Water drops over 49 m high Niagara Falls at the rate of 6.0 106 kg/s. If all the energy of the falling water could be harnessed by a hydroelectric power plant, what would be the plant's power output
hey can you please do this one step at a time, i couldn't understand the previous equation
power= work/time= mgh/time= m/time *gh
= 6E6kg/s * 9.8N/kg * 49m
but N-m/s is a watt
Sure, I'll break it down step by step to help you understand the equation.
To calculate the power output of a hydroelectric power plant, we need to find the amount of energy transferred per unit time. The formula to calculate power is:
Power = Energy / Time
In this case, the energy transferred is the gravitational potential energy of the falling water, which can be calculated using the following formula:
Potential Energy = Mass * Gravity * Height
Where:
Mass = 6.0 * 10^6 kg/s (Given)
Gravity = 9.8 m/s^2 (Acceleration due to gravity)
Height = 49 m (Given)
Now, let's calculate the potential energy:
Potential Energy = (6.0 * 10^6 kg/s) * (9.8 m/s^2) * (49 m)
Simplifying this equation, we get:
Potential Energy = 2.9412 * 10^9 J/s
Now, we can substitute the potential energy value into the power equation:
Power = 2.9412 * 10^9 J/s / Time
Since we haven't been given the time, we can't provide an exact power value. However, this equation gives you the general formula to calculate the power output of a hydroelectric power plant when you have the falling water mass flow rate and the height of the fall.