Work = Force * distance.
d = 8 m
W = 2400 J
F = ???
2400 = 8 * F
F = 2400/8
F = 300 N
Answer:
Efficiency = 52%
Explanation:
Given:
First stage
heat absorbed, Q₁ at temperature T₁ = 500 K
Heat released, Q₂ at temperature T₂ = 430 K
and the work done is W₁
Second stage
Heat released, Q₂ at temperature T₂ = 430 K
Heat released, Q₃ at temperature T₃ = 240 K
and the work done is W₂
Total work done, W = W₁ + W₂
Now,
The efficiency is given as:

or
Work done = change in heat
thus,
W₁ = Q₁ - Q₂
W₂ = Q₂ - Q₃
Thus,

or

or

also,

or

thus,

thus,

or

or
Efficiency = 52%
No. Sorry. A sound wave is a mechanical wave. There's nothing electromagnetic about it.
Are there possibly any choices to this question?
To develop this problem we will apply the concept of energy conservation. For which the work carried out must be equivalent to the potential energy stored on the capacitor. We will start by finding the capacitance to later be able to calculate the energy and therefore the work in the capacitor

Here,
C = Capacitance
V = Potential difference between the plates
Q = Charge between the capacitor plates
At the same time the energy stored in the capacitor can be defined as,

We will start by finding the value of the capacitance, so we will have to,


Finally using the expression for the energy we have that,



Therefore the minimum amount of work that must be done in charging this capacitor is 