M = 30 g = 0.03 kg, the mass of the bullet
v = 500 m/s, the velocity of the bullet
By definition, the KE (kinetic energy) of the bullet is
KE = (1/2)*m*v²
= 0.5*(0.03 kg)*(500 m/s)² = 3750 J
Because the bullet comes to rest, the change in mechanical energy is 3750 J.
The work done by the wall to stop the bullet in 12 cm is
W = (1/2)*(F N)*(0.12 m) = 0.06F J
If energy losses in the form of heat or sound waves are ignored, then
W = KE.
That is,
0.06F = 3750
F = 62500 N = 62.5 kN
Answer:
(a) 3750 J
(b) 62.5 kN
Explanation:
We have,
Semimajor axis is 
It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :

G is universal gravitational constant
M is solar mass
Plugging all the values,

Since,

So, the orbital period of a dwarf planet is 138.52 years.