Mathematical relationship due to definition in terms of work.
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A pump is used to lift 500 kg of water from a depth of 80 m in 10 s .Calculate : (a)the work done by the pump . (b)the power at which the pump works,and (c )the power rating of the pump if its efficiency is 40% (Take g =10 m s^(-2) ) [Hint :Efficiency= ("use power")/("power input")]
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The power of a pump, which can pump 500 kg of water to height 100 m in 10 s is
A pump is required to lift 800 kg of water per minute from a 10 m deep well and eject it with speed of 20 kg m//s . The required power in watts of the pump will be
The power of a motor pump is 2kW. How much water per minute the pump can raise to a heiht of 10 m ? (Given g = 10 m//s^2)
The power of a water pump is 2 kW. If g = 10 m//s^2, the amount of water it can raise in 1 min to a height of 10 m is :
The power of water pump is 4 kW. If g=10 ms^(-2), the amount of water it can raise in 1 minute to a height of 20 m is
The power of water pump is 4 kW. If g=10 ms^(-2), the amount of water it can raise in 1 minute to a height of 20 m is
Standard rating of each bulb is P, V. If total power consumption by combination is (3XP)/5 then calculate 'X'.