The electric output of the plant is 48.19 MW
First we need to calculate the water power, it is given by the formula
WP=ρQgh
Here, ρ=1000 kg/m3 is density of water,Q is the flow rate, g is the gravity, and h is the water head
Therefore, WP=1000*65*9.81*90=57388500 W=57.38 MW
Now the overall efficiency of the hydroelectric power plant is given as
η=
Plugging the values in the above equation
0.84=EP/57.38
EP=48.19 MW
Therefore, the electric output of the plant is 48.19 MW.
Answer:
E = h ν energy of electromagnetic particle
(b) has the greater frequency
Answer:
0.1667 m/s
Explanation:
m1V1 + m2V2 = m1V3 + m2V4
0.01 = ( 0.0075) + (0.015 * V4)
V4 = (0.01 - 0.0075) / (0.015)
V4= 0.1667