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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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okpalawalter8

Answer:

The frequency of the sona-pulse reflected back is  [tex]f_b = 48109.22Hz[/tex]

Explanation:

From the question we are told that

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

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

       The frequency emitted by the bat is [tex]f = 47000 \ Hz[/tex]

        The speed of sound is  [tex]v_s = 343 m/s[/tex]

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

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

Substituting values

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

                     [tex]f_a = 47556.4 Hz[/tex]

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

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

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

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

                  [tex]f_b = 48109.22Hz[/tex]

Cricetus

The frequency will be "47088.36 Hz".

According to the question,

Frequency,

  • 47000-Hz

Speed of sound,

  • 343 m/s

Now,

Frequency received by insect will be:

→ [tex]f' = f\times (\frac{343-1.1}{343-5.1} )[/tex]

By substituting the values, we get

       [tex]= 47000\times \frac{341.9}{337.9}[/tex]

       [tex]= 47000\times 1.012[/tex]

       [tex]= 47564 \ Hz[/tex]

hence,

The frequency received as echo will be:

→ [tex]f''=f'\times (\frac{343+5.1}{343+1.1} )[/tex]

       [tex]= 47564\times \frac{337.9}{341.9}[/tex]

       [tex]= 47564\times 0.99[/tex]

       [tex]= 47088.36 \ Hz[/tex]

Thus the above answer is appropriate.

Learn more:

https://brainly.com/question/16107942

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