# Really tough physics problems?????HELP!!

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#### basha

##### Senior Member
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Hi Everyone, since none of the people who viewed this post could answer these questions, I have decided that they are just too hard for the mcat. If you guys see any concepts tested below which might appear on the mcat, please let me know so I can review them. Also, a lot of the questions below aren't that hard, but the answer for them doesn't feel right. Dont you guys think? Again, please let me know. Thanks a lot for your help. Good luck to all on the mcat.

1)a) A student connects three 2-microfarad capacitors in parallel (Total Cap. = 6 microfarads), connected to a 6 V battery. What is the current in the circuit after the transients have disappeared?
b) Now, he replaces the 6 V with 4 Vrms, oscillating at a freq. of 1 MHz. What is the root mean square current in the circuit?

2)a) 50 cm long hollow glass cylinder open at both ends is held in the air. A device that produces pure tones is placed adjacent to one end. The frequency of the sound is initially set very low then increased gradually. At a frequency of 320 Hz, the tube resonates. What is the speed of sound in air during this experiment?
b) When helium replaced with air, the resonant frequency increases. How come????
c) When the temp is increased, the the freq decreases. How come???

3)a)When the plates of a capacitor are moved farther while voltage source remains attached...
The answer is: The voltage remains the same and the charge decreases. (According the EK Physics book, the voltage should increase with increasing distance, and the charge should subsequently stay the same since the capacitance will decrease as V increases)
b) When the capacitance increases, Electric field intensity remains the same. How come?? (Also, is Electric field same as electric field intensity)

4)a)An astronaut onboard the space shuttle takes a hollow metallic ball and by means of electrostatic induction, gives it a net positive charge of 2 C. The ball is placed between the plates of a capacitor. 12 V is applied across the plates. The capacitance is 1.5 farads, charge is 18 C. What is the electric field intensity at a point 4 m away from the center of the ball. k represents the const in Coloumbs law.
The answer: k/8 N/C (I have no idea how this is so, I even plugged it in the formula for E = kq/r2)
b) An astronaut adjusts the capacitor plates so that they are 3 m apart. The voltage is 12 V. The hollow metallic ball that has a negligible radius, a net positive charge of 2 C, and a mass of 100 g. It is held inside the capacitor against the positive plate. It is then released. What is the kinetic energy of the ball when it reaches the negative plate?
The answer is: 24 J ( I did this one using the U = 1/2 QV formula, but got 18 J instead)

#### J.Pearlman

##### Member
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Umm..these look like typical mcat problems basha. Let me copy this post onto the pre-med forum so someone can help you out there. I would help but I took the mcat a while back. Definitely know how to do these, I had a real difficult circuits passage when i took the mcat and wish I had done some problems like these to gear me up. Anyhow, good luck bro.

#### freakazoid

##### Guy Friend Extraordinaire
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1a) Remember, current is defined as moving charge. There is an accumulation of charge in the capacitors during the transient period, but after it's been charged, the charges don't move, so there's no "charge/time", so zero amps

1b) Not completely sure on this one, but there is the equation Q=CV. Once you figure out Q, you can multiply that by frequency to get the current. Frequency is defined as "cycles/second" and Q is "charge", so multiplying them you get "charge/second" which is current.

2a) Not too sure about this one. The equation for open-tube harmonics is f=(nv)/(2L). Since the tube is resonating, I'm assuming it's the first harmonic. Solve for velocity, you get v=2Lf/n, and since 2L = 2(.50 --converst to meters) = 1, f(1) = v, so v=320 m/s

2b) and 2c) Do they give you options on this one? If you could give me that, perhaps I could help you more.

3a) C=(kEA)/d, so when d increases, C decreases. Since the voltage source gives out a constant voltage, Q=CV gives a smaller Q. (E = permittivity of space)

3b) V=Ed, where E = electric field, so as long as V doesn't change, E doesn't change.

Have to go now, try to finish this up later.

#### freakazoid

##### Guy Friend Extraordinaire
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Thought about 1b), 2b), and 2c)

Here goes:

1b) If you go through what I said, Q=CV=(6*10-6)*(4)=24*10-6
and multiplied by the frequency, you get 24. For some reason they want to multiply that by 2pi . . . and then you'll get your answer, but I can't think of a reasonable explanation . .

2b) I'm not sure if this is right, but I'm thinking in order to make a material resonate, you have impart a certain amount of kinetic energy to the material so that they start to vibrate. So assuming you have to impart the same amount of kinetic energy (KE), and

KE = (1/2)mv^2,

Since helium has a lower density in air, it has less mass, so velocity increases.

From the equation I stated earlier: f = (n/2L)*v, frequency is proportional to velocity, so as velocity increases, so does frequency.

2c) is slightly trickier. However, raising the temperature of the material will give the molecules of glass more kinetic energy, so it needs less kinetic energy from the air paticles, and since as I mentioned earlier velocity^2 is proportional to kinetic energy, the velocity will be less, and since velocity is proportional to frequency, frequency will be less too.

For clarification on 3a), I know you're thinking of V=Ed. However, it's connected to a voltage source

For clarification on 3b), we know that Q=CV. It looks like if capacitance increases, either charge has to go up or voltage has to go down. If it is connected to a voltage source, the voltage will not change since the voltage source will keep it constant. As a result, the voltage source will increase the charge on the plates in order to keep the voltage constant, so that according to V=Ed, voltage is the same, and so the electric field is the same.

4a) The electric field of the capacitor does not matter. Imagine. The capacitor has two plates. When charged, all of the positive charges are on one plate while all of the negative charges are on the other plate. If you were to draw the electric field, all the electric field lines are straight and emanate out of the positive charges towards the negative charges. What you get is a uniform electric field, but only within the plates. The only charge that matters to the electric field is the ball. Using kq/r^2, where q= 2C and r=4m, you get (2/16)*k, which reduces to (1/8)*k.

4b) I went into the whole thing with kinematic equations and such . . . the problem with these equations is that you get so bogged down with the excess information it's hard to tell which equation to use. After a long and hard derivation, I finally got KE=(-q*delta(V)). Since q is positive, you can imagine 12V at the positive plate and 0V at the negative plate, since the positive plate will repel the positive charge. V(final) - V(initial) = -12V
so KE=-(+2)(-12) = 24.

If you want to go through the whole derivation, some equations that will be helpful are,

F=qE=ma
V=Ed
PE=mah
KE=(1/2)mv^2

#### freakazoid

##### Guy Friend Extraordinaire
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I'm not sure if anyone cares anymore, but it came to me today as I was washing dishes:

1b) Frequency is defined as cycles/second. A cycle can be defined as equal to 2pi, so once you get the charge, you multiply charge by 2pi*10^6, which will give you 48pi.