Answer:
1340.2MW
Explanation:
Hi!
To solve this problem follow the steps below!
1 finds the maximum maximum power, using the hydraulic power equation which is the product of the flow rate by height by the specific weight of fluid
W=αhQ
α=specific weight for water =9.81KN/m^3
h=height=220m
Q=flow=690m^3/s
W=(690)(220)(9.81)=1489158Kw=1489.16MW
2. Taking into account that the generator has a 90% efficiency, Find the real power by multiplying the ideal power by the efficiency of the electric generator
Wr=(0.9)(1489.16MW)=1340.2MW
the maximum possible electric power output is 1340.2MW
4A. PE = MxGxH. (You can consider g as 9.8 / 10m/s as well)
509 J = 12x10xH
509 J = 120xH
H = 509/120
H = 4.24 m
Hope u got the answer....pls rate the answer if it is helpful for u....and I'm sorry I could not understand B part so I didn't do it.
Thank you
Answer:
424.26 m/s
Explanation:
Given that Two air craft P and Q are flying at the same speed 300m/s. The direction along which P is flying is at right angles to the direction along which Q is flying. Find the magnitude of velocity of the air craft P relative to air craft Q
The relative speed will be calculated by using pythagorean theorem
Relative speed = sqrt(300^2 + 300^2)
Relative speed = sqrt( 180000 )
Relative speed = 424.26 m/s
Therefore, the magnitude of velocity of the air craft P relative to air craft Q is 424.26 m/s
Answer:
42.9 μF
Explanation:
V = 3.50 V, Q = 150 μC
C = Q/V = 150/3.50 μF = 42.9 μF
Answer:
Internal resistance = 0.545 ohm
Explanation:
As per ohm's law we know that
here we know that
i = electric current = 2.75 A
V = potential difference = 1.50 Volts
now from above equation we have
now we have
