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4) A bat is flitting around a cave navigating via ultrasonic bleeps. Assume that the sound emission frequency of the bat is 3PHYSICS!! PLEASE HELP! MY TEACHER GIVES US THE ANSWERS BUT WE HAVE TO SHOW ALL WORK! PLEASE SHOW WORK FOR EACH PART OF THE QUESTION! THANK YOU!!

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Answer #1

PLEASE GIVE IT A THUMBS UP AND LET ME KNOW IF YOU HAVE ANY DOUBT.

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First of all,we will find speed of sound for given temperature

v = 331m/s + 0.6m/s/C * T

v = 331 + 0.6*18.9

v = 342.34 m/s

Now, we will use Doppler's equation

(a) Here bat is source and wall is observer

f' = (v / v - vs)f

v = Velocity of sound or light in medium,
f = Real frequency,
f' = Apparent frequency

vs = speed of source

f' = (342.34 / 342.34 - 0.025*342.34 )39,000

f' = 40000 Hz

Therefore, frequency detected by wall is 40000 Hz

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(b) Now, the frequency which is reflected off the wall is caught by bat

so,

f = (342.34 + 0.025 * 342.34 / 342.34 ) * 40000

f = (350.9 / 342.34) * 40000

f = 41001 Hz

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