Before you start working on any motion problem, YOU decide which direction you're going to call 'positive'. Everybody almost always calls UP positive, and the acceleration of gravity points down, so it winds up negative. But you could just as well call DOWN the positive direction. Then, the cannonball is fired with a negative vertical speed, and the acceleration of gravity eventually robs all of its negative speed, and makes it start falling in the positive direction. The whole thing is your choice.
Answer:
Part(a): the capacitance is 0.013 nF.
Part(b): the radius of the inner sphere is 3.1 cm.
Part(c): the electric field just outside the surface of inner sphere is
.
Explanation:
We know that if 'a' and 'b' are the inner and outer radii of the shell respectively, 'Q' is the total charge contains by the capacitor subjected to a potential difference of 'V' and '
' be the permittivity of free space, then the capacitance (C) of the spherical shell can be written as

Part(a):
Given, charge contained by the capacitor Q = 3.00 nC and potential to which it is subjected to is V = 230V.
So the capacitance (C) of the shell is

Part(b):
Given the inner radius of the outer shell b = 4.3 cm = 0.043 m. Therefore, from equation (1), rearranging the terms,

Part(c):
If we apply Gauss' law of electrostatics, then

Answer:
-8.56V
Explanation:
Our values are given by,
e = 6.04 V
Φ = 30.3
VC = 5.32
We can calculate the voltage across the circuit with the emf formula, that is,




Now, Using Kirchoff Voltage Law,


Finally we have the potential difference across the inductor.

I’m pretty sure it’s circuit three.
The equivalent of the Newton's second law for rotational motions is:

where

is the net torque acting on the object

is its moment of inertia

is the angular acceleration of the object.
Re-arranging the formula, we get

and since we know the net torque acting on the (vase+potter's wheel) system,

, and its angular acceleration,

, we can calculate the moment of inertia of the system: