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Chapter 12: Problem 5

During a thunderstorm, you can easily estimate your distance from a lightning strike. Count the number of seconds that elapse from when you see the flash of lightning to when you hear the thunder. The rule of thumb is that 5 s clapse for each mile of distance. Verify that this rule of thumb is (approximately) correct. (One mile is \(1.6 \mathrm{km}\) and light travels at a speed of $3 \times 10^{8} \mathrm{m} / \mathrm{s} .$ )

Short Answer

Expert verified
Answer: Yes, the rule of thumb is approximately correct. Our calculation shows that it takes thunder about 4.66 seconds to travel 1 mile, which is close to the rule's estimate of 5 seconds per mile.

Step by step solution

01

Determine the speed of sound

Starting from the speed of light, we will determine the speed of sound. Sound travels at different speeds through different mediums, but we will assume it travels approximately at \(343 \mathrm{m} / \mathrm{s}\) through the air, which is a reasonable approximation for the speed of sound at room temperature and atmospheric pressure.
02

Calculate the time it takes for sound to travel 1 mile

First, we need to convert 1 mile to meters so that our unit is consistent with the given speed of sound. Since 1 mile is equal to 1.6 km, we will convert it to meters: 1 mile = \(1.6 \mathrm{km} = 1.6 \times 10^{3} \mathrm{m}\) Next, we will use the formula for distance to find out the time it takes for thunder to travel 1 mile. Distance = Speed × Time Time = Distance / Speed Time = 1 mile / speed of sound = \((1.6 \times 10^3 \mathrm{m}) / (343 \mathrm{m} / \mathrm{s})\)
03

Evaluate the result and compare with the rule

Now we can evaluate the time it takes for thunder to travel 1 mile: Time = \((1.6 \times 10^3 \mathrm{m}) / (343 \mathrm{m} / \mathrm{s}) \approx 4.66 \mathrm{s}\) The result we have calculated is approximately 4.66 seconds, which is close to the 5 seconds in the rule of thumb. This confirms that the rule of thumb given in the exercise (5 seconds per mile) is approximately correct.

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Most popular questions from this chapter

A bat emits chirping sounds of frequency \(82.0 \mathrm{kHz}\) while hunting for moths to eat. If the bat is flying toward the moth at a speed of $4.40 \mathrm{m} / \mathrm{s}\( and the moth is flying away from the bat at \)1.20 \mathrm{m} / \mathrm{s},$ what is the frequency of the sound wave reflected from the moth as observed by the bat? Assume it is a cool night with a temperature of \(10.0^{\circ} \mathrm{C} .\) [Hint: There are two Doppler shifts. Think of the moth as a receiver, which then becomes a source as it "retransmits" the reflected wave. \(]\)
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Humans can hear sounds with frequencies up to about \(20.0 \mathrm{kHz},\) but dogs can hear frequencies up to about \(40.0 \mathrm{kHz} .\) Dog whistles are made to emit sounds that dogs can hear but humans cannot. If the part of a dog whistle that actually produces the high frequency is made of a tube open at both ends, what is the longest possible length for the tube?
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