Subjects

- Accounting
- Aerospace Engineering
- Anatomy
- Anthropology
- Arts & Humanities
- Astronomy
- Biology
- Business
- Chemistry
- Civil Engineering
- Computer Science
- Communications
- Economics
- Electrical Engineering
- English
- Finance
- Geography
- Geology
- Health Science
- History
- Industrial Engineering
- Information Systems
- Law
- Linguistics
- Management
- Marketing
- Material Science
- Mathematics
- Mechanical Engineering
- Medicine
- Nursing
- Philosophy
- Physics
- Political Science
- Psychology
- Religion
- Sociology
- Statistics

HomeStudy GuidesPhysics

Menu

The Nature of Science and Physics

Kinematics

Introduction to One-Dimensional KinematicsDisplacementVectors, Scalars, and Coordinate SystemsTime, Velocity, and SpeedVideo: One-Dimensional KinematicsAccelerationMotion Equations for Constant Acceleration in One DimensionProblem-Solving Basics for One-Dimensional KinematicsFalling ObjectsGraphical Analysis of One-Dimensional Motion

Two-Dimensional Kinematics

Dynamics: Force and Newton's Laws of Motion

Introduction to Dynamics: Newton's Laws of MotionDevelopment of Force ConceptNewton's First Law of Motion: InertiaNewton's Second Law of Motion: Concept of a SystemNewton's Third Law of Motion: Symmetry in ForcesVideo: Newton's LawsNormal, Tension, and Other Examples of ForcesProblem-Solving StrategiesFurther Applications of Newton's Laws of MotionExtended Topic: The Four Basic Forces—An Introduction

Further Applications of Newton's Laws: Friction, Drag, and Elasticity

Uniform Circular Motion and Gravitation

Introduction to Uniform Circular Motion and GravitationRotation Angle and Angular VelocityCentripetal AccelerationCentripetal ForceFictitious Forces and Non-inertial Frames: The Coriolis ForceNewton's Universal Law of GravitationVideo: GravitationSatellites and Kepler's Laws: An Argument for Simplicity

Work, Energy, and Energy Resources

Introduction to Work, Energy, and Energy ResourcesWork: The Scientific DefinitionKinetic Energy and the Work-Energy TheoremGravitational Potential EnergyVideo: Potential and Kinetic EnergyConservative Forces and Potential EnergyNonconservative ForcesConservation of EnergyPowerWork, Energy, and Power in HumansWorld Energy Use

Linear Momentum and Collisions

Rotational Motion and Angular Momentum

Introduction to Rotational Motion and Angular MomentumAngular AccelerationKinematics of Rotational MotionVideo: Rotational MotionDynamics of Rotational Motion: Rotational InertiaRotational Kinetic Energy: Work and Energy RevisitedAngular Momentum and Its ConservationVideo: Angular MomentumCollisions of Extended Bodies in Two DimensionsGyroscopic Effects: Vector Aspects of Angular Momentum

Statics and Torque

Fluid Statics

Introduction to Fluid StaticsWhat Is a Fluid?DensityPressureVariation of Pressure with Depth in a FluidPascal's PrincipleGauge Pressure, Absolute Pressure, and Pressure MeasurementArchimedes' PrincipleVideo: BuoyancyCohesion and Adhesion in Liquids: Surface Tension and Capillary ActionPressures in the Body

Fluid Dynamics and Its Biological and Medical Applications

Introduction to Fluid Dynamics and Biological and Medical ApplicationsFlow Rate and Its Relation to VelocityBernoulli's EquationVideo: Fluid FlowThe Most General Applications of Bernoulli's EquationViscosity and Laminar Flow; Poiseuille's LawThe Onset of TurbulenceMotion of an Object in a Viscous FluidMolecular Transport Phenomena: Diffusion, Osmosis, and Related Processes

Temperature, Kinetic Theory, and the Gas Laws

Heat and Heat Transfer Methods

Thermodynamics

Introduction to ThermodynamicsThe First Law of ThermodynamicsThe First Law of Thermodynamics and Some Simple ProcessesIntroduction to the Second Law of Thermodynamics: Heat Engines and Their EfficiencyCarnot's Perfect Heat Engine: The Second Law of Thermodynamics RestatedApplications of Thermodynamics: Heat Pumps and RefrigeratorsEntropy and the Second Law of Thermodynamics: Disorder and the Unavailability of EnergyStatistical Interpretation of Entropy and the Second Law of Thermodynamics: The Underlying Explanation

Oscillatory Motion and Waves

Introduction to Oscillatory Motion and WavesHooke's Law: Stress and Strain RevisitedPeriod and Frequency in OscillationsSimple Harmonic Motion: A Special Periodic MotionVideo: Harmonic MotionThe Simple PendulumEnergy and the Simple Harmonic OscillatorUniform Circular Motion and Simple Harmonic MotionDamped Harmonic MotionForced Oscillations and ResonanceWavesSuperposition and InterferenceEnergy in Waves: Intensity

Physics of Hearing

- Calculate flow rate.
- Define units of volume.
- Describe incompressible fluids.
- Explain the consequences of the equation of continuity.

$Q=\frac{V}{t}\\$

,
How many cubic meters of blood does the heart pump in a 75-year lifetime, assuming the average flow rate is 5.00 L/min?

**Strategy**

Time and flow rate *Q* are given, and so the volume *V* can be calculated from the definition of flow rate.

**Solution**

Solving *Q *= *V*/*t* for volume gives

**Discussion**

This amount is about 200,000 tons of blood. For comparison, this value is equivalent to about 200 times the volume of water contained in a 6-lane 50-m lap pool.

*V = Qt.*

$\begin{array}{lll}V& =& \left(\frac{5.00\text{ L}}{\text{1 min}}\right)\left(\text{75}\text{y}\right)\left(\frac{1{\text{ m}}^{3}}{{\text{10}}^{3}\text{ L}}\right)\left(5.26\times {\text{10}}^{5}\frac{\text{min}}{\text{y}}\right)\\ \text{}& =& 2.0\times {\text{10}}^{5}{\text{m}}^{3}\end{array}\\$

.$\bar{v}\\$

is$Q=A\overline{v}\\$

,$\bar{v}\\$

is the average velocity. This equation seems logical enough. The relationship tells us that flow rate is directly proportional to both the magnitude of the average velocity (hereafter referred to as the speed) and the size of a river, pipe, or other conduit. The larger the conduit, the greater its cross-sectional area. Figure 1 illustrates how this relationship is obtained. The shaded cylinder has a volume*V = Ad,*

$\frac{V}{t}=\frac{Ad}{t}\\$

.
We note that $\overline{v}=d/t\\$

. Thus the equation becomes $Q=A\overline{v}\\$

. Figure 2 shows an incompressible fluid flowing along a pipe of decreasing radius. Because the fluid is incompressible, the same amount of fluid must flow past any point in the tube in a given time to ensure continuity of flow. In this case, because the cross-sectional area of the pipe decreases, the velocity must necessarily increase. This logic can be extended to say that the flow rate must be the same at all points along the pipe. In particular, for points 1 and 2,$\begin{cases}Q_{1} &=& Q_{2}\\ A_{1}v_{1} &=&A_{2}v_{2} \end{cases}\\$

This is called the equation of continuity and is valid for any incompressible fluid. The consequences of the equation of continuity can be observed when water flows from a hose into a narrow spray nozzle: it emerges with a large speed—that is the purpose of the nozzle. Conversely, when a river empties into one end of a reservoir, the water slows considerably, perhaps picking up speed again when it leaves the other end of the reservoir. In other words, speed increases when cross-sectional area decreases, and speed decreases when cross-sectional area increases.
A nozzle with a radius of 0.250 cm is attached to a garden hose with a radius of 0.900 cm. The flow rate through hose and nozzle is 0.500 L/s. Calculate the speed of the water (a) in the hose and (b) in the nozzle.

**Strategy**

We can use the relationship between flow rate and speed to find both velocities. We will use the subscript 1 for the hose and 2 for the nozzle.

**Solution for (a)**

First, we solve *v*_{1} and note that the cross-sectional area is *A *= *πr*^{2}, yielding

Substituting known values and making appropriate unit conversions yields

**Solution for (b)**

We could repeat this calculation to find the speed in the nozzle

solving for *πr*^{2} for the cross-sectional area yields

Substituting known values,

**Discussion**

A speed of 1.96 m/s is about right for water emerging from a nozzleless hose. The nozzle produces a considerably faster stream merely by constricting the flow to a narrower tube.

$Q=A\overline{v}\\$

for ${\overline{v}}_{1}=\frac{Q}{{A}_{1}}=\frac{Q}{{{{\pi r}}_{1}}^{2}}\\$

.Substituting known values and making appropriate unit conversions yields

$\bar{v}_{1}=\frac{\left(0.500\text{ L/s}\right)\left(10^{-3}\text{ m}^{3}\text{L}\right)}{\pi \left(9.00\times 10^{-3}\text{ m}\right)^{2}}=1.96\text{ m/s}\\$

.$\bar{v}_{2}\\$

, but we will use the equation of continuity to give a somewhat different insight. Using the equation which states${A}_{1}{\overline{v}}_{1}={A}_{2}{\overline{v}}_{2}\\$

,${\overline{v}}_{2}\\$

and substituting $\overline{v}_{2}=\frac{{A}_{1}}{{A}_{2}}\bar{v}_{1}=\frac{{\pi r_{1}}^{2}}{{\pi r_{2}}^{2}}\bar{v}_{1}=\frac{{r_{1}}^{2}}{{r_{2}}^{2}}\bar{v}_{1}\\$

.$\overline{v}_{2}=\frac{\left(0.900\text{ cm}\right)^{2}}{\left(0.250\text{ cm}\right)^{2}}1.96\text{ m/s}=25.5 \text{ m/s}\\$

.
${n}_{1}{A}_{1}{\overline{v}}_{1}={n}_{2}{A}_{2}{\overline{v}}_{2}\\$

,
where
The aorta is the principal blood vessel through which blood leaves the heart in order to circulate around the body. (a) Calculate the average speed of the blood in the aorta if the flow rate is 5.0 L/min. The aorta has a radius of 10 mm. (b) Blood also flows through smaller blood vessels known as capillaries. When the rate of blood flow in the aorta is 5.0 L/min, the speed of blood in the capillaries is about 0.33 mm/s. Given that the average diameter of a capillary is 8.0 *μ*m, calculate the number of capillaries in the blood circulatory system.

**Strategy**

We can use

**Solution for (a)**

The flow rate is given by

**Solution for (b)**

Using *n*_{2} (the number of capillaries) gives

**Discussion**

Note that the speed of flow in the capillaries is considerably reduced relative to the speed in the aorta due to the significant increase in the total cross-sectional area at the capillaries. This low speed is to allow sufficient time for effective exchange to occur although it is equally important for the flow not to become stationary in order to avoid the possibility of clotting. Does this large number of capillaries in the body seem reasonable? In active muscle, one finds about 200 capillaries per mm^{3}, or about 200 × 10^{6} per 1 kg of muscle. For 20 kg of muscle, this amounts to about 4 × 10^{9} capillaries.

$Q=A\overline{v}\\$

to calculate the speed of flow in the aorta and then use the general form of the equation of continuity to calculate the number of capillaries as all of the other variables are known.$Q=A\overline{v}\\$

or $\overline{v}=\frac{Q}{{\pi r}^{2}}\\$

for a cylindrical vessel. Substituting the known values (converted to units of meters and seconds) gives$\overline{v}=\frac{\left(5.0\text{ L/min}\right)\left(10^{-3}{\text{ m}}^{3}\text{/L}\right)\left(1\text{ min/}60\text{s}\right)}{\pi {\left(0.010\text{ m}\right)}^{2}}=0.27\text{ m/s}\\$

.${n}_{1}{A}_{1}{\overline{v}}_{1}={n}_{2}{A}_{2}{\overline{v}}_{1}\\$

, assigning the subscript 1 to the aorta and 2 to the capillaries, and solving for ${n}_{2}=\frac{{n}_{1}{A}_{1}{\overline{v}}_{1}}{{A}_{2}{\overline{v}}_{2}}\\$

. Converting all quantities to units of meters and seconds and substituting into the equation above gives${n}_{2}=\frac{\left(1\right)\left(\pi \right){\left(\text{10}\times {\text{10}}^{-3}\text{m}\right)}^{2}\left(0.27 \text{ m/s}\right)}{\left(pi \right){\left(4.0\times {\text{10}}^{-6}\text{m}\right)}^{2}\left(0.33\times {\text{10}}^{-3}\text{m/s}\right)}=5.0\times {\text{10}}^{9}\text{capillaries}\\$

.- Flow rate
*Q*is defined to be the volume*V*flowing past a point in time*t*, or$Q=\frac{V}{t}\\$where*V*is volume and*t*is time. - The SI unit of volume is m
^{3}. - Another common unit is the liter (L), which is 10
^{-3}m^{3}. - Flow rate and velocity are related by $Q=A\overline{v}\\$where
*A*is the cross-sectional area of the flow and$\overline{v}\\$is its average velocity. - For incompressible fluids, flow rate at various points is constant. That is,

$\begin{cases}Q_{1} &=& Q_{2}\\ A_{1}v_{1} &=&A_{2}v_{2}\\ n_{1}A_{1}\bar{v}_{1} &=& n_{2}A_{2}\bar{v}_{2}\end{cases}\\$

.
1. What is the difference between flow rate and fluid velocity? How are they related?

2. Many figures in the text show streamlines. Explain why fluid velocity is greatest where streamlines are closest together. (Hint: Consider the relationship between fluid velocity and the cross-sectional area through which it flows.)

3. Identify some substances that are incompressible and some that are not.

1. What is the average flow rate in cm^{3}/s of gasoline to the engine of a car traveling at 100 km/h if it averages 10.0 km/L?

2. The heart of a resting adult pumps blood at a rate of 5.00 L/min. (a) Convert this to cm^{3}/s . (b) What is this rate in m^{3}/s ?

3. Blood is pumped from the heart at a rate of 5.0 L/min into the aorta (of radius 1.0 cm). Determine the speed of blood through the aorta.

4. Blood is flowing through an artery of radius 2 mm at a rate of 40 cm/s. Determine the flow rate and the volume that passes through the artery in a period of 30 s.

5. The Huka Falls on the Waikato River is one of New Zealand’s most visited natural tourist attractions (see Figure 3). On average the river has a flow rate of about 300,000 L/s. At the gorge, the river narrows to 20 m wide and averages 20 m deep. (a) What is the average speed of the river in the gorge? (b) What is the average speed of the water in the river downstream of the falls when it widens to 60 m and its depth increases to an average of 40 m?

6. A major artery with a cross-sectional area of 1.00 cm^{2} branches into 18 smaller arteries, each with an average cross-sectional area of 0.400 cm^{2}. By what factor is the average velocity of the blood reduced when it passes into these branches?

7. (a) As blood passes through the capillary bed in an organ, the capillaries join to form venules (small veins). If the blood speed increases by a factor of 4.00 and the total cross-sectional area of the venules is 10.0 cm^{2}, what is the total cross-sectional area of the capillaries feeding these venules? (b) How many capillaries are involved if their average diameter is 10.0 *μ*m?

8. The human circulation system has approximately 1 × 10^{9} capillary vessels. Each vessel has a diameter of about 8 *μ*m. Assuming cardiac output is 5 L/min, determine the average velocity of blood flow through each capillary vessel.

9. (a) Estimate the time it would take to fill a private swimming pool with a capacity of 80,000 L using a garden hose delivering 60 L/min. (b) How long would it take to fill if you could divert a moderate size river, flowing at 5000 m^{3}/s, into it?

10. The flow rate of blood through a 2.00 × 10^{-6}-radius capillary is 3.80 × 10^{9}. (a) What is the speed of the blood flow? (This small speed allows time for diffusion of materials to and from the blood.) (b) Assuming all the blood in the body passes through capillaries, how many of them must there be to carry a total flow of 90.0 cm^{3}/s? (The large number obtained is an overestimate, but it is still reasonable.)

11. (a) What is the fluid speed in a fire hose with a 9.00-cm diameter carrying 80.0 L of water per second? (b) What is the flow rate in cubic meters per second? (c) Would your answers be different if salt water replaced the fresh water in the fire hose?

12. The main uptake air duct of a forced air gas heater is 0.300 m in diameter. What is the average speed of air in the duct if it carries a volume equal to that of the house’s interior every 15 min? The inside volume of the house is equivalent to a rectangular solid 13.0 m wide by 20.0 m long by 2.75 m high.

13. Water is moving at a velocity of 2.00 m/s through a hose with an internal diameter of 1.60 cm. (a) What is the flow rate in liters per second? (b) The fluid velocity in this hose’s nozzle is 15.0 m/s. What is the nozzle’s inside diameter?

14. Prove that the speed of an incompressible fluid through a constriction, such as in a Venturi tube, increases by a factor equal to the square of the factor by which the diameter decreases. (The converse applies for flow out of a constriction into a larger-diameter region.)

15. Water emerges straight down from a faucet with a 1.80-cm diameter at a speed of 0.500 m/s. (Because of the construction of the faucet, there is no variation in speed across the stream.) (a) What is the flow rate in cm^{3}/s? (b) What is the diameter of the stream 0.200 m below the faucet? Neglect any effects due to surface tension.

16.** Unreasonable Results **A mountain stream is 10.0 m wide and averages 2.00 m in depth. During the spring runoff, the flow in the stream reaches 100,000 m^{3}/s. (a) What is the average velocity of the stream under these conditions? (b) What is unreasonable about this velocity? (c) What is unreasonable or inconsistent about the premises?

- flow rate:
- abbreviated
*Q*, it is the volume*V*that flows past a particular point during a time*t*, or*Q = V/t*

- liter:
- a unit of volume, equal to 10
^{−3}m^{3}

3. 27 cm/s

5. (a) 0.75 m/s (b) 0.13 m/s

7. (a) 40.0 cm

11. (a) 12.6 m/s (b) 0.0800 m

13. (a) 0.402 L/s (b) 0.584 cm

15. (a) 128 cm