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A bird is flying directly toward a stationary bird-watcher and emits a frequency of 1390 Hz. The bird-watcher, however, hears a frequency of 1430 Hz. What is the speed of the bird, expressed as a percentage of the speed of sound?

  • Physics - ,

    f =f₀/(1-v/u)
    f₀=1390 Hz
    f=1430 Hz
    v/u=(f-f₀)/f= (1430-1390) /1430=2.8•10⁻² m/s
    v/u = 2.8%

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