Una ambulancia se desplaza hacia el norte. a 23, 5 m/s. su sirena emite un sonido de 500 hz de frecuencia . calcular la frecuencia aparente que percibe. un observador. que viaja en su auto, también hacia el norte. a 18, 6 m/s cuando : el auto se aleja de la ambulancia

The apparent frquency would be

f' =  493.26 Hz

Step-by-step explanation:

This a Doppler's effect problem. This effects produces a slight change in the frequency of a traveling wave, due to the effect of the velocity of the receiver (observer).

The formula used to describe the change, as both vehicles are moving away of each other:

f'  =   [ (Vwave  +  Vobserver) / (Vwave + Vsource ) ]*f

Where

Vwave = 340 m/s             ( speed of sound)

Vobserver = 18.6 m/s       (observer = car)

Vsource  = 23.5 m/s          (speed of the ambulance = source)

f = 500 Hz                          (frequency of the sound)

We just need to substitute in the equation

f'  =    [(340+  18.6) / (340+ 23.5)] * 500 Hz

f'  =    [0.9865] * 500 Hz = 493.26 Hz

f' =  493.26 Hz

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐

angle addition postulate

2x=56

step-by-step explanation: