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Expert-verified Found in: Page 443 ### Physics For Scientists & Engineers

Book edition 9th Edition
Author(s) Raymond A. Serway, John W. Jewett
Pages 1624 pages
ISBN 9781133947271 # (a.) Calculate the absolute pressure at an ocean depth of 1 000 m. Assume the density of seawater is ${\mathbf{1030}}{\mathbf{}}{\mathbf{}}{\mathbit{k}}{\mathbit{g}}{\mathbf{/}}{{\mathbit{m}}}^{{\mathbf{3}}}$and the air above exerts a pressure of ${\mathbf{101}}{\mathbf{.}}{\mathbf{3}}{\mathbf{}}{\mathbf{}}{\mathbit{k}}{\mathbit{p}}{\mathbit{a}}$. (b.) At this depth, what is the buoyant force on a spherical submarine having a diameter of 5.00 m?

(a) The absolute pressure at an ocean depth is $P=1.00pa$

(b) The buoyant force on a spherical submarine is $B\simeq 19.5×{10}^{9}N$

See the step by step solution

## Step 1:

The pressure in a fluid at rest varies with depth h in the fluid according to the expression:

$P={P}_{0}+\rho gh$

Where, ${P}_{0}$ is the pressure at $h=0$ and $\rho$ the density of the fluid is assumed uniform. Pascal’s law states that when pressure is applied to an enclosed fluid, the pressure is transmitted undiminished to every point in the fluid and to every point on the walls of the container.

When an object is partially or fully submerged in a fluid, the fluid exerts on the object an upward force called the buoyant force. According to Archimedes’s principle, the magnitude of the buoyant force is equal to the weight of the fluid displaced by the object:

$B={\rho }_{fluid}g{V}_{fluid}$

## Step 2:

Given:

Part (a):

Depth of ocean $h=1000m$

Density of water $\rho =1030kg/{m}^{3}$

Pressure of air ${P}_{0}=101.3kpa$

By using concept and the formula from step (1), we get

$P={P}_{0}+\rho gh\phantom{\rule{0ex}{0ex}}P=101.3×{10}^{3}+1000×1030×9.8\phantom{\rule{0ex}{0ex}}P=1.00pa\phantom{\rule{0ex}{0ex}}$

## Step 3:

Given:

Part (b):

Depth of ocean $h=1000m$

Density of water $\rho =1030kg/{m}^{3}$

Diameter of submarine $D=5m$

Radius of submarine $r=\frac{D}{2}=2.5m$

By using concept and the formula from step (1), we get

$B={\rho }_{fluid}g{V}_{fluid}\phantom{\rule{0ex}{0ex}}B={\rho }_{fluid}g×Ah\phantom{\rule{0ex}{0ex}}B={\rho }_{fluid}g×\pi ×{r}^{2}×h\phantom{\rule{0ex}{0ex}}B=101.3×{10}^{3}×9.8×3.14×{\left(2.5\right)}^{2}×1000\phantom{\rule{0ex}{0ex}}B\simeq 19.5×{10}^{9}N\phantom{\rule{0ex}{0ex}}$ ### Want to see more solutions like these? 