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 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 ) )

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))

A bolt of mass 0.2 kg falls from the ceiling of an elevator moving down with a uniform speed of 5 m s^(-1) . It hits the floor of the elevator (length of the elevator = 5 m) and does not rebound. The amount of heat produced by the impact is (Take g = 10 m s^(-2) )

A bolt of mass 0.2 kg falls from the ceiling of an elevator moving down with a uniform speed of 5 m s^(-1) . It hits the floor of the elevator (length of the elevator = 5 m) and does not rebound. The amount of heat produced by the impact is (Take g = 10 m s^(-2) )