Water flows over a section of a waterfall at the rate of 5.4 104 kg/s and falls 38 m. How much power is generated by the falling water?
(Mass flow rate)*(energy release per mass)
= g*H* (5.4*10^4 kg/s)
= 20.1 Megawatts
20130120
I don't understand how you got 20.1. What did you divide that first number by?
dr wellos is a genius
To calculate the power generated by the falling water, we can use the formula:
Power = Force × Velocity
In this case, the force is equal to the weight of the water, which is given by the formula:
Force = mass × acceleration due to gravity
where the mass is the rate at which water flows (5.4 × 10^4 kg/s), and the acceleration due to gravity is approximately 9.8 m/s^2.
So, the force exerted by the falling water is:
Force = (5.4 × 10^4 kg/s) × (9.8 m/s^2)
Now, we need to calculate the velocity of the falling water. The velocity of an object in free fall can be determined using the equation:
v = √(2gh)
where v is the velocity, g is the acceleration due to gravity, and h is the height of the fall (38 m in this case).
Plugging in the values, we have:
v = √(2 × 9.8 m/s^2 × 38 m)
Now that we have the velocity, we can calculate the power using the formula:
Power = Force × Velocity
Substituting the values:
Power = ((5.4 × 10^4 kg/s) × (9.8 m/s^2)) × √(2 × 9.8 m/s^2 × 38 m)
Calculating this expression will give us the power generated by the falling water.