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

At the race track, one race car starts its engine with a resulting intensity level of \(98.0 \mathrm{dB}\) at point \(P .\) Then seven more cars start their engines. If the other seven cars each produce the same intensity level at point \(P\) as the first car, what is the new intensity level with all eight cars running?
An aluminum rod, \(1.0 \mathrm{m}\) long, is held lightly in the middle. One end is struck head-on with a rubber mallet so that a longitudinal pulse-a sound wave- -travels down the rod. The fundamental frequency of the longitudinal vibration is \(2.55 \mathrm{kHz}\). (a) Describe the location of the node(s) and antinode(s) for the fundamental mode of vibration. Use either displacement or pressure nodes and antinodes. (b) Calculate the speed of sound in aluminum from the information given in the problem. (c) The vibration of the rod produces a sound wave in air that can be heard. What is the wavelength of the sound wave in the air? Take the speed of sound in air to be \(334 \mathrm{m} / \mathrm{s}\). (d) Do the two ends of the rod vibrate longitudinally in phase or out of phase with each other? That is, at any given instant, do they move in the same direction or in opposite directions?
A geological survey ship mapping the floor of the ocean sends sound pulses down from the surface and measures the time taken for the echo to return. How deep is the ocean at a point where the echo time (down and back) is $7.07 \mathrm{s} ?\( The temperature of the seawater is \)25^{\circ} \mathrm{C}$
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?
When playing fortissimo (very loudly), a trumpet emits sound energy at a rate of \(0.800 \mathrm{W}\) out of a bell (opening) of diameter \(12.7 \mathrm{cm} .\) (a) What is the sound intensity level right in front of the trumpet? (b) If the trumpet radiates sound waves uniformly in all directions, what is the sound intensity level at a distance of \(10.0 \mathrm{m} ?\)
See all solutions

Recommended explanations on Physics Textbooks

Fields in Physics

Read Explanation

Solid State Physics

Read Explanation

Oscillations

Read Explanation

Physics of Motion

Read Explanation

Electrostatics

Read Explanation

Kinematics Physics

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