Physics question thread

[Shrike]

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

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

Acceptable topics:

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

Unacceptable topics:

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

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

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

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

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

 

[QofQuimica]

I have a question about which way the magnetic field curls for a electron. the the direction of an electron is one direction, it means current is in the other, so then i should point thumb in the other direction and curl right?

You can do one of two things:

1) Pretend the electron is a proton, figure out the direction of B, and then take the reverse.

2) Use your left hand instead of your right.

 

[QofQuimica]

I don't know how to deal with capacitors and resistors that are in the same circuit. Unfortunately we did not cover this in my physics class (of course we covered them severly, but not together). Any thoughts on this?

Thanks

You mean you are trying to figure out whether their values are additive or something like you do with two resistors in series? They have different units, so you won't be able to add them together like that. You will have to consider each one separately.

 

[Lests55]

Yeah, I did a poor job of "asking" a question, more like I told you I didn't know anything...

My question is ....if I were calculating equivalent resistance or voltage drop through a resistor for example (maybe on a test here in a few weeks... 😱 )how would this be affected if there was a capacitor in series/parallel with the resistor? Do I just ignore the "other units" sources when doing this? It just seems like a capacitor in parallel/series with a resistor would somehow affect that resistors current and pot. diff. properties.

 

[Anastasis]

Okay so I apologize because I'm sure this question has been asked a million times before: What kinds of potential energies can be included in work?

For instance, if I pick a brick up and place it on a table, haven't I done work on it equivalent to its change in gravitational potential energy? (ie w=mgh)

Also, since Gravity is a conservative force, then it technically can't do work right?

Or am I just tragically confused? 😕

 

[QofQuimica]

Yeah, I did a poor job of "asking" a question, more like I told you I didn't know anything...

My question is ....if I were calculating equivalent resistance or voltage drop through a resistor for example (maybe on a test here in a few weeks... 😱 )how would this be affected if there was a capacitor in series/parallel with the resistor? Do I just ignore the "other units" sources when doing this? It just seems like a capacitor in parallel/series with a resistor would somehow affect that resistors current and pot. diff. properties.

Putting a resistor in series with a capacitor will affect how long it takes the capacitor to charge, but not the current. Things in series can't have different values for current without violating Kirchoff's Laws. Likewise, two things in parallel should have the same potential across them. DrChandy or Nutmeg, if you're reading this, can one of you give a more thorough explanation?

 

[QofQuimica]

Okay so I apologize because I'm sure this question has been asked a million times before: What kinds of potential energies can be included in work?

For instance, if I pick a brick up and place it on a table, haven't I done work on it equivalent to its change in gravitational potential energy? (ie w=mgh)

Also, since Gravity is a conservative force, then it technically can't do work right?

Or am I just tragically confused? 😕

No, I don't think you're tragically confused. Conservative forces can do work as long as you don't have the same ending point as your starting point. In other words, if I throw something up in the air, and it falls back down into my hand, then no net work was done. But if I could throw the object up onto a ledge and it doesn't fall back down, then I *have* done net work. So in your example, as long as the brick doesn't fall back down off the table, then you did do net work on it by picking it up.

Non-conservative forces like friction and air resistance always do non-zero work, even if the object ends up where it started.

 

[Anastasis]

No, I don't think you're tragically confused. Conservative forces can do work as long as you don't have the same ending point as your starting point. In other words, if I throw something up in the air, and it falls back down into my hand, then no net work was done. But if I could throw the object up onto a ledge and it doesn't fall back down, then I *have* done net work. So in your example, as long as the brick doesn't fall back down off the table, then you did do net work on it by picking it up.

Non-conservative forces like friction and air resistance always do non-zero work, even if the object ends up where it started.

(I think I may have found another error with Kaplan's material then... oh well.)

About conservative forces doing work - so if that brick fell off the ledge, Gravity would be doing work to make it fall down? But I thought because all the potential energy is converted to kinetic energy then no work was done.

Thanks for the reply, btw!

 

[QofQuimica]

(I think I may have found another error with Kaplan's material then... oh well.)

About conservative forces doing work - so if that brick fell off the ledge, Gravity would be doing work to make it fall down? But I thought because all the potential energy is converted to kinetic energy then no work was done.

Thanks for the reply, btw!

Well, you had to perform some amount of work to put the brick up on the table, and gravity performs the same amount of work when the brick falls back down again. So then you end up having no NET work. I think the confusion here is that it's next to impossible to have an object with potential energy that did not have to be "lifted" into position first. Maybe meteorites would be an exception? But any earthly object like a brick that falls to the ground had to be lifted up above the ground in the first place. Once the brick falls, it has completed a cycle, and so there is zero net work.

 

[Anastasis]

Well, you had to perform some amount of work to put the brick up on the table, and gravity performs the same amount of work when the brick falls back down again. So then you end up having no NET work. I think the confusion here is that it's next to impossible to have an object with potential energy that did not have to be "lifted" into position first. Maybe meteorites would be an exception? But any earthly object like a brick that falls to the ground had to be lifted up above the ground in the first place. Once the brick falls, it has completed a cycle, and so there is zero net work.

Ah okay - that makes more sense now. What about electric fields? Same logic hold true?

(btw, even with this new info, Kaplan is wrong. Makes me really frustrated and wondering if this happens more often than I realize)

 

[QofQuimica]

Ah okay - that makes more sense now. What about electric fields? Same logic hold true?

(btw, even with this new info, Kaplan is wrong. Makes me really frustrated and wondering if this happens more often than I realize)

Electric fields are a bit trickier, because you can have negative charges, and they do the opposite of positive charges in terms of potential increasing or decreasing. (There is no equivalent negative mass in gravitational energy considerations!) But electric fields are also conservative forces, so again, if your charge ends up where it starts, no net work is performed.

BTW, where in your Kaplan materials do you think you saw the mistake? I'd like to take a look at it.

 

[Anastasis]

Electric fields are a bit trickier, because you can have negative charges, and they do the opposite of positive charges in terms of potential increasing or decreasing. (There is no equivalent negative mass in gravitational energy considerations!) But electric fields are also conservative forces, so again, if your charge ends up where it starts, no net work is performed.

BTW, where in your Kaplan materials do you think you saw the mistake? I'd like to take a look at it.

Kaplan FL #7, PS section (obviously 😉 ) #50:
A crane lifts a 1000kg steel beam off the ground, and sets it down on scaffolding 100 meters off the ground. All of the following are true EXCEPT:
a. The net work done on the steel beam is 9.8x10^5 J.
b. The net work done on the steel beam is 0 J
c. The magnitude of the work done on the steel beam by gravity is 9.8x10^5J
d. The magnitude of the work done on the steel beam by the crane is 9.8x10^5 J

I answered B. But now that I read back through it I think perhaps I was being confused because I thought gravity could not do work. The answer sheet says the answer is A. It says because the beam's kinetic energy is 0 then no work is done, which seems to me obvious bull**** since work can increase potential energy as well.

But I'm starting to overly confuse myself; I'll defer to your opinion on this.

BTW, if me posting this somehow violates copyright, I'm sorry 😳

 

[QofQuimica]

Kaplan FL #7, PS section (obviously 😉 ) #50:
A crane lifts a 1000kg steel beam off the ground, and sets it down on scaffolding 100 meters off the ground. All of the following are true EXCEPT:
a. The net work done on the steel beam is 9.8x10^5 J.
b. The net work done on the steel beam is 0 J
c. The magnitude of the work done on the steel beam by gravity is 9.8x10^5J
d. The magnitude of the work done on the steel beam by the crane is 9.8x10^5 J

I answered B. But now that I read back through it I think perhaps I was being confused because I thought gravity could not do work. The answer sheet says the answer is A. It says because the beam's kinetic energy is 0 then no work is done, which seems to me obvious bull**** since work can increase potential energy as well.

But I'm starting to overly confuse myself; I'll defer to your opinion on this.

BTW, if me posting this somehow violates copyright, I'm sorry 😳

Technically I am sure we aren't supposed to post any test prep company questions, but the big no-no is that we can't post any real MCAT questions. So I'll leave it.

Ok, so here you have a situation like I was describing before, where you pick an object up and set it down somewhere on a ledge. So the net work isn't zero, because the object didn't end up where it started (on the ground). Thus, there must have been some net work performed, and you can eliminate B. The object is raised to a height of 100 m, and its mass is 1000 kg, so the final PE would be mgh or 1000 x 10 x 100 (never use 9.8 for g on the MCAT; always round it to 10 to simplify calculations). C could not be true because the object remains up on the scaffold. (I think maybe the explanation is trying to tell you that if gravity had done some work on the beam, it should have fallen and there WOULD be a non-zero KE.) So now what we have to consider is whether this value (about 1 x 10^6) is the total work done by the crane or not. The answer to this is that we don't know for sure (and it's probably not, because presumably the crane must have lifted the object higher than 100 m first before "setting it down.") So D may be true, but it's not as good of an answer as A, which MUST be true no matter how high the crane originally lifted the object.

I think you should go back and draw this sequence of events step by step. It seems like you are unclear about the difference between work done by each thing (gravity versus the crane) and NET work. You have to keep track of which thing is doing what, as well as the initial and final locations of the object. If the crane picks up the object, it has done work equal to mgh. If the object then falls back to the ground, gravity has done work equal to the same amount. But the NET work in that case will be zero if the object falls all the way back to the ground even though gravity did work to make it fall; this is because the work done by the gravity and the crane cancel out. On the other hand, if the object remains on the ledge like in this problem, there IS net work done of mgh, because the crane has picked it up, but gravity has not done work on it to make it fall again. Does that make sense, or is it even more confusing???

 

[Lests55]

I am hung up on the same question.

If I am reading correctly the question is saying "all of the following are TRUE EXCEPT" or "Which one is false?" Here is my rational for them saying it is A:

They are saying that A is false: "net work done is 9.8x10^5"
The work by the crane is W=Fd, which is positive.
The work done by gravity W=-change(PE) is thus negative.
So C&D are ok....
Sum them together and you get no net work. Wnet=change(KE)=0

Is this correct?

 

[Anastasis]

Technically I am sure we aren't supposed to post any test prep company questions, but the big no-no is that we can't post any real MCAT questions. So I'll leave it.

Ok, so here you have a situation like I was describing before, where you pick an object up and set it down somewhere on a ledge. So the net work isn't zero, because the object didn't end up where it started (on the ground). Thus, there must have been some net work performed, and you can eliminate B. The object is raised to a height of 100 m, and its mass is 1000 kg, so the final PE would be mgh or 1000 x 10 x 100 (never use 9.8 for g on the MCAT; always round it to 10 to simplify calculations). C could not be true because the object remains up on the scaffold. (I think maybe the explanation is trying to tell you that if gravity had done some work on the beam, it should have fallen and there WOULD be a non-zero KE.) So now what we have to consider is whether this value (about 1 x 10^6) is the total work done by the crane or not. The answer to this is that we don't know for sure (and it's probably not, because presumably the crane must have lifted the object higher than 100 m first before "setting it down.") So D may be true, but it's not as good of an answer as A, which MUST be true no matter how high the crane originally lifted the object.

I think you should go back and draw this sequence of events step by step. It seems like you are unclear about the difference between work done by each thing (gravity versus the crane) and NET work. You have to keep track of which thing is doing what, as well as the initial and final locations of the object. If the crane picks up the object, it has done work equal to mgh. If the object then falls back to the ground, gravity has done work equal to the same amount. But the NET work in that case will be zero if the object falls all the way back to the ground even though gravity did work to make it fall; this is because the work done by the gravity and the crane cancel out. On the other hand, if the object remains on the ledge like in this problem, there IS net work done of mgh, because the crane has picked it up, but gravity has not done work on it to make it fall again. Does that make sense, or is it even more confusing???

It's an except question. I followed your same reasoning so b is the untrue statement. Make sense?

 

[QofQuimica]

I am hung up on the same question.

If I am reading correctly the question is saying "all of the following are TRUE EXCEPT" or "Which one is false?" Here is my rational for them saying it is A:

They are saying that A is false: "net work done is 9.8x10^5"
The work by the crane is W=Fd, which is positive.
The work done by gravity W=-change(PE) is thus negative.
So C&D are ok....
Sum them together and you get no net work. Wnet=change(KE)=0

Is this correct?

Oops, I actually missed that "except" part the first time through. 😳 (Which is another good lesson: always read the question thoroughly and make sure you are answering the question that you've actually been asked! 😛 )

In that case, hmm, it looks like this is supposed to be a work-energy theorem problem. I was considering a beam-earth system, not a beam in isolation. If you ignore the potential energy and use W = KE(final) - KE(initial), you will get their answer. This is because the final and initial kinetic energies are both zero. So yes, what you said is correct.

Don't feel bad, though, anastasis; this question is pretty ambiguous. The questions on the MCAT should not be ambiguous like that.

 

[Lests55]

I thought your book said that A was the "except" and thus the false answer?

 

[Anastasis]

Oops, I actually missed that "except" part the first time through. 😳 (Which is another good lesson: always read the question thoroughly and make sure you are answering the question that you've actually been asked! 😛 )

In that case, hmm, it looks like this is supposed to be a work-energy theorem problem. I was considering a beam-earth system, not a beam in isolation. If you ignore the potential energy and use W = KE(final) - KE(initial), you will get their answer. This is because the final and initial kinetic energies are both zero. So yes, what you said is correct.

Don't feel bad, though, anastasis; this question is pretty ambiguous. The questions on the MCAT should not be ambiguous like that.

So on the real MCAT will they specify when a question uses that system (which, btw, I don't understand now, 😳 )

 

[QofQuimica]

So on the real MCAT will they specify when a question uses that system (which, btw, I don't understand now, 😳 )

They should always specify what the system under consideration comprises. I just asked a physicist friend of mine about your question, and his response was that no real physicist would ever write a question like that.

Anyway, I wouldn't stress about it too much.

 

[Nutmeg]

Putting a resistor in series with a capacitor will affect how long it takes the capacitor to charge, but not the current. Things in series can't have different values for current without violating Kirchoff's Laws. Likewise, two things in parallel should have the same potential across them. DrChandy or Nutmeg, if you're reading this, can one of you give a more thorough explanation?

Putting a capacitor in the mix likely gives you a problem that isn't steady-state, and since the MCAT doesn't require differential equations, it seems unlikely that this would come up. A capacitor basically starts with no resistance, and the resistance increases to infinity as the capacitor charges. So in the series, you initially ignore it, but as the capacitor charges, current stops flowing, until the capacitor becomes fully charged and the resistance of the capacitor becomes infinity, bring all current flow through the resistor to a dead stop. The state with a charged capacitor and no current is the steady state, with the full voltage drop being over the capacitor and no voltage drop over the resistor (since V = i*R and i = 0).

Conversely, in parallel, the fact that the uncharged capacitor acts as a short means initially no current goes through the resistor. As the capacitor charges, the resistor takes more and more current. Eventually, when the capacitor becomes fully charged, you ignore it. The steady state is that the voltage drop is set by the resistor, and you can assume that the capacitor in parallel has the same voltage drop, and this can be used to find the charge on the capacitor.

 

[Lests55]

Does an increase in frequency correspond to an increase in "loudness"? How about "pitch"? 😍

 

[Anastasis]

Does an increase in frequency correspond to an increase in "loudness"? How about "pitch"? 😍

Frequency directly influences pitch; Loudness, I assume you mean dB, is a relative measure of air pressure or intensity. (It's on a logarithmic scale, so remember that when doing comparision problems). Increase in frequency, increase in pitch. Decrease in frequency, decrease in pitch.

Air pressure or intensity is directly related to amplitude, so your amplitude affects your "loudness".

If I remember right, frequency affects intensity in electromagnetic waves but I think it affects sounds waves only in a minor way. Someone want to correct me if I'm wrong?

 

[Nutmeg]

Frequency directly influences pitch; Loudness, I assume you mean dB, is a relative measure of air pressure or intensity. (It's on a logarithmic scale, so remember that when doing comparision problems). Increase in frequency, increase in pitch. Decrease in frequency, decrease in pitch.

Air pressure or intensity is directly related to amplitude, so your amplitude affects your "loudness".

If I remember right, frequency affects intensity in electromagnetic waves but I think it affects sounds waves only in a minor way. Someone want to correct me if I'm wrong?

You're right about sound, but frequency in electromagnetic waves effects the color (blue is high frequency, red is low) and energy (high frequency = high energy--that's why UV rays and X-rays are so much more destructive than radio waves and microwaves, which are low frequency).

Amplitude is the number of photons, and frequency is the energy per photon. So a high amplitude wave still has more energy than a low amplitude wave, but for a given frequency/color, the energy per photon is set by the frequency, not the amplitude.

 

[Lanced]

Quick question about capacitors and dielectrics:

Do we need to know the detail presented in the TPR science review book? I.e. what happens if the dielectric is inserted/removed when the battery is connected or disconnected?

On a side topic, should we memorize the Arrhenius equation?

Thanks all

 

[gujuDoc]

Quick question about capacitors and dielectrics:

Do we need to know the detail presented in the TPR science review book? I.e. what happens if the dielectric is inserted/removed when the battery is connected or disconnected?

On a side topic, should we memorize the Arrhenius equation?

Thanks all

It is not so much about memorizing so much as being put in there to help with understanding where some people may not understand conceptually what happens in different situations. I'd just get a general idea of what happens in different situations. On my real MCAT, I never had to know any equations that required logarthims, such as the Arrenius equation or the equation for 1/2 lifes in radioactive decay. Knowing the conceptual aspects was enough. But don't worry about memorizing that eqn.

 

[ladpm]

Hi -

Sorry about if this was posted earlier...could someone give a good explanation of diffraction? Also, show proportionalities of variables (ie. that effect the dark and bright fringes, number of slits, distance from the diffraction, etc?) Sorry for the loaded question. Thanks

 

[Lests55]

(1) In a perf. inelastic collision, it is assumed that no energy is conserved (in the system at least). I know this energy is likely turned into sound (and other stuff?) but why is all of the energy gone from the system? It still moves afterword. It just seems odd that when two objects collide, stick together, and then move off they have not retained any of the energy.

(2) If a substance was described as "highly elastic" does that mean it has a high or low Young's modulus? Or does it simply mean that it has a high yield point?

(3) In the friction portion of the EK series there is a discussion about how friction opposes relative motion and they give an example of a car. Friction acts in the direction of motion of the car, and I don't understand their explanation.

THANKS!

 

[QofQuimica]

(1) In a perf. inelastic collision, it is assumed that no energy is conserved (in the system at least). I know this energy is likely turned into sound (and other stuff?) but why is all of the energy gone from the system? It still moves afterword. It just seems odd that when two objects collide, stick together, and then move off they have not retained any of the energy.

(2) If a substance was described as "highly elastic" does that mean it has a high or low Young's modulus? Or does it simply mean that it has a high yield point?

(3) In the friction portion of the EK series there is a discussion about how friction opposes relative motion and they give an example of a car. Friction acts in the direction of motion of the car, and I don't understand their explanation.

THANKS!

1) KE is conserved for elastic collisions, but not inelastic. The energy is lost as heat.

2) Hmm, I would call that low, assuming that by "highly elastic," you mean that the substance is greatly deformed by the stress. Remember that stronger stuff has a smaller strain for a given stress, so its Young's modulus should be higher.

3) They are probably referring to the friction on the wheels. The friction points in the opposite direction of the wheel's spin, so it will have to point forward.

 

[Lests55]

You're right about sound, but frequency in electromagnetic waves effects the color (blue is high frequency, red is low) and energy (high frequency = high energy--that's why UV rays and X-rays are so much more destructive than radio waves and microwaves, which are low frequency).

Amplitude is the number of photons, and frequency is the energy per photon. So a high amplitude wave still has more energy than a low amplitude wave, but for a given frequency/color, the energy per photon is set by the frequency, not the amplitude.

My EK books say that intensity is propprtion to the square of frequency and amplitude. So by combining what you guys said and this, an increase in frequency via Doppler, the resulting increase in intensity should lead to something being louder?

 

[j-med]

what are the types of materials for magnets?

There are diamagnetic, paramagnetic, and ferromagnetic.
But what about permanent magnets? (Is that a category of its own? vs. temporary magnets or something?)

How does compass fit in these categories?

 

[DrChandy]

what are the types of materials for magnets?

There are diamagnetic, paramagnetic, and ferromagnetic.
But what about permanent magnets? (Is that a category of its own? vs. temporary magnets or something?)

How does compass fit in these categories?

Permanent magnets are ferromagnetic. Compasses are made up of ferromagnetic metals which align to the earth's magnetic field.

 

[SeminoleFan3]

I was working through the AAMC 3R test, and the last question is about work. It has a force (oriented at a 60 degree angle) pulling a mass off the ground through a pulley system. When giving the answer, it doesn't appear that the angle came into play at any point. Why not? I though it would be W=Fd(cos60).

 

[QofQuimica]

I was working through the AAMC 3R test, and the last question is about work. It has a force (oriented at a 60 degree angle) pulling a mass off the ground through a pulley system. When giving the answer, it doesn't appear that the angle came into play at any point. Why not? I though it would be W=Fd(cos60).

Because it's a pulley. If you draw a free-body diagram of the mass, you will see that the tension in the string is going to be 180 degrees away from the direction of the object's weight. Since the tension is the same everywhere on the rope, then the angle that the end of the rope is being pulled at doesn't matter. They do give you extra info on the MCAT, so don't assume that you have to use all of the info they provide.

 

[DrChandy]

I was working through the AAMC 3R test, and the last question is about work. It has a force (oriented at a 60 degree angle) pulling a mass off the ground through a pulley system. When giving the answer, it doesn't appear that the angle came into play at any point. Why not? I though it would be W=Fd(cos60).

It's an mgh problem. Note that the force to lift the block would have to be:

F-mg>0, if the block moves upwards and is not static.

Where F is the composite of both the horizontal and vertical components of the force.


Since the value of F is not given, all you have is the mass and the distance (height) the block has moved. When you see mass and height, think mgh. This will give you the PE, which is equal to the work done in lifting the object.

 

[Anastasis]

I was having a hard time with simple harmonic motion and I can accross this website: http://ww2.unime.it/weblab/mirror/ExplrSci/dswmedia/harmonic.htm

It really helps me to manipulate things to see the relationship. I hope this helps someone!

(btw, Q if this is in the wrong thread, feel free to delete/move, I won't be offended. I wasn't sure where it should go.)

 

[Lests55]

What is the source of the bouyant force? Is it electromagnetic in origin?

Also, does it do work when moving an object "up" just as gravity does in pulling it "down"?

 

[Anastasis]

What is the source of the bouyant force? Is it electromagnetic in origin?

Also, does it do work when moving an object "up" just as gravity does in pulling it "down"?

The bouyant force is basically gravity. It's the force of the displaced water wanting to come to the point of lowest potential gravitational energy. I hope that makes sense.

As for the work, this is one of my weak areas - but I think it does do work (because It's gravity) but I'm sketchy with conservative forces and work.

 

[QofQuimica]

What is the source of the bouyant force? Is it electromagnetic in origin?

Also, does it do work when moving an object "up" just as gravity does in pulling it "down"?

The buoyant force is due to the hydrostatic pressure of the fluid. I don't understand what connection you are making with buoyancy and electromagnetics. Maybe you mean a gravitational field? 😕

Yes, the buoyant force can do work if the object rises.

 

[Lests55]

The buoyant force is due to the hydrostatic pressure of the fluid. I don't understand what connection you are making with buoyancy and electromagnetics. Maybe you mean a gravitational field? 😕

Yes, the buoyant force can do work if the object rises.

I was just trying to relate the Bouyanct Force to one of the fundamental forces. It seems to be a "contact" force and my physics text describes the source of the normal force as electromagnetic (I think through a repulsion explanation).

Question regarding EM Waves...Do they require a medium to travel through (if so, I guess 'vacuum' counts as a medium)? Do they transmit matter as well as energy (i.e. does a photon count as "matter")?

 

[Anastasis]

I was just trying to relate the Bouyanct Force to one of the fundamental forces. It seems to be a "contact" force and my physics text describes the source of the normal force as electromagnetic (I think through a repulsion explanation).

Question regarding EM Waves...Do they require a medium to travel through (if so, I guess 'vacuum' counts as a medium)? Do they transmit matter as well as energy (i.e. does a photon count as "matter")?

EM waves do not need a medium to travel through (that's how they can travel through the vacuum of space, no a vacuum is not a medium). No they do not transmit matter; photons may be particles but in theory, they have no mass. (I've seen the math that proves this but it lost me 1/4 of the way through... 😕 )

As for whether it is a contact force, that is the force that causes the force of the water to transfer to the object to force it up but were the force originates is the force of gravity on the water.

(It's kind of like this distinction: say you push down on a lever and lift a box up, what force is moving the box? Well in the context of that question it is the force you place on the lever but that force takes different forms: mechanical force of your arm, electrostatic force of the atoms of your hand and atoms of the lever, electrostatic force of atoms of lever and atoms of box. I dunno, this made sense in my head, maybe it's just confusing?)

 

[Lests55]

Thanks!

 

[93106]

does ohms law apply for ac circuits?

 

[QofQuimica]

does ohms law apply for ac circuits?

Yes, but the equation is different. You don't have to know about this for the MCAT.

 

[Anastasis]

Can someone explain the difference between anode/cathodes in galvanic cells and anode/cathodes in electrolytic cells?

I know oxidation occurs at the anode and reduction at the cathode in electrolytic cells but I thought it was reversed in galvantic cells - or is only the relative charge of each reversed?

 

[PBandJ]

for all types of cells: An Ox...Red Cat (oxidation occurs at the anode, and reduction occurs at the cathode)
however, in galvanic cells, since they are spontaneous, the anode is labeled negative and the cathode is labeled positive. in electrolytic cells, which are nonspontaneous, the signs are reversed [anode is labeled positive and the cathode is labeled negative].

hope that helps.

 

[Anastasis]

for all types of cells: An Ox...Red Cat (oxidation occurs at the anode, and reduction occurs at the cathode)
however, in galvanic cells, since they are spontaneous, the anode is labeled negative and the cathode is labeled positive. in electrolytic cells, which are nonspontaneous, the signs are reversed [anode is labeled positive and the cathode is labeled negative].

hope that helps.

Yeah it does, a bit. So where red/oxid takes place is the same for both types of cells but the charge of each node is reversed? Does this have to do with galvanic having an applied voltage? sorry to be annoying, just want to get my head around this before the exam.

Basically I'm trying to follow the flow of electrons in my head and having a hard time.

 

[Anastasis]

Okay - so maybe if I type this out you can tell me if my logic follows:
In an electrolytic cell, at the anode, oxidation is occuring which means the electrons flow away in the wire (leaving the anode positive) to the cathode where reduction occurs (more negative).

In a galvanic cell, there is applied voltage (right?), at the cathode reduction is still occuring - taking up electrons and making it more positive. At the anode, oxidation is occuring releasing electons which makes it negative.

Argh - I dunno if that makes any sense, am I right?

 

[QofQuimica]

Okay - so maybe if I type this out you can tell me if my logic follows:
In an electrolytic cell, at the anode, oxidation is occuring which means the electrons flow away in the wire (leaving the anode positive) to the cathode where reduction occurs (more negative).

In a galvanic cell, there is applied voltage (right?), at the cathode reduction is still occuring - taking up electrons and making it more positive. At the anode, oxidation is occuring releasing electons which makes it negative.

Argh - I dunno if that makes any sense, am I right?

It doesn't make sense to me; I think you're mixing up galvanic and electrolytic. Electrolytic cells are the ones where you have an applied voltage, while galvanic cells are spontaneous and have no applied voltage. Like scentimint said, electrons always go to the cathode, which is where reduction occurs. This is always true, in every cell.

I think that the easiest way to think about it is to remember that electrons are all charged negatively. Since they are negative, they would prefer, if given their choice, to go to a positive electrode. In a spontaneous (galvanic) cell, they will therefore go to a positively charged cathode. But in an electrolytic cell, you are applying an EMF to force them to go "backwards" to the electrode that has the same charge as they do. So in this case, the cathode is charged negatively. The only way to get negatively charged electrons to go to a negatively charged cathode is to "force" them to do so by applying an EMF. Electrolytic cells then can never be spontaneous, because electrons always spontaneously go to a positively charged electrode. Does that make sense to you?

 

[Anastasis]

It doesn't make sense to me; I think you're mixing up galvanic and electrolytic. Electrolytic cells are the ones where you have an applied voltage, while galvanic cells are spontaneous and have no applied voltage. Like scentimint said, electrons always go to the cathode, which is where reduction occurs. This is always true, in every cell.

I think that the easiest way to think about it is to remember that electrons are all charged negatively. Since they are negative, they would prefer, if given their choice, to go to a positive electrode. In a spontaneous (galvanic) cell, they will therefore go to a positively charged cathode. But in an electrolytic cell, you are applying an EMF to force them to go "backwards" to the electrode that has the same charge as they do. So in this case, the cathode is charged negatively. The only way to get negatively charged electrons to go to a negatively charged cathode is to "force" them to do so by applying an EMF. Electrolytic cells then can never be spontaneous, because electrons always spontaneously go to a positively charged electrode. Does that make sense to you?

Okay - I'm kinda embarassed 😳 Yes, the whole problem was that I was mixing up the types of cells. I get it now.

 

[PRamos]

My EK books say that intensity is propprtion to the square of frequency and amplitude. So by combining what you guys said and this, an increase in frequency via Doppler, the resulting increase in intensity should lead to something being louder?

There are two effects going on simultaneously. If a wave is being compressed by the Doppler effect, it is because the sender and receiver are getting closer to one another as the wave propagates. Each subsequent wave that the receivers detects was sent from a closer point, so it will appear more intense. This is why a sound from a car coming towards you gets louder as it apporaches while all the time giving off a higher perceived frequency.

It's not a Doppler effect that increases the intensity, but a higher perceived frequency is caused by a scenario that is also associated with increasing sound intensity.

BTW, it's just my opinion, but EK was good for mental pictures, but was often either over-simplified or incorrect. If you want a way to think about something, reference EK. If you want a definite answer you can trust, use another book.

 

[kevin86]

question about the torque equations for both center of mass problems and physical pendulum. So the F*R, what is the R. For a center of mass problem, it's the pivot point to the center of that mass, but what about physical pedulum, is it still the same? If the pendulum was a uniform stick with one end tied to a beam, then R would be half the stick length right? Is the true for all the torque eqations?