Respuesta :
m = mass of water pumped = 3 kg
t = time taken to pump mass "m" of water = 1 sec
h = depth from which the water is pumped = 50 m
P = electrical power used by the pump = Input power = 2000 Watt
P' = output power
U = potential energy gained by water
η = efficiency of the pump
potential energy gained by water is given as
U = mgh
output power is given as
Output power = Potential energy gained by water / time
P' = mgh/t
inserting the values
P' = (3) (9.8) (50)/1
P' = 1470 Watt
efficiency of the pump is given as
efficiency = output power/Input power
η = P'/P
η = 1470/2000
η = 0.735
Answer:
70% efficiency
Explanation:
Power out:
P=W/t
P=mgh/t
P=[(3kg)*(9.8m/s^2)*(50m)]/1sec
P=1470W
Power used: 2000W
efficiency=work out/work in
eff=1470/2000W
eff=0.735=73.5%
When accounting for significant figures:
eff=70%