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.

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.

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... :eek: )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.

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? :confused:

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... :eek: )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?

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? :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.

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!

(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.

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)

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.

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 bullsh!t 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 :o

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 bullsh!t 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 :o
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???

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?

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???

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

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. :o (Which is another good lesson: always read the question thoroughly and make sure you are answering the question that you've actually been asked! :p )

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.

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

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

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, :o )

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

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

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.

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

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

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.

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

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.

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

(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) 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.

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?

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?

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.

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).

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.

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.

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.)

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"?

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.

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? :confused:

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

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? :confused:

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")?

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... :confused: )

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?)

Thanks!

does ohms law apply for ac circuits?

does ohms law apply for ac circuits?
Yes, but the equation is different. You don't have to know about this for the MCAT.

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?

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.

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.

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?

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?

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 :o Yes, the whole problem was that I was mixing up the types of cells. I get it now.

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.

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?

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?
If you mean the T = F x R x sin theta equation, then R is the distance from the fulcrum to the mass. Assuming that the pendulum was uniform, then its center of mass should be in the center, and the fulcrum is on one end, so I think you're right. DrChandy or Nutmeg, if you're reading this, will you confirm?

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.


shrike you are awsome!! wow, i am amazed at all your posts here! thanks for all your help! :)

Circuits and light bulbs....

Circuit A: Let's say you have a 12 V battery, and a resistor (light bulb) with 2 Ohms resistance.

Circuit B: 12V battery with two 2 Ohm light bulbs connected in series.

Circuit C: 12V battery with two 2 Ohm light bulbs connected in parallel.


Could you please discuss the overall brightness of each circuit, and also the brightness of each light bulb?

Thanks!

Circuits and light bulbs....

Circuit A: Let's say you have a 12 V battery, and a resistor (light bulb) with 2 Ohms resistance.

Circuit B: 12V battery with two 2 Ohm light bulbs connected in series.

Circuit C: 12V battery with two 2 Ohm light bulbs connected in parallel.


Could you please discuss the overall brightness of each circuit, and also the brightness of each light bulb?

Thanks!

Well lets look at the equations you'd need to solve this. They are

P=IV
V=IR

solving for I in the first equation we get P=V^2/R.

Ok, so what's the total resistance? In the first one we're given it, it's 2 Ohms. Plug the values in and we get

P=12^2/2 = 72Watts

What about the second? The rule for resisters in series is just add them up and that's the total resistance. So

Rtotal= R1 + R2 = 2 Ohms + 2 Ohms = 4 Ohms

So just plug that in we get

P=12^2/4 = 36 Watts

So together the bulbs put out a total of 36 watts, or 18 watts each

The last one remember the rule for getting the total resistance with parallel circuits. It's

1/(Rtotal) = 1/R1 + 1/R2 = 1/2 + 1/2 = 1


So 1/Rtotal = 1. So we take the inverse to get Rtotal = 1 Ohms. BTW, be very careful when doing parallel circuits that you remember to take the inverse, otherwise you'll get the wrong total resistance. So plug in the values.

P=12^2/1=144Watts.

So the bulbs put out a total of 144 Watts, or each one puts out 72 Watts.

So lets sum up. In A the bulb puts out 72 Watts, in B each puts out 18 Watts, in C each bulb each puts out 72 watts. So in A and C the bulbs are the same brightness.(Which is to be expected, they're in parallel and can draw current separately. So the bulbs should affect each others brightness.) However in B since we link them in series we affect the bulbs brightness and find the bulb now puts out 1/4 the energy as before. As for total brightness B is the least bright and is half as bright as A. C is twice as bright as A.

Anyway hope I got that right.(Haven't done circuits in a few months so I'm doing this off the top of my head. Take what I say with a grain of salt.)

so brightness is just another way to say power, would intensity be power as well
? my book only mentions intensity, never brightness.

so brightness is just another way to say power, would intensity be power as well
? my book only mentions intensity, never brightness.

Well in my physics course brightness was synonymous with intensity which was just power.(Watch me be wrong though.) I think they might mention "intensity" because people might think brightness = how bright it looks to me.(Which brings up the issue in what spectrum is the light being emitted in. I'll clarify, suppose some of these bulbs emitted mostly in infrared and others mostly in visible. So to the naked eye we might see the infrared bulb dimmer than expected even though it's emitting as much energy as the equations suggest it should. These questions never really care about that, they just care about how much energy total is coming out, regardless of frequency or anything. Anyway more than you need to know.)

Just wondering if anyone knew when momentum is conserved and when total energy is conserved. Also, when the momentum is conserved that means that m1v1=m2v2 and that v1 doesn't necessarily have to equal v2 right? They just need to balance out in the end?

Just wondering if anyone knew when momentum is conserved and when total energy is conserved. Also, when the momentum is conserved that means that m1v1=m2v2 and that v1 doesn't necessarily have to equal v2 right? They just need to balance out in the end?
Momentum will always be conserved on the MCAT. KE is only conserved for elastic collisions, not for inelastic. You are right that the final momentum has to equal the initial momentum in order for momentum to be conserved.

Just wondering if anyone knew when momentum is conserved and when total energy is conserved. Also, when the momentum is conserved that means that m1v1=m2v2 and that v1 doesn't necessarily have to equal v2 right? They just need to balance out in the end?

Wait, are you asking is total energy in a system conserved or kinetic? Q has it right, KE is conserved in totally elastic collision but not in inelastic. However in general the total energy (and I mean EVERYTHING) is always conserved as far as the MCAT is concerned.(It's just transformed. So for example in an inelastic collision that lost KE gets turned into thermal energy if I remember correctly.) Of course this is probably more than you need to know.

wow you are guys are fast...thanks! that really helped! i also had another question (sorry I just realized that there was this discussion board and i am super stoked!) i am confused when to answer if a particle is accelerating or not. for instance, two charges with opposite signs would accelerate towards each other because they have a force acting on them, right (kqq/r2)? But, how about like an object rolling down an incline plane...

For example, after a block begins to slide, how does it speed vary with time? (assuming that tension and kinetic frictional forces are constant in magnitude)

Momentum will always be conserved on the MCAT.


Sorry Q, beg to differ. Momentum is conserved when there are no external forces. If you have external forces, you're not going to be able to conserve momentum. I seem to remember this being an issue in at least one MCAT question.

Remember, the second law says the net force on a particle is equal to the product of it's mass and acceleration (F=ma). Another way Newton's 2nd is presented, and another way you can think of it, is that the net force on a particle is equal to the rate of change in that particles momentum (F=[p2-p1]/[t2-t1])

For the curious:
If you don't know calculus, just change the little d's to delta signs. I don't know how to make delta signs here.

Remember, F=ma
since a=dv/dt, we can sub in the def of acceleration.
F=m(dv/dt)
recalling p=mv, we see now how net force = rate of change in p.
F=dp/dt

wow you are guys are fast...thanks! that really helped! i also had another question (sorry I just realized that there was this discussion board and i am super stoked!) i am confused when to answer if a particle is accelerating or not. for instance, two charges with opposite signs would accelerate towards each other because they have a force acting on them, right (kqq/r2)? But, how about like an object rolling down an incline plane...

For example, after a block begins to slide, how does it speed vary with time? (assuming that tension and kinetic frictional forces are constant in magnitude)

Newton again. Sum the force vectors. That will be equal to the magnitude of the mass times the acceleration and point in the direction of the accel.

F=ma Live it, breathe it, be it.

Count the forces. Things you're touching, gravity if you need it, friction if you need it, electric or mag if you need it.

So with the block
Things you're touching: the plane (normal force)
Gravity: check
Friction: check (kinetic).
There's no tension unless it's touching something like a string.

If those forces cancel out, no accel. If they don't cancel out, you're accelerating.

Edit: You said "after" the block starts to slide. Realize the "just beginning to slide" is code for static friction at the breaking piont (set F=ma to 0, because you haven't started moving yet).

Q, is this allowed? Me answering these questions? You mentioned something, so I just kind of dove in...

I seem to remember this being an issue in at least one MCAT question.

Really? Was it a test prep company question, or an AAMC one? Every MCAT-like momentum question I've ever seen has been a scenario where there are no external forces acting on the system, like two hockey pucks colliding or something like that. Anyway, it's a good point that what I said in my last post assumes that there are NOT any external forces acting on the system. Thanks for clarifying.

Q, is this allowed? Me answering these questions? You mentioned something, so I just kind of dove in...
Definitely. We're always looking for good people to help answer questions. That's why I invited you. If you feel that physics is your strong point, I will more or less leave this thread to you, and I'll take the chem ones.

Edit: You said "after" the block starts to slide. Realize the "just beginning to slide" is code for static friction at the breaking piont (set F=ma to 0, because you haven't started moving yet).



Actually I think you should clarify this a little. Once it "just begins to slide" we stop worrying about static friction and start looking at dynamic friction.(Static friction> dynamic) So we have to do all the work, calculating the normal force, force down the plane, etc to see what the net force is. (Just an aside to the original asker, do you know how to calculate a normal with trig? It's not hard, I can run through the thought process if you want but then again loads of people here can as well.)

Oh BTW (again to the original asker.) you need to be careful to remember that acceleration and velocity are vectors.(dbhvt is being careful to point this out but it's really important to remember since it can get students far too often.) I'm just bringing it up because dbhvt points out that the acceleration in directed in some direction but he is not saying the velocity is in any particular direction.(Because he doesn't have enough info to say what direction the velocity is in, it could be in a completely different direction than acceleration.)

(Just an aside to the original asker, do you know how to calculate a normal with trig? It's not hard, I can run through the thought process if you want but then again loads of people here can as well.)

Dave, if you want to write up a general explanation of how to determine the forces for blocks sliding down inclines, that would be great. We'll add it to the explanations thread.

Well, the answer to that block question is that its speed increased linearly with time....I thought that since there was a consant force that acceleration would be constant leading to no change in speed?

Well, the answer to that block question is that its speed increased linearly with time....I thought that since there was a consant force that acceleration would be constant leading to no change in speed?

ACCELERATION AND VELOCITY
(this is a key point, so let me know if you got it)

Acceleration is defined as a change in velocity over some time.

So if the net force is not zero, then every single time without fail no matter what, the acceleration is not zero and the velocity is changing.

If the net force is constant, the acceleartion is constant and the velocity is changing at a constant rate.

Say you have a constant acceleration of 10 m/s^2 down.

Every second, you add 10 m/s down to your previous velocity vector. If you start out going 5 m/s down, in 1 second you'll be going 15 m/s down, in two you'll be going 25 m/s down, etc....

If you have non-zero accelaration, but it is NOT constant, your velocity vector will change, but it will change different amounts every second. Eg, start out at 5 m/s down, next second your at 40 m/s down, the next second you'r at 2 m/s to the left.


SPEED vs VELOCITY

As far as speed is concerned, remember that speed is a scalar (just a number) and velocity is a vector (a number and a direction). You could come up with a situation where there is no change in speed but a change in velocity. That situation would be a particle moving in a circle without speeding up or slowing down (ie uniform circular motion). The velocity will be changing, because the particle is not moving in a straight line. Even though the number part of the velocity isn't changing, the direction part is changing.

I'm getting confused about waves. Is this correct?

Sound waves: needs a medium to travel. Velocity increases as it travels through a medium, while f stays the same, thus wavelength increases? But the Doppler effect can change just the observed frequency?

EM waves: does not need a medium to travel. Velocity decreases depending on the medium, frequency stays the same, so wavelength changes?

Light (EM): Higher frequency waves take longer to travel through a medium and are thus dispersed more than lower frequency wavelengths. So does this mean that velocity and wavelength are the only things that change also?

Lastly, in reference to the galvanic/electrolytic cell, electrons always go to the anode. This might be the wrong thread, but in o-chem, anions always go to the anode, correct? (regarding lab techniques)

Thanks all!

Actually I think you should clarify this a little. Once it "just begins to slide" we stop worrying about static friction and start looking at dynamic friction.(Static friction> dynamic)

FINE POINTS:

This is what I meant by 'code for...'

If it's actually moving, and you're worried about what is happening while it's moving, then yes you do stop worrying about static friction and start looking at dynamic friction (or kinetic friction, whichever term you want to use).

If the problem says "JUST BEGINS to slides", almost always, the moment of interest is the change from sitting there and not moving to moving. This moment is best described in equations as static friction maxed out.

(I wish I knew how to do greek letters, subtext and supertext)

Recall that F(static friction) is < or = (the coefficient of static friction)*(the normal force).

Because it's an inequality, you generally can't solve the equation definitively if you are told the block is just sitting there and not moving.

But if they tell you it is at the point where it JUST STARTS moving, they're interested in the moment when the net force vector overcomes the force of static friction. Set F=ma to zero, and F(static friction) to it's maximum value in the inequality.

There are a couple of buzz words in physics problems that are like this, where we go from one state to another state, each with different force vectors. They can be a little tricky and counter intuitive, but generally want to look at the state BEFORE the change.

Definitely. We're always looking for good people to help answer questions. That's why I invited you. If you feel that physics is your strong point, I will more or less leave this thread to you, and I'll take the chem ones.


Q-

How about if I just answer things here when I'm bored?

Now that I know where I'm going, I'm in shedding responsibility mode.

Lastly, in reference to the galvanic/electrolytic cell, electrons always go to the anode. This might be the wrong thread, but in o-chem, anions always go to the anode, correct? (regarding lab techniques)

Thanks all!

regarding your electrochemistry question (which probably belongs in the general chemistry thread):

electrons always flow from anode to cathode. the cathode is where reduction occurs (gaining electrons); the anode is where oxidation occurs (losing electrons). in a galvanic cell (spontaneous), the anode is negative while the cathode is positive. in an electrolytic cell (non-spontaneous), the anode is positive while the cathode is negative (hence why an applied voltage is needed to move the electrons TOWARD the negative charge).

hi scentimint, thanks, I was afraid I wrote that. I meant, electrons go to the cathode. The thing I wanted to clarify was that in regards to o-chem lab techniques (i.e. electrophoresis), the anions migrate to the anode, right? I've been keeping that in my head, and then I read the posts about electrolytic/galvanic cells and realized that I should separate these thoughts in my head. Hope that made sense.

ACCELERATION AND VELOCITY
(this is a key point, so let me know if you got it)

Acceleration is defined as a change in velocity over some time.

So if the net force is not zero, then every single time without fail no matter what, the acceleration is not zero and the velocity is changing.

If the net force is constant, the acceleartion is constant and the velocity is changing at a constant rate.

Say you have a constant acceleration of 10 m/s^2 down.

Every second, you add 10 m/s down to your previous velocity vector. If you start out going 5 m/s down, in 1 second you'll be going 15 m/s down, in two you'll be going 25 m/s down, etc....

If you have non-zero accelaration, but it is NOT constant, your velocity vector will change, but it will change different amounts every second. Eg, start out at 5 m/s down, next second your at 40 m/s down, the next second you'r at 2 m/s to the left.


SPEED vs VELOCITY

As far as speed is concerned, remember that speed is a scalar (just a number) and velocity is a vector (a number and a direction). You could come up with a situation where there is no change in speed but a change in velocity. That situation would be a particle moving in a circle without speeding up or slowing down (ie uniform circular motion). The velocity will be changing, because the particle is not moving in a straight line. Even though the number part of the velocity isn't changing, the direction part is changing.


Oh okay, I think I got it...does this make sense? Since the force is constant...the acceleration is constant, meaning no change in velocity. But, velocity is a change in either speed or direction and only one of them have to remain constant. So, since the acceleration is constant, a=F/m...its speed will be equal to the equation v=at. While t increases v (its speed) must also increase with time since acceleration remains constant. It kinda reminds me of those V=IR questions because we know R remains constant (depends on pL/A)...so if V increases so does I because R is constant.

Thanks so much for your help! One more quick question...If you throw an object up in the air the time it takes for it to reach the top is longer than the time it takes during its descent right? Is it because on the way down it has the force of gravity pulling it down so its velocity is greater than the way going up? If not, can you tell me a scenario when this would be true?

...the acceleration is constant, meaning no change in velocity.

Repeat this mantra to yourself in the bathroom mirror for 10 minutes every morning untill Saturday:

"Acceleration IS a change in velocity. If there is acceleration, the velocity is changing."

If the acceleration is constant, and it is not zero, there IS a change in velocity. Always, every single time, no matter what. It could be a change in the direction, the speed, or both. On the MCAT, acceleration problems will generally be constant non-zero acceleration. All of those problems will involve a change in velocity. Everyone, everytime, every year, every administration. Now repeat after me:

"Acceleration IS a change in velocity. If there is acceleration, the velocity is changing."

HINT:

Try to stop thinking about speed. Speed doesn't really make sense. If you're going 50 mph, do you care what direction you are going? Yes, of course you do. Doing your thing around town do you ever care about speed without direction? No.

Speed is an abstract silly thing that is used when bragging about the Porsche they give you if you get a 45 on the MCAT. In physics and real life, you always deal with direction at the same time, so you want to think about velocity.

...Thanks so much for your help! One more quick question...If you throw an object up in the air the time it takes for it to reach the top is longer than the time it takes during its descent right? Is it because on the way down it has the force of gravity pulling it down so its velocity is greater than the way going up? If not, can you tell me a scenario when this would be true?

FAST AND EASY ANSWER:
It takes the same amount of time to go up as come down.


EXPLANATION:


Any object near the surface of the earth (ie, not the moon or a spaceship--closer than 300km for the quantitative folks), say a ball, or a block, or a person, will have the same force of gravity acting on it all the time everywhere.

If you throw a ball in the air and are worried about the time when the ball is in the air, what forces are acting on it?

Things your touching: nothing
Gravity if you need it: check
Friction if you need it: lets say you don't need air resistance (the ball is small and moving slowly)

So, you have a constant force acting on a particle. That force is a vector pointing down with a magnitude of (mass of the ball)*10m/s^2. That force is constant.

The force is constant, the acceleration is constant, the velocity is changing at a constant rate for as long as the ball can be described by the force diagram we just talked through (as long as nothing else is touching it--like the ground, for instance).

So describe the motion of the particle starting with upward velocity, but under constant downward acceleration:

The ball's velocity will change by 10m/s down every second of it's flight. Constant force, constant acceleration, constant rate of change in it's velocity. As far as the speed is concerned, the ball will keep moving up for some amount of time, but slow down at a constant rate untill it stops. Then it will start moving down. Because the rate it slows down on the way up is equal to the rate it speeds up on the way down, it takes the same amount of time to go up as it does to come down, and at any given height, it will have the same speed on it's way down as it did earlier on it's way up. In the up path, at 5 ft, the balls velocity might be 15 m/s up. In the down path, at 5 ft, the balls velocity would then be 15 m/s down.

Thanks so much for your help dbhvt ...that really clears things up! I really appreciate you taking the time to help me with these questions! :)

Thanks so much for your help dbhvt ...that really clears things up! I really appreciate you taking the time to help me with these questions! :)


No problem.
:luck:

Dave, if you want to write up a general explanation of how to determine the forces for blocks sliding down inclines, that would be great. We'll add it to the explanations thread.

You mean just how to calculate a normal or a force down a ramp? It's pretty straight forward but the normal is F=mg*cos(x) and the force down the ramp is F=mg*sin(x) where x is the incline of the ramp. I can write something up later today about what all that means.(Or do you mean how do you go about attacking the problem in the first place?)

FINE POINTS:

This is what I meant by 'code for...'

If it's actually moving, and you're worried about what is happening while it's moving, then yes you do stop worrying about static friction and start looking at dynamic friction (or kinetic friction, whichever term you want to use).

If the problem says "JUST BEGINS to slides", almost always, the moment of interest is the change from sitting there and not moving to moving. This moment is best described in equations as static friction maxed out.

(I wish I knew how to do greek letters, subtext and supertext)

Recall that F(static friction) is < or = (the coefficient of static friction)*(the normal force).

Because it's an inequality, you generally can't solve the equation definitively if you are told the block is just sitting there and not moving.

But if they tell you it is at the point where it JUST STARTS moving, they're interested in the moment when the net force vector overcomes the force of static friction. Set F=ma to zero, and F(static friction) to it's maximum value in the inequality.

There are a couple of buzz words in physics problems that are like this, where we go from one state to another state, each with different force vectors. They can be a little tricky and counter intuitive, but generally want to look at the state BEFORE the change.

Oh point taken. I agree, it can be a bit tricky since wording can make a world of difference.(Yes, been burnt by that before.)

I'm getting confused about waves. Is this correct?

Sound waves: needs a medium to travel. Velocity increases as it travels through a medium, while f stays the same, thus wavelength increases? But the Doppler effect can change just the observed frequency?


I'm not sure what you're asking here. Yes sound requires a medium and it travels at a constant speed through a given medium. It doesn't speed up or slow down traveling through a given medium. It will change speed going from one medium to another but frequency stays the same. Also the doppler effect can change the frequency.(Relative motion between observer and emitter will do that.)


EM waves: does not need a medium to travel. Velocity decreases depending on the medium, frequency stays the same, so wavelength changes?

Light (EM): Higher frequency waves take longer to travel through a medium and are thus dispersed more than lower frequency wavelengths. So does this mean that velocity and wavelength are the only things that change also?

Lastly, in reference to the galvanic/electrolytic cell, electrons always go to the anode. This might be the wrong thread, but in o-chem, anions always go to the anode, correct? (regarding lab techniques)

Thanks all!
Yes, light does not need a medium to travel. Velocity does depend on the medium although frequency stays the same. I hate to admit but I'm a little iffy on how refraction relates to frequency.

Actually here's a rule of thumb, waves going from one medium to another don't change frequency. Speed and wavelength change.(There's an example of a wave going down a string crossing from a light to heavy string that really illustrates this. I think I can find a picture of the set up and I can give the explaination of why it's true. It's apparently is true of any wave.) Do you want me to go through why frequency doesn't change between mediums?

DaveD,
Thank you for taking the time to answer my questions. You don't have to go into the explanation of why freq. doesn't change. I was just having a hard time keeping the relationships straight. I was looking through EK physics and it seems that they state some steadfast rules and then add to it saying something along the lines of 'except for light,' or 'except for sound'.
So i will just keep it at: frequency does not change depending on the medium, only velocity and wavelength, for all types of waves. (unless there is a doppler shift).

Stupid question:

If the SI unit for elec pot is volts, how is it equal to an amount of Work? Work has units of Joules, right?


Ridiculous question:

Will it help me for the mcat to be very familiar with the SI units and their derivatives for potential electricity and magnetism questions?


Last question:

What would you say is the most often misunderstood sort of question about elec and magn on the mcat? :idea:

Let's say someone is standing in a frictionless ice rink and someone tosses them a ball. Who would experience a greater impulse, someone that catches it and holds on or someone that catches it starts to move and then drops it? Please explain in detail using F, dt, and dp. Thanks! Assume that both times the balls were thrown with the same constant speed.

Q-

How about if I just answer things here when I'm bored?

Now that I know where I'm going, I'm in shedding responsibility mode.
Well, I'll take whatever I can get; this forum is a lot of work, as I'm sure you can imagine. I don't know what it is with you physics people, but I sure go through a lot of you. :p

BTW, I'm going to use your a vs. v post for the explanations thread. Thanks for writing that.

Bernoulli's equation:

Okay, so it says that where when velocity of the fluid increases, the pressure decreases.

Would it also be correct to say that pressure is directly proportional to area?

EK says that pressure is inversely proportional to area, and this just doesn't make sense to me...

Thanks!

Pressure = Force/Area. An increase in area results in a decrease in pressure; a decrease in area results in an increase in pressure.

Pressure = Force/Area. An increase in area results in a decrease in pressure; a decrease in area results in an increase in pressure.


But what about fluid flow going from a fat pipe into a thin pipe. The velocity increases when fluid enters the thin pipe, and the pressure decreases. Hasn't the cross sectional area also decreased?

You see, the reason for my confusion is that in the EK bio review book (edition 5, page 142), they try to make the point that blood flow does NOT follow Bernoulli's equation. In trying to make this point, they say, "Bernoulli's equation tells us that pressure is inversely proportional to cross-sectional area, but in blood vessels this is not the case..."
But it IS, like that in blood, right? Capillaries have big overall cross-sectional area, low pressure.
I can't quite make the connection though because Pressure is low in capillaries, which have the greatest area...


EDIT: ohhh but the capillaries do have more pressure than venules.. I get it.. So it kinda follows Bernoulli's equation, but not completely...

But what about fluid flow going from a fat pipe into a thin pipe. The velocity increases when fluid enters the thin pipe, and the pressure decreases. Hasn't the cross sectional area also decreased?

You see, the reason for my confusion is that in the EK bio review book (edition 5, page 142), they try to make the point that blood flow does NOT follow Bernoulli's equation. In trying to make this point, they say, "Bernoulli's equation tells us that pressure is inversely proportional to cross-sectional area, but in blood vessels this is not the case..."
But it IS, like that in blood, right? Capillaries have big overall cross-sectional area, low pressure.
I can't quite make the connection though because Pressure is low in capillaries, which have the greatest area...


EDIT: ohhh but the capillaries do have more pressure than venules.. I get it.. So it kinda follows Bernoulli's equation, but not completely...

I am sorry if I am not supposed to answer this, but from what I understand Bernoulli's equation applies to ideal fluid flow(there is a variation to apply to non-ideal system, but that is beyond scope of MCAT). Blood is a non-ideal fluid, it is considered Non-newtonian fluid and hence, that bernoulli's equation for ideal flow does not apply to blood. If anyone wants to add or correct me feel free to do so. I hope that clarifies a little bit

m2sin(theta2) = m1sin(theta1)

Which equation is NOT used in the derivation?


A. F = mg
B. F = Fsin(theta)
C. F = ma
D. F = Fcos(theta)

I have two physics-related questions:

why does a falling body take some time to reach terminal velocity? Aren't the weight of the falling body and the air resistance constant throughout the fall?


The work-energy theorem states Work = ΔKE. (how) would this be changed/affected when there is ΔPE or friction involved?


Thanks in advance.

I have two physics-related questions:

why does a falling body take some time to reach terminal velocity? Aren't the weight of the falling body and the air resistance constant throughout the fall?


The work-energy theorem states Work = ?KE. (how) would this be changed/affected when there is ?PE or friction involved?


Thanks in advance.
air resistance is a force in the opposite direction to which the body is falling...a force can be dependent on several things but the important one here is acceleration which is increasing as a body is falling.

1st Total Energy is conserved in the universe...so a change in KE will usually be accompanied by a change in PE or friction. Don't forget that work is also Force time distance.
hope that helps

I have two physics-related questions:

why does a falling body take some time to reach terminal velocity? Aren't the weight of the falling body and the air resistance constant throughout the fall?


The work-energy theorem states Work = ΔKE. (how) would this be changed/affected when there is ΔPE or friction involved?


Thanks in advance.
1) No. Air resistance is proportional to the velocity of the falling object. At t=0, before the object begins to fall, the resistance is zero. As the object begins to fall faster and faster, the resistance increases until it reaches the point where it is equal and opposite to the weight of the object. This happens fairly quickly (within several seconds), but your object is still falling plenty fast!

Next time you're driving in your car, try this experiment:
While your car is stopped, stick your hand out the window. You won't feel any resistance, because the car isn't moving. If you start driving slowly, and you stick your hand out the window, you will feel some resistance. If you were to go 80 mph on the highway and stick your hand out the window (I don't recommend actually doing this!!!), you would feel A LOT of resistance.

2) Whenever you have friction, the total energy is equal to the sum of the energy lost to friction plus the kinetic energy. So the kinetic energy will be lower in the presence of friction assuming a constant total energy, because some of that total energy must be expended overcoming the friction.

1) No. Air resistance is proportional to the velocity of the falling object. At t=0, before the object begins to fall, the resistance is zero. As the object begins to fall faster and faster, the resistance increases until it reaches the point where it is equal and opposite to the weight of the object. This happens fairly quickly (within several seconds), but your object is still falling plenty fast!

Next time you're driving in your car, try this experiment:
While your car is stopped, stick your hand out the window. You won't feel any resistance, because the car isn't moving. If you start driving slowly, and you stick your hand out the window, you will feel some resistance. If you were to go 80 mph on the highway and stick your hand out the window (I don't recommend actually doing this!!!), you would feel A LOT of resistance.

2) Whenever you have friction, the total energy is equal to the sum of the energy lost to friction plus the kinetic energy. So the kinetic energy will be lower in the presence of friction assuming a constant total energy, because some of that total energy must be expended overcoming the friction.


thanks.

Please explain this to me:

Accelerating in a car, the friction force acts in the direction of motion of the car.

Please explain this to me:

Accelerating in a car, the friction force acts in the direction of motion of the car.
I know this question has come up before, but I can't find it right now. Basically it has to do with the fact that the friction is on the tires, not the car, and the tires are rolling backward at the point where they hit the road. Thus, the friction points forward. Look at how the wheels move on a toy car and you will see what I mean.

Please explain this to me:

Accelerating in a car, the friction force acts in the direction of motion of the car.
if we assume the car is moving to the right, the wheels are rotating clockwise. So at the point where the wheels touch the ground, the tire is exerting a force to the left, making the friction push the car to the right. Also remember at this point its all static fricition! If you lock the wheels and basically begin skidding, then the tire is not exerting any force and kinetic friction starts slowing the car down by having a force pointing to the left. Does that make sense?

I'm looking at EK Physics manual (5th) page 5. It says that since distance traveled by one is unknown, his speed or instantaneous velocity at any moment during the trip is unknown. I see why the speed is unknown, since speed=distance/time, but I don't understand why IV is also unknown. Can't we get the instantaneous velocity by using eq. v^2 = v0^2 + 2ax, where x is displacement, not distance. Since the IV, according to the formula, is independent of distance, I don't understand why EK book said we cannot calculate IV without knowing distance traveled.

Any help would be appreciated.

I'm looking at EK Physics manual (5th) page 5. It says that since distance traveled by one is unknown, his speed or instantaneous velocity at any moment during the trip is unknown. I see why the speed is unknown, since speed=distance/time, but I don't understand why IV is also unknown. Can't we get the instantaneous velocity by using eq. v^2 = v0^2 + 2ax, where x is displacement, not distance. Since the IV, according to the formula, is independent of distance, I don't understand why EK book said we cannot calculate IV without knowing distance traveled.

Any help would be appreciated.

For example, if you start at point zero on a 400 meter track, run around and return to point 0, than your distance travelled is 400 meters but the displacement is 0. Displacement is by defintion the final position minus the intial position or the net distance travelled. If you travel 3 mile north than 4 miles east, your displacement is 5 and not 7 because you take into the consideration the final point and initial point (drawing the figure helps--it is a triangle). Velocity in general is displacement over time. If you do not know the overall distance travelled, you cannot find displacement. If you cannot find displacement, you cannot find velocity. Hope that helps!

For example, ...

So if I don't know the distance traveled, but know the
starting point and the ending point, then I can calculate
velocity since EP-SP = displacement. Am I right?
And thanks for your reply.

So if I don't know the distance traveled, but know the
starting point and the ending point, then I can calculate
velocity since EP-SP = displacement. Am I right?
And thanks for your reply.
In terms of velocity, you don't need to worry about distance traveled. Velocity is defined as the change in position (displacement) over the change in time. So if you have the starting point (x1), the ending point (x2), and the time it took to get from x1 to x2 (delT), your velocity is simply:

v=(x2-x1)/delT

So, to answer your question, yes, you can calculate velocity without knowing distance traveled.

In terms of velocity, you don't need to worry about distance traveled. Velocity is defined as the change in position (displacement) over the change in time. So if you have the starting point (x1), the ending point (x2), and the time it took to get from x1 to x2 (delT), your velocity is simply:

v=(x2-x1)/delT

So, to answer your question, yes, you can calculate velocity without knowing distance traveled.

Thanks!

So if I don't know the distance traveled, but know the
starting point and the ending point, then I can calculate
velocity since EP-SP = displacement. Am I right?
And thanks for your reply.

Yes you are. But be careful with the EP-SP notion. For example, if you travel 3 miles north, and then 4 miles east, you would think the displacement is 4-0=4. However, this is not true. You have to take into consideration the net displacement. Drawing this on paper, you get a displacement of 5 and 5. (Draw the triangle out...it helps to see it better). So displacement is final minus intial but it is representative of net quantity.

Yes you are. But be careful with the EP-SP notion. For example, if you travel 3 miles north, and then 4 miles east, you would think the displacement is 4-0=4. However, this is not true. You have to take into consideration the net displacement. Drawing this on paper, you get a displacement of 5 and 5. (Draw the triangle out...it helps to see it better). So displacement is final minus intial but it is representative of net quantity.
Yes, this is a good point as well. With my explanation above, I assumed that the motion all took place on one line. You have to take into account that distance and velocity are vectors, with both magnitude and direction.

In an ideal fluid through a horizontal pipe, if the radius of the pipe is doubled, shouldn't the pressure be halved by two according to the equation P = F/A? Why is this approach wrong? Thanks for your help in advance. :D

In an ideal fluid through a horizontal pipe, if the radius of the pipe is doubled, shouldn't the pressure be halved by two according to the equation P = F/A? Why is this approach wrong? Thanks for your help in advance. :D
So you have to think about what A is in this equation. It's related to the radius, so your are on the right track in your approach, but you need to know that the area A refers to the cross-sectional area where the fluid is flowing through the pipe. Since the cross-sectional area is a circle in this case, the area A is related to the radius R by the following equation:

A = PI*(R^2)

So, if you plug that back into your P = F/A equation, you get:

P = F/A = F/(PI*(R^2))

From that point, you can see that pressure P is proportional to 1/(R^2). So, as your question asks, if the radius of the pipe is doubled (R' = 2R), then the pressure would in fact be quartered (P' = P/4) because of the R^2 in the denominator.

I hope that makes sense. :)

So you have to think about what A is in this equation. It's related to the radius, so your are on the right track in your approach, but you need to know that the area A refers to the cross-sectional area where the fluid is flowing through the pipe. Since the cross-sectional area is a circle in this case, the area A is related to the radius R by the following equation:

A = PI*(R^2)

So, if you plug that back into your P = F/A equation, you get:

P = F/A = F/(PI*(R^2))

From that point, you can see that pressure P is proportional to 1/(R^2). So, as your question asks, if the radius of the pipe is doubled (R' = 2R), then the pressure would in fact be quartered (P' = P/4) because of the R^2 in the denominator.

I hope that makes sense. :)

Hi, I completely agree with you, but the answer is "the new pressure cannot be determined without more info" Perhaps I have briefed the question too much. So here is the actual Q:
An ideal fluid with pressure P flows through a horizontal pipe with a radis r. If the radius of the pipe is increased by a factor of 2, which of the following more likely gives the new pressure?
a.P
b.4P
c.16P
d."the new pressure cannot be determined without more info

So what's wrong with the way you described above in the last post about your approach to this question and why is the answer d? Thanks

Here is another physics question that's bothering me. Please help...:)
Q: Two particles are held in equilibrium by the gravitational and electrostatic forces between them. Particle A has mass m(a) and charge q(a). Particle B has mass m(b) and charge q(b). The distance between the charges is d. Which of the following changes will cause the charges to accelerate toward one another?

A. m(a) is doubled and m(b) is doubled
B. m(a) is doubled and m(b) is halved
C. q(a) is doubled and q(b) is doubled
D. d is doubled

The answer is A, but I think A works only if the heavier particle is located above the lighter one as both their forces will be doubled but the actual magnitude of increase in forces will be proportional to the original mass of the particles as shown in the eq. F = ma.
How is this so? Thanks again!!!!!

Here is another physics question that's bothering me. Please help...:)
Q: Two particles are held in equilibrium by the gravitational and electrostatic forces between them. Particle A has mass m(a) and charge q(a). Particle B has mass m(b) and charge q(b). The distance between the charges is d. Which of the following changes will cause the charges to accelerate toward one another?

A. m(a) is doubled and m(b) is doubled
B. m(a) is doubled and m(b) is halved
C. q(a) is doubled and q(b) is doubled
D. d is doubled

The answer is A, but I think A works only if the heavier particle is located above the lighter one as both their forces will be doubled but the actual magnitude of increase in forces will be proportional to the original mass of the particles as shown in the eq. F = ma.
How is this so? Thanks again!!!!!

Newton's law of gravitation states that each object exerts a equal but opposite pull for one another (it is a action-reaction pair as Newton's third law states). On the other hand, the electrostatic force between two static charges depends on the magnitude of the charge. Opposite charges attract and like charges repel. In this case, since the charges are in equilibrium, I believe the charges are like in magnitude since the system is in equilibrium and the law of gravitation relies on the pull for one another. That means:
F net = (GM1M2/r^2) = KQ1Q2/r^2. The r in relation cancels. Therefore you can knock out answer choices c and d. Now, since it seems that both charges are of the same sign and magnitude, doubling the magnitude will cause for further repulsion, and F net will have a value. In the question, F net is equal to zero since the system is in equilibrium and the charges are static. Therefore, b can be eliminated and that leaves choice a. Choice a is correct because doubling the mass causes for a four times greater pull of the objects for one another. I hope this helps. Remember, the law of gravitation means the pull of the an object for another object. Good luck!!

Newton's law of gravitation states that each object exerts a equal but opposite pull for one another (it is a action-reaction pair as Newton's third law states). On the other hand, the electrostatic force between two static charges depends on the magnitude of the charge. Opposite charges attract and like charges repel. In this case, since the charges are in equilibrium, I believe the charges are like in magnitude since the system is in equilibrium and the law of gravitation relies on the pull for one another. That means:
F net = (GM1M2/r^2) = KQ1Q2/r^2. The r in relation cancels. Therefore you can knock out answer choices c and d. Now, since it seems that both charges are of the same sign and magnitude, doubling the magnitude will cause for further repulsion, and F net will have a value. In the question, F net is equal to zero since the system is in equilibrium and the charges are static. Therefore, b can be eliminated and that leaves choice a. Choice a is correct because doubling the mass causes for a four times greater pull of the objects for one another. I hope this helps. Remember, the law of gravitation means the pull of the an object for another object. Good luck!!

BioMedEngineer, Thanks so much for your help. But I'm not clear how the signs are the same for the two particles. I would really appreciate if you can explain how you proved that teh charges for the two particles are the same. Thanks!!

BioMedEngineer, Thanks so much for your help. But I'm not clear how the signs are the same for the two particles. I would really appreciate if you can explain how you proved that teh charges for the two particles are the same. Thanks!!

Hey! The law of gravitation states the pull objects exert on one another and is a action-reaction force pair. The electrostatic force is the force two charges feel on one another. The system in the question is in equilibrium meaning the charges are static and fixed in place. If the charges were opposite in magnitude and charge, they would feel a natural pull for one another. But in this case, the law of gravitation balances the electrostatic force. This means that the charges have to be the same in charge and magnitude and. Thus, Fnet is equal to zero and there is no acceleration. However, if the mass is doubled, the gravitational force will slightly overcome the electrostatic force and the objects will pull on one another. Drawing a diagram helps. If the charges are like, than there will be one vector towards each charge representing the gravitational force and one vector away from each charge, they are like so they repel. Since we are dealing with vectors, we assign one as negative and one as positive, depends on the preferred sign convention, and the vectors will cancel out. Thus, the only way to disturb equilibrium is to deal with the gravitational force. I hope this helps. Good luck!! :luck:

I am having some trouble with harmonics (by trouble I mean I did not get a single question correct in the practice test I took concerning waves and harmonics).
Is it correct to say:
"A higher harmonic will have a higher frequency. Also the frequencies double as the harmonic increases."
For instance if the first harmonic has a frequency of 50hz then the second harmonic will have a frequency of 100 hz, and the third harmonic will have a frequency of 150 hz?
Also any tips or places to look to review harmonics would be appreciated. When I looked on the SDN Physics FAQ thread it only listed a wave topic which focused on longitudinal versus transverse waves (nothing about harmonics).

Thanks in advance

I am having some trouble with harmonics (by trouble I mean I did not get a single question correct in the practice test I took concerning waves and harmonics).
Is it correct to say:
"A higher harmonic will have a higher frequency. Also the frequencies double as the harmonic increases."
For instance if the first harmonic has a frequency of 50hz then the second harmonic will have a frequency of 100 hz, and the third harmonic will have a frequency of 150 hz?
Also any tips or places to look to review harmonics would be appreciated. When I looked on the SDN Physics FAQ thread it only listed a wave topic which focused on longitudinal versus transverse waves (nothing about harmonics).

Thanks in advance

Hey. If you are talking about waves, than the frequency is given by the following relation (velocity of the wave in the medium*harmonic number)/(2*the length of the string). The harmonic frequency is given by the following relationship (f(n)=nf1) where n is the harmonic number and f1 is the first frequency---frequency of the first harmonic. So, your analysis is correct--if the first harmonic is 50 than the second harmonic is 100. But beware, the second harmonic can mean that there are 3 nodes on the string. For good review of harmonics, I suggest looking at any physics textbook--or in a review book. If you have any other questions, I am glad to help or I can post more information on harmonics--provided I have the permission first to do so. Good luck :luck:

Hey. If you are talking about waves, than the frequency is given by the following relation (velocity of the wave in the medium*harmonic number)/(2*the length of the string). The harmonic frequency is given by the following relationship (f(n)=nf1) where n is the harmonic number and f1 is the first frequency---frequency of the first harmonic. So, your analysis is correct--if the first harmonic is 50 than the second harmonic is 100. But beware, the second harmonic can mean that there are 3 nodes on the string. For good review of harmonics, I suggest looking at any physics textbook--or in a review book. If you have any other questions, I am glad to help or I can post more information on harmonics--provided I have the permission first to do so. Good luck :luck:
Yeah, I think harmonics would be a great topic to write an explanations post for if you have the time and inclination, BME. :thumbup:

let's see if i remember this, because I think this showed up on my MCAT...

make sure you understand the difference between harmonic # and overtones in a standing wave.

the harmonic number corresponds to the # of antinodes. The number of nodes in a standing wave is n - 1.

All harmonic frequencies are integer multiples of the fundamental frequency.... i.e. F(n) = n*f1. The fundamental frequency has harmonic # n = 1, but not all harmonic frequencies are necessarily valid for the boundary conditions of a standing wave.

This is where the terminology 'overtone' comes from... the fact that some harmonic #s cannot satisfy the boundary conditions. Overtones depend on the boundary conditions... i.e. for bondary conditions of one node and one antinode, the valid harmonic numbers will be n = 1, 3, 5.... and the three standing waves are considered the fundamental wave, the first overtone, and the second overtone.

Someone correct me if i'm telling lies :meanie:

I've been studying centrepetal motion today, and I'm stuck at curved plane question.

https://netfiles.uiuc.edu/dylee1/shared/curvedplane.bmp?uniq=-e3fhxs

Here's the picture of the problem. I understand that there is a gravitational force that acts on the ball vertically downward with the magnitude of mg.
And the component of the force that is perpendicular to the surface is mgcos(phata). Since the ball is not jumping toward the center or moving away from the center, I understand that there should be a force that counters mgcos(phata). In case of noncurved surface, that would be normal force with the same magnitude. But there is just too much complication involved with this. And I read from EK that the components for the normal force for curved surface case is mgcos(phata) and mv^2/r, but did not tell me which directinos of the two vectors are.

So my question is,
1) What is the magnitude of the centrepetal force and how did you calculate it?
2) What is the direction of each vector that makes up the normal force vector in this curved surface problem?
3) If you used normal force as a part of calculating the centrepetal force, is that really valid? I mean centrepetal force that we are calculating here is that of the ball, and the normal force is that of the surface that ball has contact with, so shouldn't we use only the forces acted directly on the ball (i.e. mg) when we calculate the CF of the ball? or is it one of those tricky determining "system" question?

I appreciate in advance.

I've been studying centrepetal motion today, and I'm stuck at curved plane question.

https://netfiles.uiuc.edu/dylee1/shared/curvedplane.bmp?uniq=-e3fhxs

Here's the picture of the problem. I understand that there is a gravitational force that acts on the ball vertically downward with the magnitude of mg.
And the component of the force that is perpendicular to the surface is mgcos(phata). Since the ball is not jumping toward the center or moving away from the center, I understand that there should be a force that counters mgcos(phata). In case of noncurved surface, that would be normal force with the same magnitude. But there is just too much complication involved with this. And I read from EK that the components for the normal force for curved surface case is mgcos(phata) and mv^2/r, but did not tell me which directinos of the two vectors are.

So my question is,
1) What is the magnitude of the centrepetal force and how did you calculate it?
2) What is the direction of each vector that makes up the normal force vector in this curved surface problem?
3) If you used normal force as a part of calculating the centrepetal force, is that really valid? I mean centrepetal force that we are calculating here is that of the ball, and the normal force is that of the surface that ball has contact with, so shouldn't we use only the forces acted directly on the ball (i.e. mg) when we calculate the CF of the ball? or is it one of those tricky determining "system" question?

I appreciate in advance.


Let me take a shot at this bad boy....the ball is in somewhat circular motion. As it is moving down the curved slide per say, it feels forces: The normal force due to the force the curved slide exerts on the ball, and two components of the gravity force: mgsin(theta) and mgcos(theta). As the ball is moving, it will feel a centripetal force inwards. Since these are vectors, you have to choose a convention to associate the vectors with signs--i.e. the force pointing in will be positive and out will be negative. The force balance will then yield: F normal - mgcos(theta) = force centripetal. Depending on the values given in the problem, this should be enough to solve for the normal force, or the velocity of the ball.
You have to use the normal force when solving this problem because the curved surface is exerting a force on the ball the same way the ball is exerting a force on the surface. However, the force exerted by the surface is directing inwards, toward the center of the circle, and the force of the ball is broken down into components.
If there is not enough information provided to solve the problem, you can substitute a value of velocity in terms of the angular velocity and radius. Remember, the ball is rotating, so it exhibits a angular velocity. The formula for this is velocity = angular velocity times the radius of the path. In most physics classes, you will be dealing with angular motion, period, and simple harmonic motion because simple harmonic motion can be represented as circular motion since it is one-dimensional.
The gravitational component, mgsin(theta), is perpendicular to the motion and keeps the object moving in a circle. It's purpose can be clarified as follows: If you tie a string to a rock and then hurl it in a circle over your head (don't try this), with constant speed (does not mean constant velocity), and cut the string at a point, then the rock will fly away parallel to the path of the velocity vector.
I hope this helps and good luck. Someone correct me if I am wrong!!!

Thanks, that really helped.
We decided that the force pointing inward is positive and outward is negative, we know that the centrepetal force that is pointed inward should be positive. And from the equation Fnormal + (-mg*cos(theta)) = centrepetal force,
we see that Fnormal should be larger than mg*cos(theta). Let's suppose that Fnormal be 5 unit and -mg*cos(theta) be -4 unit, and thus centrepetal force be 1 unit.

If I draw a force diagram, I get something like
https://netfiles.uiuc.edu/dylee1/shared/curvedplane_2.bmp?uniq=lnfsj3
this.

As you can see the sum of forces that are perpendicular to the surface
is not zero (with my example, we have Fnet = 5+1+(-4) = 2 is not 0).
Is it because there must be some non zero force acting on the ball
toward the center for the circular motion to take place?
If that is true, I finally understood the concept. :D

Which of the following has the greatest inertia?

A. a 5kg mass at rest
B. a 5KG mass moving at 10 m/s
C. a 5 kg mass acceerating at 10 m/s^2
D. all have the same inertia

Ans:mass is a quantitative measure of linear inertia. I dont understand what this statement means. Does it mean that Inertia depends only on mass regardless of what happens to any of the other factors?

Thanks

Which of the following has the greatest inertia?

A. a 5kg mass at rest
B. a 5KG mass moving at 10 m/s
C. a 5 kg mass acceerating at 10 m/s^2
D. all have the same inertia

Ans:mass is a quantitative measure of linear inertia. I dont understand what this statement means. Does it mean that Inertia depends only on mass regardless of what happens to any of the other factors?

Thanks

I would probably guess D because of Newton's first law of motion which states that a object at rest tends to stay at rest and a object in motion tends to stay in motion. Resistance to a change in motion is inertia. Not too sure...

I would probably guess D because of Newton's first law of motion which states that a object at rest tends to stay at rest and a object in motion tends to stay in motion. Resistance to a change in motion is inertia. Not too sure...


Their ans is D, But the explanation they give for that is what i wrote above. and i m nore too sure what it means..
Thanks :)

Their ans is D, But the explanation they give for that is what i wrote above. and i m nore too sure what it means..
Thanks :)

Inertia is a objects ability to resist changes in motion. Newton's first law states that a object at rest will stay at rest or an object in motion will stay in motion unless an external force acts on the system. Thus, all objects will display equal amounts of inertia--all will want to stay in their respective states unless a force acts on it. That is why no object from the list has the greater inertia. Choice will want to stay in rest and B anc C will want to stay in motion. I hope this helps and good luck :luck:

I meant to say the objects in this question...not all objects in general...oops!

All objects do not display equal amounts of inertia; it depends on mass. Answer D is correct because the masses are all the same.

Inertia is resistance to acceleration (changes in speed or direction of motion). Look at F=ma; the greater the mass, the more force you need to apply to cause a given acceleration. For example, if I want to move my cat, it's very easy. If I want to move my boyfriend, it's significantly harder. My boyfriend's greater mass means that his body has a greater resistance to changes in motion.

That's what the answer explanation is trying to say.

(For another explanation, see: http://en.wikipedia.org/wiki/Inertia#Mass_and_.27inertia.27 )

EDIT: on looking back at the wikipedia article, it appears that this is considered something of an abuse of the term, and the better term for the concept the question is asking about is "inertial mass." But anyway, if they're asking you to compare inertias, that's what they're talking about.

This might be out of scope for the MCAT, and if it isn't, I'd bet they'd ask the question as "which has the greater resistance to changes in velocity" because the real MCAT does quite well at having correct questions.

For example, if I want to move my cat, it's very easy. If I want to move my boyfriend, it's significantly harder. My boyfriend's greater mass means that his body has a greater resistance to changes in motion.

Yeah, but I'll bet you have means other than force to produce accelerations in your boyfriend.

I have one more question:

Planets A and B have the same mass. planet A has a radius half as large as planet B. A 5 kg mass is dropped 10 m above the surgace of the planet A and at the same time a 5 kg mass is dropped 10 m above the surgavce of the planet B. If