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[" A "60" HP electric motor lifts an ele...

[" A "60" HP electric motor lifts an elevator "],[" having a maximum total load capacity of "],[2000kg" .If the frictional force on the "],[" elevator is "4000N" ,the speed of the elevator "],[" at full load is close to : "(1HP=746W" ,"],[g=10ms^(-2)" ) "]

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A 60 HP electric motor lifts an elevator having a maximum total load capacity of 2000 kg. if the frictional force on the elevator is 4000 N, the speed of the elevator at full load is close to : ( 1 HP = 746 W, g = 10 ms ^( - 2 ) )

A 60 HP electric motor lifts an elevator having a maximum total load capacity of 2000 kg. if the frictional force on the elevator is 4000 N, the speed of the elevator at full load is close to : ( 1 HP = 746 W, g = 10 ms ^( - 2 ) )

A 120 HP electric motor lifts an elevator having a maximum total load capacity of 3000 kg. If the frictional force on the elevator is 6000N, the speed of the elevator at full load is close to : (1 HP = 746 W, g = 10 ms^(-2) ).

A 120 HP electric motor lifts an elevator having a maximum total load capacity of 3000 kg. If the frictional force on the elevator is 6000N, the speed of the elevator at full load is close to : (1 HP = 746 W, g = 10 ms^(-2) ).

A 120 HP electric motor lifts an elevator having a maximum total load capacity of 3000 kg. If the frictional force on the elevator is 6000N, the speed of the elevator at full load is close to : (1 HP = 746 W, g = 10 ms^(-2) ).

An elevator has full load capacity 800 kg. Frictional force on the elevator is 4000 N. If elevator is going down with constant speed 2.0 m/s, then the power applied by electric motor at full load is lose to : (1 HP = 746 W, g = 10 m//s^2 )

An elevator has full load capacity 800 kg. Frictional force on the elevator is 4000 N. If elevator is going down with constant speed 2.0 m/s, then the power applied by electric motor at full load is lose to : (1 HP = 746 W, g = 10 m//s^2 )

An elevator has full load capacity 800 kg. Frictional force on the elevator is 4000 N. If elevator is going down with constant speed 2.0 m/s, then the power applied by electric motor at full load is lose to : (1 HP = 746 W, g = 10 m//s^2 )

An elevator that can carry a maximum load of 1500 kg (elvator+ passengers) is moving up with a constant speed of 2 ms^(-1). The frictional force opposing the motion is 3000 N. Find the minmum power delvered by the motor to the elevator in watts as well as in horse power (g=10ms^(-2))

An elevator that can carry a maximum load of 1500 kg (elvator+ passengers) is moving up with a constant speed of 2 ms^(-1). The frictional force opposing the motion is 3000 N. Find the minmum power delvered by the motor to the elevator in watts as well as in horse power (g=10ms^(-2))