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In audio and communications systems, the gain, $\…

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Problem 84 Hard Difficulty

Workers around jet aircraft typically wear protective devices over their ears. Assume that the sound level
of a jet airplane engine, at a distance of $30 \mathrm{m},$ is 130 $\mathrm{dB}$ ,and that the average human ear has an effective radius of 2.0 $\mathrm{cm} .$ What would be the power intercepted by
an unprotected ear at a distance of 30 $\mathrm{m}$ from a jet airplane engine?

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Physics 101 Mechanics

Physics for Scientists and Engineers with Modern Physics

Chapter 16

Sound

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Video Transcript

our question that workers around jet aircrafts typically wear protective gear devices over their ears. Assuming the sound level of a jet airplane engine at a distance of 30 meters is 130 decibels and that the average human ear has, um, effective radius of two centimeters. What would be the power interpreted by an unprotected here at a distance of 30 meters from a jet airplane? So we're told here right now what we're told, we're told that the distance from the jet airplane is 30 meters, which I read his d. The decibel level, which I read his beta is 130 decibels and the effective radius of an ear which I call our is two centimetres. Every right that meter terms of S I units and two centimeters is, uh, two times 10 to the minus two meters. And that's because there are 100 centimeters in every meter. Okay, So to find the power we can use, the equation for the power output is equal to the intensity times the area being affected. So this would be the intensity of the sound times four in this case, times pi times the radius of the ear squared. So our swear. Okay, well, we don't know I but we can find I using the definition of decibels, which says that beta is equal to 10 times log base 10 says log base 10 of the ratio of the intensity to I not. Okay. Well, I know it is just a constant, um, in this definition, and it's equal to 1.0 times 10 to the minus 12. And intensity has Eunice of Watts per meter squared in my side. Okay, so let's solve this equation for I. We can divide both sides by 10 to get rid of the 10. So this goes away, and this is now beta over 10. To get rid of the long you raise both sides to the 10th power. Okay? And they will multiply both sides by I Not after doing that to you, I by itself. So you have 10. This will be 10 to the beta. Over 10 multiplied by. I know. Okay. Well, we know beta. We know, I know. We'll plug those values into this expression. We find that the power is equal to 10 once for meter squared. Okay? I'm starting out the power of the intensity intensity is 10 less per meter squared. So now we'll go back to finding the power which is equal to begin I the intensity that we just found times four times pi times the radius of the ear Swear we plug all those values and you find this is equal to zero. Wait. 013 What box set in your solution, Okay?

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Physics for Scientists and Engineers with Modern Physics

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Problem

In audio and communications systems, the gain, $\…

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Physics for Scientists and Engineers with Modern Physics

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Watch More Solved Questions in Chapter 16

Video Transcript

Douglas C. Giancoli

Physics for Scientists and Engineers with Modern Physics

Elyse Gonzalez

Marshall Styczinski

Farnaz Mohseni

Meghan Miholics

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Physics for Scientists and Engineers with Modern Physics

Elyse Gonzalez

Marshall Styczinski

Farnaz Mohseni

Meghan Miholics

Douglas C. GiancoliPhysics for Scientists and Engineers with Modern Physics

Elyse Gonzalez

Marshall Styczinski

Farnaz Mohseni

Meghan Miholics

Douglas C. Giancoli

Physics for Scientists and Engineers with Modern Physics