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:
13,750 N
Yes
Explanation:
Given:
v₀ = 90 km/h = 25 m/s
v = 0 m/s
t = 4 s
Find: a and Δx
a = Δv / Δt
a = (0 m/s − 25 m/s) / (4 s)
a = -6.25 m/s²
F = ma
F = (2200 kg) (-6.25 m/s²)
F = -13,750 N
Δx = ½ (v + v₀) t
Δx = ½ (0 m/s + 25 m/s) (4 s)
Δx = 50 m
They both are two different elements.SO the answer is Element
Pretty sure that it is 0.
Answer:
a) When moving towards a high pressure center the pressure values increase in the equipment
b) This area is called high prison since the weight of the atmosphere on top is maximum
Explanation:
A) A high atmospheric pressure system is an area where the pressure is increasing the maximum value is close to 107 Kpa, the other side as low pressure can have small values 85.5 kPa.
When moving towards a high pressure center the pressure values increase in the equipment
B) This area is called high prison since the weight of the atmosphere on top is maximum
in general they are areas of good weather
