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Each minute, 73,800 m3 of water passes over a waterfall 96.3 m high. Assuming that 58.0% of the kinetic energy gained by the water in falling is converted to electrical energy by a hydroelectric generator, calculate the power output of the generator. (The density of water is 1000 kg/m3 .)

Respuesta :

Given,

V= 73800[tex]m^{3}[/tex]

H= 96.3 m

D= 1000 kg/[tex]m^{3}[/tex]

So, mass m=V*D=73800*1000=738*[tex]10^{5}[/tex] kg

Power= Energy/time= mgH/t = 0.58*(738*[tex]10^{5}[/tex]  *9.81*96.3)/60

=6.74*[tex]10^{8}[/tex] W

So the power output is 6.74*[tex]10^{8}[/tex] Watt.

Power

Power in physics is the quantity of energy transported or converted per unit time. The watt, which is equal to one joule per second in the International System of Units, serves as the unit of power. Power is occasionally referred to as activity in earlier writings. Power has relationships with other factors. For instance, the power required to move a ground vehicle is equal to the product of the traction force on the wheels and the velocity of the vehicle. A motor's output power is calculated by multiplying its torque output by the angle of rotation of its output shaft. Similar to this, the power lost by an electrical component in a circuit is a function of both its voltage and current.

Each minute, 73,800 m3 of water passes over a waterfall 96.3 m high. Assuming that 58.0% of the kinetic energy gained by the water in falling is converted to electrical energy by a hydroelectric generator, calculate the power output of the generator. (The density of water is 1000 kg/m3 .)

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