Ahot-air balloonist, rising vertically with a constant speed of 5.00 m/s releases a sandbag at the instant the balloon is 40.0 m above the ground. after it is released, the sandbag encounters no appreciable air drag. compute the position of the sandbag at 0.250 s after its release.

Y = 40.94m

Explanation:

The initial speed of the sandbag is the same as the balloon and so is its position, so:

Replacing these values:

Yo = 40m     Vo = 5m/s     g = 9.81m/s^2     t = 0.25s

We get the position of the sandbag:

Y = 40.94m

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