Log out Go to App
Go to App

Learning Materials

Features

Discover

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The escape speed from the surface of Planet Zoroaster is $12.0 \mathrm{km} / \mathrm{s} .$ The planet has no atmosphere. A meteor far away from the planet moves at speed \(5.0 \mathrm{km} / \mathrm{s}\) on a collision course with Zoroaster. How fast is the meteor going when it hits the surface of the planet?
A planet with a radius of \(6.00 \times 10^{7} \mathrm{m}\) has a gravitational field of magnitude \(30.0 \mathrm{m} / \mathrm{s}^{2}\) at the surface. What is the escape speed from the planet?
Juana slides a crate along the floor of the moving van. The coefficient of kinetic friction between the crate and the van floor is \(0.120 .\) The crate has a mass of \(56.8 \mathrm{kg}\) and Juana pushes with a horizontal force of \(124 \mathrm{N}\). If \(74.4 \mathrm{J}\) of total work are done on the crate, how far along the van floor does it move?
Emil is tossing an orange of mass \(0.30 \mathrm{kg}\) into the air. (a) Emil throws the orange straight up and then catches it, throwing and catching it at the same point in space. What is the change in the potential energy of the orange during its trajectory? Ignore air resistance. (b) Emil throws the orange straight up, starting \(1.0 \mathrm{m}\) above the ground. He fails to catch it. What is the change in the potential energy of the orange during this flight?
A \(62-\mathrm{kg}\) woman takes \(6.0 \mathrm{s}\) to run up a flight of stairs. The landing at the top of the stairs is \(5.0 \mathrm{m}\) above her starting place. (a) What is the woman's average power output while she is running? (b) Would that be equal to her average power input-the rate at which chemical energy in food or stored fat is used? Why or why not?
See all solutions

Recommended explanations on Physics Textbooks

Energy Physics

Read Explanation

Physics of Motion

Read Explanation

Magnetism

Read Explanation

Wave Optics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Electricity

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free