Answer:
![\Delta f= 144.71KHz](https://img.qammunity.org/2020/formulas/physics/college/jx39cjfet0i06kodywtxhpa2f407ddxwlt.png)
Step-by-step explanation:
We define two formulas. Firt for the stationary object and then for the shpo,
The frequency formula for the stationary is,
![f_(Lr)=(V+V_w)/(V)f^1\\f_(S)=(V)/(V-V_w)f^2](https://img.qammunity.org/2020/formulas/physics/college/snr8fuoa5a65rzh1e4o7fvlvja39q63z4r.png)
We need to find the difference or the rate of the relevant frequencies, that is
![\Delta f= f_(Lr)-f_S\\\Delta=f(2V_w)/(V-V_w)](https://img.qammunity.org/2020/formulas/physics/college/od4dv9kxibm5oqinjegob7tsthgjwt8vvu.png)
With the dates that we have, we only make a substitution,
![f=22 KHz\\V_w = 4.95m/s\\V= 1510m/s](https://img.qammunity.org/2020/formulas/physics/college/otabf1xg0i8kveijz033yvqp7f5cys1smg.png)
![\Delta f= (22.0KHz)((2(4.95))/(1510-4.95))](https://img.qammunity.org/2020/formulas/physics/college/e1ciisclxotszdtrvvupcdsyxfl1nwnudx.png)
![\delta f= 144.71KHz](https://img.qammunity.org/2020/formulas/physics/college/wgojfpwz7q7z4ni3bnlww4li1i9bsu2ztw.png)