/*! 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 34 An office window has dimensions ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 34

An office window has dimensions \(3.4 \mathrm{~m}\) by \(2.1 \mathrm{~m}\). As a result of the passage of a storm, the outside air pressure drops to \(0.93 \mathrm{~atm}\), but inside the pressure is held at \(1.0 \mathrm{~atm}\). What net force pushes out on the window?

Short Answer

Expert verified
The net force pushing out on the window is approximately 50,546.47 N.

Step by step solution

01

Understand the problem

We need to find the net force exerted on the window due to the difference in atmospheric pressure outside and inside the office.
02

Identify given values

The problem states the dimensions of the window are \(3.4\) meters by \(2.1\) meters. The outside pressure is \(0.93 \, \text{atm}\) and the inside pressure is \(1.0 \, \text{atm}\).
03

Convert atmospheric pressure to Pascals

1 atmosphere (atm) is equivalent to \(101,325 \, \text{Pa}\). We convert the given pressures to Pascals:- Inside pressure: \(P_{\text{inside}} = 1.0 \, \text{atm} = 101,325 \, \text{Pa}\)- Outside pressure: \(P_{\text{outside}} = 0.93 \, \text{atm} = 0.93 \times 101,325 \, \text{Pa} = 94,233.25 \, \text{Pa}\)
04

Calculate the area of the window

The area of the window is the product of its height and width:\[A = 3.4 \, \text{m} \times 2.1 \, \text{m} = 7.14 \, \text{m}^2\]
05

Calculate the net force

The net force on the window is given by the difference in pressure multiplied by the area of the window:\[F = (P_{\text{inside}} - P_{\text{outside}}) \times A\]Substitute the known values:\[F = (101,325 \, \text{Pa} - 94,233.25 \, \text{Pa}) \times 7.14 \, \text{m}^2 = 50,546.4715 \, \text{N}\]
06

Provide the final answer

The net force pushing outward on the window is approximately \(50,546.47 \, \text{N}\).

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Atmospheric Pressure
When we talk about atmospheric pressure, we refer to the force exerted by the weight of the atmosphere. This force acts on every surface it touches. At sea level, this pressure averages around 101,325 Pascals (Pa), equivalent to 1 atmosphere (atm). Atmospheric pressure can significantly affect weather patterns, like in the scenario of a storm mentioned in the problem, causing pressure differences between external and internal environments. Such differences are essential in determining the net force on objects, like in our case with the office window.
Pressure Conversion
Pressure is often measured in different units, such as atmospheres (atm), Pascals (Pa), and others like bar or PSI. However, scientists commonly use Pascals in calculations due to its SI unit status.
  • 1 atm is equivalent to 101,325 Pa.
When solving problems, it's essential to convert all quantities to the same unit for consistency. In the problem at hand, the internal pressure is given as 1 atm and converted to Pascals as 101,325 Pa. Similarly, the external pressure is 0.93 atm, which converts to 94,233.25 Pa.
Area Calculation
Area calculation is a critical component in determining the force exerted on the window. For a rectangle such as our window, the area is found by multiplying its length by its width. In this case, the height is 3.4 meters, and the width is 2.1 meters.
The formula to find this area is: \[ A = ext{height} imes ext{width} \] This gives us: \[ A = 3.4 imes 2.1 = 7.14 ext{ m}^2 \] This area is used in calculating the net force by determining how much pressure acts over the entire surface of the window.
Force Calculation
In physics, force results from interactions between objects, and in this exercise, it arises due to pressure differences. The force calculation involves using the pressure difference and the already calculated area of the window.
The force exerted by a pressure difference can be calculated with: \[ F = (P_\text{inside} - P_\text{outside}) \times A \] Where:
  • \(P_\text{inside} = 101,325 \, \text{Pa}\)
  • \(P_\text{outside} = 94,233.25 \, \text{Pa}\)
  • \(A = 7.14 \, \text{m}^2\)
Substituting these values into the formula gives us: \[F = (101,325 - 94,233.25) \times 7.14 \] Which results in a force of approximately 50,546.47 Newtons acting outward on the window, illustrating how pressure differentials can lead to tangible forces on structures.

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

Blood pressure in Argentinosaurus. (a) If this long-necked, gigantic sauropod had a head height of \(18 \mathrm{~m}\) and a heart height of \(8.0 \mathrm{~m}\), what (hydrostatic) gauge pressure in its blood was required at the heart such that the blood pressure at the brain was 80 torr (just enough to perfuse the brain with blood)? Assume the blood had a density of \(1.06 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\). (b) What was the blood pressure (in torr or \(\mathrm{mm} \mathrm{Hg}\) ) at the feet?

A spring of spring constant \(3.75 \times 10^{4} \mathrm{~N} / \mathrm{m}\) is between a rigid beam and the output piston of a hydraulic lever. An empty container with negligible mass sits on the input piston. The input piston has area \(A_{i}\), and the output piston has area \(18.0 A_{2}\). Initially the spring is at its rest length. How many kilograms of sand must be (slowly) poured into the container to compress the spring by \(5.00 \mathrm{~cm}\) ?

Giraffe bending to drink. In a giraffe with its head \(1.8 \mathrm{~m}\) above its heart, and its heart \(2.0 \mathrm{~m}\) above its feet, the (hydrostatic) gauge pressure in the blood at its heart is 250 torr. Assume that the giraffe stands upright and the blood density is \(1.06 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\).In torr (or \(\mathrm{mm} \mathrm{Hg}\) ), find the (gauge) blood pressure (a) at the brain (the pressure is enough to perfuse the brain with blood, to keep the giraffe from fainting) and (b) at the feet (the pressure must be countered by tight-fitting skin acting like a pressure stocking). (c) If the giraffe were to lower its head to drink from a pond without splaying its legs and moving slowly, what would be the increase in the blood pressure in the brain? (Such action would probably be lethal.)

Two identical cylindrical vessels with their bases at the same level each contain a liquid of density \(1.30 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\). The area of each base is \(4.25 \mathrm{~cm}^{2}\), but in one vessel the liquid height is \(0.854 \mathrm{~m}\) and in the other it is \(1.560 \mathrm{~m}\). Find the work done by the gravitational force in equalizing the levels when the two vessels are connected.

A hollow spherical iron shell floats almost completely submerged in water. The outer diameter is \(50.0 \mathrm{~cm}\), and the density of iron is \(7.87 \mathrm{~g} / \mathrm{cm}^{3}\). Find the inner diameter.
See all solutions

Recommended explanations on Physics Textbooks

Physical Quantities And Units

Read Explanation

Energy Physics

Read Explanation

Oscillations

Read Explanation

Circular Motion and Gravitation

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Translational Dynamics

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.