<span>Last choice on the list:
Object A has a net charge of 0 because the positive and negative
charges are balanced.
Object B has a net charge of –2 because there is an imbalance of
charged particles (2 more negative electrons than positive protons).</span>
Answer: The skier has potential and kinetic energy.
Explanation: This is what I found from a different user on this website
The concept required to solve this problem is associated with potential energy. Recall that potential energy is defined as the product between mass, gravity, and change in height. Mathematically it can be described as

Here,
= Change in height
m = mass of super heroine
g = Acceleration due to gravity
The change in height will be,

The final position of the heroin is below the ground level,

The initial height will be the zero point of our system of reference,


Replacing all this values we have,



Since the final position of the heroine is located below the ground, there will net loss of gravitational potential energy of 10744.81J
1) At the moment of being at the top, the piston will not only tend to push the penny up but will also descend at a faster rate at which the penny can reach in 'free fall', in that short distance. Therefore, at the highest point, the penny will lose contact with the piston. Therefore the correct answer is C.
2) To solve this problem we will apply the equations related to the simple harmonic movement, hence we have that the acceleration can be defined as

Where,
a = Acceleration
A = Amplitude
= Angular velocity
From a reference system in which the downward acceleration is negative due to the force of gravity we will have to



From the definition of frequency and angular velocity we have to




Therefore the maximum frequency for which the penny just barely remains in place for the full cycle is 2.5Hz