Question

15. What is the kinetic energy of a particle whose momentum is 0.02 kg.m.s-l? 2x10-16 2x104J © 2x10-5 J @2x106​

Answer 1

Explanation:

Kinetic energy of bullet = K.E=12mv2 where, m is the mass of the bullet and v is its velocity .

Given, m=0.02Kg,v=300ms−1

So,K.E=12⋅(0.02)(300)2=900J