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Chapter 6: Problem 65

Show that 1 kilowatt-hour (kW-h) is equal to 3.6 MJ.

Short Answer

Expert verified
Answer: 1 kilowatt-hour is equal to 3.6 mega-joules.

Step by step solution

01

Convert kilowatt-hours to watts-seconds

Using the given information, we first convert 1 kilowatt-hour (kW-h) into watt-seconds (W-s): 1 kW-h = 1 kW * 1 hour = 1000 W * 3600 s.
02

Convert watts-seconds to joules

Now, we convert watt-seconds (W-s) into joules (J) using the conversion 1 W = 1 J/s: 1000 W * 3600 s = 1000 J/s * 3600 s.
03

Simplify the expression

The expression in Step 2 becomes: 1000 J/s * 3600 s = 3600000 J.
04

Convert joules to mega-joules

Finally, we convert joules (J) into mega-joules (MJ): 3600000 J = 3.6 MJ. So, 1 kilowatt-hour is equal to 3.6 mega-joules.

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Most popular questions from this chapter

The orbit of Halley's comet around the Sun is a long thin ellipse. At its aphelion (point farthest from the Sun), the comet is $5.3 \times 10^{12} \mathrm{m}\( from the Sun and moves with a speed of \)10.0 \mathrm{km} / \mathrm{s} .$ What is the comet's speed at its perihelion (closest approach to the Sun) where its distance from the Sun is \(8.9 \times 10^{10} \mathrm{m} ?\)
Rachel is on the roof of a building, \(h\) meters above ground. She throws a heavy ball into the air with a speed \(v,\) at an angle \(\theta\) with respect to the horizontal. Ignore air resistance. (a) Find the speed of the ball when it hits the ground in terms of \(h, v, \theta,\) and \(g .\) (b) For what value(s) of \(\theta\) is the speed of the ball greatest when it hits the ground?
(a) If the length of the Achilles tendon increases \(0.50 \mathrm{cm}\) when the force exerted on it by the muscle increases from \(3200 \mathrm{N}\) to $4800 \mathrm{N},$ what is the "spring constant" of the tendon? (b) How much work is done by the muscle in stretching the tendon \(0.50 \mathrm{cm}\) as the force increases from \(3200 \mathrm{N}\) to \(4800 \mathrm{N} ?\)
Human feet and legs store elastic energy when walking or running. They are not nearly as efficient at doing so as kangaroo legs, but the effect is significant nonetheless. If not for the storage of elastic energy, a \(70-\mathrm{kg}\) man running at \(4 \mathrm{m} / \mathrm{s}\) would lose about \(100 \mathrm{J}\) of mechanical energy each time he sets down a foot. Some of this energy is stored as elastic energy in the Achilles tendon and in the arch of the foot; the elastic energy is then converted back into the kinetic and gravitational potential energy of the leg, reducing the expenditure of metabolic energy. If the maximum tension in the Achilles tendon when the foot is set down is \(4.7 \mathrm{kN}\) and the tendon's spring constant is $350 \mathrm{kN} / \mathrm{m},$ calculate how far the tendon stretches and how much elastic energy is stored in it.
(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 \(1 \mathrm{kW} .\) )
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