A machine which is `75%` efficient uses 12 J of energy in lifting up a 2 kg mass through a cerfain distance. The mass is then allowed to fall through that distance. The velocity at the end of its fall is (in m/s)

A machine which is 75 percent efficient, uses 12 joules of energy in lifting up a 1 kg mass through a certain distance. The mass is then allowed to fall through that distance. The velocity at the end of its fall is (in ms^(-1) )

A machine which is 75 percent efficient, uses 12 joules of energy in lifting up a 1 kg mass through a certain distance. The mass is then allowed to fall through that distance. The velocity at the end of its fall is (in ms^(-1) )

A machine which is 75 % efficient uses 12 J of energy lifting up a 1 kg mass through a certain distance . The mass is then allowed to fall through that distance. What will be its velocity at end of fall:

A machine, which is 72 percent efficient, uses 36 joules of energy in lifting up 1kg mass through a certain distance. The mass is the allowed to fall through that distance. The velocity at the end of its fall is

A machine, which is 72 percent efficient, uses 36 joules of energy in lifting up 1kg mass through a certain distance. The mass is the allowed to fall through that distance. The velocity at the end of its fall is

A lifting machine, having an efficiency of 80 % uses 2500 J of energy in lifting a 10 kg load over a certain height. If the load is now allowed to fall through that height freely, its velocity at the end of the fall will be (g =10 m//s^(2))

An electron of mass m, charge e falls through a distance h metre in a uniform electric field E. Then time of fall