Question

Calculate the power of an engine required to lift 10^5 Kg of Coal per hour from a mine 360m deep, take g as 10 m/s^2.​

Answer 1

Answer:

100kW

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

Answer 2

Answer:

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