Answer:
(a) 83475 MW
(b) 85.8 %
Explanation:
Output power = 716 MW = 716 x 10^6 W
Amount of water flows, V = 1.35 x 10^8 L = 1.35 x 10^8 x 10^-3 m^3
mass of water, m = Volume x density = 1.35 x 10^8 x 10^-3 x 1000
= 1.35 x 10^8 kg
Time, t = 1 hr = 3600 second
T1 = 25.4° C, T2 = 30.7° C
Specific heat of water, c = 4200 J/kg°C
(a) Total energy, Q = m x c x ΔT
Q = 1.35 x 10^8 x 4200 x (30.7 - 25.4) = 3 x 10^12 J
Power = Energy / time
Power input =
Power input = 83475 MW
(b) The efficiency of the plant is defined as the ratio of output power to the input power.
Thus, the efficiency is 85.8 %.
Answer:
v = 73.75 m/s
Explanation:
It is given that,
A rocket rises vertically, from rest, with an acceleration of 3.2 m/s² until it runs out of fuel at an altitude of 850 m.
Let us assume we need to find the velocity of the rocket when it runs out of fuel.
Let v is the final speed. Using the third equation of kinematics as :
u = 0
So, the velocity of the rocket when it runs out of the fuel is 73.75 m/s
There is no factor on your list of choices that has any effect.
Area of parallelogram is b*h=0.952. There are 9500 of them so 0.952*9500=904.4 (answer c)