Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 12: Problem 11

A sound wave with an intensity level of \(80.0 \mathrm{dB}\) is incident on an eardrum of area \(0.600 \times 10^{-4} \mathrm{m}^{2} .\) How much energy is absorbed by the eardrum in 3.0 min?

Short Answer

Expert verified
Answer: The eardrum absorbs 108 x 10⁻⁸ Joules of energy in 3 minutes.

Step by step solution

01

Convert the intensity level to intensity

Given the intensity level (L) of the sound wave is \(80.0 \mathrm{dB}\). To convert this value to intensity (I) in watts per square meter, we use the decibel formula: $$L = 10 \cdot \log_{10}\left(\frac{I}{I_0}\right)$$ where \(I_0=10^{-12} \mathrm{W/m^2}\) is the reference intensity. We need to solve this equation for I: $$I = I_0 \cdot 10 ^{\frac{L}{10}}$$ Now, substituting the given values: $$I = 10^{-12} \cdot 10 ^{\frac{80}{10}}$$
02

calculate the intensity

Performing the calculation of intensity: $$I = 10^{-12} \cdot 10^{8} = 10^{-4} \mathrm{W/m^2}$$
03

Find the power absorbed by the eardrum

The power absorbed by the eardrum (P) can be calculated by multiplying the intensity (I) by the area of the eardrum (A): $$P = I \cdot A$$ Given the area of the eardrum, A is \(0.600 \times 10^{-4} \mathrm{m^2}\). Now substituting the intensity and area of the eardrum to find the power: $$P = 10^{-4} \cdot 0.600 \times 10^{-4}$$
04

Calculate the power absorbed by the eardrum

Performing the calculation of power absorbed: $$P = 10^{-4} \cdot 0.600 \times 10^{-4} = 0.6 \times 10^{-8} \mathrm{W}$$
05

Calculate the energy absorbed by the eardrum

To find the energy absorbed by the eardrum in 3 minutes, we can use the formula: $$E = P \cdot t$$ where t is the time in seconds. Given the time t is 3 minutes, convert it to seconds: $$t = 3 \cdot 60 = 180 \,\text{seconds}$$ Now substituting the power and time to find the energy absorbed: $$E = 0.6 \times 10^{-8} \cdot 180$$
06

Calculate the energy absorbed

Performing the calculation of energy absorbed: $$E = 0.6 \times 10^{-8} \cdot 180 = 108 \times 10^{-8} \mathrm{Joules}$$ So, the eardrum absorbs \(108 \times 10^{-8} \mathrm{J}\) of energy in 3 minutes.

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

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?
A musician plays a string on a guitar that has a fundamental frequency of \(330.0 \mathrm{Hz}\). The string is \(65.5 \mathrm{cm}\) long and has a mass of \(0.300 \mathrm{g} .\) (a) What is the tension in the string? (b) At what speed do the waves travel on the string? (c) While the guitar string is still being plucked, another musician plays a slide whistle that is closed at one end and open at the other. He starts at a very high frequency and slowly lowers the frequency until beats, with a frequency of \(5 \mathrm{Hz}\), are heard with the guitar. What is the fundamental frequency of the slide whistle with the slide in this position? (d) How long is the open tube in the slide whistle for this frequency?
Find the speed of sound in mercury, which has a bulk modulus of $2.8 \times 10^{10} \mathrm{Pa}\( and a density of \)1.36 \times\( \)10^{4} \mathrm{kg} / \mathrm{m}^{3}.$
A source and an observer are each traveling at 0.50 times the speed of sound. The source emits sound waves at \(1.0 \mathrm{kHz}\). Find the observed frequency if (a) the source and observer are moving towand each other, (b) the source and observer are moving away from each other; (c) the source and observer are moving in the same direction.
(a) What is the pressure amplitude of a sound wave with an intensity level of \(120.0 \mathrm{dB}\) in air? (b) What force does this exert on an eardrum of area \(0.550 \times 10^{-4} \mathrm{m}^{2} ?\)
See all solutions

Recommended explanations on Physics Textbooks

Rotational Dynamics

Read Explanation

Particle Model of Matter

Read Explanation

Electrostatics

Read Explanation

Turning Points in Physics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Oscillations

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