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
EXPLANATION
Parameters given:
Mass of the student, M = 70 kg
Mass of the textbook, m = 1 kg
Distance, r = 1 m
To find the gravitational force acting between the student and the textbook, apply the formula for gravitational force:
where G = gravitational constant
Therefore, the gravitational force acting between the student and the textbook is:
That is the answer.
The formula for the period of wave is: wave period is equals to 1 over the frequency.
To get the value of period of wave you need to divide 1 by 200 Hz. However, beforehand, you have to convert 200 Hz to cycles per second. So that would be, 200 cyles per second or 200/s.
By then, you can start the computation by dividing 1 by 200/s. Since 200/s is in fractional form, you have to find its reciprocal form and multiply it to one which would give you 1 (one) second over 200. This would then lead us to the value
0.005 seconds as the wave period.
wave period= 1/200 Hz
Convert Hz to cycles per second first
200 Hz x 1/s= 200/second
Make 200/second as your divisor, so:
wave period= 1/ 200/s
get the reciprocal form of 200/s which is s/200
then you can start the actual computation:
wave period= 1 x s divided by 200
this would give us an answer of
0.005 s.
Answer:
1.school and community helps in growth and development of you social health as when u learn how to be compatible with the people around you, you will do better in life
2. by trying and solving
by not giving
by innovating
by being motivated
BY KNOWING THAT U CAN DO IT
