Question

Each second, 1250 m3 of water passes over a waterfall 150 m high. Three-fourths of the kinetic energy gained by the water in falling is transferred to electrical energy by a hydroelectric generator. At what rate does the generator produce electrical energy? (The mass of 1 m3 of water is 1000 kg.)

Answer 1

The generator produces electrical energy at a rate of 1378125000 J per second.

Step-by-step explanation:

volume of water falling each second is 1250
`m^(3)`

height through which it falls, h is 150 m

mass of 1
`m^(3)` of water is 1000 kg

⇒mass of 1250
`m^(3)` of water, m = 1250×1000 = 1250000 kg

acceleration due to gravity, g = 9.8
`(m)/(sec^(2) )`

in falling through 150 m in each second, by Work-Energy Theorem:

Kinetic Energy(KE) gained by it = Potential Energy(PE) lost by it

⇒KE = mgh

= 1250000×9.8×150 J

= 1837500000 J

Electrical Energy =
`(3)/(4)`(KE)

=
`(3)/(4)`×1837500000

= 1378125000 J per second