Question

Question #5ca90

Answer 1

`"the receiver frequency is 9907.68 Hz."`

Explanation:

`v_("air"):"velocity of sound in the air ,"v_("air")=340 " "m/s`

`v_s:"velocity of source ,"v_s=140 (km)/(hr)=(140.000)/(3600)=38.89" "m/s`

`v_r:"velocity of receiver ,"v_r=150" "(km)/(hr)=(150.000)/(3600)=41.67" "m/s`

`f_s:"frequency of source ,"f_s=10kHz=10000Hz`

`f_r:"frequency of receiver "`

`f_r=f_s * (v_("air")+-v_r)/(v_("air")+-v_s )`

`"Since the distance between receiver and source increase,"`
`"you should write the sign of " v_r " as negative and the sign of the " v_s " as positive."`

`f_r=f_s *(v_("air")-v_r)/(v_("air")+v_s)`

`f_r=10000 *((340-41.67))/((340+38.89))`

`f_r=9907.68 " "Hz`