Question

Approximately 3.7 x 10⁷ kg of water falls 53 m over a waterfall each second. (a) What is the decrease in the gravitational potential energy of the water-Earth system each second? (b) If all this energy could be converted to electrical energy (it cannot be), at what rate would electrical energy be supplied? (The mass of 1 m³ of water is 1000 kg.) (c) If the electrical energy were sold at 4.2 cent/kW·h, what would be the yearly income in dollars?

Answer 1

a)

Given that

Mass (m) of the water = 3.7 × 10^7 kg

Height (h) from it is falling = 53 m

Decrease in the gravitational potential energy is

                  U = mgh

                      = (3.7 × 10^7 kg )(9.8m/s²)(53m) =19217.8*10⁶J

b)

1 Joule = 2.78*10^-7 kWh

According to the law of conservation of energy, the decreasein the gravitational potential enrgy is equal to electricalenergy.

Then  electrical energy at what rate
would electrical energy be supplied = 5342.55 kWh

3

and that is being produced every second.
so we get:

0.042*5342.55*3600*24*365

$7076 mil /year