Based on the law of conservation of energy, we know that we can't create energy, machines can only convert one type of energy into another. So, if we want to improve a machines's ability then we need to reduce it's loss energy (part of energy which is useless). Out of all the options only Option C fits best with it.
In short, Your Answer would be Option C
Hope this helps!
Answer:
Explanation:
Energy of signal being radiated per second on all sides = 71 x 10³ J .
At a distance of 220 m it is spread over an area of 4 π x (220)² because it is spreading uniformly on all sides.
So energy crossing per unit area
= 
= 11.67 x 10⁻² Wm⁻²s⁻¹.
This is the intensity of the signal.
At 2200 m this intensity will further reduce by 100 times
So there it becomes equal to
11.67 x 10⁻⁴ Wm⁻² s⁻¹.
The answer is 266. Hope that helps!
1) The electric potential at a distance r from a single point charge is given by

where k is the Coulomb's constant, q is the charge and r is the distance from the charge.
The charge in this problem is

So the potential at distance

is

2) By using the same formula as before, we can find the electric potential at distance r=99 m from the charge: