calculate the power of an engine required to lift 900 kg of coal per hour from a mine whose depth is 300 g=10
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Explanation:
Hey there!
Mistakes in the question!
Question will be ---->
Calculate the power of an engine required to lift 900 kg of coal per hour from a mine 300m deep.
Answer----->
Mass(m) = 900 kg = 30² kg
Height (h) = 300m
Acceleration due to gravity = 10 m/s^2
Time(t) = 1 hour = 3600 s
Then,
P.E = mgh
= 30² * 10 * 300 J
Work done = P.E = 30² * 10 * 300 J
We know,
Power = Work done / Time
= 30² x 10 x 300/3000
= 30² watt.
Hope this helps you
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