Part A:
A capacitor is completely charged with 650 nC by a voltagesourcethat has 275 V.
What is its capacitance?

C=Q/V = (550e-9 C)/375V = 1/47e-9 F (1.47 nF)ConfirmedRight

Part B:
Now the plates of the charged capacitor are pushed together withthevoltage source already disconnected:

- The voltage drop between the plates decreases.
- The energy stored in the capacitor remains the same.
- The capacitance increases.
- The voltage drop between the plates increases.
- None of the above.

Part C:
The initial air gap of the capacitor above was 7 mm. What isthestored energy if the air gap is now 3 mm?

1) We know Capacitance when the air gap is 7 mm (d1=7e-3 m), butnotthe area, A. So use:

C= (Îµ0A)/(d1)
or
A=(Cd)/(Îµ0) = ((1.47E-9 F)*(7e-3 m))/(8.85e-12) = 1.163m^2

2) Using the area, A, that you find, find the capacitance, C'withan air gap of 3 mm. (d2=3e-3 m)

C'= (Îµ0A)/(d2) = ((8.85e-12)*(1.163 m^2))/(3e-3) = 3.431e-9F

3) The potential energy with this air gap of 3 mm is:

U = 0.5C'V^2 = 0.5(3.431e-9 F)(375^2) = 2.41e-4 JNotRight

Alright, as you can see, I worked out the last part completelyandI'm pretty sure my work is right. However, the answer is showingupas wrong. I'm not sure what I did wrong here. Also, I'm notsurewhat to do for Part B and I only have one try. Wouldn't theenergyremain the same (or nothing happen at all)?

Need some help!

A playground ride consists of a disk of mass M = 56 kg and radius R= 2.0 m mounted on a low-friction axle. A child of mass m = 21 kgruns at speed v = 2.8 m/s on a line tangential to the disk andjumps onto the outer edge of the disk.

ANGULAR MOMENTUM
(a) Consider the system consisting of the child and the disk, butnot including the axle. Which of the following statements are true,from just before to just after the collision?
The momentum of the system doesn't change.
The momentum of the system changes.
The axle exerts a force on the system but nearly zero torque.
The torque exerted by the axle is zero because the force exerted bythe axle is very small.
The angular momentum of the system about the axle changes.
The angular momentum of the system about the axle hardlychanges.
The torque exerted by the axle is nearly zero even though the forceis large, because || is nearly zero.

(b) Relative to the axle, what was the magnitude of the angularmomentum of the child before the collision?
|C| = kgÂ·m2/s
(c) Relative to the axle, what was the angular momentum of thesystem of child plus disk just after the collision?
|C| = kgÂ·m2/s
(d) If the disk was initially at rest, now how fast is it rotating?That is, what is its angular speed? (The moment of inertia of auniform disk is Â½MR2.)
= radians/s
(e) How long does it take for the disk to go around once?
Time to go around once = s
ENERGY
(f) If you were to do a lot of algebra to calculate the kineticenergies before and after the collision, you would find that thetotal kinetic energy just after the collision is less than thetotal kinetic energy just before the collision. Where has most ofthis energy gone?
Increased chemical energy in the child.
Increased thermal energy of the disk and child.
Increased translational kinetic energy of the disk.

MOMENTUM
(g) What was the speed of the child just after the collision?
v = m/s
(h) What was the speed of the center of mass of the disk just afterthe collision?
vcm = m/s
(i) What was the magnitude of the linear momentum of the disk justafter the collision?
|p| = kgÂ·m/s
(j) Calculate the change in linear momentum of the systemconsisting of the child plus the disk (but not including the axle),from just before to just after impact, due to the impulse appliedby the axle. Take the x axis to be in the direction of the initialvelocity of the child.
px = px,f - px,i = kgÂ·m/s

ANGULAR MOMENTUM
(k) The child on the disk walks inward on the disk and ends upstanding at a new location a distance R/2 = 1 m from the axle. Nowwhat is the angular speed? (It helps to do this analysisalgebraically and plug in numbers at the end.)
= radians/s
ENERGY
(l) If you were to do a lot of algebra to calculate the kineticenergies in part (k), you would find that the total kinetic energyafter the move is greater than the total kinetic energy before themove. Where has this energy come from?
Decreased chemical energy in the child.
Decreased momentum of the disk.
Decreased thermal energy of the disk and child.

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