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15 January, 21:39

A bird is flying directly toward a stationary bird-watcher and emits a frequency of 1180 Hz. The bird-watcher, however, hears a frequency of 1260 Hz. What is the speed of the bird, expressed as a percentage of the speed of sound?

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  1. 15 January, 22:01
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    The speed of the bird expressed as a percentage of the speed of sound are %V = 6.39%.

    Explanation:

    V = 343.2 m/s

    Vo = 0

    Vf=?

    f' = 1260 Hz

    f = 1180 Hz

    %V=?

    f' = f * ((V / V-Vf))

    Clearing Vf:

    Vf = 21.94 m/s

    %V = (Vf/V) * 100

    %V = 6.39%
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