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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

A 402 -kg pile driver is raised 12 \(\mathrm{m}\) above ground. (a) How much work must be done to raise the pile driver? (b) How much work does gravity do on the driver as it is raised? (c) The driver is now dropped. How much work does gravity do on the driver as it falls?
The power output of a cyclist moving at a constant speed of $6.0 \mathrm{m} / \mathrm{s}\( on a level road is \)120 \mathrm{W} .$ (a) What is the force exerted on the cyclist and the bicycle by the air? (b) By bending low over the handlebars, the cyclist reduces the air resistance to \(18 \mathrm{N} .\) If she maintains a power output of \(120 \mathrm{W},\) what will her speed be?
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?
Bruce stands on a bank beside a pond, grasps the end of a 20.0 -m-long rope attached to a nearby tree and swings out to drop into the water. If the rope starts at an angle of \(35.0^{\circ}\) with the vertical, what is Bruce's speed at the bottom of the swing?
(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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