I can guarantee you that it is not
C.<span>the angle that the incident ray makes with a line drawn perpendicular to the reflecting surface I hope this somewhat helps</span>
The elastic potential energy of a spring is given by

where k is the spring's constant and x is the displacement with respect to the relaxed position of the spring.
The work done by the spring is the negative of the potential energy difference between the final and initial condition of the spring:

In our problem, initially the spring is uncompressed, so

. Therefore, the work done by the spring when it is compressed until

is

And this value is actually negative, because the box is responsible for the spring's compression, so the work is done by the box.
Answer:

Explanation:
The mass of the man can be found by using the formula

f is the force
a is the acceleration
From the question we have

We have the final answer as
<h3>50 kg</h3>
Hope this helps you
The tension in the string corresponds to the gravitational attraction between the Sun and any planet.
The wavelength decreases to roughly half.
(The frequency roughly doubles.)