Abird is flying directly toward a stationary bird-watcher and emits a frequency of 1180 hz. the bird-watcher, however, hears a frequency of 1260 hz. what is the speed of the bird, expressed as a percentage of the speed of sound?

The speed of the bird expressed as a percentage of the speed of sound are %V= 6.39%.

Explanation:

V= 343.2 m/s

Vo= 0

Vf= ?

f'= 1260 Hz

f= 1180 Hz

%V= ?

f'= f* ( (V/ V-Vf))

Clearing Vf:

Vf= 21.94 m/s

%V= (Vf/V)*100

%V= 6.39%

when jerome paddle in flowing river towards east the net speed of jerome will increase

so the net speed of jerome will be given as

now the speed of paul will be same as it is given as he is running on the ground

so

so here we can say since the speed of jerome is more so he will win the race

now as we know the distance so the time difference is given as

so jerome will reach first and the time difference will be 203.55 s or 3.4 min

= initial pressure

= final pressure = 2

= initial volume

= final volume = /4

= initial temperature = 20 c = 20 + 273 = 293 k

= final temperature = ?

using the equation