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Intensity of sound wave & energy passes into ear

  1. Apr 30, 2013 #1
    1. The problem statement, all variables and given/known data

    Please refer to the image. Question (a).

    2. Relevant equations
    I= P/(4∏r^2)
    P=E/t


    3. The attempt at a solution

    After a few attempts, I found the way to get the answer:

    (6.3 x 10^-6) x (1.5 x 10^-4) x 60 = 5.67 x 10^-8

    But I don't understand why the area of the ear canal is used, instead of the area which the energy carried by the wave passes through. From what I understand, intensity depends on the distance from the source, so how come we don't need to take into account the distance of 5.0m?
     

    Attached Files:

  2. jcsd
  3. Apr 30, 2013 #2

    SammyS

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    attachment.php?attachmentid=58367&d=1367331297.jpg

    The sound intensity at 5 m from the sound source is given. The ear canal is also at 5 m from the source. Therefore the distance of 5 m is taken into account.
     
  4. May 1, 2013 #3
    Thank you.
     
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