Physics question thread

[Shrike]

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

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

Acceptable topics:

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

Unacceptable topics:

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

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

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

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

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

 

[ladida]

hello, i haf a few questions about waves if the wavelength of a water is 8m and its speed is 2m/s. how many waves will pass a fixed point in the water. the answer is 15 waves but i can't get it i got 240😕

2) wavelength of water wave is 3.7m and period is 1.5s calculate
a)speed of wave i get this one 2.5m/s
need help on
b)the time required for teh wave to travel 100m.
c) the distance travelled by wave in 1 min<-- i don get these questions does it mean 60s is time wavelenth is 3.7 therefore i can find veloctiy and i times it by 1.5s to find the distance?_?

answer for b is 41s
answer for c is 1.5*10^2m/s

Thanks a bunch

 

[heymanooh1]

hello, i haf a few questions about waves if the wavelength of a water is 8m and its speed is 2m/s. how many waves will pass a fixed point in the water. the answer is 15 waves but i can't get it i got 240😕

2) wavelength of water wave is 3.7m and period is 1.5s calculate
a)speed of wave i get this one 2.5m/s
need help on
b)the time required for teh wave to travel 100m.
c) the distance travelled by wave in 1 min<-- i don get these questions does it mean 60s is time wavelenth is 3.7 therefore i can find veloctiy and i times it by 1.5s to find the distance?_?

answer for b is 41s
answer for c is 1.5*10^2m/s

Thanks a bunch

For 2b and 2c, use the equation: distance = velocity x time. I think there's some information missing in your initial question. You can calculate frequency and period from the data given but not much else beyond that.

 

[commuter9]

One of the conversion factors you need is 1 eV = 1.6 x 10 ^-19 J.

The change in energy, delta U, is given as the potential difference, delta V, between two points multiplied by the charge of the particle, q. Therefore, delta U = q x (delta V) = Work done on the particle W.

delta U has units of Joules, J. delta V has units of Joules per coulomb, delta V = J/C

W = delta U = [1.6 x 10 ^-19] x [10, 0000 J/C] = 1.6 x 10^-15 J

Using the 1 eV conversion factor,
delta U = [1.6 x 10^-15 J] x [1eV/(1.6 x 10^-19 J)] = 10,000 eV = W.

The change in electric potential, delta V, is work done per unit charge, not work done per coulomb.

Also, U = delta P.E. + delta K.E. = 0 because of energy conservation, keeping in mind that energy is on a relative scale. Therefore there's always a reference point which will be either stated or implied based on the information given in the passage.

Thanks heymanooh1. This makes sense (although I don't think I would have ever thought of doing it this way!...twhich is a bit worrisome).

 

[ladida]

For 2b and 2c, use the equation: distance = velocity x time. I think there's some information missing in your initial question. You can calculate frequency and period from the data given but not much else beyond that.

thanks!!

 

[rcd]

I'm confused about how this problem was solved.

An electron is accelerated through a distance of .1m by a potential difference of 10,000 volts. What is the electron's energy as it strikes the anode?
A. 100ev
B. 1,000 eV
C. 10,000 eV
D. 1 J

The answer is C. This question is supposedly asking about change in KE and so W=qV is used. This is what I don't get. They say q = e so it is just equal to one. Why isn't it equal to X coulombs?

1eV = 1e * 1V

W = q v
= e * 10,000 V
W = 10,000 eV

 

[lisichka]

Hey guys,
today i encountered this question in kaplan: here it goes
PLEASE HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
204 Po undergoes radioactive decay with 1/2 life of 3.8 hours emitting alpha particle.

2 moles of pure 204 polonium are isolated and put on scale. After 7.6 hours, a reading of sample mass is made. Its mass is:
apparantly a correct answer is 402 grams
incorrect ones are 408g 400g 201g 203 g

the explanation is in terms of alpha particle. Since we have 2 half lives, 3/4 of alpha particle has decayed, and 1/4 of alpha particles are still in polonium. So since we have 2 moles of polonium, we also have 2 moles of decayed alpha particle, 3/4 *2mol=3/2 moles of alpha particle decay. then we convert moles to mass-->4g*3/2=6 grams of alpha paricle has decayed. So since we have 2 moles of polonium originally, that means we originally had 408 grams of polonium, but since 6 grams of alpha particle has decayed out, we now have 408g-6g=402 grams of sample left, and is a correct answer.


ok, so my reaction to this explanations was whaaaaaaaaaaaaa...xyz,,,,,WTF ?

i thought it was a simple 408-->(1/2 life)200g-->(1/2 life)100g left? why not solve it like this? how do we know when to use alpha decay vs. my method above.

i understand how we have 2 half lives and the original 408 grams, but... everything else is beyond me. please help....!
.

 

[heymanooh1]

Hey guys,
today i encountered this question in kaplan: here it goes
PLEASE HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
204 Po undergoes radioactive decay with 1/2 life of 3.8 hours emitting alpha particle.

2 moles of pure 204 polonium are isolated and put on scale. After 7.6 hours, a reading of sample mass is made. Its mass is:
apparantly a correct answer is 402 grams
incorrect ones are 408g 400g 201g 203 g

the explanation is in terms of alpha particle. Since we have 2 half lives, 3/4 of alpha particle has decayed, and 1/4 of alpha particles are still in polonium. So since we have 2 moles of polonium, we also have 2 moles of decayed alpha particle, 3/4 *2mol=3/2 moles of alpha particle decay. then we convert moles to mass-->4g*3/2=6 grams of alpha paricle has decayed. So since we have 2 moles of polonium originally, that means we originally had 408 grams of polonium, but since 6 grams of alpha particle has decayed out, we now have 408g-6g=402 grams of sample left, and is a correct answer.


ok, so my reaction to this explanations was whaaaaaaaaaaaaa...xyz,,,,,WTF ?

i thought it was a simple 408-->(1/2 life)200g-->(1/2 life)100g left? why not solve it like this? how do we know when to use alpha decay vs. my method above.

i understand how we have 2 half lives and the original 408 grams, but... everything else is beyond me. please help....!
.

General equation for alpha decay of Po: (a/z) Po -> Po + (a-4/z-2)X + (4/2)He
a=mass #; z=atomic #. For alpha-decay, it's helpful to keep mass conservation in mind. In the alpha process, each particle that decays generates a Helium atom which is the mass "lost", another atom (X) and Po.

In the 1st alpha process , only 1 mole of Po will decay to give 1 mol Po, 1 mol X and 1mol He (this is the mass "lost") as products. In the 2nd decay, out of the 1 mole Po remaining, half of this will be decay to give 0.5 mol Po, 0.5 mol X and 0.5 mol He. Now you might be asking, how much does X weigh in terms of mass? Well that's what mass conservation is for, in order to keep track of the atoms that are actually undergoing alpha decay. Nothing further happens to X, it's is still in the sample as is the remaining 0.5 moles of Po after 2 half lives. But notice that you exactly know how many moles of He were "lost". If you add up the masses for all the atoms, after the 1st decay it'll be 408. The same after the 2nd decay and so on.

You are correct that there will be about ~100 g of Po remaining but the question was asking how much will be left of the sample = Po + X

 

[Dr Durden]

Your mistake lisichka is thinking decay means the half the substance completely disappears in a half-life. What really happens is that after one half-life, half the amount of your given element decays and turns into a different element, usually one with a slightly smaller atomic number.

Think of an alpha particle as a helium atom with two protons and two neutrons, for a total atomic mass of four. After 3.8 hours, half of the original Po atoms will have spit out this alpha particle. Thus Po-204 becomes Pb-200. The atomic number drops by two (for the proton loss) and the mass by four (for the proton and neutron loss).

For the first half life, half of your two moles have created an alpha particle. Thus (1/2*2=1) one mole of alpha particles have been created. You now have a single mole of Po-204 and a single mole of Pb-200. After another half life, half the remaining mole of Po-204 spits out another alpha particle. This creates another half a mole of alpha particles, giving you a total of 1.5 moles of alpha particles, or six grams (1.5*atomic wt of 4). Subtract the six grams from your original 408 and you get 402.

 

[rcd]

Hey guys,
today i encountered this question in kaplan: here it goes
PLEASE HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
204 Po undergoes radioactive decay with 1/2 life of 3.8 hours emitting alpha particle.

2 moles of pure 204 polonium are isolated and put on scale. After 7.6 hours, a reading of sample mass is made. Its mass is:
apparantly a correct answer is 402 grams
incorrect ones are 408g 400g 201g 203 g

the explanation is in terms of alpha particle. Since we have 2 half lives, 3/4 of alpha particle has decayed, and 1/4 of alpha particles are still in polonium. So since we have 2 moles of polonium, we also have 2 moles of decayed alpha particle, 3/4 *2mol=3/2 moles of alpha particle decay. then we convert moles to mass-->4g*3/2=6 grams of alpha paricle has decayed. So since we have 2 moles of polonium originally, that means we originally had 408 grams of polonium, but since 6 grams of alpha particle has decayed out, we now have 408g-6g=402 grams of sample left, and is a correct answer.


ok, so my reaction to this explanations was whaaaaaaaaaaaaa...xyz,,,,,WTF ?

i thought it was a simple 408-->(1/2 life)200g-->(1/2 life)100g left? why not solve it like this? how do we know when to use alpha decay vs. my method above.

i understand how we have 2 half lives and the original 408 grams, but... everything else is beyond me. please help....!
.

2 moles * half * half = .5 moles, 2 - .5 = 1.5 moles of decay rxn

"Sample" in this case means the mixture of Po plus the things it just decayed to minus the alpha particles which presumably flew away. 408g originally - 1.5 moles of decay * mass of He per mole = 408 - 1.5 * 4 = 402.

 

[rcd]

if the wavelength of a water is 8m and its speed is 2m/s. how many waves will pass a fixed point in the water. the answer is 15 waves but i can't get it i got 240😕

1 wave / 8m * 2m / s = .25 waves / s. = 15 waves / min.

 

[lisichka]

oh thank you, everyone soooo much. i see now. you guys are great
😍 😍
p.s. as i was solving this problem, i was thinking about Russian spy polonium poisoning, and how basically alpha particles penetrated his body tissues.
wow, mcat physics goes a loooong way. how ironic 😉

 

[RAD11]

Let's say you have a parallel plate, which has a electric field going across it, where + charge is on the left side of the plate and - charge is on the right side of the plate.

If you put a proton into the field, from the + plate end:

1. p+ travels from higher to lower potential
2. field does + work (since p+ and field move in the same direction)
3. PE decreases, which leads to an increase in KE

** Now here's the part where I get confused** .....

If you put an electron into the field, from the - plate end:

1. e- travels from low to high potential (shouldn't e- travel from high to low still? I don't get why it's from low to high)
2. field does + work (not sure why the field is doing + work here)
3. PE decreases, which leads to an increase in KE (no clue about this since I don't get part 1)

Thanks in advance for your help.

 

[gridiron]

Let's say you have a parallel plate, which has a electric field going across it, where + charge is on the left side of the plate and - charge is on the right side of the plate.

If you put a proton into the field, from the + plate end:

1. p+ travels from higher to lower potential
2. field does + work (since p+ and field move in the same direction)
3. PE decreases, which leads to an increase in KE

** Now here's the part where I get confused** .....

If you put an electron into the field, from the - plate end:

1. e- travels from low to high potential (shouldn't e- travel from high to low still? I don't get why it's from low to high)
2. field does + work (not sure why the field is doing + work here)
3. PE decreases, which leads to an increase in KE (no clue about this since I don't get part 1)

Thanks in advance for your help.

Hey! The first part of your analysis is correct. In physics, the positive end is designated the high potential and the negative end the low potential. The reason for this is because the potential difference between the plates, the voltage, can be found from the electric field between the plates. Using electric field lines, the field lines run from the positive end of the plate to the negative end of the plate. You can then calculate the potential using two different methods: the line integral of the electric field or the potential energy per unit charge. Since you don't need to know integrals for the MCAT, the latter method is more appropriate. The potential difference is equal to the change in potential energy per unit charge. The potential energy is equal to negative work so you have two representations of the equation:

change in U/ q

or

-W/q

The potential difference will be positive for positive charges and negative for negative charges because of the magnitude of q. That is why as a charge moves from the negative plate to the positive plate it is moving to a region of higher potential--more positive potential.

To understand the concept of work, consider the macroscopic world scenario of rigid bodies. Gravitational field lines run earth to the ground and point downward--thus keeping us "grounded." When you lift an object it is common to say that you do negative work and the object gains potential energy. Consider the equation for work:

W = Fdcostheta.

In this case, you are doing work against the gravitational field and therefore your angle theta is 180 degrees. Thus the work done is negative in lifting the object and the potential energy increases (change in potential energy is equal to -W). Thus, the generalization can be made that when an object does something it doesn't want to do, it will gain potential energy and negative work will be done. However, when the object does something it normally will do (when the object falls), positive work is done and the object loses potential energy. This generalization can be applied to the microscopic world as well---now the gravitational field is the electric field and the object is some charge of a given magnitude. To relate this to your question, as the negative charge is placed on the negative plate what does it naturally want to do? It will want to move toward the positive plate due to repulsion of like negative charge. Because it does what it naturally wants to do, the work is positive and the potential energy decreases. If instead the charge were placed on the positive plate, external work would be required to move the negative charge toward the negative plate. Similar to lifting an object against an gravitational field, the work done is negative and the potential energy will increase.

Here is something to remember when you take the MCAT: work and potential are opposite in magnitude of one another: if one increases the other will decrease because -W = change in potential energy. When you encounter a question on the test, you can automatically cross out answer choices where both quantities are of similar magnitude. I hope this helps and good .

 

[RAD11]

Hey! The first part of your analysis is correct. In physics, the positive end is designated the high potential and the negative end the low potential. The reason for this is because the potential difference between the plates, the voltage, can be found from the electric field between the plates. Using electric field lines, the field lines run from the positive end of the plate to the negative end of the plate. You can then calculate the potential using two different methods: the line integral of the electric field or the potential energy per unit charge. Since you don't need to know integrals for the MCAT, the latter method is more appropriate. The potential difference is equal to the change in potential energy per unit charge. The potential energy is equal to negative work so you have two representations of the equation:

change in U/ q

or

-W/q

The potential difference will be positive for positive charges and negative for negative charges because of the magnitude of q. That is why as a charge moves from the negative plate to the positive plate it is moving to a region of higher potential--more positive potential.

To understand the concept of work, consider the macroscopic world scenario of rigid bodies. Gravitational field lines run earth to the ground and point downward--thus keeping us "grounded." When you lift an object it is common to say that you do negative work and the object gains potential energy. Consider the equation for work:

W = Fdcostheta.

In this case, you are doing work against the gravitational field and therefore your angle theta is 180 degrees. Thus the work done is negative in lifting the object and the potential energy increases (change in potential energy is equal to -W). Thus, the generalization can be made that when an object does something it doesn't want to do, it will gain potential energy and negative work will be done. However, when the object does something it normally will do (when the object falls), positive work is done and the object loses potential energy. This generalization can be applied to the microscopic world as well---now the gravitational field is the electric field and the object is some charge of a given magnitude. To relate this to your question, as the negative charge is placed on the negative plate what does it naturally want to do? It will want to move toward the positive plate due to repulsion of like negative charge. Because it does what it naturally wants to do, the work is positive and the potential energy decreases. If instead the charge were placed on the positive plate, external work would be required to move the negative charge toward the negative plate. Similar to lifting an object against an gravitational field, the work done is negative and the potential energy will increase.

Here is something to remember when you take the MCAT: work and potential are opposite in magnitude of one another: if one increases the other will decrease because -W = change in potential energy. When you encounter a question on the test, you can automatically cross out answer choices where both quantities are of similar magnitude. I hope this helps and good .

So in the case of the e- placed on the - end plate, it will move from high potential to low potential resulting in a decrease in potential energy.

Okay, I guess I just got confused because I thought that the + end plate is always high potential and the - end plate is always low potential, but just from reading your explanation, it depends on the scenario of whether we are placing a p+ or an e- in the field and on what location within the field (whether next to the + end plate or the - end plate).

Thanks for much for your explanation! 🙂

 

[rcd]

So in the case of the e- placed on the - end plate, it will move from high potential to low potential resulting in a decrease in potential energy.

Okay, I guess I just got confused because I thought that the + end plate is always high potential and the - end plate is always low potential, but just from reading your explanation, it depends on the scenario of whether we are placing a p+ or an e- in the field and on what location within the field (whether next to the + end plate or the - end plate).

Thanks for much for your explanation! 🙂

Hold on. You're not grasping the difference between potential and potential energy. The + plate is always high potential. It's just low potential energy for electrons, and high potential energy for protons.

 

[RAD11]

Hold on. You're not grasping the difference between potential and potential energy. The + plate is always high potential. It's just low potential energy for electrons, and high potential energy for protons.

I think I got it now! Just clarify this with me, though. So basically, what you are saying is that when an e- is placed at the - end plate, is goes from low potential (- end plate) to high potential (+ end plate), which results in a decrease in potential energy, right? (e- is attracted to + charge)

....and when a p+ is placed on a + end plate, it goes from high potential (+ end plate) to low potential (- end plate), which also results in a decrease in potential energy (p+ is attracted to - charge). Is my rationalization correct?

 

[gridiron]

I think I got it now! Just clarify this with me, though. So basically, what you are saying is that when an e- is placed at the - end plate, is goes from low potential (- end plate) to high potential (+ end plate), which results in a decrease in potential energy, right?

....and when a p+ is placed on a + end plate, it goes from high potential (+ end plate) to low potential (- end plate), which also results in a decrease in potential energy. Is my rationalization correct?

Yes. I didn't mean to imply that a decrease in potential energy means low potential--I was trying to develop the equation and concept using an analogy. If you want, I can derive the reasoning behind high and low potential using calculus?

 

[RAD11]

Yes. I didn't mean to imply that a decrease in potential energy means low potential--I was trying to develop the equation and concept using an analogy. If you want, I can derive the reasoning behind high and low potential using calculus?

😳

I got it now! 🙂

 

[gridiron]

😳

I got it now! 🙂

Sorry if I wasn't clear earlier. Good .

 

[killinsound]

Why are the charges on two capacitors in parallel equivalent? (conceptually)


P.S. I thought the Electric field everywhere between a pair of capacitors were the same? So why does PE vary? PE = 1/2 VQ and V = Ed where d is the distance between capacitors. Is the electrical potential the one that is staying constant, and potential energy changing? (I don't see how potential energy could change when electrical potential is constant).

 

[killinsound]

Why are the charges on two capacitors in parallel equivalent? (conceptually)


P.S. I thought the Electric field everywhere between a pair of capacitors were the same? So why does PE vary? PE = 1/2 VQ and V = Ed where d is the distance between capacitors. Is the electrical potential the one that is staying constant, and potential energy changing? (I don't see how potential energy could change when electrical potential is constant).

edit: i think i figured out my answer to the P.S. question, but I want to make sure that my reasoning is correct.

The electric field is the same everwhere within the capacitor. The electric potential k(Q/d) is the distance from the (+) plate. Electric potential is large closest to the (+) plate, and conversely, largest farthest away from the (-) plate. This means that when you put a positive charge in a constant field, it is going to move towards the (-) plate and towards lower electric potential.

The change in electric potential times the charge is the electric potential energy. So by decreasing your electric potential, you are also decreasing your Electrical PE.

I think what was/and still is throwing me off is since PE = q(change in electric potential)

and electric potential = k(q/r)
that means electric potential = Ed

so PE = qEd... and since E is constant, how can PE change?

Also, I'm still confused on "Why are the charges on two capacitors in parallel equivalent?"

 

[gridiron]

Why are the charges on two capacitors in parallel equivalent? (conceptually)


P.S. I thought the Electric field everywhere between a pair of capacitors were the same? So why does PE vary? PE = 1/2 VQ and V = Ed where d is the distance between capacitors. Is the electrical potential the one that is staying constant, and potential energy changing? (I don't see how potential energy could change when electrical potential is constant).

Hey! When capacitors are in parallel, it means that the capacitors are wired directly together at one plate and wired directly at the other plate. Since they are wired directly to one another, a potential difference is applied across all the wired plates--resulting in each capacitor having the same potential difference. A generalization can thus be made:

--When a potential difference is applied across capacitors wired in parallel, the potential difference is the same across each capacitor. The total charge q is the sum of the charges stored in all the capacitors.

As a proof, assume you have capacitors C1, C2 and C3 wired in parallel with a voltage source of V. To find an expression of C equivalent:

q1 = C1V
q2 = C2V
q3 = C3V
Total charge is:
q = q1+q2+q3 = (C1+C2+C3)V
C equivalent = q/V = C1+C2+C3

Thus, capacitors connected in parallel can be replaced with an equivalent capacitor that has the same total charge q and the same potential difference V as the actual individual capacitors. Remember, the potential difference (Voltage) is the same across each capacitor in parallel because they are all wired directly with one another at each plate which is connected to the voltage source. A good way to remember a rule for capacitors in series and parallel is:

for parallel: "par-V" --which means the potential difference is the same across each capacitor
for series: "ser-q" --which means the charge across each capacitor in series is the same.

Potential energy can vary because depending on the strength of the capacitor, different charge can be stored. The greater the value of C the more charge that can be stored.

 

[heymanooh1]

edit: i think i figured out my answer to the P.S. question, but I want to make sure that my reasoning is correct.

The electric field is the same everwhere within the capacitor. The electric potential k(Q/d) is the distance from the (+) plate. Electric potential is large closest to the (+) plate, and conversely, largest farthest away from the (-) plate. This means that when you put a positive charge in a constant field, it is going to move towards the (-) plate and towards lower electric potential.

The change in electric potential times the charge is the electric potential energy. So by decreasing your electric potential, you are also decreasing your Electrical PE.

I think what was/and still is throwing me off is since PE = q(change in electric potential)

and electric potential = k(q/r)
that means electric potential = Ed

so PE = qEd... and since E is constant, how can PE change?

Also, I'm still confused on "Why are the charges on two capacitors in parallel equivalent?"

Using V = k(q/r) is the general equation for point charges or a source charge with spherical geometry. Not quite appropriate for parallel plates

For parallel plates:
C = Q/V ; V = E dot d; C = epsilon (Area/d)

 

[killinsound]

Thanks so much for your explanation.

now you might hate me, but what i meant to say was why the charge is equal for capicators in series.

 

[gridiron]

Thanks so much for your explanation.

now you might hate me, but what i meant to say was why the charge is equal for capicators in series.

When capacitors are wired in series, they are wired one after another. Because of this, a potential difference is applied across the two ends of the series. The potential difference which then exists across the capacitors produces the same charge q across each capacitor. Assume you have three capacitors wired in series: C1, C2, C3 to a battery of voltage V. Capacitor1 is at the top and C3 is at the bottom. The top plate of C1 is wired to the positive terminal of the battery and the bottom plate of C3 is wired to the negative terminal of the battery. That makes the bottom plate of C1 negative, top plate of C2 positive, bottom plate of C2 negative and top plate of C3 positive. When the battery is first connected to bottom plate of C3 is produces negative charge on the plate. That charge is repels negative from the top plate of C3 giving it a positve charge. The repelled negative charge then moves to the bottom plate of C2 and gives it negative charge. The charge on the bottom plate of C2 then repels negative charge from the top plate of C2 thus giving it a positive charge. The repelled negative charge moves to the bottom plate of C1 and gives it a negative charge. The negative charge at the bottom plate of C1 repels negative charge on the top plate and moves negative charge from the top plate to the battery. Charges produced on the plates are a direct result of shifting of charges from one plate to another. Therefore the charging of each capacitor causes the charging of the next capacitor--the charges will be the same in each capacitor because of the shifting of charge from one plate to another in the same magnitude.

To calculate C equivalent:​

V1 = q/C1
V2 = q/C2
V3 = q/C3
V = V1+V2+V3 = q(1/C1 + 1/C2 + 1/C3)
C equivalent = q/v
1/C equivalent = 1/C1 + 1/C2 + 1/C3

​

Therefore, capacitors connected in series can be replaced with an equivalent capacitor that has the same charge and the same total voltage (potential) as the capacitors in series.

 

[akinf]

In laminar and turbulent flow, can you explain the basis of the relationships in the equation for critical velocity.

Vc = (Nr*n)/(p*D)

Nr: dimensionless constant called Reynolds number
n: viscosity of the fluid
p: density of the fluid
D: diamter of the tube

Basically, why does critical velocity increase with an increase in viscosity, and decrease with an increase in both density and diamter of the tube?

 

[akinf]

Why does the velocity of sound decrease with decreasing temperature? Does it have to do with the amount of motion of the particle needed to propograte the sound?

 

[commuter9]

A canon shoots an orange (3kg) straight up in the air with initial velocity 5m/s. A horizontal wind exerts a force of 6N on the orange while it is in the air.

1. What is the horizontal component of the acceleration of the orange while it is in the air?
A. 0m/s^2
B. 2 m/s^2

Answer is B. Why is there acceleration in the X direction (I think I'm getting confused with projectile motion problems)? Is there acceleration in the X direction because there is a force in the X direction (and I guess that there is no force in the X direction during projectile motion...just the force of gravity).


2. What is the horizontal velocity of the orange at the top of its path?
A. 0 m/s^2
B 1m/s^2

Answer is B. In the previous questions we had figured out that t=.5sec and from the question above that a=2 m/s^2. The answer is saying that we know that v0=0. Why is that? Is it due to the fact that the orange only had vertical velocity at first and then at some point in time it gains some horizontal velocity so we can assume initially it didn't have any and once the horizontal force kicked in, there was this 2 m/s^2 constant acceleration?

I think that I kind of see why these are the answers, but it is still hazy for me and I can see how I could make this mistake again. Please help.

 

[NontradICUdoc]

A canon shoots an orange (3kg) straight up in the air with initial velocity 5m/s. A horizontal wind exerts a force of 6N on the orange while it is in the air.

1. What is the horizontal component of the acceleration of the orange while it is in the air?
A. 0m/s^2
B. 2 m/s^2

Answer is B. Why is there acceleration in the X direction (I think I'm getting confused with projectile motion problems)? Is there acceleration in the X direction because there is a force in the X direction (and I guess that there is no force in the X direction during projectile motion...just the force of gravity).


2. What is the horizontal velocity of the orange at the top of its path?
A. 0 m/s^2
B 1m/s^2

Answer is B. In the previous questions we had figured out that t=.5sec and from the question above that a=2 m/s^2. The answer is saying that we know that v0=0. Why is that? Is it due to the fact that the orange only had vertical velocity at first and then at some point in time it gains some horizontal velocity so we can assume initially it didn't have any and once the horizontal force kicked in, there was this 2 m/s^2 constant acceleration?

I think that I kind of see why these are the answers, but it is still hazy for me and I can see how I could make this mistake again. Please help.

1. There is an acceleration in the x direction because the question tells you that the wind exerts a FORCE of 6N in the x-direction. Force=Mass*Acceleration. Now, since you are told that the orange is shot vertically, you can calculate the vertical force because you have acceleration due to gravity pointing down. If you draw your force vectors, you will see that the net force is the hypotenuse of the triangle. So you can calculate the net force on the orange of 31N. Net acceleration is 31=3(Anet) , Anet=10.2 m/s^2. Anet^2=A👍^2+A(x)^2. A👍=10, therefor 10.2^2=10^2+A(x)^2 therefor 104=100+A(x)^2, 4=A(x)^2, so A(x)=2 and the answer is B.

2. Since you calculated that t=0.5, and we know that at the top of its path, there is no velocity in the vertical direction so the only velocity is in the horizontal direction. Using the formula A=(vf-vi)/(Tf-Ti) we plug in the folowing values: A=2, Vf=X, Vi=0, Tf=0.5, Ti=0. So: 2=(X-0)/(0.5-0), 2=Vf/0.5 finally, Vf=1m/s.

I hope this helped

 

[killinsound]

Thanks biomedengineer!

 

[gridiron]

In laminar and turbulent flow, can you explain the basis of the relationships in the equation for critical velocity.

Vc = (Nr*n)/(p*D)

Nr: dimensionless constant called Reynolds number
n: viscosity of the fluid
p: density of the fluid
D: diamter of the tube

Basically, why does critical velocity increase with an increase in viscosity, and decrease with an increase in both density and diamter of the tube?

According to the equation provided, the parameter viscosity is directly proportional to the critical velocity--thus an increase in viscosity will lead to an increase in critical velocity. The critical velocity is when flow will become turbulent. Turbulence by definition is an increase in the resistance of flow--which is also an increase in viscosity. When concerned with fluid flow in a tube, you must make one very important assumption: The liquid touching the surface of the tube is stationary and the liquid moves fastest at the center of the tube--this is the basis behind the derivation of any fluids equation (Flow, Poiseuille's law...etc). The force causing the fluid flow is a pressure gradient multiplied by the area over which flows occurs: F=-changeP*A (don't worry about why there is a negative sign). If you were to draw the laminar stream lines from the surface of the tube to the center of the tube, you can see that the velocity profile is determined by the radial distance from the surface of the tube. If you were to think of each laminar stream line as a cylinder with radius r, you can verify that the velocity profile is fastest at the center of the tube. As the diameter of the tube is increased, the critical velocity decreases because there is more area for fluid to travel through, thus resulting in lower shear forces the fluid encounters. A good example of this is the aorta--it has a very large cross sectional area. Accordingly, it also has a high reynolds number--a dimensionless quantity which is the ratio of inertial forces to viscous forces. This means, that at high speed of flow, fluid flow will become turbulent in the aorta. Density is the change in differential mass over differential volume. An increase in density leads to a decrease in critical velocity because viscous forces predominate. This leads to a greater resistance in flow--meaning it will become turbulent faster. For the MCAT, know how to manipulate the above equation: density and diameter are inversely proportional to critical velocity--so an increase in density or diameter will lead to a decrease in the the critical velocity. Don't worry too much behind the derivation of the equation--it is beyond the scope of the MCAT.

 

[akinf]

BioMedEngineer....I am not following your explanation. I am not seeing why an increase in viscosity would increase the critical velocity. Shouldn't the "thicker" the fluid is make it easier for turbulence to occur? Or is it the other way around? Since the fluid is so thick, it is harder for the flow to become turbulent?

I want to try to get this straight in my head. Could you almost compare turbulence to movement similar to that of an out of control gas? So, if density is low for a fluid, it must move to a higher velocity before it gets "out of control"? This is a weird analogy but I don't know if it works.

 

[killinsound]

When a capacitor and a resistor are in parallel, I'm confused on how you figure out the voltage going through the capacitor.

Same thing applies with a capcitor and resistor in series.

 

[akinf]

I'm taking a look at the MCAT Interpretive Manual and it doesn't seem to mention the Lensmakers equation. Is that equation categorized somewhere else or is it beyond the scope of the MCAT?

 

[akinf]

Umm...could someone explain the main concepts and ideas we need to pay attention for regarding diffreaction and interference (especially Young's double slit experiment). I would really appreciate this. I am reading the Kaplan books and some things just aren't registering.

A note that Kaplan mentions about diffraction is that "it is really the interference of an infinite number of waves where each point along the slit acts as a wave source." .....umm...huh? I don't get that line.

Also, I checked the Physics General Topics before coming here and noticed that it wasn't there. So, someone highly knowledgable may want to consider putting something up...just a suggestion. TY.

 

[commuter9]

1. There is an acceleration in the x direction because the question tells you that the wind exerts a FORCE of 6N in the x-direction. Force=Mass*Acceleration. Now, since you are told that the orange is shot vertically, you can calculate the vertical force because you have acceleration due to gravity pointing down. If you draw your force vectors, you will see that the net force is the hypotenuse of the triangle. So you can calculate the net force on the orange of 31N. Net acceleration is 31=3(Anet) , Anet=10.2 m/s^2. Anet^2=A👍^2+A(x)^2. A👍=10, therefor 10.2^2=10^2+A(x)^2 therefor 104=100+A(x)^2, 4=A(x)^2, so A(x)=2 and the answer is B.

2. Since you calculated that t=0.5, and we know that at the top of its path, there is no velocity in the vertical direction so the only velocity is in the horizontal direction. Using the formula A=(vf-vi)/(Tf-Ti) we plug in the folowing values: A=2, Vf=X, Vi=0, Tf=0.5, Ti=0. So: 2=(X-0)/(0.5-0), 2=Vf/0.5 finally, Vf=1m/s.

I hope this helped

Thanks EMT2ER-Doc. So I guess for the second question I just have to look at it like at some point the orange didn't have horizontal velocity and then when the force was applied it did so then that is where the different in velocities occur.

 

[NontradICUdoc]

Thanks EMT2ER-Doc. So I guess for the second question I just have to look at it like at some point the orange didn't have horizontal velocity and then when the force was applied it did so then that is where the different in velocities occur.

With any physics question you need to pick a system and draw your force/acceleration/velocity vectors for THAT system and never mind anything else.

In this example, one system is when the orange immediately leaves the cannon. Another system is when the orange is at its highest point.

Remember, just pick one system and don't mind the rest. Do one system at a time.

 

[gridiron]

BioMedEngineer....I am not following your explanation. I am not seeing why an increase in viscosity would increase the critical velocity. Shouldn't the "thicker" the fluid is make it easier for turbulence to occur? Or is it the other way around? Since the fluid is so thick, it is harder for the flow to become turbulent?

I want to try to get this straight in my head. Could you almost compare turbulence to movement similar to that of an out of control gas? So, if density is low for a fluid, it must move to a higher velocity before it gets "out of control"? This is a weird analogy but I don't know if it works.

By definition, critical velocity of a fluid is the point at which flow becomes turbulent--resulting in chaotic motion. This usually occurs when the velocity of the fluid is high. Viscosity by definition is the resistance to flow for a liquid and laminar flow is the smooth flow of a viscous liquid. Fluids with high viscosity will have a greater resistance to flow but at low shear will still exhibit laminar flow. Since turbulence is usually the result of high shear speeds, a greater velocity will need to be attained by the more viscous fluid in order to exhibit turbulence.

 

[gridiron]

When a capacitor and a resistor are in parallel, I'm confused on how you figure out the voltage going through the capacitor.

Same thing applies with a capcitor and resistor in series.

When the capacitor and resistor are in parallel to a voltage source, both the capacitor and resistor are at the same electric potential. Thus, the following equations for current would apply:

for resistor: I = Vsource/R
for capacitor: I = C (dVsouce/dt) of the form dv/dt because current is a function of time.

For the MCAT, you do not need to know derivatives or integrals, so don't worry too much how to solve for the voltage for the capacitor because you will need to use a integral.

The series case is way too complicated for the MCAT--it involves using the laplace transform to solve for the voltage. I wouldn't worry about learning how to solve for the voltage because if it is on the MCAT, which is doubtful, then equations will be provided.

Interestingly, the RC circuit is used in electronics for filtering. The series RC circuit is used more often then the parallel RC circuit because the parallel circuit doesn't act as a filter unless it is fed by a current source. This type of filter is readily employed in common medical devices like the EKG and EEG. Don't worry too much about the analysis of such circuits--it involves using calculus and that isn't tested on the MCAT.

 

[gridiron]

Why does the velocity of sound decrease with decreasing temperature? Does it have to do with the amount of motion of the particle needed to propograte the sound?

The velocity of sound depends on the proporties of the medium through which it travels. The two properties are: inertial and elastic. Inertial properties deal with the mass density of the particles of the medium and elastic properties are the tendency of a medium to change shape or deform due to an applied stress. The effect of temperature on the speed of sound can be explained using the elastic properties of the medium. The temperature of air effects the strength of particle interactions in air. A decrease in temperature leads to a decrease in the elastic properties of the air thus leading to a decrease in speed. In general, the greater the elasticity of the medium, the faster the speed of sound will travel--so the strength of particle interactions within medium dictate the speed of sound. The stronger the particle interactions, the faster the speed of sound because the disturbance (a pressure wave for sound) is transmitted faster from particle to particle.

 

[akinf]

Thank you.

 

[killinsound]

When the capacitor and resistor are in parallel to a voltage source, both the capacitor and resistor are at the same electric potential. Thus, the following equations for current would apply:

for resistor: I = Vsource/R
for capacitor: I = C (dVsouce/dt) of the form dv/dt because current is a function of time.

For the MCAT, you do not need to know derivatives or integrals, so don't worry too much how to solve for the voltage for the capacitor because you will need to use a integral.

The series case is way too complicated for the MCAT--it involves using the laplace transform to solve for the voltage. I wouldn't worry about learning how to solve for the voltage because if it is on the MCAT, which is doubtful, then equations will be provided.

Interestingly, the RC circuit is used in electronics for filtering. The series RC circuit is used more often then the parallel RC circuit because the parallel circuit doesn't act as a filter unless it is fed by a current source. This type of filter is readily employed in common medical devices like the EKG and EEG. Don't worry too much about the analysis of such circuits--it involves using calculus and that isn't tested on the MCAT.

Thanks!!! I asked my dad (electrial engineer) and he pretty much explained it the same way.

 

[5moreminutes]

neverimmind

 

[midn]

There is a paragraph in my EK Physics book that doesn't make any sense to me and I was hoping somebody here could explain it:

"If a question asks, 'How much work is done by gravity?' (or any other conservative force), the question itself implies that gravity is not part of the system. There are three methods to answer such a questions: 1) Use Fdcos(theta); 2) simply calculate the change in deltaU(sub g); 3) use W= deltaK + deltaU + deltaE(sub i) but do not include gravitational potential energy in your calculation of deltaU. Technically speaking, a conservative force doesn't do work because energy is never lost nor gained by the system."

I understand using 1) and 2), but 3) makes no sense. How do you calculate the work done and not consider the change in gravitational potential energy? The last sentence is even more confusing because we are asked how much work is done by the conservative force gravity, but it says that conservative forces do no work.

😕

If you have the EK 6th ed book, it is p. 47 at the very top.

 

[NontradICUdoc]

There is a paragraph in my EK Physics book that doesn't make any sense to me and I was hoping somebody here could explain it:

"If a question asks, 'How much work is done by gravity?' (or any other conservative force), the question itself implies that gravity is not part of the system. There are three methods to answer such a questions: 1) Use Fdcos(theta); 2) simply calculate the change in deltaU(sub g); 3) use W= deltaK + deltaU + deltaE(sub i) but do not include gravitational potential energy in your calculation of deltaU. Technically speaking, a conservative force doesn't do work because energy is never lost nor gained by the system."

I understand using 1) and 2), but 3) makes no sense. How do you calculate the work done and not consider the change in gravitational potential energy? The last sentence is even more confusing because we are asked how much work is done by the conservative force gravity, but it says that conservative forces do no work.

😕

If you have the EK 6th ed book, it is p. 47 at the very top.

With a question like this, I would suggest going to the EK website and post this question. Jordan will clarify it for you.

 

[midn]

Oh yeah, sorry I forgot to check that. Turns out there is a discussion on that topic already so I'll post there.

Anyways, I have ANOTHER physics question 😀 .

So I'm reading my old physics books to get a bit of clarification on energy (kinetic and potential).

I understand the work-kinetic energy theorem which basically says the change in kinetic energy is merely the sum of all work done on or by the system (I'm only restating it here so you can correct me if I'm wrong) if all the work goes into changing the velocity of a particle.

Now, when we are talking about work done by kinetic friction, we say the change in kinetic energy is equal to the net force on an object times its displacement. If the object slows down due to friction, the net force is equal to the kinetic friction force. My book ("Physics for Scientists and Engineers" by Serway and Jewett, p. 200) makes the point in saying that the net force times the displacement of the object is NOT work although it has the same units. Its reason for this statement is that the displacement is that of the object and not the point of application of the friction force. This is where I am confused.

Here is the actual quote:
"sum of F(sub x) * delta(x) = K(sub f) - K(sub i)

This looks like the work-kinetic energy theorem, but the left hand side has not been called work. The quantity delta(x) is the displacement of the book - not the displacement of the point of application of the friction force."

(This refers to a book sliding allowing a surface with friction that ends up slowing down due to the force of kinetic friction)

Why is it not work? Is it because the friction force is working against the actual displacement? If so, does that mean in case for a force to do work, it must be displacing the object in the same direction as the force?

I'm no good at physics.

 

[rcd]

Oh yeah, sorry I forgot to check that. Turns out there is a discussion on that topic already so I'll post there.

Anyways, I have ANOTHER physics question 😀 .

So I'm reading my old physics books to get a bit of clarification on energy (kinetic and potential).

I understand the work-kinetic energy theorem which basically says the change in kinetic energy is merely the sum of all work done on or by the system (I'm only restating it here so you can correct me if I'm wrong) if all the work goes into changing the velocity of a particle.

Now, when we are talking about work done by kinetic friction, we say the change in kinetic energy is equal to the net force on an object times its displacement. If the object slows down due to friction, the net force is equal to the kinetic friction force. My book ("Physics for Scientists and Engineers" by Serway and Jewett, p. 200) makes the point in saying that the net force times the displacement of the object is NOT work although it has the same units. Its reason for this statement is that the displacement is that of the object and not the point of application of the friction force. This is where I am confused.

Here is the actual quote:
"sum of F(sub x) * delta(x) = K(sub f) - K(sub i)

This looks like the work-kinetic energy theorem, but the left hand side has not been called work. The quantity delta(x) is the displacement of the book - not the displacement of the point of application of the friction force."

(This refers to a book sliding allowing a surface with friction that ends up slowing down due to the force of kinetic friction)

Why is it not work? Is it because the friction force is working against the actual displacement? If so, does that mean in case for a force to do work, it must be displacing the object in the same direction as the force?

I'm no good at physics.

A lot of the energy of a friction force is lost to heat, and not work. That's the only thing I can imagine happening. ie) F * d = E = q + w.

 

[gridiron]

Oh yeah, sorry I forgot to check that. Turns out there is a discussion on that topic already so I'll post there.

Anyways, I have ANOTHER physics question 😀 .

So I'm reading my old physics books to get a bit of clarification on energy (kinetic and potential).

I understand the work-kinetic energy theorem which basically says the change in kinetic energy is merely the sum of all work done on or by the system (I'm only restating it here so you can correct me if I'm wrong) if all the work goes into changing the velocity of a particle.

Now, when we are talking about work done by kinetic friction, we say the change in kinetic energy is equal to the net force on an object times its displacement. If the object slows down due to friction, the net force is equal to the kinetic friction force. My book ("Physics for Scientists and Engineers" by Serway and Jewett, p. 200) makes the point in saying that the net force times the displacement of the object is NOT work although it has the same units. Its reason for this statement is that the displacement is that of the object and not the point of application of the friction force. This is where I am confused.

Here is the actual quote:
"sum of F(sub x) * delta(x) = K(sub f) - K(sub i)

This looks like the work-kinetic energy theorem, but the left hand side has not been called work. The quantity delta(x) is the displacement of the book - not the displacement of the point of application of the friction force."

(This refers to a book sliding allowing a surface with friction that ends up slowing down due to the force of kinetic friction)

Why is it not work? Is it because the friction force is working against the actual displacement? If so, does that mean in case for a force to do work, it must be displacing the object in the same direction as the force?

I'm no good at physics.

Work is defined as the energy transferred to or from an object by means of a force acting on the object. Positive energy transferred would be positive work and energy transferred from the object would be negative work. The work-kinetic energy theorem states that the change in the kinetic energy of a particle is equal to the net work done on the particle. When friction is involved, changes in thermal energy need to be taken into consideration. For example, consider a constant horizontal force pulls a block along a table through a displacement of magnitude d. This motion increases the block's velocity from some initial velocity to some final velocity . During this process, a constant kinetic frictional force from the table acts on the block. Applying Newton's second law on the system, which is the block, you would get the following:

F - fk(force kinetic friction) = ma​

The forces are constant and thus the acceleration is also constant, so you can use the following equation:

(vfinal)^2 = (vinitial)^2 + 2ad

If you solve the above equation for a and then subsitute the value into the second law expression, you will obtain:

Fd = 0.5m(vfinal)^2 - 0.5m(vinitial)^2 +(fk)d

From the work energy theorem:

0.5m(vfinal)^2 - 0.5m(vinitial)^2 = change in kinetic energy for the block (K)

So, you have:

Fd = K + (fk)d

As the block slides along the table, the table will become warmer. This temperature change in the table is equivalent to the change in thermal energy of the object--the block. The sliding of the block increases the total thermal energy. In physics, changes in thermal energy can be denoted as:

change in thermal energy = (fk)d

The change in thermal energy arises due to the fact there is friction acting between the block and table as the block slides. Now, we can say that Fd is the work done by an external force on the system. But you might ask, on which system is this external force done on? To answer this question, you have to see the energies that change as a result. The mechanical energy of the block changes (its velocity changes) and the thermal energy of the block and the table changes. Changes in mechanical energy of the block would be equal to the change in kinetic energy plus the change in the potential energy of the block. Therefore, considering friction, work is equal to:

W = change in mechanical energy + change in thermal energy

This is the work done on a system by an external agent or force and when friction is taken into consideration.

Another defintion of work is: the energy transferred to or from a system as a result of an external agent or force acting upon the system.

 

[midn]

Thanks a lot guys.

The way I am going to think of it from now on is:
delta(E)=q + W=delta(E;internal) + delta(E;mechanical)

In the case of the sliding book, temperature stays constant so q = 0 thus all change in delta(E) is done by work. I now see that the f (force of kinetic friction)*d contributes to mechanical work and thus is not necessarily directly related to work. The work done by the frictional force depends on how much energy is lost by the book and gained by the table.

I'm not going to think too hard about this since I found out that my problem is over-thinking and not solving questions in a timely manner.

👍

 

[commuter9]

A car is driving on a level road at a contant speed 8m/s when it attempts to execute a turn about curve of effective radius 10m. For the following questions, we will assume the turn is successful, that is, the car performs the turns as the driver intends. The static coeff of friction between the tires and the road is .9, the kinetic coeff of friction is .7.

What force provides the centripetal force?
A. Gravity
B. The normal force
C. Static friction
D. Kinetic friction

Answer is C. I thought that it would be kinetic friction since the car is moving. The explanation for static friction is: Since the tires are not slipping on the road, the appropriate friction is static.