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Sagot :
Given:
The force is
[tex]F=1.90\text{ N}[/tex]The area of the eardrum is
[tex]\begin{gathered} A=1.14\text{ cm}^2 \\ =1.14\times10^{-4}\text{ m}^2 \end{gathered}[/tex]To find:
The maximum tolerable gauge pressure inside the eardrum
a) the pressure in mm of Hg
b) At what depth in freshwater would this person's eardrum rupture
Explanation:
The pressure at the eardrum is
[tex]\begin{gathered} P=\frac{F}{A} \\ =\frac{1.90}{1.14\times10^{-4}} \\ =16.67\times10^3\text{ N/m}^2 \end{gathered}[/tex]Hence, the pressure is
[tex]16.67\times10^3\text{ N/m}^2[/tex]a)
We know,
[tex]1\text{ N/m}^2=0.0075\text{ mm of Hg}[/tex]So,
[tex]\begin{gathered} 16.67\times10^3\text{ N/m}^2=0.0075\times16.67\times10^3\text{ mm of Hg} \\ =125.02\text{ mm of Hg} \end{gathered}[/tex]Hence, the pressure is 125.02 mm of Hg.
b)
The depth of fresh water is,
[tex]\begin{gathered} h=\frac{P}{dg} \\ Here,\text{ d=1000 kg/m}^3 \\ g=9.8\text{ m/s}^2 \end{gathered}[/tex]So,
[tex]\begin{gathered} h=\frac{16.67\times10^3}{1000\times9.8} \\ =1.70\text{ m} \end{gathered}[/tex]Hence, the depth of water is 1.70 m.
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