/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 At the surface of a freshwater l... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

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 \(5.37\,\text{kPa}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A fish uses a swim bladder to change its density so it is equal to that of water, enabling it to remain suspended under water. If a fish has an average density of \(1080 \mathrm{kg} / \mathrm{m}^{3}\) and mass \(10.0 \mathrm{g}\) with the bladder completely deflated, to what volume must the fish inflate the swim bladder in order to remain suspended in seawater of density $1060 \mathrm{kg} / \mathrm{m}^{3} ?$
A container is filled with gas at a pressure of $4.0 \times 10^{5} \mathrm{Pa}\( The container is a cube, \)0.10 \mathrm{m}$ on a side, with one side facing south. What is the magnitude and direction of the force on the south side of the container due to the gas inside?
(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?
In a hydraulic lift, the radii of the pistons are \(2.50 \mathrm{cm}\) and $10.0 \mathrm{cm} .\( A car weighing \)W=10.0 \mathrm{kN}$ is to be lifted by the force of the large piston. (a) What force \(F_{\mathrm{a}}\) must be applied to the small piston? (b) When the small piston is pushed in by $10.0 \mathrm{cm},$ how far is the car lifted?(c) Find the mechanical advantage of the lift, which is the ratio \(W / F_{\mathrm{a}}\).
A hydraulic lift is lifting a car that weighs \(12 \mathrm{kN}\). The area of the piston supporting the car is \(A\), the area of the other piston is \(a,\) and the ratio \(A / a\) is \(100.0 .\) How far must the small piston be pushed down to raise the car a distance of \(1.0 \mathrm{cm} ?[\text {Hint}:\) Consider the work to be done.]
See all solutions

Recommended explanations on Physics Textbooks

Fluids

Read Explanation

Modern Physics

Read Explanation

Electricity and Magnetism

Read Explanation

Space Physics

Read Explanation

Engineering Physics

Read Explanation

Conservation of Energy and Momentum

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.