Kathleen Rausch

2021-12-19

The deepest point in the ocean is 11 km below sea level, deeper than MT. Everest is tall. What is the pressure in atmospheres at this depth?

vrangett

Beginner2021-12-20Added 36 answers

Density of sea water $p=1029k\frac{g}{{m}^{3}}$

Atmospheric pressure${P}_{0}=1.01\times {10}^{5}Pa$

Depth$h=11km=11\times {10}^{3}m$

Pressure of water at the deepest point

$P=pgh=1029\times 9.81\times 11\times {10}^{3}=1.11\times {10}^{8}Pa$

$\frac{P}{{P}_{0}}=1099$

Therefore net pressure$=P+{P}_{0}=1099+1atm=1100atm$

Atmospheric pressure

Depth

Pressure of water at the deepest point

Therefore net pressure

zesponderyd

Beginner2021-12-21Added 41 answers

The following provides the pressure equation:

$P={P}_{0}+pgd=1.013\times {10}^{5}+1030\left(9.8\right)\left(11000\right)=1.1\times {10}^{8}=\frac{1.1\times {10}^{8}}{1.013\times {10}^{5}}=1096atm$

Don Sumner

Skilled2021-12-27Added 184 answers

Concepts and reason

The concepts required to solve the given questions is the pressure in the atmosphere at the depth.

Initially, calculate the depth form km to m. Later, write an expression for the pressure at the depth. Finally, calculate the pressure in the atmosphere at this depth.

Fundamentals

The following is the term for the atmosphere at a certain depth:

Here, is the initial atmospheric pressure, D is the density of water, g is the acceleration due to gravity, and h is the deepest point in the ocean.

The depth in km is equal to,

$h=11\text{}km$

Then, change it to m as follows:

$h=11\text{}km(\frac{{10}^{3}\text{}m}{1\text{}km})$

$=11000\text{}m$

Solution:

The depth of the ocean is converted from km to m on multiplying by a factor of ${10}^{3}$

Substitute 1.00 atm for ${P}_{0}$, $1030\text{}kg/{m}^{3}$ for D, $9.8m/{s}^{2}$ for g, and 11000 m for h in the equation $P={P}_{0}+{D}_{gh}$.

$P=1.00\text{}atm+(1030\text{}kg/{m}^{3})(9.8\text{}m/{s}^{2})(11000\text{}m)$

$=1.00\text{}atm+1096.09\text{}atm$

$=1097.09\text{}atm$

The pressure in the atmosphere at the depth is equal to 1097.09 atm.

Explanation:

The density of the water is equal to $1000\text{}kg/{m}^{3}$. However, we must determine the depth of the ocean at that particular point. The ocean contains salt water so, we have to use density of the salt water.

xleb123

Skilled2023-06-19Added 181 answers

Andre BalkonE

Skilled2023-06-19Added 110 answers

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