/*! 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 100 When one person shouts at a foot... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 16: Problem 100

When one person shouts at a football game, the sound intensity level at the center of the field is \(60.0 \mathrm{~dB}\). When all the people shout together, the intensity level increases to \(109 \mathrm{~dB}\). Assuming that each person generates the same sound intensity at the center of the field, how many people are at the game?

Short Answer

Expert verified
Approximately 7943 people are at the game.

Step by step solution

01

Understanding Decibel Levels

The sound intensity level in decibels (dB) relates to the intensity of the sound by the formula: \[ \beta = 10 \log_{10} \left( \frac{I}{I_0} \right) \]where \( \beta \) is the sound level in dB, \( I \) is the intensity of the sound, and \( I_0 \) is a reference intensity, typically \( 10^{-12} \text{ W/m}^2 \). Here, we have individual intensity level, \( \beta_1 = 60 \text{ dB} \), and total intensity level, \( \beta_2 = 109 \text{ dB} \).
02

Calculate Individual Intensity

Recognize that the intensity \( I_1 \) for one person can be calculated from the sound level \( \beta_1 = 60 \text{ dB} \):\[ 60 = 10 \log_{10} \left( \frac{I_1}{I_0} \right) \]Rewriting gives:\[ \frac{I_1}{I_0} = 10^{6} \]Thus,\[ I_1 = 10^{6} \cdot I_0 \]
03

Calculate Total Intensity

For the total intensity \( I_t \) corresponding to \( \beta_2 = 109 \text{ dB} \), we have:\[ 109 = 10 \log_{10} \left( \frac{I_t}{I_0} \right) \]Rewriting gives:\[ \frac{I_t}{I_0} = 10^{10.9} \]Thus,\[ I_t = 10^{10.9} \cdot I_0 \]
04

Determine Number of People

Since each person produces the same intensity \( I_1 \), the total intensity is the sum of the intensities of all people shouting. Therefore,\[ I_t = N \cdot I_1 \]Substitute \( I_t = 10^{10.9} \cdot I_0 \) and \( I_1 = 10^{6} \cdot I_0 \) into the equation:\[ 10^{10.9} \cdot I_0 = N \cdot 10^{6} \cdot I_0 \]Cancel \( I_0 \) and solve for \( N \):\[ N = 10^{10.9 - 6} = 10^{4.9} \approx 7943 \]
05

Conclusion

Therefore, approximately 7943 people are present at the game shouting together to achieve an intensity level of 109 dB.

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.

Decibel Scale
The decibel scale is a logarithmic unit used to measure sound intensity level. It's a representation that allows us to express very large or very small power quantities in an easily understandable form.
With the decibel scale, a tenfold increase in sound intensity corresponds to an increase of 10 dB. For example:
  • A sound intensity of 60 dB is 10 times more intense than 50 dB.
  • Similarly, a sound of 70 dB is 100 times more intense than 50 dB, as each 10 dB increment represents a tenfold increase.
This scale is crucial because our ears perceive sound intensity logarithmically. We actually perceive changes on this scale as linear changes in loudness, which is why it's so effective in sound measurement.
Sound Intensity Level
Sound Intensity Level, measured in decibels (dB), quantifies the intensity of sound over a given area. Intensity refers to the power per unit area carried by a wave.
  • Sound intensity is mathematically expressed with the formula: \( \beta = 10 \log_{10} \left( \frac{I}{I_0} \right) \)
  • Here, \(I\) is the sound intensity, and \(I_0\) is a reference intensity, typically \( 10^{-12} \text{ W/m}^2 \).
This threshold of hearing represents the quietest sound that can be heard by the average human ear. Sound intensity level helps us understand not only the loudness but also the impact sound can have on surroundings.
Logarithmic Scale
The logarithmic scale is crucial in plotting sound intensity levels because it transforms multiplicative processes into additive ones. This is particularly useful in sound measurements where intensity levels can vary exponentially.
  • The key advantage is ease of use when dealing with a wide range of values, allowing compression of exponentially varying data.
  • In sound, using a logarithmic scale like decibels makes it easier to compare different levels, as multiplication becomes simple additions.
Without the logarithmic scale, working with sound levels spanning so many orders of magnitude would be overwhelmingly complex and unintuitive for most users.
Physics Problem Solving
Physics problem-solving often requires clear steps and an organized approach. Dealing with concepts such as sound intensity involves understanding the equations and what they represent.
  • Understanding the formula: grasp what each variable signifies.
  • Solve systematically: break down the problem into digestible parts, as was done in the original exercise through several steps.
  • Check assumptions: e.g., assuming all people produce identical sound intensity.
Careful calculation and simplification turn complex physical equations into manageable solutions, helping verify results and assumptions in the given scenario.

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

The sonar unit on a boat is designed to measure the depth of fresh water \(\left(\rho=1.00 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}, B_{\mathrm{ad}}=2.20 \times 10^{9} \mathrm{~Pa}\right) .\) When the boat moves into seawater \(\left(\rho=1025 \mathrm{~kg} / \mathrm{m}^{3}, B_{\mathrm{ad}}=2.37 \times 10^{9} \mathrm{~Pa}\right),\) the sonar unit is no longer calibrated properly. In seawater, the sonar unit indicates the water depth to be \(10.0 \mathrm{~m}\). What is the actual depth of the water?

A sound wave travels twice as far in neon (Ne) as it does in krypton (Kr) in the same time interval. Both neon and krypton can be treated as monatomic ideal gases. The atomic mass of neon is \(20.2 \mathrm{u}\), and that of krypton is \(83.8 \mathrm{u}\). The temperature of the krypton is \(293 \mathrm{~K}\). What is the temperature of the neon?

A wave traveling in the \(+x\) direction has an amplitude of \(0.35 \mathrm{~m},\) a speed of \(5.2 \mathrm{~m} /\) \(\mathrm{s},\) and a frequency of \(14 \mathrm{~Hz}\). Write the equation of the wave in the form given by either Equation 16.3 or 16.4

Two sources of sound are located on the \(x\) axis, and each emits power uniformly in all directions. There are no reflections. One source is positioned at the origin and the other at \(x=+123 \mathrm{~m}\). The source at the origin emits four times as much power as the other source. Where on the \(x\) axis are the two sounds equal in intensity? Note that there are two answers.

Review at for one approach to this problem. A dish of lasagna is being heated in a microwave oven. The effective area of the lasagna that is exposed to the microwaves is \(2.2 \times 10^{-2} \mathrm{~m}^{2}\). The mass of the lasagna is \(0.35 \mathrm{~kg}\), and its specific heat capacity is \(3200 \mathrm{~J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right)\). The temperature rises by \(72 \mathrm{C}^{\circ}\) in \(8.0\) minutes. What is the intensity of the microwaves in the oven?
See all solutions

Recommended explanations on Physics Textbooks

Force

Read Explanation

Torque and Rotational Motion

Read Explanation

Engineering Physics

Read Explanation

Measurements

Read Explanation

Rotational Dynamics

Read Explanation

Fluids

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.