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Sagot :
Answer:
_s = 37.77 m / s
Explanation:
This is an exercise of the Doppler effect that the change in the frequency of the sound due to the relative speed of the source and the observer, in this case the observer is still and the source is the one that moves closer to the observer, for which relation that describes the process is
f ’= f₀ [tex]\frac{v}{v - v_s}[/tex]
where d ’= 530 Make
when the ambulance passes away from the observer the relationship is
f ’’ = f₀ [tex]\frac{v}{v + v_s}[/tex]
where d ’’ = 424 beam
let's write the two expressions
f ’ (v-v_s) = fo v
f ’’ (v + v_s) = fo v
let's solve the system, subtract the two equations
v (f ’- f’ ’) - v_s (f’ + f ’’) = 0
v_s = v [tex]\frac{ f' - f''}{ f' + f''}[/tex]
the speed of sound is v = 340 m / s
let's calculate
v_s = 340 [tex](\frac{ 530 -424}{530+424} )[/tex]
v_s = 340 [tex](\frac{106}{954}[/tex])
v_s = 37.77 m / s
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