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The average sound intensity inside a busy neighborhood restaurant is \(3.2 \times 10^{-5} \mathrm{W} / \mathrm{m}^{2} .\) How much energy goes into each ear (area \(\left.=2.1 \times 10^{-3} \mathrm{m}^{2}\right)\) during a one-hour meal?

Short Answer

Expert verified
Each ear receives about \( 2.42 \times 10^{-4} \text{ J} \) of energy during a one-hour meal.

Step by step solution

01

Review the given variables

Identify the average sound intensity (I) and the area of each ear (A). Here, \( I = 3.2 \times 10^{-5} \text{ W/m}^2 \) and \( A = 2.1 \times 10^{-3} \text{ m}^2 \).
02

Calculate the power received by an ear

The power (P) incident on the ear can be calculated using the formula: \( P = I \times A \). Substitute the given values: \( P = (3.2 \times 10^{-5}) \times (2.1 \times 10^{-3}) \).
03

Calculate the power

Perform the multiplication to find the power: \( P = 6.72 \times 10^{-8} \text{ W} \).
04

Determine the total energy over time

The energy (E) absorbed is calculated by multiplying the power by the total time: \( E = P \times T \). Here, \( T = 3600 \text{ seconds} \) (since 1 hour is 3600 seconds).
05

Calculate the final energy

Substitute the values into the energy formula: \( E = (6.72 \times 10^{-8}) \times 3600 \). This results in \( E = 2.4192 \times 10^{-4} \text{ J} \).
06

Interpret the result

The total energy going into each ear during the one-hour period is approximately \( 2.42 \times 10^{-4} \text{ J} \).

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Key Concepts

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

Energy Absorption
When we talk about energy absorption in the context of sound, we are referring to the amount of sound energy that is absorbed by an object over a certain period. In the scenario of a busy neighborhood restaurant, this is the sound energy that enters your ear over an hour. 
To understand this process, consider the role of sound intensity. Sound intensity is measured in watts per square meter \(\left(\text{W/m}^2\right)\). It indicates the power of the sound per unit area. Here, the average sound intensity \( I \) is given as \(3.2 \times 10^{-5} \text{W/m}^2\). Each ear has an effective area of \(2.1 \times 10^{-3} \text{m}^2\). 
  • First, calculate the power that has been incident on the ear by multiplying the sound intensity \( I \) with the area \( A \).
  • To find out the total energy absorbed, multiply this power \( P \) by the time period \( T \) (here it's 1 hour or 3600 seconds).
This gives us the total energy in joules \( J \) that the ear absorbs during the meal.
Power Calculation
Power calculation is a crucial element in determining how much energy an object receives over a time period. Power, in terms of sound, can be understood as the rate at which sound energy is being transferred to a specific area, such as an ear. 
The formula for power \( P \) is quite straightforward: \( P = I \times A \), where \( I \) is the sound intensity, and \( A \) is the area that the sound is hitting.  

Example Calculation:

In the given example, the average sound intensity is \(3.2 \times 10^{-5} \text{W/m}^2\), and the area of the ear is \(2.1 \times 10^{-3} \text{m}^2\). 
  • Substitute these values into the formula: \( P = (3.2 \times 10^{-5}) \times (2.1 \times 10^{-3}) \).
  • This results in power \( P = 6.72 \times 10^{-8} \text{W} \).
Power provides a snapshot of the amount of sound energy being absorbed by the ear at any one moment.
Sound Energy
Sound energy is a form of energy associated with the vibrations of matter, such as air particles. It travels in waves and can vary significantly in intensity, which affects how we perceive volume. 
Specifically, sound energy can be described as a transfer of energy from a sound source through a medium (like air) to an area (such as the human ear). The formula \( E = P \times T \) is used to calculate the total sound energy absorbed over time, where \( P \) is the power of the sound and \( T \) is the time. 

Practical Application:

For the given example, to find the total sound energy absorbed by an ear during the meal, you multiply the power received by the ear \(6.72 \times 10^{-8} \text{W} \) by the duration of the meal in seconds \(3600 \text{s}\). 
  • The energy \( E \) then becomes \( E = (6.72 \times 10^{-8}) \times 3600 = 2.4192 \times 10^{-4} \text{J} \).
  • This gives the total sound energy absorbed, which in practical terms, is the perceived sound exposure during the meal.
Understanding sound energy can help us manage our exposure to varying sound levels, especially in busy environments.

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Most popular questions from this chapter

The security alarm on a parked car goes off and produces a frequency of \(960 \mathrm{Hz}\). The speed of sound is \(343 \mathrm{m} / \mathrm{s}\). As you drive toward this parked car, pass it, and drive away, you observe the frequency to change by \(95 \mathrm{Hz}\) Al what speed are you driving?

A loudspeaker has a circular opening with a radius of \(0.0950 \mathrm{m} .\) The electrical power needed to operate the speaker is \(25.0 \mathrm{W}\). The average sound intensity at the opening is \(17.5 \mathrm{W} / \mathrm{m}^{2}\). What percentage of the electrical power is converted by the speaker into sound power?

As the drawing shows, one microphone is located at the origin, and a second microphone is located on the \(+y\) axis. The microphones are separated by a distance of \(D=1.50 \mathrm{m} .\) A source of sound is located on the \(+x\) axis, its distances from microphones 1 and 2 being \(L_{1}\) and \(L_{2}\), respectively. The speed of sound is \(343 \mathrm{m} / \mathrm{s} .\) The sound reaches microphone 1 first, and then, \(1.46 \mathrm{ms}\) later, it reaches microphone \(2 .\) Find the distances \(L_{1}\) and \(L_{2}\).

The mass of a string is \(5.0 \times 10^{-3} \mathrm{kg}\), and it is stretched so that the tension in it is \(180 \mathrm{N}\). A transverse wave traveling on this string has a frequency of \(260 \mathrm{Hz}\) and a wavelength of \(0.60 \mathrm{m} .\) What is the length of the string?

Light is an electromagnetic wave and travels at a speed of \(3.00 \times\) \(10^{8} \mathrm{m} / \mathrm{s} .\) The human eye is most sensitive to yellow- green light, which has a wavelength of \(5.45 \times 10^{-7} \mathrm{m} .\) What is the frequency of this light?

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