Question

14.
Calculate the horse power of the engine required to lift
1.08 x 10^6 kg of coal per hour from a mine 74.6 m deep.​

Answer 1

Answer:

Explanation:

m=10  

5

kg,g=10ms  

−2

,h=360m,t=1h=3600s

The work needed in lifting a mass m to a height h against the force due to gravity is W=mg×h

And, power, P=  

t

W

​

=  

t

mgh

​

 

∴P=  

3600

10  

5

×10×360

​

W=10  

5

W=100kW