. . . water).¹

Hint: you must first find the pressure and density at that depth. Look at sections 9.2, 9.2.1 and 9.2.9 in the textbook.

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. . . \/.²

This would be good for long-distance communication, since lower frequencies are more weakly attenuated.

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. . . that?³

The tone sf A has the frequency $f_{\sf A} = 440$ Hz. One octave corresponds to a factor of 2 in frequency. Each of the 12 semitones within one octave differs from its neighbor by the frequency ratio 2^1/12. For instance, $f_{\sf F}/f_{\sf E} = 2^{1/12}$ and $f_{\sf A}/f_{\sf G} = 2^{2/12}$.

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. . . \/.⁴

This may seem rather short, but there are several reasons: first, the lungs are not a fixed-volume container - they are elastic, and the effective volume has been increased to take this into account; second, the "piston" air column is not the whole mouth and throat, but just the region with the small area A specified - which corresponds to a diameter of only about 1.4 cm, which in turn is (not so surprisingly) about the same size as our calculated L. The vibrating "mass on a spring" is just the air between the baritone's teeth and lips.

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. . . land.⁵

Actually this question could lead to a nice project: How do they make the sound (PHYSICS), and why do they do it (ZOOLOGY)?

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