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Sagot :
Lets calculate the wavelength of the sound in water. But before we do that, we must transform the unit of the speed to m/s. We get:
5220 / 3,6 = 1450 m/s
So
[tex]\begin{gathered} v=\lambda f \\ 1450=\lambda\times500 \\ \lambda=\frac{1450}{500} \\ \lambda=2.9\text{ m} \end{gathered}[/tex]The wave of sound does not change the frequency as it goes through different mediums. So f = 500 Hz. The speed of sound in air is 340 m/s. Then we can say that:
[tex]\begin{gathered} v=\lambda f \\ 340=\lambda\times500 \\ \lambda=\frac{340}{500} \\ \lambda=0.68\text{ m} \end{gathered}[/tex]We can see how many times more wavelengths occur in air by calculating the ratio:
[tex]\begin{gathered} R=\frac{2.9}{0.68} \\ R\cong4 \end{gathered}[/tex]So wavelengths in air occur approximately 4 times more than in water.
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