Q52P During a test dive in 1939, prio... [FREE SOLUTION]

91影视

University Physics with Modern Physics

Hugh D. Young, Roger A. Freedman

Physics

14th edition Edition 路 1596 pages

ISBN: 9780321973610

Chapter 3: Q52P (page 613)

During a test dive in 1939, prior to being accepted by the U.S. Navy, the submarine Squalus sank at a point where the depth of water was 73.0 m. The temperature was $27.0 掳 C$ at the surface and at the bottom. The density of seawater is 1030 kg/m. (a) A diving bell was used to rescue 33 trapped crewmen from the Squalus. The diving bell was in the form of a circular cylinder high, open at the bottom and closed at the top. When the diving bell was lowered to the bottom of the sea, to what height did water rise within the diving bell? (Hint: Ignore the relatively small variation in water pressure between the bottom of the bell and the surface of the water within the bell.) (b) At what gauge pressure must compressed air have been supplied to the bell while on the bottom to expel all the water from it?

Short Answer

Expert verified

(a) The height of the water is 2.04 m.

(b) The pressure is 7 $7.37 脳 10^{5} Pa$ .

Step by step solution

Definition of Density

The term density is defined as the ratio of mass and volume.

Determine height and pressure for water

There ae two cases first when the bell at the surface and second when the bell at h = 73 m under the surface.

For first stage

The temperature, $T_{1} = 300 K$

The length, $L_{1} = 230 m$

The volume, $V_{1} = {AL}_{1}$

The pressure, $p_{1} = 1.013 脳 10^{5} Pa$

For second stage

The temperature, $T_{2} = 280 K$

The length is $L_{1}$ .

The volume, $V_{1} = {AL}_{1}$

The pressure is $p_{2}$ .

So, the pressure can be calculated as

$\begin{aligned} p_{2} = p_{1} + 螖 p \\ = p_{1} + 蚁 g h \end{aligned}$

Put all the values in the above equation.

$\begin{aligned} p_{2} = 1.013 脳 10^{5} + (1030 脳 9.8 脳 73 \\ = 8.38162 脳 10^{5} Pa \end{aligned}$

Now using ideal law

pV = nRT

Here n and R are constants, than

$\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}}$

Put ${AL}_{1}$ for $V_{1}$ for ${AL}_{2}$ $V_{2}$ in the above equation.

$\begin{aligned} \frac{p_{1} A L_{1}}{T_{1}} = \frac{p_{2} A L_{2}}{T_{1}} \\ \frac{p_{1} L_{1}}{T_{1}} = \frac{p_{2} L_{2}}{T_{2}} \\ L_{2} = \frac{p_{1} L_{1} T_{2}}{p_{2} T_{1}} \end{aligned}$

Find $L_{2}$ by substituting all known values.

$\begin{aligned} L_{2} = \frac{1 .013 脳 10^{5} 脳 2 .30 脳 280}{8 .38162 脳 10^{5} 脳 300} \\ = 0 .260 m \\ The column of the air at the bottom and the height of the water raised with in the diving bell is, the height of the water is calculated as \\ L_{1} 鈭� L_{2} = 2 .30 m 鈭� 0 .260 m \\ = 2 .04 m \end{aligned}$