Chapter 1: Q14E (page 391)

You are designing a diving bell to withstand the pressure of seawater at a depth of 250 m. (a) What is the gauge pressure at this depth? (You can ignore changes in the density of the water with depth.) (b) At this depth, what is the net force due to the water outside and the air inside the bell on a circular glass window 30.0 cm in diameter if the pressure inside the diving bell equals the pressure at the surface of the water? (Ignore the small variation of pressure over the surface of the window.)

Short Answer

Expert verified

(a) The gauge pressure is $2.5 \times 10^{6} N / m^{2}$ and (b) the net force due to the water and the air is $1.7 \times 10^{5} N$ .

Step by step solution

Given Data

The depth is h=250m .

The diameter of glass window is d=30cm .

Understanding the net force on the glass window

The net force on the circular glass window can be calculated by using the relationship between force, pressure and cross-sectional area.

Determining the gauge pressure

The relation of gauge pressure can be written as:

$P = p g h$

Here, gis the gravitational acceleration and $p$ is the density of seawater.

Substitute 250m for $h$ ,1030 $k g / m^{3}$ for p, and $9.80 m / s^{2}$ for g in the above relation.

$P = (1030 kg / m^{3}) (9.80 m / s^{2}) (250 m) P = 2.5 \times 10^{6} N / m^{2}$

Thus, the required pressure is $2.5 \times 10^{6} N / m^{2}$ .

Determining the net force on the window

The relation of net force can be written as:

F=PA

$F = P (\frac{{πd}^{2}}{4})$

Here, Ais the cross-sectional area.

Substitute $2.5 \times 10^{6} N / m^{2}$ for $P$ , and 30cm for d in the above relation.

$F = (2.5 \times 10^{6} N / m^{2}) (\frac{π {(30 cm \times \frac{1 m}{100 cm})}^{2}}{4}) F = 1.7 \times 10^{5} N$

Thus, the net force is $1.7 \times 10^{5} N$ .

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Short Answer

Step by step solution

Given Data

Understanding the net force on the glass window

Determining the gauge pressure

Determining the net force on the window

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