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Approximately \(5.5 \times 10^{6} \mathrm{~kg}\) of water falls \(50 \mathrm{~m}\) over Niagara Falls each second. (a) What is the decrease in the gravitational potential energy of the water-Earth system each second? (b) If all this energy could be converted to electrical energy (it cannot be), at what rate would electrical energy be supplied? (The mass of \(1 \mathrm{~m}^{3}\) of water is \(1000 \mathrm{~kg} .\) ) (c) If the electrical energy were sold at 1 cent \(/ \mathrm{kW} \cdot \mathrm{h},\) what would be the yearly income?

Short Answer

Expert verified
a) 2.695 脳 10鈦 J, b) 2.695 GW, c) Income: 2.36 脳 10鹿鹿 cents/year or 2.36 billion dollars/year.

Step by step solution

01

Calculating Potential Energy Decrease

The decrease in gravitational potential energy can be calculated using the formula:\[\Delta U = m g h\]where \(m\) is the mass of the water, \(g\) is the acceleration due to gravity (9.8 m/s虏), and \(h\) is the height the water falls. Here, \(m = 5.5 \times 10^6\) kg and \(h = 50\) m.Substituting the values, we get:\[\Delta U = 5.5 \times 10^6 \times 9.8 \times 50 = 2.695 \times 10^9 \text{ J}\]This is the decrease in potential energy per second.
02

Converting to Electrical Power

To find the rate of electrical energy production, assume all potential energy is converted to electrical energy. This rate, called power, is equal to the potential energy change per second, which is the same as the calculated decrease in potential energy.The power output is:\[P = 2.695 \times 10^9 \text{ Watts} \]
03

Calculating Yearly Income

The power output is converted from watts to kilowatt-hours, which is the unit used in energy billing. 1 watt is equal to 1 joule per second, so:1 kW = 1000 W1 kWh = 3.6 \times 10^6 JLet's find the kilowatt-hours generated in one year:\[\text{Energy per year} = 2.695 \times 10^9 \text{ W} \times \frac{1}{1000} \text{ kW/W} \times 3600 \text{ s/hour} \times 24 \text{ hours/day} \times 365 \text{ days/year}\]Which simplifies to:\[2.695 \times 10^9 \times 8760 \text{ kWh} = 2.36052 \times 10^{13} \text{ kWh/year}\]Assuming electricity is sold at 1 cent per kWh, we have:\[\text{Yearly Income} = 2.36052 \times 10^{13} \text{ kWh} \times 0.01 \text{ \$/kWh} = 2.36052 \times 10^{11} \text{ cents/year}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power Conversion
When we discuss power conversion in the context of Niagara Falls or any other hydroelectric system, we're essentially talking about transforming the energy of falling water into usable electrical power. In our example, the potential energy of water is converted to electricity. This potential energy originates from gravity. As the water falls from a height, it loses gravitational potential energy.

To calculate the amount of electrical power that could be produced, we use the concept that power is the rate at which energy is converted. In ideal conditions, all potential energy could be transformed into electrical energy. However, in reality, some energy is always lost to friction, resistance, and other inefficiencies, making complete conversion impossible. The power output is still calculated as if 100% conversion occurs, resulting in a power output of:
  • \[ P = 2.695 \times 10^9 \text{ Watts} \]
This powerful conversion showcases the immense energy of natural phenomena like waterfalls, highlighting why hydroelectric plants are common around such sites.
Energy Calculation
Energy calculation is a critical step to understand how much energy we are dealing with or can generate. In the case of Niagara Falls, the potential energy is initially calculated using the gravitational potential energy formula:
  • \[ \Delta U = mgh \]
Here, mass (\( m \)), gravitational force (\( g = 9.8 \text{ m/s}^2 \)), and height (\( h \)) are used to find the potential energy. For each second, the potential energy converted is \( 2.695 \times 10^9 \) Joules.

This calculation is foundational because energy billing and efficiency measurements depend on it. By converting these energy units into kilowatt-hours, a standard unit in electricity billing, it becomes easier to equate to other forms of energy or costs of production. The transition from energy to power helps in understanding how continuous or repetitive processes, like water flowing over a waterfall, can consistently produce energy for consumption.
Electricity Billing
Electricity billing might sound complex, but it's based on straightforward calculations of energy consumed over time. In our case with Niagara Falls, the key unit of measurement is the kilowatt-hour (\( ext{kWh} \)). This is how energy is typically billed to consumers.

From our power conversion step, the energy per second is obtained in watts. When we convert this into annual energy production, the value is massive:
  • \[ 2.36052 \times 10^{13} \text{ kWh/year} \]
Electricity bills usually involve pricing per kilowatt-hour. Here, if electricity is sold at 1 cent per \( ext{kWh} \), that's \( 0.01 \) dollars. Calculating the yearly income involves multiplying the total annual kilowatt-hours by this rate, giving:
  • \[ 2.36052 \times 10^{11} \text{ cents/year} \]
This straightforward approach helps in understanding not just the potential revenue from such energy conversions, but also sets a practical context for individuals managing or investing in hydroelectric power resources.

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Most popular questions from this chapter

131 Fasten one end of a vertical spring to a ceiling, attach a cabbage to the other end, and then slowly lower the cabbage until the upward force on it from the spring balances the gravitational force on it. Show that the loss of gravitational potential energy of the cabbage-Earth system equals twice the gain in the spring's potential energy.

A \(20 \mathrm{~kg}\) block on a horizontal surface is attached to a horizontal spring of spring constant \(k=4.0 \mathrm{kN} / \mathrm{m} .\) The block is pulled to the right so that the spring is stretched \(10 \mathrm{~cm}\) beyond its relaxed length, and the block is then released from rest. The frictional force between the sliding block and the surface has a magnitude of \(80 \mathrm{~N}\). (a) What is the kinetic energy of the block when it has moved \(2.0 \mathrm{~cm}\) from its point of release? (b) What is the kinetic energy of the block when it first slides back through the point at which the spring is relaxed? (c) What is the maximum kinetic energy attained by the block as it slides from its point of release to the point at which the spring is relaxed?

A locomotive with a power capability of \(1.5 \mathrm{MW}\) can accelerate a train from a speed of \(10 \mathrm{~m} / \mathrm{s}\) to \(25 \mathrm{~m} / \mathrm{s}\) in \(6.0 \mathrm{~min} .\) (a) Calculate the mass of the train. Find (b) the speed of the train and (c) the force accelerating the train as functions of time (in seconds) during the 6.0 min interval. (d) Find the distance moved by the train during the interval.

To make a pendulum, a \(300 \mathrm{~g}\) ball is attached to one end of a string that has a length of \(1.4 \mathrm{~m}\) and negligible mass. (The other end of the string is fixed.) The ball is pulled to one side until the string makes an angle of \(30.0^{\circ}\) with the vertical; then (with the string taut ) the ball is released from rest. Find (a) the speed of the ball when the string makes an angle of \(20.0^{\circ}\) with the vertical and (b) the maximum speed of the ball. (c) What is the angle between the string and the vertical when the speed of the ball is one-third its maximum value?

What is the spring constant of a spring that stores \(25 \mathrm{~J}\) of elastic potential energy when compressed by \(7.5 \mathrm{~cm} ?\)

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