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:
A)s = 104.16 m
b)s= 104.16 m
Explanation:
Given that
u = 25 m/s
μ = 0.3
The friction force will act opposite to the direction of motion.
Fr= μ m g
Fr= - m a
a=acceleration
μ m g = - m a
a= - μ g
a= - 0.3 x 10 m/s² ( take g= 10 m/s²)
a= - 3 m/s²
The final speed of the mass is zero ,v= 0
We know that
v² = u² +2 a s
s=distance
0² = 25² - 2 x 3 x s
625 = 6 s
s = 104.16 m
By using energy conservation
Work done by all the forces =Change in the kinetic energy
Negative sign because force act opposite to the displacement.
- 3 x 2 x s = - 625
s= 104.16 m