Answer:
W = 7.06 J
Explanation:
From the given information the spring constant 'k' can be calculated using the Hooke's Law.

Now, using this spring constant the additional work required by F to stretch the spring can be found.
The work energy theorem tells us that the work done on the spring is equal to the change in the energy. Therefore,
![W = U_2 - U_1\\W = \frac{1}{2}kx_2^2 - \frac{1}{2}kx_1^2 = \frac{1}{2}(275.13)[0.29^2 - 0.18^2] = 7.06~J](https://tex.z-dn.net/?f=W%20%3D%20U_2%20-%20U_1%5C%5CW%20%3D%20%5Cfrac%7B1%7D%7B2%7Dkx_2%5E2%20-%20%5Cfrac%7B1%7D%7B2%7Dkx_1%5E2%20%3D%20%5Cfrac%7B1%7D%7B2%7D%28275.13%29%5B0.29%5E2%20-%200.18%5E2%5D%20%3D%207.06~J)
I may be wrong, but I think you're trying to say that Planet-A is
<em>3 times as far from the sun</em> as Planet-C is.
If that's the real question, then the answer is that the period of Orbit-A
is about<em> 5.2</em> times as long as the period of Orbit-C .
Orbital period ≈ (proportional to) (the orbital distance) ^ 3/2 power.
This was empirically demonstrated about 350 years ago by Johannes
and his brilliant Kepple, and derived about 100 years later by Newton
from his formula for the forces of gravity.
I think it’s ultraviolet, if not i’m sorry!
To solve the problem we will apply the concepts related to the Intensity as a function of the power and the area, as well as the electric field as a function of the current, the speed of light and the permeability in free space, as shown below.
The intensity of the wave at the receiver is




The amplitude of electric field at the receiver is


The amplitude of induced emf by this signal between the ends of the receiving antenna is


Here,
I = Current
= Permeability at free space
c = Light speed
d = Distance
Replacing,


Thus, the amplitude of induced emf by this signal between the ends of the receiving antenna is 0.0543V
D is the wrong answer. New information does often completely change the theory. Its hard to change something and leave the major theory intact.