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Chapter 15: Problem 8

Find the total pressure, in pascal, \(118 \mathrm{~m}\) below the surface of the ocean. The density of seawater is \(1.024 \mathrm{~g} / \mathrm{cm}^{3}\) and the atmospheric pressure at sea level is \(1.013 \times 10^{5} \mathrm{~Pa}\).

Short Answer

Expert verified
The total pressure 118 m below the surface of the ocean is \(1287518.4 Pa\).

Step by step solution

01

Converting the units

We first need to convert the density of seawater from \(g / cm^3\) to \(kg / m^3\) as the other units in the formula are in the metric system. Since \(1g = 0.001kg\) and \(1cm^3 = 0.000001m^3\), the conversion gives \(\rho = 1.024 * 0.001 / 0.000001 = 1024 kg/m^3\).
02

Finding the pressure due to water

Next, we can find the pressure at the given depth due to the seawater using the formula \( P = \rho g h \). Let's use \(\rho = 1024 kg/m^3\), \( g = 9.8 m/s^2 \) and \( h = 118 m \). This gives the pressure as \( P = 1024 * 9.8 * 118 = 1184518.4 Pa \).
03

Finding the total pressure

The total pressure at the given depth in the ocean is the sum of the pressure due to the water and the atmospheric pressure at sea level. Given that \( P_0 = 1.013 * 10^{5} Pa \), the total pressure is \( P = 1184518.4 + 1.013 * 10^{5} = 1287518.4 Pa \).

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