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Chapter 6: Problem 76

(a) Calculate the change in potential energy of \(1 \mathrm{kg}\) of water as it passes over Niagara Falls (a vertical descent of \(50 \mathrm{m}\) ). (b) \(\mathrm{At}\) what rate is gravitational potential energy lost by the water of the Niagara River? The rate of flow is $5.5 \times 10^{6} \mathrm{kg} / \mathrm{s} .\( (c) If \)10 \%$ of this energy can be converted into electric energy, how many households would the electricity supply? (An average household uses an average electrical power of about \(1 \mathrm{kW} .\) )

Short Answer

Expert verified
Answer: The change in potential energy of 1 kg of water over Niagara Falls is approximately 490.5 J. If 10% of this energy is converted into electric energy, it can supply approximately 269,800 households.

Step by step solution

01

Calculate the change in potential energy

To find the change in potential energy our equation is: $$\Delta U = mgh$$ where, $$m = 1 \mathrm{kg}$$, $$g = 9.81 \mathrm{m/s^2}$$ (acceleration due to gravity), and $$h = 50 \mathrm{m}$$ (height). Calculating the change in potential energy: $$\Delta U = 1 \mathrm{kg} \times 9.81 \mathrm{m/s^2} \times 50 \mathrm{m} \approx 490.5 \mathrm{J}$$
02

Calculate the rate of loss of gravitational potential energy

The rate of flow is given as \(5.5 \times 10^{6} \mathrm{kg/s}\). We can find the rate of gravitational potential energy loss by multiplying the change in potential energy by the rate of flow: $$P = \Delta U \times \text{rate of flow}$$ $$P = 490.5 \mathrm{J} \times 5.5 \times 10^{6} \mathrm{kg/s} = 2.698 \times 10^{9} \mathrm{W}$$
03

Calculate the amount of electric energy generated and the number of households it can supply

If 10% of the gravitational potential energy is converted into electric energy, then: $$\text{Electric energy} = 0.1 \times 2.698 \times 10^{9} \mathrm{W} = 2.698 \times 10^{8} \mathrm{W}$$ An average household uses an average electrical power of 1 kW or 1000 W. Therefore, the number of households that the generated electric energy can supply is: $$\text{Number of households} = \frac{\text{Electric energy}}{\text{Energy per household}}$$ $$\text{Number of households} = \frac{2.698\times 10^{8} \mathrm{W}}{1000 \mathrm{W/household}} = 2.698\times 10^{5}$$ The generated electric energy can supply approximately \(269,800\) households.

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