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An ambulance driver traveling at 31.0 m/s (69.3 mph) honks his horn as he sees a motorist ahead on the highway traveling in the same direction. The motorist hears a frequency of 392 Hz and notices that his own speedometer reads 44.7 mph (20.0 m/s). Calculate the frequency that the ambulance driver hears. (Speed of sound is 340 m/s) (in Hz)

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lcmendozaf

Answer:

fo = 378.52Hz

Explanation:

Using Doppler effect formula:

[tex]f'=\frac{C-Vb}{C-Va}*fo[/tex]

where

f' = 392 Hz

C = 340m/s

Vb = 20m/s

Va = 31m/s

Replacing these values and solving for fo:

fo = 378.52Hz

Lanuel

The frequency that the ambulance driver (observer) hears is equal to 468.84 Hz.

Given the following data:

  • Observer velocity = 44.7 mph (20.0 m/s)
  • Frequency of sound = 392 Hz
  • Source velocity = 31.0 m/s (69.3 mph)
  • Speed of sound = 340 m/s

To calculate the frequency that the ambulance driver (observer) hears, we would apply Doppler's effect of sound waves:

Mathematically, Doppler's effect of sound waves is given by the formula:

[tex]F_o = \frac{V \;+ \;V_o}{V\; - \;V_s} F[/tex]

Where:

  • V is the speed of a sound wave.
  • F is the actual frequency of sound.
  • [tex]V_o[/tex] is the observer velocity.
  • [tex]V_s[/tex] is the source velocity.
  • [tex]F_o[/tex] is the observer frequency.

Substituting the given parameters into the formula, we have;

[tex]F_o = \frac{340 \;+ \;20}{340\; - \;31} \times 392\\\\F_o = \frac{360}{301} \times 392\\\\F_o = 1.20 \times 392\\\\F_o =468.84 \;Hz[/tex]

Ambulance driver (observer) frequency = 468.84 Hz

Read more: https://brainly.com/question/23460034

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