A pump of 200 W power is lifting 2 kg water from an average depth of 10 meters per second. Find the velocity of water delivered by the pump?
Power= work done/time
=( mgh+0.5mv^2)/t
200= (2*9.8*10 + 0.5*2*v^2)/1
200=196+ v^2
v= 2 m/s
2m/s
Excellent
Can anyone say how t=1 sec??
To find the velocity of water delivered by the pump, we can apply the principle of conservation of energy. The power output of the pump is equal to the work done by the pump in lifting the water.
The power (P) is given as 200 Watts. Power is defined as the rate at which work is done, i.e., P = Work/Time. In this case, the work done is the energy required to lift the water.
The work done (W) can be calculated by multiplying the force exerted by the pump (F) with the distance traveled by the water (d).
The force exerted by the pump can be calculated using Newton's second law, which states that F = m * g, where m is the mass and g is the acceleration due to gravity.
In this case, the mass (m) of the water is given as 2 kg, and the acceleration due to gravity (g) is approximately 9.8 m/s^2.
The distance traveled by the water (d) is given as 10 meters.
The work done (W) is given by the equation W = F * d.
Now, we can substitute the values in the equation:
W = (m * g) * d
W = (2 kg * 9.8 m/s^2) * 10 m
W = 196 N * 10 m
W = 1960 N*m or 1960 Joules
Since power (P) is defined as the work done per unit time, we can rearrange the equation P = W / t to get the value of time (t).
t = W / P
t = 1960 J / 200 W
t = 9.8 s
Now, the velocity of water delivered by the pump (v) can be calculated by dividing the distance traveled by the water (d) by the time taken to cover that distance (t).
v = d / t
v = 10 m / 9.8 s
v ≈ 1.02 m/s
Therefore, the velocity of water delivered by the pump is approximately 1.02 m/s.
how can you have a
depth of 10 meters per second
???