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Question

The greatest ocean depths on Earth are found in the Marianas Trench near the Philippines. Calculate the pressure due to the ocean at the bottom of this trench, given its depth is 11.0 km and assuming the density of seawater is constant all the way down.

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From the definition of Absolute Pressure\textbf{Absolute Pressure} we know that :

The absolute pressure, or total pressure, is the sum of gauge pressure and atmospheric pressure

pabs=pg+patmsp_\text{abs} = p_\text{g} + p_\text{atms}

,

pabs=ρgh+patmsp_\text{abs} = \rho g h + p_\text{atms}

Where:

  • pabsp_\text{abs} is the absolute pressure .
  • patmsp_\text{atms} is the atmospheric pressure .
  • pgp_\text{g} is gauge pressure .
  • ρ\rho is the density of the fluid .
  • gg is the acceleration of the gravity .
  • hh is the height of fluid .

 Givens:\textbf{ Givens:} g=9.8m/s2g = 9.8 \mathrm{m/s^2} , h=11×103kmh = 11 \times 10^3 \text{km} , ρ=1030kg/m3\rho = 1030 \mathrm{kg/m^3}

Plugging\textbf{Plugging} known information to get :

pabs=ρgh+patms=1030×9.8×11×103+101325=1.1×109\begin{align*} p_\text{abs}& = \rho g h + p_\text{atms} \\ &= 1030 \times 9.8 \times 11 \times 10^3 + 101325 \\ &= 1.1 \times 10^9 \end{align*}

pabs=1.1×109pa\boxed{ p_\text{abs} = 1.1 \times 10^9 \text{pa} }

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