The water that powers the generators enters and leaves the system at a low speed (thus we can neglect its change of kinetic energy) and loses 160 m of altitude during the process. What is the water flow required, in cubic meters per second, to produce this power, assuming 83.5 % efficiency?

Explanation:

Let density of water be ρ .

During flow , volume of water flowing per second is constant

loss of P. E per unit volume  = ρ gh , 83.5 % is lost

Gain of K E per unit volume = 1/2  ρ v²

83.5 % of mgh =   ρ 1/2  ρ v²

1/2  ρ v² =  .835 x 9.8

v² =  2 x .835 x 9.8

= 16.366

v = 4.04 m /s

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards