Una central hidroeléctrica tiene 2,5 Hm3 de agua embalsada a una altura media de 120 m con relación a la turbina. a) ¿Cuál es la energía potencial en kWh? b) Si el rendimiento de las instalaciones es del 65% ¿Qué energía producirá en una hora si el agua cae con un caudal de 2 m3 /s)

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W= fd
where f is an applied force and d is the distance over which it is applied. when a parachuting skydiver is at constant velocity, he is experiencing "terminal velocity", the point at which his acceleration down is canceled out by air resistance. that means that the resisting force of the air is equal to his weight, but in the opposite direction. with this, we can solve for the work done by the force.
w = fd
w = (92.0)(9.81 m/s^2) (325)
w= 2.93 x 10^5 j
the value for work in this case is positive because it is pushing up on the skydiver, who has a downward force of gravity.
therefore the answer is 2.93 x 10^5 j

Answer the above statement is ( false );