/*! 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 26 To suck lemonade of density \(10... [FREE SOLUTION] | 91Ó°ÊÓ

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To suck lemonade of density \(1000 \mathrm{~kg} / \mathrm{m}^{3}\) up a straw to a maximum height of \(4.0 \mathrm{~cm}\), what minimum gauge pressure (in atmospheres) must you produce in your lungs?

Short Answer

Expert verified
The minimum gauge pressure is approximately 0.00387 atm.

Step by step solution

01

Understand the Problem

To find the minimum gauge pressure needed to suck lemonade up to a height of 4.0 cm using a straw, we must consider the relationship between pressure, fluid density, gravity, and height.
02

Use the Pressure Formula

The pressure required to lift a liquid to a certain height is given by the formula: \[ P = \rho g h \]where \( P \) is the pressure, \( \rho \) is the density of the fluid (1000 \( \mathrm{kg/m^3} \) for lemonade), \( g \) is the acceleration due to gravity (approximately 9.81 \( \mathrm{m/s^2} \)), and \( h \) is the height (0.04 m).
03

Calculate the Pressure in Pascals

Substitute the known values into the pressure formula:\[ P = 1000 \, \mathrm{kg/m^3} \times 9.81 \, \mathrm{m/s^2} \times 0.04 \, \mathrm{m} \]\[ P = 392.4 \, \mathrm{Pa} \]
04

Convert Pascals to Atmospheres

To convert the pressure from Pascals to atmospheres, use the conversion factor \( 1 \, \mathrm{atm} = 101325 \, \mathrm{Pa} \):\[ P = \frac{392.4 \, \mathrm{Pa}}{101325 \, \mathrm{Pa/atm}} \approx 0.00387 \, \mathrm{atm} \]
05

Determine Minimum Gauge Pressure

Gauge pressure is the pressure relative to atmospheric pressure. Since this calculation already considers the pressure required to move the lemonade up the straw, 0.00387 atm is the minimum gauge pressure.

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Key Concepts

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

Pressure Calculation
Pressure in fluid mechanics refers to the force exerted per unit area within the fluid. It's crucial for understanding how fluids behave in various conditions. When dealing with a fluid in a vertical tube, such as lemonade in a straw, there's a pressure difference due to gravity. This difference is crucial for moving the fluid to a specific height. To calculate this pressure, the formula \[ P = \rho g h \]allows us to determine how much force per area is needed. Here,
  • \(P\) is pressure in pascals (Pa).
  • \(\rho\) is the fluid's density in kilograms per cubic meter \(\mathrm{kg/m^3}\).
  • \(g\) is the acceleration due to gravity, typically \(9.81 \mathrm{m/s^2}\).
  • \(h\) is the height the fluid must be lifted, in meters (\(\mathrm{m}\)).
Knowing how to calculate pressure helps predict how much force is needed to move fluids in various applications.
Gauge Pressure
Gauge pressure measures the pressure relative to atmospheric pressure. This is different from absolute pressure, which includes atmospheric pressure in its measurement. When sucking lemonade through a straw, gauge pressure tells us how much pressure is needed, excluding what's already exerted by the atmosphere.
For many practical applications, gauge pressure is more useful than absolute pressure, as it gives a clearer picture of the force needed for specific tasks, such as moving a fluid. In fluid mechanics, calculating the minimum gauge pressure helps in determining the least amount of additional force needed to lift a liquid to a required height, using the suction power of one's lungs or a pump.
Fluid Density
Fluid density, denoted \(\rho\), is a measure of how much mass a fluid has per unit volume. The density of a fluid influences how it behaves under pressure. For lemonade, the density was given as \(1000 \mathrm{~kg/m^3}\), similar to water.
Fluid density plays an essential role in calculating pressure, as a denser fluid will require more force to move. The formula \[P = \rho g h\] incorporates density to help determine the pressure needed to lift the fluid. Understanding a fluid's density is key in predicting how it'll react in a scenario where it's moved or subjected to forces, such as when using a straw to sip a beverage.
Pressure Conversion
Pressure conversion is important when comparing or utilizing different units of measurement. In the step-by-step solution, the pressure was initially calculated in pascals, a standard unit for pressure. To convert this to atmospheres, which is often used in everyday settings, we use the conversion factor \[1 \, \mathrm{atm} = 101325 \, \mathrm{Pa}.\] Applying this allows us to convert \[392.4 \, \mathrm{Pa}\] to approximately \[0.00387 \, \mathrm{atm}.\] This process of conversion ensures fluid mechanics calculations are consistent, and understandable across different contexts, making theoretical discussions and practical applications easier to follow.

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Most popular questions from this chapter

Three liquids that will not mix are poured into a cylindrical container. The volumes and densities of the liquids are \(0.50 \mathrm{~L}, 2.6 \mathrm{~g} / \mathrm{cm}^{3}\); \(0.25 \mathrm{~L}, 1.0 \mathrm{~g} / \mathrm{cm}^{3} ;\) and \(0.40 \mathrm{~L}, 0.80 \mathrm{~g} / \mathrm{cm}^{3} .\) What is the force on the bot- tom of the container due to these liquids? One liter \(=1 \mathrm{~L}=1000\) \(\mathrm{cm}^{3}\). (Ignore the contribution due to the atmosphere.)

Models of torpedoes are sometimes tested in a horizontal pipe of flowing water, much as a wind tunnel is used to test model airplanes. Consider a circular pipe of internal diameter \(25.0 \mathrm{~cm}\) and a torpedo model aligned along the long axis of the pipe. The model has a \(5.00 \mathrm{~cm}\) diameter and is to be tested with water flowing past it at \(2.50 \mathrm{~m} / \mathrm{s}\). (a) With what speed must the water flow in the part of the pipe that is unconstricted by the model? (b) What will the pressure difference be between the constricted and unconstricted parts of the pipe?

About one-third of the body of a person floating in the Dead Sea will be above the waterline. Assuming that the human body density is \(0.98 \mathrm{~g} / \mathrm{cm}^{3}\), find the density of the water in the Dead Sea. (Why is it so much greater than \(1.0 \mathrm{~g} / \mathrm{cm}^{3} ?\) )

A large aquarium of height \(5.00\) \(\mathrm{m}\) is filled with fresh water to a depth of \(2.00\) \(\mathrm{m}\). One wall of the aquarium consists of thick plastic \(8.00 \mathrm{~m}\) wide. By how much does the total force on that wall increase if the aquarium is next filled to a depth of \(4.00 \mathrm{~m}\) ?

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 \(0.96\) atm, but inside the pressure is held at \(1.0 \mathrm{~atm} .\) What net force pushes out on the window?

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