Answer:
The first eagle hear a frequency of 4.43kHz and the second eagle hear a frequency of 3.80kHz
Explanation:
Given that
both eagle are flying towards one another
speed of the first eagle v1 = 14.5m/s
speed of the second eagle v2 = 22.5m/s
frequency emitted by the first eagle f1= 3400Hz
frequency emitted by the second eagle f2 = 3950Hz
speed of sound v = 330m/s
first part
[tex]F_1 = (\frac{v + v_2}{v - v_1} )f_1\\F_1 = (\frac{330 + 22.5}{330 - 14.5} )(3400)\\= 3.80 \times 10^3Hz\\= 3.80kHz\\[/tex]
second part
[tex]F_1 = (\frac{v + v_2}{v - v_1} )f_1\\F_1 = (\frac{330 + 14.5}{330 - 22.5} )(3950)\\= 4.43 \times 10^3Hz\\= 4.43kHz\\[/tex]
The first eagle hear a frequency of 4.43kHz and the second eagle hear a frequency of 3.80kHz
