Answer:
The first eagle hear a frequency of 4.23kHz and the second eagle hear a frequency of 3.56kHz
Explanation:
Given that
both eagle are flying towards one another
speed of the first eagle v1 = 15m/s
speed of the second eagle v2 = 20m/s
frequency emitted by the first eagle f1= 3200Hz
frequency emitted by the second eagle f2 = 3800Hz
speed of sound v = 330m/s
[tex]F_1 = (\frac{v + v_2}{v - v_1} )f_1\\F_1 = (\frac{330 + 20}{330 - 15} )(3200)\\= 3.56 \times 10^3Hz\\= 3.56kHz\\[/tex]
the second part
[tex]F_1 = (\frac{v + v_2}{v - v_1} )f_1\\F_1 = (\frac{330 + 15}{330 - 20} )(3800)\\= 4.23 \times 10^3Hz\\= 4.23kHz\\[/tex]
The first eagle hear a frequency of 4.23kHz and the second eagle hear a frequency of 3.56kHz