Answer:
<h2>
4.25m/s</h2><h2>
E. None of the option is correct</h2>
Explanation:
Using the law of conservation of momentum to solve the problem. According to the law, the sum of momentum of the bodies before collision is equal to the sum of the bodies after collision. The bodies move with the same velocity after collision.
Mathematically.
mu + MU = (m+M)v
m and M are the masses of the bullet and the block respectively
u and U are their respective velocities
v is their common velocity
from the question, the following parameters are given;
m = 20g = 0.02kg
u = 960m/s
M = 4.5kg
U =0m/s (block is at rest)
Substituting this values into the formula above to get v;
0.02(960)+4.5(0) = (0.02+4.5)v
19.2+0 = 4.52v
4.52v = 19.2
Dividing both sides by 4.52
4.52v/4.52 = 19.2/4.52
v = 4.25m/s
Since they have the same velocity after collision, then the speed of the block immediately after the collision is also 4.25m/s
Answer:

Explanation:
From the question we are told that:
coefficient of static friction 
Velocity 
Generally the equation for Conservation of energy is mathematically given by



Answer:
Power=720[watt]
Explanation:
We need to remember the definition of mechanical work which is equal to the product of the force applied by the distance traveled.
In this problem, we have to find the power which is defined as the work divided into the time in which such work is performed. This way if we have the displacement and the time, this will be the speed with which this work is done.
![Power= W/T\\T=time [s]\\W=work [J]\\Power = F*V\\where\\V=velocity [m/s]\\Power=1800 * 0.4 = 720 [watt]](https://tex.z-dn.net/?f=Power%3D%20W%2FT%5C%5CT%3Dtime%20%5Bs%5D%5C%5CW%3Dwork%20%5BJ%5D%5C%5CPower%20%3D%20F%2AV%5C%5Cwhere%5C%5CV%3Dvelocity%20%5Bm%2Fs%5D%5C%5CPower%3D1800%20%2A%200.4%20%3D%20720%20%5Bwatt%5D)