sand is falling on a conveyor belt at rate of 5kg/s. the extra power required to move the belt with a velocity of 6 m/s is _____ ?
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Answered by
44
The force required to keep the belt moving is equal to the rate of increase of horizontal momentum of the sand.
Or, F = rate of change of mass × velocity change
=> F = (5 kg/s)(6 m/s)
=> F = 30 N
Now,
Power = work done per second
=> P = force × rate of displacement
=> P = Fv = (30 N)(6 m/s)
=> P = 180 W
Or, F = rate of change of mass × velocity change
=> F = (5 kg/s)(6 m/s)
=> F = 30 N
Now,
Power = work done per second
=> P = force × rate of displacement
=> P = Fv = (30 N)(6 m/s)
=> P = 180 W
Answered by
7
Answer:
P = 180 W
Explanation:
F = rate of change of mass × velocity change
F = (5 kg/s)(6 m/s)
F = 30 N
P = force × rate of displacement
P = Fv = (30 N)(6 m/s)
P = 180 W
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