<h2>Answer:</h2><h3>(A) the positively charged surface increases and the energy stored in the capacitor increases.</h3>
When charging a capacitor transferring charge from one surface to the other, the first surface becomes negatively charged while the second surface becomes positively charged. As you transfer the charge, the voltage of the positively charged surface increases and the energy stored in the capacitor also increases. We can solve this by the definition of <em>capacitance</em><em> </em>that is <em>a measure of the ability of a capacitor to store energy. </em>For any capacitor, the capacitance is a constant defined as:

To maintain
constant, if Q increases V also increases.
On the other hand, the potential energy
can be expressed as:

In conclusion, as Q increases the potential energy also increases.
Answer:
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Answer:
110.9 m/s²
Explanation:
Given:
Distance of the tack from the rotational axis (r) = 37.7 cm
Constant rate of rotation (N) = 2.73 revolutions per second
Now, we know that,
1 revolution =
radians
So, 2.73 revolutions = 
Therefore, the angular velocity of the tack is, 
Now, radial acceleration of the tack is given as:

Plug in the given values and solve for
. This gives,
![a_r=(17.153\ rad/s)^2\times 37.7\ cm\\a_r=294.225\times 37.7\ cm/s^2\\a_r=11092.28\ cm/s^2\\a_r=110.9\ m/s^2\ \ \ \ \ \ \ [1\ cm = 0.01\ m]](https://tex.z-dn.net/?f=a_r%3D%2817.153%5C%20rad%2Fs%29%5E2%5Ctimes%2037.7%5C%20cm%5C%5Ca_r%3D294.225%5Ctimes%2037.7%5C%20cm%2Fs%5E2%5C%5Ca_r%3D11092.28%5C%20cm%2Fs%5E2%5C%5Ca_r%3D110.9%5C%20m%2Fs%5E2%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5B1%5C%20cm%20%3D%200.01%5C%20m%5D)
Therefore, the radial acceleration of the tack is 110.9 m/s².
Momentum = mv
= .15 *20
= 3 kgm/s^2