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?
it says they want the answer in terms of GW
mass/time * g* h= power
To calculate the power output of the hydroelectric power plant, we need to understand the concept of power. Power is the rate at which energy is transferred or the rate at which work is done. The formula for power is given by:
Power = Energy / Time
In this case, the energy transferred is the potential energy of the falling water, and the time is not explicitly given. However, we can assume that time is not a limiting factor, as the amount of water falling per second (6.0 × 10^6 kg/s) remains constant.
The potential energy of an object is given by the equation:
Potential Energy = mass × gravitational acceleration × height
Given that the height of Niagara Falls is 49 m and the mass of water falling per second is 6.0 × 10^6 kg/s, we can calculate the potential energy per second.
Potential Energy per second = (6.0 × 10^6 kg/s) × (9.8 m/s^2) × (49 m)
Next, we need to convert the potential energy per second to power. To do that, we divide the potential energy per second by the unit of time (1 second). This will give us the power output in watts (W).
Power in watts = Potential Energy per second / 1 second
Finally, we can convert the power from watts to gigawatts (GW). Since 1 GW is equal to 1 billion watts, we divide the power in watts by 1 billion.
Power in gigawatts = Power in watts / 1,000,000,000
By following these steps, we can determine the power output of the hydroelectric power plant in terms of gigawatts.