Question

An engine of power 10 Kw can lift 1000Kg of water to a height of 10m in 1min. find useful output energy​

Answer 1

Answer:

5*10⁶joules

Explanation:

given, p=10kw=10000W;  m=1000kg ; d=10m.

intput energy :

p=w/t

10⁴=w/600

∴w=6*10⁶J

 

output energy:

w=m*a*d

  = 1000kg*10ms⁻²*10m

  = 10⁶J.

useful output energy : (input -output) energy

            = 6*10⁶-10⁶

             = 5*10⁶J