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alexdok [17]
2 years ago
13

Elephants can communicate over distances as far as 6 km by using very low-frequency sound waves. What is the wavelength of a 10

Hz sound wave emitted by an elephant? Assume air temperature is 20∘C.
Physics
1 answer:
harkovskaia [24]2 years ago
3 0

The characteristics of the sound waves allow to find the result for the wavelength is:

         λ = 34.29 m

The sound wave is a longitudinal traveling wave, where the speed of the wave depends on the air temperature.

         v = v_o \sqrt{1 + \frac{T}{273} }  

where v₀ is the speed of sound for T = 0ºc v₀ = 331 m / s.

indicate that the temperature is T = 20ºC, let's find the speed of the traveling wave.

        v = 331 \sqrt{1 + \frac{20}{273} }  

        v = 342.9 m / s

The speed of sound is related to wavelength and frequency.

       v = λ f

        λ = \frac{v}{f}

Let's calculate

        \lambda = \frac{342.29}{10}  

        λ  = 34.29 m

In conclusion using the characteristics of sound waves we can find the result for the wavelength is:

          λ = 34.29 m

Learn more here:  brainly.com/question/21995826

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Answer:

0.087 m

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the formula for the time period of the pendulum is given by

T = 2\pi \sqrt{\frac{I}{mgd}}    .... (1)

where, I is the moment of inertia of the rod about the pivot point and g is the acceleration due to gravity.

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By using the parallel axis theorem, the moment of inertia of the rod about the pivot is

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I = \frac{mL^{2}}{12}+ md^{2}

Substituting the values in equation (1)

3 = 2 \pi \sqrt{\frac{\frac{mL^{2}}{12}+ md^{2}}{mgd}}

9=4\pi^{2}\times \left ( \frac{\frac{L^{2}}{12}+d^{2}}{gd} \right )

12d² -26.84 d + 2.25 =  0

d=\frac{26.84\pm \sqrt{26.84^{2}-4\times 12\times 2.25}}{24}

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Answer:

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8 0
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