Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 12: Problem 3

At a baseball game, a spectator is \(60.0 \mathrm{m}\) away from the batter. How long does it take the sound of the bat connecting with the ball to travel to the spectator's ears? The air temperature is \(27.0^{\circ} \mathrm{C}\)

Short Answer

Expert verified
Answer: Approximately 0.1725 seconds.

Step by step solution

01

Find the speed of sound in air at the given temperature

To find the speed of sound at the given temperature, we can use the formula: \(v = 331.4 + 0.6T\) where \(v\) is the speed of sound in meters per second and \(T\) is the temperature in degrees Celsius.
02

Substitute the temperature value into the formula

In this case, the air temperature is \(27.0^{\circ} \mathrm{C}\). So, substitute the value of \(T\) in the formula: \(v = 331.4 + 0.6(27)\).
03

Calculate the speed of sound

Calculate the right-hand side of the equation: \(v = 331.4 + 0.6(27) = 331.4 + 16.2 = 347.6 \mathrm{m/s}\). The speed of sound in air at the given temperature is \(347.6\mathrm{m/s}\).
04

Use the distance and speed of sound to find the time

The spectator is \(60.0 \mathrm{m}\) away from the batter, and the speed of sound in air at the given temperature is \(347.6 \mathrm{m/s}\). To find the time it takes for the sound to travel this distance, we can use the formula: \(t = \frac{d}{v}\), where \(t\) is the time, \(d\) is the distance, and \(v\) is the speed.
05

Substitute the distance and speed values into the formula

Substitute the values of \(d\) and \(v\) in the formula: \(t = \frac{60.0}{347.6}\).
06

Calculate the time it takes for the sound to reach the spectator

Calculate the right-hand side of the equation: \(t = \frac{60.0}{347.6} \approx 0.1725 \mathrm{s}\). It takes approximately \(0.1725 \mathrm{s}\) for the sound of the bat connecting with the ball to reach the spectator's ears at the given conditions.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

A 30.0 -cm-long string has a mass of \(0.230 \mathrm{g}\) and is vibrating at its next-to-lowest natural frequency \(f_{2} .\) The tension in the string is \(7.00 \mathrm{N} .\) (a) What is \(f_{2} ?\) (b) What are the frequency and wavelength of the sound in the surrounding air if the speed of sound is $350 \mathrm{m} / \mathrm{s} ?$
(a) Show that if \(I_{2}=10.01_{1},\) then $\beta_{2}=\beta_{1}+10.0 \mathrm{dB} .$ (A factor of 10 increase in intensity corresponds to a 10.0 dB increase in intensity level.) (b) Show that if \(I_{2}=2.0 I_{1}\) then \(\beta_{2}=\beta_{1}+3.0 \mathrm{dB} .\) (A factor of 2 increase in intensity corresponds to a 3.0 -dB increase in intensity level. tutorial: decibels)
Some bats determine their distance to an object by detecting the difference in intensity between cchoes.(a) If intensity falls off at a rate that is inversely proportional to the distance squared, show that the echo intensity is inversely proportional to the fourth power of distance. (b) The bat was originally \(0.60 \mathrm{m}\) from one object and \(1.10 \mathrm{m}\) from another. After flying closer, it is now \(0.50 \mathrm{m}\) from the first and at \(1.00 \mathrm{m}\) from the second object. What is the percentage increase in the intensity of the ccho from each object?
An organ pipe that is open at both ends has a fundamental frequency of $382 \mathrm{Hz}\( at \)0.0^{\circ} \mathrm{C}$. What is the fundamental frequency for this pipe at \(20.0^{\circ} \mathrm{C} ?\)
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. \(]\)
See all solutions

Recommended explanations on Physics Textbooks

Physics of Motion

Read Explanation

Circular Motion and Gravitation

Read Explanation

Geometrical and Physical Optics

Read Explanation

Measurements

Read Explanation

Dynamics

Read Explanation

Physical Quantities And Units

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.

Sign-up for free