/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 To find how far away a lightning... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 23: Problem 2

To find how far away a lightning flash is, a rough rule is the following: "Divide the time in seconds between the flash and the sound, by three. The result equals the distance in \(\mathrm{km}\) to the flash." Justify this. The speed of sound is \(v \approx 333 \mathrm{~m} / \mathrm{s} \approx \frac{1}{3} \mathrm{~km} / \mathrm{s}\), and so the distance to the flash is approximately $$ s=v t \approx \frac{t}{3} $$ where \(t\), the travel time of the sound, is in seconds and \(s\) is in kilometers. The light from the flash travels so fast, \(3 \times 10^{8} \mathrm{~m} / \mathrm{s}\), that it reaches the observer almost instantaneously. Hence, \(t\) is essentially equal to the time between seeing the flash and hearing the thunder. The rule works.

Short Answer

Expert verified
Dividing time in seconds by 3 gives the distance in kilometers because sound travels 1 km in about 3 seconds.

Step by step solution

01

Understand the Speed of Sound

The speed of sound is approximately \( v \approx 333 \mathrm{~m/s} \). If we convert this into kilometers per second, we have \( v \approx \frac{333}{1000} \approx \frac{1}{3} \mathrm{~km/s} \). This means that sound travels roughly 1 kilometer for every 3 seconds.
02

Calculate Distance Using the Formula

The distance \( s \), in kilometers, covered by sound traveling for \( t \) seconds is given by \( s = vt \). Substituting \( v \approx \frac{1}{3} \mathrm{~km/s} \) into the formula gives us \( s \approx \frac{1}{3}t \). This simplifies to \( s \approx \frac{t}{3} \), showing that the distance in kilometers is approximately the time in seconds divided by three.
03

Justify Using Light Speed

Light travels at approximately \( 3 \times 10^8 \mathrm{~m/s} \), much faster than sound. Therefore, when you see a flash, the light reaches you almost instantly compared to the sound. This means that \( t \), the time interval you measure between seeing the flash and hearing the thunder, is effectively the travel time of the sound.
04

Confirm the Rule

Combining the information, the rule "divide the time in seconds between the flash and the sound by three" stems from the relationship \( s \approx \frac{t}{3} \). By calculating \( \frac{t}{3} \), you are finding the distance to the lightning in kilometers using the speed of sound approximation.

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Speed of Sound
The speed of sound is an important concept in understanding how we perceive lightning and thunder. Sound moves quite slower than light, which is why we often see lightning first, and then hear thunder afterward. In the given problem, the speed of sound is approximated as \(v \approx 333 \,\text{m/s}\). Converting this to kilometers per second, we have \(v \approx \frac{1}{3} \,\text{km/s}\).

This conversion highlights that sound travels about 1 kilometer every 3 seconds. The fact that sound moves through the air at this velocity allows us to calculate distances from thunder using time intervals.
Distance Calculation
To calculate the distance based on the time you hear a sound, the key formula used is \(s = vt\), where \(s\) is the distance, \(v\) is the speed of sound, and \(t\) is the time in seconds it takes for the sound to reach you. Substituting the speed of sound, we have \(s \approx \frac{1}{3}t\). This simplifies further to \(s \approx \frac{t}{3}\).

This formula shows that by dividing the time in seconds by 3, you directly get the distance in kilometers. It's a simple calculation that makes use of the constant speed at which sound travels through the air, allowing for quick distance assessments in situations like determining how far away a lightning strike occurred.
Speed of Light
Light travels at an astonishing speed of approximately \(3 \times 10^8 \text{ m/s}\). This rapid pace means light reaches observers virtually instantly. In contrast to sound, light practically doesn't make us consider travel time over short distances, like when observing a lightning flash.

This contrast in speed is crucial when calculating the distance to lightning. The rapidity of light ensures that by the time we see a flash, it has already happened, allowing us to use the delay in sound as a primary measure. This makes the time \(t\) between seeing and hearing the event a reliable indicator of the distance to the strike. This concept explains why we rely on sound to gauge distances, as light doesn鈥檛 present a noticeable delay.

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

An increase in pressure of \(100 \mathrm{kPa}\) causes a certain volume of water to decrease by \(5 \times 10^{-3}\) percent of its original volume. ( \(a\) ) What is the bulk modulus of water? (b) What is the speed of sound (compression waves) in water?

A car moving at \(20 \mathrm{~m} / \mathrm{s}\) with its horn blowing \((f=1200 \mathrm{~Hz})\) is chasing another car going at \(15 \mathrm{~m} / \mathrm{s}\) in the same direction. What is the apparent frequency of the horn as heard by the driver being chased? Take the speed of sound to be \(340 \mathrm{~m} / \mathrm{s}\). This is a Doppler problem. Draw the observer-to-source arrow; that's the positive direction (see Fig. 23-1). Both the source and the observer are moving in the negative direction. Hence, we use \(-v_{o}\) and \(-v_{s}\) $$ f_{o}=f_{s} \frac{v \pm v_{o}}{v \mp v_{s}}=(1200 \mathrm{~Hz}) \frac{340-15}{340-20}=1.22 \mathrm{kHz} $$ Because the source is approaching the observer, the latter will measure an increase in frequency.

Two sound waves have intensities of \(10 \mu \mathrm{W} / \mathrm{cm}^{2}\) and \(500 \mu \mathrm{W} / \mathrm{cm}^{2}\). What is the difference in their intensity levels? Call the \(10 \mu \mathrm{W} / \mathrm{cm}^{2}\) sound \(A\), and the other \(B\). Then $$ \begin{array}{l} \beta_{A}=10 \log _{10}\left(\frac{I_{A}}{I_{0}}\right)=10\left(\log _{10} I_{A}-\log _{10} I_{0}\right) \\ \beta_{B}=10 \log _{10}\left(\frac{I_{B}}{I_{0}}\right)=10\left(\log _{10} I_{B}-\log _{10} I_{0}\right) \end{array} $$ Subtracting \(\beta_{A}\) from \(\beta_{B}\), $$ \begin{aligned} \beta_{B}-\beta_{A} &=10\left(\log _{10} I_{B}-\log _{10} I_{A}\right)=10 \log _{10}\left(\frac{I_{B}}{I_{A}}\right) \\ &=10 \log _{10}\left(\frac{500}{10}\right)=10 \log _{10} 50=(10)(1.70) \\ &=17 \mathrm{~dB} \end{aligned} $$

The two sound sources in Fig. 23-3 vibrate in-phase. A loud sound is heard at \(P\) when \(L_{1}=L_{2}\). As \(L_{1}\) is slowly increased, the weakest sound is heard when \(L_{1}-L_{2}\) has the values \(20.0 \mathrm{~cm}, 60.0\) \(\mathrm{cm}\), and \(100 \mathrm{~cm}\). What is the frequency of the sound source if the speed of sound is \(340 \mathrm{~m} / \mathrm{s}\) ? The waves coming down directly from the fork toward the guy must be a little longer (more spaced) than the waves going up from the fork and back down from the wall which have same spacings. The weakest sound will be heard at \(P\) when a crest from \(S_{1}\) and a trough from \(S_{2}\) reach there at the same time. This will happen if \(L_{1}-L_{2}\) is \(\frac{1}{2} \lambda\), or \(\lambda+\frac{1}{2} \lambda\), or \(2 \lambda+\frac{1}{2} \lambda\), and so on. Hence, the increase in \(L_{1}\) between weakest sounds is \(\lambda_{h}\) and from the data we see that \(\lambda=0.400 \mathrm{~m}\). Then, from \(\lambda=v / f\), $$ f=\frac{v}{\lambda}=\frac{340 \mathrm{~m} / \mathrm{s}}{0.400 \mathrm{~m}}=850 \mathrm{~Hz} $$

Three seconds after a gun is fired, the person who fired the gun hears an echo. How far away was the surface that reflected the sound of the shot? Use \(340 \mathrm{~m} / \mathrm{s}\) for the speed of sound.
See all solutions

Recommended explanations on Physics Textbooks

Radiation

Read Explanation

Electricity and Magnetism

Read Explanation

Quantum Physics

Read Explanation

Physical Quantities And Units

Read Explanation

Electric Charge Field and Potential

Read Explanation

Particle Model of Matter

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.