Now if you ask me about the dysregulation of apoptosis in renal cell carcinoma and my thoughts on whether or not this is a funtion of the extrinsic, intrinsic, or both pathways, I could help you. Unfortunately, this ain't on the MCAT!!

So is this the thread of physics MCAT questions or physics MCAT questions YOU guys think are "good/smart" questions? :confused:

And by the way, light being ONLY a wave is from the physics book I used at the time: Physics: Concepts and Connections by Art Hobson.

Wondering why that physics book said that !!!! It must have been a pretty old book at the time. Einstein proved light has the characteristics of particles and waves before you took physics...

Would someone explain what the normal force is supposed to be? Where does it come from, and what does it represent? I never got a clear definition in Physics class, and still don't really understand what it is. I did a search, and didn't find anything defining it. Thanks!

The normal force is just the reaction force in Newton's Third Law. Remember that if one object exerts a force on another object, the second object will exert an equal and opposite force against the first object. Lets say you placed an object on a table. The object exerts a downward force on the table, which is in this case is the weight of the object. The normal force is the force exerted by the table on the object (the reaction force) and is equal to the weight of the object. Lets say you placed a second object on top of the first. The normal force is then the sum of the weights of the two objects. Weight is also a force, keep that in mind.

Is light a wave, particle or both? When I first took physics 100 years ago, it was a wave only. EK says it's a wave AND particle. Which is right? Thanks!

It can be either one. This phenonemon is called wave-particle duality. In simple terms, light can behave as a stream of particles (photon). But it also possesses properties that are characteristic of a wave - it reflects and refracts like a wave would. Interference and the Doppler Effect are further proof light can exhibit wave-like characteristics.

I SAID I took physics many,many years ago (14 to be exacrt) so dude, gimmie a break? :rolleyes: Now if you ask me about the dysregulation of apoptosis in renal cell carcinoma and my thoughts on whether or not this is a funtion of the extrinsic, intrinsic, or both pathways, I could help you. Unfortunately, this ain't on the MCAT!!

So is this the thread of physics MCAT questions or physics MCAT questions YOU guys think are "good/smart" questions? :confused:

And by the way, light being ONLY a wave is from the physics book I used at the time: Physics: Concepts and Connections by Art Hobson.


i thought u were being sarcastic cuz u said u took physics "100 years ago," and they found out the light as wave/particle thing less than 100 years ago. take a chill pill :eek:

The normal force is just the reaction force in Newton's Third Law. . .

Thanks so much! That makes sense now, and I remember it now. :D

Well, like I said, you can "ALWAYS" say work is equal to change in kinetic energy but you can "SOMETIMES" say work is equal to the change in potential energy. In other words, work is equal to the change in potential energy only when delta KE = delta PE.

I suppose you are right, since a force and displacement are always necessary to perform some work. But I would also think its important to know when to use what equation.

If you have a converging two lense system and the distance between the two lenses is less then the distance needed to form an image do you still go about the problem the same way. To rephrase, if the second lens intercepts the light from the first lens before it forms an image will this change anything?

That should be 100 cm x 100 cm x 100cm. Just thought I'd help. I'm pretty stoked, it's 4th of July weekend and Shrike's supposed to be posting some new writeups this weekend :D


Ohh cool, that's awesome that he's gonna be posting. I can't wait

The function of the dielectric is to create an opposing electric field by assuming a polarity in the opposite direction. For a given charge, the electric field would end up lower with the dielectric than without. But in the set up you describe, the voltage drop is controlled by the battery--the voltage drop of the capacitor must remain equal to the voltage drop of the battery (remember Kirchhoff's voltage law: The directed sum of the electrical potential differences around a circuit must sum to zero). Hence the electric field is fixed because the voltage is fixed. The dielectric merely acts to increase the charge on the plates.

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?

It can be either one. This phenonemon is called wave-particle duality. In simple terms, light can behave as a stream of particles (photon). But it also possesses properties that are characteristic of a wave - it reflects and refracts like a wave would. Interference and the Doppler Effect are further proof light can exhibit wave-like characteristics.


Ugh.....this reminds me of the first damned passage I had on my MCAT this past April. I hated it. It was one of those more verbal type of passages that I kept wanting to change my answers to.

It started with a paragraph about the wave theory, then a paragraph about the particle theory, and finally the dual theory.

Greetings, all. I am on vacation: pretty much away from computers, among other things. I logged on for a quick check this morning, only to find fighting. This is not what I wanted to see before the coffee was done brewing.

Let's be clear about a few ground rules for this:

1. There is absolutely no fighting, no name calling, no unpleasantness permitted. Whether someone is being sarcastic or rude should never be an issue.

2. As one needs permission from the moderators of this thread even to post answers, rule 1 should never be a problem; nor should there be issues with posters correcting each other at all. The reason for rule 2 is that we wanted to have a place for test takers to come for answers they knew they could trust; if we ever have disagreements about either the physics or the way to apply it on the MCAT, that goal will not be achieved.

3. I still need help with the thread; if you want to be approved to answer, please PM me.

4. I say in the thread header that questions will be answered in a reasonable amount of time. I never defined reasonable. I'm not sure what amount of time is reasonable. But I'm trying, as are the other moderators and approved answerers.

Now, have a hot dog, watch a firework, and then gear up for the crucial six weeks.

Thank you,

Shrike.

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?
You're missing a small thing: this relationship holds only if the capacitor with and without the dielectric are being compared while being hooked up to the same-voltage battery (or other source). When there's a fixed voltage source, the voltage across the capacitor is fixed, too, and that means so is electric field. Now, in the equation Q = CV, if voltage is fixed, and capacitace C changes (because that's what a dielectric does -- it raises capacitance) then Q has to go up. Frankly, for the MCAT the reason this happens is irreleveant, but it relates to energy stored in dipoles within the dielectric.

If voltage is not held constant, but charge is (because the dielectric is inserted after the capacitor is charged, then removed from the voltage source), then voltage across the capacitor, and thus field, goes down.

If you have a converging two lense system and the distance between the two lenses is less then the distance needed to form an image do you still go about the problem the same way. To rephrase, if the second lens intercepts the light from the first lens before it forms an image will this change anything?
Basically, you find the lens power of the resulting system by taking the powers of each of the component lenses, and adding them. You then solve normally. Power if defined as 1/f; use the power of the system to get the effective focal length, and pretend the lens has that focal length.

For a more detailed answer about what comes next, see the Physics FAQs thread.

What is the relation between the force applied on the rope to tug it so that a weigh gets lifted on a pulley?
Look at how many sections of rope are pulling up on the weight: how many of them would have to shorten for the weight to move? Divide the weight of the object by that number, and you get force required on the rope (also known as tension).

For more on this topic, see my writeup on simple machines in the Physics FAQs thread.

so how come exankrackers said that...

when its humid, water vapor goes into the air. air is made up of co2 and n2 that has a greater mass than water. so if water vapor goes into the air, the mass decreases (18g/mol).. which decreases the density. ok, that makes sense. but then they said that sound travels faster in that.. it increases speed in *less* dense material... why?
In the two gases being considered, density -- in terms of molecules per cubic whatever -- is the same. I know, you don't usually think of density that way, but sound does. Sound also considers, for gases, how often the molecules are running into each other. In the case described, the addition of water vapor increases the average molecular speed (because it decreases average mass while, presumably, keeping temperature [and therefore average KE] constant), so it increases the number of collisions. Sound will therefore travel faster.

The original question was too vague for the MCAT; don't worry too much about it.

I was looking at the examkrackers book and it doesn't do go into much detail about the center of mass and center of gravity. what do we have to knwo about the center of mass and the center of gravity?
You have to know that, on MCAT problems, they're the same.

You have to know how to find center of mass; the answer is found in the FAQ thread. Basically, you reduce everything to point masses, pick a zero, and use CM = (sum of x * m)/(sum of m). Although any zero works, you should always pput it under the mass on the far left.

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.

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. :D

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.

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

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.

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.

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???

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

Yes.

Cool thanks.

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 :)

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.

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

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.

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!

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.

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.

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.

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.

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.

... 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.

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???

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?

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!

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).

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

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.

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.

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 :love:

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!

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 :love:

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

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

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

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.

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 :eek:

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?

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 :eek:

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! :)

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.

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.

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.

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. :confused:

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..?

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.

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.

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).

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!
:confused:

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.

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 θ = Ω·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 θ = 360 degrees is 360/(500/(60 x 60)) = 360 x 60 x (60/500) centuries. Thus, answer choice C is the correct answer.

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!
:confused:

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

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.

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)

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. :)

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.

Note that the charges are not balancing. The fields generated by +Q and -Q act in the same direction.
http://en.wikipedia.org/math/d99d651b881fd3fe915dcfc9dabb4c2f.png

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:

http://www.physics.unc.edu/~fisherb/dipole.gif

Am I missing something?? :confused:

http://en.wikipedia.org/math/d99d651b881fd3fe915dcfc9dabb4c2f.png

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?? :confused:
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.

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. :)

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

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.