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Chapter 14: Qn :1 (page 408)

Question: To suck lemonade of density 1000 kg / m3up a straw to a maximum height of 4.0 cm , what minimum gauge pressure (in atmospheres) must you produce in your lungs?

Short Answer

Expert verified

Answer:

Gauge pressure produced in the lungs in atmospheric unit is-3.9×10-3atm

Step by step solution

01

The given data 

Maximum height,h=4cmor0.04m

Density of lemonade, ρ=1000kg/m3

Acceleration due to gravity,g=9.8m/s2g=9.8m/s2

02

Understanding the concept of pressure 

To find gauge pressure in theatm, we can use the formula for pressure in terms of density, acceleration due to gravity, and height. Using the conversion factor, we can convert the pressure into atmospheric pressure.

Formula:

Pressure applied on a body, p=ρgh (i)

03

Calculation of gauge pressure in atm 

The pressure produced in the lungs in SI unit using equation (i) can be given as follows:

p=1000kg/m3×0.04m×9.8m/s2=-392pa=-392pa×1atm1.01×105pa(1atm=1.01×105pa)=-3.9×10-3atm

So, the pressure value is-3.9×10-3atm

Negative sign indicates that the pressure inside the lung is less than the outside pressure.

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Most popular questions from this chapter

A 8.60kgsphere of radius 6.22cmis at a depth of 2.22kmin seawater that has an average density of 1025kg/m3.What are the (a) gauge pressure, (b) total pressure, and (c) corresponding total force compressing the sphere’s surface? What are (d) the magnitude of the buoyant force on the sphere and (e) the magnitude of the sphere’s acceleration if it is free to move? Take atmospheric pressure to be 1.01×105Pa.

Figure shows two sections of an old pipe system that runs through a hill, with distances dA=dB=30mand D=110 m. On each side of the hill, the pipe radius is 2.00 cm. However, the radius of the pipe inside the hill is no longer known. To determine it, hydraulic engineers first establish that water flows through the left and right sections at 2.50 m/s. Then they release a dye in the water at point A and find that it takes 88.8 sto reach point B.

What is the average radius of the pipe within the hill?

Water is moving with a speed of 5.0msthrough a pipe with a cross-sectional area of 4.0cm2. The water gradually descends as the pipe cross-sectional area increases to 8.0cm2.

(a) What is the speed at the lower level?

(b) If the pressure at the upper level is 1.5×105Pa, What is the pressure at the lower level?

When a person snorkels, the lungs are connected directly to the atmosphere through the snorkel tube and thus are at atmospheric pressure 1) In atmospheres, what is the difference∆pbetween this internal air pressure and the water pressure against the body if the length of the snorkel tube is 20 cm (standard situation) &2) In atmospheres, what is the difference֏pbetween this internal air pressure and the water pressure against the body if the length of the snorkel tube is 4.0m(probably lethal situation)? In the latter, the pressure difference causes blood vessels on the walls of the lungs to rupture, releasing blood into the lungs. As depicted in Figure, an elephant can safely snorkel through its trunk while swimming with its lungsbelow the water surface because the membrane around its lungs contains connective tissue that holds and protects the blood vessels, preventing rupturing.

Fresh water flows horizontally from pipe section 1 of cross-sectional area A1into pipe section 2 of cross-sectional area A2. Figure gives a plot of the pressure difference p2-p1versus the inverse area squared A,-2that would be expected for a volume flow rate of a certain value if the water flow were laminar under all circumstances. The scale on the vertical axis is set by ps=300kN/m2. For the conditions of the figure,

(a) what is the value of A2and

(b) what is the value of the volume flow rate?
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