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.)

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

Explanation:

volume of water falling each second is 1250

height through which it falls, h is 150 m

mass of 1 of water is 1000 kg

⇒mass of 1250 of water, m = 1250×1000 = 1250000 kg

acceleration due to gravity, g = 9.8

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 = (KE)

                            = ×1837500000

                            = 1378125000 J per second

explanation:

v=uf/(u-f)

= 5.-10/(5+10) convex lens has negative focal length

t = 1/400 h   ⇔ 0.0025 h ⇔ 9s ⇔ 0.15 m

any of the answers are correct.