1. A bag dropped from a helicopter falls with an acceleration of 9.8m/s2. What is its velocity after 5s?

The answer is 49m/s.

Explanation:

Given,

acceleration (a) =9.8m/s^2

initial velocity (u)=0 (as it was in rest condition in helicopter)

time (t) =5 s

now, final velocity (v) after 5s =?

we have,

or, v = 49m/s

Therefore, the velocity after 5s is 49m/s.

Hope it helps..

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j