The answer would be:
Doubling the height will increase the amount of Joules produced.
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The capacitive reactance is reduced by a factor of 2.
<h3>Calculation:</h3>
We know the capacitive reactance is given as,

where,
= capacitive reactance
f = frequency
C = capacitance
It is given that frequency is doubled, i.e.,
f' = 2f
To find,
=?




Therefore, the capacitive reactance is reduced by a factor of 2.
I understand the question you are looking for is this:
A capacitor is connected across an AC source. Suppose the frequency of the source is doubled. What happens to the capacitive reactant of the inductor?
- The capacitive reactance is doubled.
- The capacitive reactance is traduced by a factor of 4.
- The capacitive reactance remains constant.
- The capacitive reactance is quadrupled.
- The capacitive reactance is reduced by a factor of 2.
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Answer:
The height reached is 20m, The time taken to reach 20m is 2 seconds
Explanation:
Observing the equations of motion we can see that the following equation will be most helpful for this question.

We are given initial velocity, u
We know that the stone will stop at its maximum height, so final velocity, v
Acceleration, a
And we are looking for the displacement (height reached), s
Substitute the values we are given into the equation

Rearrange for s



s = -20 (The negative is just showing direction, it can be ignored for now)
The height reached is 20m
Use a different equation to find the time taken

Substitute in the values we have

Rearrange for t



t = 2s
The time taken to reach 20m is 2 seconds
Answer:
539 kPa
Explanation:
Pressure equals density times acceleration of gravity times depth.
P = ρgh
Water has a density of 1000 kg/m³, and acceleration of gravity is 9.8 m/s².
P = (1000 kg/m³) (9.8 m/s²) (55.0 m)
P = 539,000 Pa
P = 539 kPa