Physics 101

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A 90-dB sound is absorbed by an eardrum 0.75 cm in diameter for 2 hours. How much energy in joules does the eardrum absorb at that time?

  • Physics 101 -

    (Power/area)*(area)*(time)= Energy

    Convert 90 dB to power/area. Use SI units. See
    http://www.sengpielaudio.com/TableOfSoundPressureLevels.htm
    if you need help with that.

  • Physics 101 -

    I figured it out already, thanks. I was just having trouble with converting dB to W/m^2.

    Anyway, here's my answer. Feel free to correct it.

    I - Intensity
    P - Power
    E - Energy
    t - time
    A - Area

    I = P/A ; I = E/t / A ; E = IAt

    Io = 1x10^-12 W/m^2 (treshold of hearing)

    Conversion of dB to W/m^2 :

    B = 10 log I/Io

    90 = 10log I/Io
    9 = log I/Io
    I = 10^9 Io

    I = 10^9 (10^-12)
    I = 1x10^-3 W/m^2

    E = IAt
    = 1x10^-3 W/m^2 [1/4 pi (0.0075m)^2] (7200 sec)
    = 3.18086 x 10^-4 Joules

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