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Suppose a 1.90 N force can rupture an eardrum having an area of 1.14 cm².(a) Calculate the maximum tolerable gauge pressure inside the eardrum (in the middle ear) in N/m². (Pressures in themiddle ear may rise when an infection causes a fluid buildup. Use 13.6 x 10³ kg/m³ as the density of mercury.) submit answer in N/m²(a) part 2: Convert this value to mm Hg.mm Hg(b) At what depth in fresh water would this person's eardrum rupture, assuming the gauge pressure in the middle ear iszero?Submit Answer in m

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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