What is the efficiency of a 0.60 kg basketball that, one dropped from a 2.0 m height (from basket rim) , rebounds to 1.2 m in height (G= 10 N/KG)

Explanation:

we know :

efficiency = output/input

We also know that :

output = m x g x h

where :

m = mass of body

g = acceleration due to gravity

h = height of body from floor

Thus, output = 0.6 x 10 x 1.2 = 7.2J

Similarly ,input = 0.6 x 10 x 2 = 12J

Thus efficiency = 7.2/12 = 0.6

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j