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Chapter 9: Problem 19

At the surface of a freshwater lake the pressure is \(105 \mathrm{kPa} .\) (a) What is the pressure increase in going \(35.0 \mathrm{m}\) below the surface? (b) What is the approximate pressure decrease in going \(35 \mathrm{m}\) above the surface? Air at \(20^{\circ} \mathrm{C}\) has density of $1.20 \mathrm{kg} / \mathrm{m}^{3} .$

Short Answer

Expert verified
Answer: The pressure increase in going 35.0 m below the surface of a freshwater lake is approximately 343.35 kPa, while the pressure decrease in going 35.0 m above the surface is approximately 5.37 kPa.

Step by step solution

01

Identify known values

The known values in this problem are: - Surface pressure \(P_0 = 105\,\text{kPa}\) - Depth \(h = 35.0\,\text{m}\) - Water density \(\rho_w = 1000\,\text{kg/m}^3\) (freshwater) - Gravitational acceleration \(g = 9.81\,\text{m/s}^2\)
02

Calculate hydrostatic pressure

To find the pressure increase, we will use the hydrostatic pressure formula: \(\Delta P = \rho_w gh\) Plug in the known values: \(\Delta P = (1000\,\text{kg/m}^3)(9.81\,\text{m/s}^2)(35.0\,\text{m})\)
03

Calculate the pressure increase

Solve for \(\Delta P\): \(\Delta P = 343,350\,\text{Pa} = 343.35\,\text{kPa}\) The pressure increase in going \(35.0\,\text{m}\) below the surface is \(343.35\,\text{kPa}\). #b) Pressure decrease above the surface#
04

Identify known values

The known values in this problem are: - Surface pressure \(P_0 = 105\,\text{kPa}\) - Altitude \(h = 35.0\,\text{m}\) - Air density \(\rho_a = 1.20\,\text{kg/m}^3\) at \(20^{\circ} \mathrm{C}\) - Gravitational acceleration \(g = 9.81\,\text{m/s}^2\) - Gas constant for air \(R_a = 287\,\text{J/kg}\cdot\text{K}\) - Temperature \(T = 20 + 273.15 = 293.15\,\text{K}\) (convert from Celsius to Kelvin)
05

Calculate the pressure decrease

To find the pressure decrease, we will use the barometric pressure formula: \(\Delta P = P_0[1 - (\frac{h}{h_0})]^{(gM/RT)}\) , where \(h_0 = \frac{RT}{Mg}\) First, calculate \(h_0\): \(h_0 = \frac{(287\,\text{J/kg}\cdot\text{K})(293.15\text{K})}{(1.20\,\text{kg/m}^3)(9.81\,\text{m/s}^2)}\)
06

Calculate the pressure decrease

Solve for \(\Delta P\): \(\Delta P = 105\,\text{kPa}\left[1 - \left(\frac{35.0\,\text{m}}{6787.4\,\text{m}}\right)^{5.26}\right]\) \(\Delta P \approx 5.37\,\text{kPa}\) The approximate pressure decrease in going \(35\,\text{m}\) above the surface is \(5.37\,\text{kPa}\).

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Most popular questions from this chapter

A dinoflagellate takes 5.0 s to travel 1.0 mm. Approximate a dinoflagellate as a sphere of radius \(35.0 \mu \mathrm{m}\) (ignoring the flagellum). (a) What is the drag force on the dinoflagellate in seawater of viscosity $0.0010 \mathrm{Pa} \cdot \mathrm{s} ?$ (b) What is the power output of the flagellate?
This table gives the terminal speeds of various spheres falling through the same fluid. The spheres all have the same radius. $$\begin{array}{llllllll}\hline m= & 5.0 & 11.3 & 20.0 & 31.3 & 45.0 & 80.0 & (\mathrm{g}) \\\ \hline v_{1}= & 1.0 & 1.5 & 2.0 & 2.5 & 3.0 & 4.0 & (\mathrm{cm} / \mathrm{s}) \\\\\hline\end{array}$$, Is the drag force primarily viscous or turbulent? Explain your reasoning.
(a) When ice floats in water at \(0^{\circ} \mathrm{C},\) what percent of its volume is submerged? (b) What is the specific gravity of ice?
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(a) What is the pressure difference required to make blood flow through an artery of inner radius \(2.0 \mathrm{mm}\) and length \(0.20 \mathrm{m}\) at a speed of \(6.0 \mathrm{cm} / \mathrm{s} ?\) (b) What is the pressure difference required to make blood flow at \(0.60 \mathrm{mm} / \mathrm{s}\) through a capillary of radius \(3.0 \mu \mathrm{m}\) and length \(1.0 \mathrm{mm} ?\) (c) Compare both answers to your average blood pressure, about 100 torr.
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