Answer:
The conservation of energy should be used to answer this question.
a)
At the position where the spring is unstretched, the elastic potential energy of the spring is zero.

since
and
is equal to zero.

The roots of this quadratic equation can be solved by using discriminant.


We should use the positive root, so
x = 0.292 m.
b)
We should use energy conservation between the point where the spring is momentarily at rest, and the point where the spring is unstretched.

since the kinetic energy at point 2 and the potential energy at point 3 is equal to zero.

Explanation:
In questions with springs, the important thing is to figure out the points where kinetic or potential energy terms would be zero. When the spring is unstretched, the elastic potential energy is zero. And when the spring is at rest, naturally the kinetic energy is equal to zero.
In part b) the cookie slides back to its original position, so the distance traveled, x, is equal to the distance in part a). The frictional force is constant in the system, so it is quite simple to solve part b) after solving part a).
Answer:
Gamma rays occupy the short-wavelength end of the spectrum; they can have wavelengths smaller than the nucleus of an atom. Visible light wavesare one-thousandths the width of human hair--about a million times longer than gamma rays. Radio waves, at the long-wavelength end of the spectrum, can be many meters long.
Yes......................