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3 years ago

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Two eagles fly directly toward one another, the first at 15.0 m/s and the second at 20.0 m/s. Both screech, the first one emitti

ng a frequency of 3200 Hz and the second one emitting a frequency of 3800 Hz. What frequencies do they receive if the speed of sound is 330 m/s?

1 answer:

Vinil7 [7]3 years ago

8 0

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

$F_1 = (\frac{v + v_2}{v - v_1} )f_1\\F_1 = (\frac{330 + 20}{330 - 15} )(3200)\\= 3.56 \times 10^3Hz\\= 3.56kHz\\$

the second part

$F_1 = (\frac{v + v_2}{v - v_1} )f_1\\F_1 = (\frac{330 + 15}{330 - 20} )(3800)\\= 4.23 \times 10^3Hz\\= 4.23kHz\\$

The first eagle hear a frequency of 4.23kHz and the second eagle hear a frequency of 3.56kHz

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Answer:

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Explanation:

Unit conversions:

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With elastic modulus being E = 200 GPa, then its maximum stress must be

$\sigma = E\epsilon = 200\times10^9*0.000545455 = 109090909 Pa$

Knowing the tension force being F = 8500 N, we can calculate the appropriate cross section area

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And its corresponding diameter is

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$7.79\times10^{-5} = \pi d^2/4$

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$\frac{mv^2}{2} = FS$

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