Question

Calculate the power of an electric motor that can lift 800 kg of water to store in a tank at a height of 1500 cm in 20 s.(g =10m/s sq)

Answer 1

HERE IS YOUR ANSWER :

mass = 800Kg
tank's height = 1500cm=15m
time = 20s
g = 10m/s²

Power = W/t
= F.s/t
= ma.s/t
= mgs/t
= 800*15*10/20
= 6000W

Answer 2

The power of an electric motor to store the water is 6000 W.

Explanation:

The power is defined as

$P=\frac{W}{t}$

Here W is the work done and t is the time taken.

The work done by motor to lift water,

W = mgh

Here, m is the mass and h is the height.

Given m= 800 kg, h = 1500 cm = 15 m, g = 10 m/s^2 and t = 20 s.

substitute the values in work done formula, we get  

W = 800 kg x 15 m x 10 m/s^2

W = 120000 J.

Therefore, power is

$P=\frac{120000J}{20s}$

P  =  6000 W.

Thus, the power of an electric motor to store the water is 6000 W.

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