A bat, flying at 5.1 m/s, pursues an insect that is flying at 1.1 m/s in the same direction. The bat emits a 47000-Hz sonar pulse. Take the

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A bat, flying at 5.1 m/s, pursues an insect that is flying at 1.1 m/s in the same direction. The bat emits a 47000-Hz sonar pulse. Take the speed of sound to be 343 m/s. show answer No Attempt At what frequency, in hertz, does the bat hear the pulse reflected back from the insect? f = | sin() cos() tan() cotan() asin() acos() atan() acotan() sinh() cosh() tanh() cotanh() Degrees Radians π ( ) 7 8 9 HOME E ↑^ ^↓ 4 5 6 ← / * 1 2 3 → + – 0 . END √() BACKSPACE DEL CLEAR Grade Summary Deductions 0% Potential 100% Submissions Attempts remaining: 3 (4% per attempt) detailed view Hints: 4% deduction per hint. Hints remaining: 2 Feedback: 5% deduction per feedback.

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Nho 4 years 2021-07-24T13:22:26+00:00 1 Answers 38 views 0

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  1. King
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    2021-07-24T13:23:33+00:00
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    Answer:

    The frequency of the sona-pulse reflected back is  f_b = 48109.22Hz

    Explanation:

    From the question we are told that

         The speed of the bat is  v = 5.1 m/s

          The speed of the insect is v_i = 1.1 m/s

           The frequency emitted by the bat is f = 47000 \ Hz

            The speed of sound is  v_s = 343 m/s

    Let look at this question in this manner

    At the first instant the that the bat  emits the sonar pulse

         Let the bat be the source of sound

          Let the insect be the observer

    This implies that the frequency of sound the the insect would receive is mathematically represented as

                        f_a = [\frac{v_s - v_i}{v_s - v}] f

    Substituting values

                     f_a = [\frac{343 - 1.1}{345 -5.1} ] * 47000

                         f_a = 47556.4 Hz

    Now at the instant the sonar pules reaches the insect

                Let the bat be the observer

                Let the insect be the source of the sound

    Here the sound wave is reflected back to the bat

    This implies that the frequency of sound the the bat  would receive is mathematically represented as

                       f_b =[ \frac{x}{y} ] * f_a

                       f_b =[ \frac{v_s + v}{v_s + v_i} ] * f_a

                      f_b =[ \frac{343 + 5.1}{343 + 1.1} ] * 47556.4

                      f_b = 48109.22Hz

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