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Chapter 9: Problem 1

Someone steps on your toe, exerting a force of \(500 \mathrm{N}\) on an area of \(1.0 \mathrm{cm}^{2} .\) What is the average pressure on that area in atm?

Short Answer

Expert verified
Express your answer in atmospheres (atm). Answer: The average pressure on the toe area is about 49.3 atm.

Step by step solution

01

Convert area to SI units (square meters, m²)

We need all values in SI units to perform calculations. Given area is in \(\mathrm{cm}^2\), let's convert it to \(\mathrm{m}^2\). We know that 1m = 100cm. Area in \(\mathrm{m}^2\) = Area in \(\mathrm{cm}^2\) × \((\frac{0.01\,\mathrm{m}}{1\,\mathrm{cm}})^2\) Area in \(\mathrm{m}^2\) = \(1.0\,\mathrm{cm}^2\) × \((\frac{0.01\,\mathrm{m}}{1\,\mathrm{cm}})^2\) = \(1.0 \times 10^{-4} \,\mathrm{m}^2\)
02

Calculate the pressure in Pascals (Pa)

Pressure (P) = Force (F) / Area (A) P = \(\frac{500\,\mathrm{N}}{1.0 \times 10^{-4}\,\mathrm{m}^2}\) = \(5000000\,\mathrm{Pa}\) or \(5 \times 10^6\,\mathrm{Pa}\)
03

Convert pressure from Pascals to atmospheres (atm)

To convert pressure from Pascals to atmospheres, we will use the following conversion factor: 1 atm = 101325 Pa. Pressure in atm = Pressure in Pa / 101325 Pressure in atm = \(\frac{5\times10^{6}\,\mathrm{Pa}}{101325\,\mathrm{Pa/atm}} \approx 49.3\,\mathrm{atm}\) So, the average pressure on that area is about 49.3 atm.

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Most popular questions from this chapter

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.]
What keeps a cloud from falling? A cumulus (fair-weather) cloud consists of tiny water droplets of average radius \(5.0 \mu \mathrm{m} .\) Find the terminal velocity for these droplets at \(20^{\circ} \mathrm{C},\) assuming viscous drag. (Besides the viscous drag force, there are also upward air currents called thermals that push the droplets upward. (tutorial: rain drop)
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A viscous liquid is flowing steadily through a pipe of diameter \(D .\) Suppose you replace it by two parallel pipes, each of diameter \(D / 2,\) but the same length as the original pipe. If the pressure difference between the ends of these two pipes is the same as for the original pipe, what is the total rate of flow in the two pipes compared to the original flow rate?
A cube that is \(4.00 \mathrm{cm}\) on a side and of density $8.00 \times 10^{2} \mathrm{kg} / \mathrm{m}^{3}$ is attached to one end of a spring. The other end of the spring is attached to the base of a beaker. When the beaker is filled with water until the entire cube is submerged, the spring is stretched by \(1.00 \mathrm{cm} .\) What is the spring constant?
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