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Ch11. Fluids Mass Density Fluids are materials that can flow, and they include both gases and liquids. DEFINITION OF MASS DENSITY The mass density is the mass m of a substance divided by its volume V: SI Unit of Mass Density: kg/m3

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**Mass Densities of Common Substances**

(Unit: kg/m3) Solids Aluminum 2 700 Brass 8 470 Concrete 2 200 Copper 8 890 Diamond 3 520 Gold 19 300 Ice Iron (steel) 7 860 Lead 11300 Quartz 2 660 Silver 10 500 Wood (yellow pine) 550

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**Gases have the smallest densities.**

Air 1.29 Carbon dioxide 1.98 Helium 0.179 Hydrogen 0.0899 Nitrogen 1.25 Oxygen 1.43 Liquids Blood (whole, 37°C) 1060 Ethyl alcohol 806 Mercury 13600 Oil (hydraulic) 800 Water (4 °C) 1 × 103 Gases have the smallest densities.

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**Example 1. Blood as a Fraction of Body Weight**

The body of a man whose weight is 690 N contains about 5.2 × 10–3 m3 (5.5 qt) of blood. (a) Find the blood’s weight and (b) express it as a percentage of the body weight. (11.1) (a) (b)

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**A convenient way to compare densities: specific gravity**

Specific gravity has no units.

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Pressure The SI unit for pressure is a newton/meter2 (N/m2), a combination that is referred to as a pascal (Pa). 105 Pa = 1 bar Another unit for pressure is pounds per square inch (lb/in.2), often abbreviated as “psi.”

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**In general, a static fluid cannot produce a force parallel to a surface.**

Pressure has no directional characteristic. The force generated by the pressure of a static fluid is always perpendicular to the surface that the fluid contacts.

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**Example 2. The Force on a Swimmer**

Suppose the pressure acting on the back of a swimmer’s hand is 1.2 × 105 Pa, a realistic value near the bottom of the diving end of a pool. The surface area of the back of the hand is 8.4 × 10–3 m2 (a) Determine the magnitude of the force that acts on it. (b) Discuss the direction of the force. (a) (b) The hand (palm downward) is oriented parallel to the bottom of the pool. Since the water pushes perpendicularly against the back of the hand, the force F is directed downward in the drawing. This downward-acting force is balanced by an upward-acting force on the palm, so that the hand is in equilibrium. If the hand were rotated by 90°, the directions of these forces would also be rotated by 90°, always being perpendicular to the hand.

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**Pressure and Depth in a Static Fluid**

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**Conceptual Example 3. The Hoover Dam**

Lake Mead is the largest wholly artificial reservoir in the United States and was formed after the completion of the Hoover Dam in The water in the reservoir backs up behind the dam for a considerable distance (about 200 km or 120 miles). Suppose that all the water were removed, except for a relatively narrow vertical column in contact with the dam.

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A side view of this hypothetical situation, in which the water against the dam has the same depth as in Figure part a. Would the Hoover Dam still be needed to contain the water in this hypothetical reservoir, or could a much less massive structure do the job? The dam for our imaginary reservoir would sustain the same forces that the Hoover Dam sustains and would need to be equally large.

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**Example 4. The Swimming Hole**

The cross section of a swimming hole. Points A and B are both located at a distance of h = 5.50 m below the surface of the water. Find the pressure at each of these two points.

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Since points A, B, C, and D are at the same distance h beneath the liquid surface, the pressure at each of them is the same.

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**Example 5. Blood Pressure**

Blood in the arteries is flowing, but as a first approximation, the effects of this flow can be ignored and the blood can be treated as a static fluid. Estimate the amount by which the blood pressure P2 in the anterior tibial artery at the foot exceeds the blood pressure P1 in the aorta at the heart when a person is (a) reclining horizontally and (b) standing.

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(a) h = 0 m, (b)

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**Conceptual Example 6. Pumping Water**

Two methods for pumping water from a well. In one method, the pump is submerged in the water at the bottom of the well, while in the other, it is located at ground level. If the well is shallow, either technique can be used. However, if the well is very deep, only one of the methods works. Which is it? A ground-level (on top of the ground) pump can only cause water to rise to a certain maximum height, so it cannot be used for very deep wells.

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**Check Your Understanding 1**

A scuba diver is swimming under water, and the graph shows a plot of the water pressure acting on the diver as a function of time. In each of the three regions, A B, B C, and C D, does the depth of the diver increase, decrease or remain constant? A B: Increase B C: Decrease C D: Remain constant

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Pressure Gauges The atmospheric pressure can be determined from the height h of the mercury in the tube, the density of mercury, and the acceleration due to gravity. PA = PB (PB is Atmospheric pressure.)

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If P1=0, then If liquid is mercury, (760mm) What if water

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The height h is proportional to P2 – Patm, which is called the gauge pressure. The gauge pressure is the amount by which the container pressure differs from atmospheric pressure. The actual value for P2 is called the absolute pressure.

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**A sphygmomanometer is used to measure blood pressure.**

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**Pascal's Principle PASCAL’S PRINCIPLE**

Any change in the pressure applied to a completely enclosed fluid is transmitted undiminished to all parts of the fluid and the enclosing walls.

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Example 7. A Car Lift In a hydraulic car lift, the input piston has a radius of r1 = m and a negligible weight. The output plunger has a radius of r2 = m. The combined weight of the car and the plunger is F2 = N. The lift uses hydraulic oil that has a density of 8.00 × 102 kg/m3. What input force F1 is needed to support the car and the output plunger when the bottom surfaces of the piston and plunger are at (a) the same level and (b) the levels with h = 1.10 m? (a)

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(b) P2 = P gh P2 =F2/(pr22) and P1 =F1/(pr12):

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Archimedes' Principle Buoyant force exists because fluid pressure is larger at greater depths.

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**FB = Wfluid ARCHIMEDES’ PRINCIPLE**

Any fluid applies a buoyant force to an object that is partially or completely immersed in it; the magnitude of the buoyant force equals the weight of the fluid that the object displaces: FB = Wfluid Magnitude of buoyant force Weight of displaced fluid

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The effect that the buoyant force has depends on its strength compared with the strengths of the other forces that are acting.

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**Example 8. A Swimming Raft**

A solid, square pinewood raft measures 4.0 m on a side and is 0.30 m thick. (a) Determine whether the raft floats in water, and (b) if so, how much of the raft is beneath the surface (see the distance h in the figure).

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(a) pine = 550 kg/m3 , Vpine = 4.0 m × 4.0 m × 0.30 m = 4.8 m3

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**FB = 26 000 N = Wfluid , Vwater = 4.0 m × 4.0 m × h**

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**Conceptual Example 9. How Much Water Is Needed to Float a Ship?**

A ship floating in the ocean is a familiar sight. But is all that water really necessary? Can an ocean vessel float in the amount of water that a swimming pool contains, for instance? All that is needed, in principle, is a thin section of water that separates the hull of the floating ship from the sides of the canal.

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**A useful application of Archimedes’ principle can be found in car batteries.**

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**Example 10. A Goodyear Airship**

Normally, a Goodyear airship, contains about 5.40 × 103 m3 of helium (He) whose density is kg/m3. Find the weight of the load WL that the airship can carry in equilibrium at an altitude where the density of air is 1.20 kg/m3.

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**Check Your Understanding 2**

A glass is filled to the brim with water and has an ice cube floating in it. When the ice cube melts, what happens? a. Water spills out of the glass. b. The water level in the glass drops. c. The water level in the glass does not change. (c)

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Fluids in Motion In steady flow the velocity of the fluid particles at any point is constant as time passes. Unsteady flow exists whenever the velocity at a point in the fluid changes as time passes. Turbulent flow is an extreme kind of unsteady flow and occurs when there are sharp obstacles or bends in the path of a fast-moving fluid.

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**Fluid flow can be compressible or incompressible, viscous or nonviscous.**

An incompressible, nonviscous fluid is called an ideal fluid. A streamline is a line drawn in the fluid such that a tangent to the streamline at any point is parallel to the fluid velocity at that point.

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**Steady flow is often called streamline flow.**

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**The Equation of Continuity**

Equation of continuity: If a fluid enters one end of a pipe at a certain rate (e.g., 5 kilograms per second), then fluid must also leave at the same rate, assuming that there are no places between the entry and exit points to add or remove fluid. The mass of fluid per second (e.g., 5 kg/s) that flows through a tube is called the mass flow rate.

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**Mass flow rate at position 2**

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**EQUATION OF CONTINUITY**

The mass flow rate ( Av) has the same value at every position along a tube that has a single entry and a single exit point for fluid flow. For two positions along such a tube = fluid density (kg/m3) A = cross-sectional area of tube (m2) v = fluid speed (m2) SI Unit of Mass Flow Rate: kg/s

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**Incompressible fluid A1v1=A2v2**

Q = Volume flow rate = Av

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Example 11. A Garden Hose A garden hose has an unobstructed opening with a cross-sectional area of 2.85 × 10–4 m2, from which water fills a bucket in 30.0 s. The volume of the bucket is 8.00 × 10–3 m3 (about two gallons). Find the speed of the water that leaves the hose through (a) the unobstructed opening and (b) an obstructed opening with half as much area.

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**Example 12. A Clogged Artery**

In the condition known as atherosclerosis, a deposit or atheroma forms on the arterial wall and reduces the opening through which blood can flow. In the carotid artery in the neck, blood flows three times faster through a partially blocked region than it does through an unobstructed region. Determine the ratio of the effective radii of the artery at the two places.

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**Check Your Understanding 3**

Water flows from left to right through the five sections (A, B, C, D, E) of the pipe shown in the drawing. In which section(s) does the water speed increase, decrease, and remain constant? Treat the water as an incompressible fluid.

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**d Speed Increases Speed Decreases Speed is Constant a. A, B D, E C b. **

Speed Increases Speed Decreases Speed is Constant a. A, B D, E C b. D B A, C, E c. d. e. C, D E d

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Bernoulli's Equation

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In this horizontal pipe, the pressure in region 2 is greater than that in region 1. The difference in pressures leads to the net force that accelerates the fluid to the right. When the fluid changes elevation, the pressure at the bottom is greater than that at the top, assuming the cross-sectional area of the pipe is constant.

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BERNOULLI’S EQUATION In the steady flow of a nonviscous, incompressible fluid of density , the pressure P, the fluid speed v, and the elevation y at any two points (1 and 2) are related by

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**Applications of Bernoulli's Equation**

When a moving fluid is contained in a horizontal pipe, all parts of it have the same elevation (y1 = y2) .

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**Conceptual Example 13. Tarpaulins and Bernoulli’s Equation**

A tarpaulin is a piece of canvas that is used to cover a cargo, like that pulled by the truck. When the truck is stationary the tarpaulin lies flat, but it bulges outward when the truck is speeding down the highway. Account for this behavior. The greater inside pressure generates a greater force on the inner surface of the canvas, and the tarpaulin bulges outward.

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**Example 14. An Enlarged Blood Vessel**

An aneurysm is an abnormal enlargement of a blood vessel such as the aorta. Suppose that, because of an aneurysm, the cross-sectional area A1 of the aorta increases to a value A2 = 1.7A1. The speed of the blood ( = 1060 kg/m3 ) through a normal portion of the aorta is v1 = 0.40 m/s. Assuming that the aorta is horizontal (the person is lying down), determine the amount by which the pressure P2 in the enlarged region exceeds the pressure P1 in the normal region

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Example 15. Efflux Speed The tank is open to the atmosphere at the top. Find an expression for the speed of the liquid leaving the pipe at the bottom.

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**Check Your Understanding 4**

Fluid is flowing from left to right through a pipe (see the drawing). Points A and B are at the same elevation, but the cross-sectional areas of the pipe are different. Points B and C are at different elevations, but the cross-sectional areas are the same. Rank the pressures at the three points, highest to lowest. a. A and B (a tie), C b. C, A and B (a tie) c. B, C, A d. C, B, A e. A, B, C e

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Viscous Flow (a) In ideal (nonviscous) fluid flow, all fluid particles across the pipe have the same velocity. (b) In viscous flow, the speed of the fluid is zero at the surface of the pipe and increases to a maximum along the center axis.

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For a highly viscous fluid, like thick honey, a large force is needed; for a less viscous fluid, like water, a smaller one will do. As part b of the drawing suggests, we may imagine the fluid to be composed of many thin horizontal layers. When the top plate moves, the intermediate fluid layers slide over each other. The velocity of each layer is different, changing uniformly from v at the top plate to zero at the bottom plate. The resulting flow is called laminar flow, since a thin layer is often referred to as a lamina.

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**FORCE NEEDED TO MOVE A LAYER OF VISCOUS FLUID WITH A CONSTANT VELOCITY**

The magnitude of the tangential force F required to move a fluid layer at a constant speed v, when the layer has an area A and is located a perpendicular distance y from an immobile surface, is given by where is the coefficient of viscosity. SI Unit of Viscosity: Pa·s Common Unit of Viscosity: poise (P)

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**Values of viscosity depend on the nature of the fluid**

Values of viscosity depend on the nature of the fluid. Under ordinary conditions, the viscosities of liquids are significantly larger than those of gases. Moreover, the viscosities of either liquids or gases depend markedly on temperature. Usually, the viscosities of liquids decrease as the temperature is increased. In contrast, the viscosities of gases increase as the temperature is raised. An ideal fluid has = 0 P.

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**Factors that determine the volume flow rate Q (in m3/s) of the fluid.**

First, a difference in pressures P2 – P1 must be maintained between any two locations along the pipe for the fluid to flow. In fact, Q is proportional to P2 – P1, a greater pressure difference leading to a larger flow rate. Second, a long pipe offers greater resistance to the flow than a short pipe does, and Q is inversely proportional to the length L. Because of this fact, long pipelines, such as the Alaskan pipeline, have pumping stations at various places along the line to compensate for a drop in pressure (see Figure 11.40). Third, high-viscosity fluids flow less readily than low-viscosity fluids, and Q is inversely proportional to the viscosity Finally, the volume flow rate is larger in a pipe of larger radius, other things being equal.

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POISEUILLE’S LAW A fluid whose viscosity is , flowing through a pipe of radius R and length L, has a volume flow rate Q given by where P1 and P2 are the pressures at the ends of the pipe.

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**Example 16. Giving an Injection**

A hypodermic syringe is filled with a solution whose viscosity is 1.5 × 10–3 Pa·s. As the figure shows, the plunger area of the syringe is 8.0 × 10–5 m2, and the length of the needle is m. The internal radius of the needle is 4.0 × 10–4 m. The gauge pressure in a vein is 1900 Pa (14 mm of mercury). What force must be applied to the plunger, so that 1.0 × 10–6 m3 of solution can be injected in 3.0 s?

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.

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**Concepts & Calculations Example 17. Pressure and Force**

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The figure shows a rear view of a loaded two-wheeled wheelbarrow on a horizontal surface. It has balloon tires and a weight of W = 684 N, which is uniformly distributed. The left tire has a contact area with the ground of AL = 6.6×10–4 m2, whereas the right tire is underinflated and has a contact area of AR = 9.9×10–4 m2. Find the force and the pressure that each tire applies to the ground.

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**Concepts & Calculations Example 18. The Buoyant Force**

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A father (weight W = 830 N) and his daughter (weight W = 340 N) are spending the day at the lake. They are each sitting on a beach ball that is just submerged beneath the water. Ignoring the weight of the air within the balls and the parts of their legs that are underwater, find the radius of each ball.

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Conceptual Question 19 REASONING AND SOLUTION Consider a stream of water that falls from a faucet. According to the equation of continuity, the mass flow rate must be the same at every point along the stream ( = water density, A = of the stream, v = speed of stream). As the water falls, it is accelerated due to gravity; therefore, the speed of the water increases as it falls. Since the density of water is uniform throughout the stream, when v increases, the cross-sectional area A of the stream must decrease in order to maintain a constant mass flow rate. Therefore, the cross-sectional area of the stream becomes smaller as the water falls. If the water is shot upward, as it is in a fountain, the velocity of the stream is upward, while the acceleration due to gravity is directed downward. Therefore, the speed of the stream decreases as the stream rises. Since v decreases, the cross-sectional area A must increase in order to maintain a constant mass flow rate Therefore, the cross-sectional area of the stream becomes larger as the water is shot upward.

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**Problem 25 P = F/A = gh = (1.00 *103 kg/m3)(9.80 m/s2)(71 m) =**

REASONING AND SOLUTION The pump must generate an upward force to counteract the weight of the column of water above it. Therefore, F = mg = ( hA)g. The required pressure is then P = F/A = gh = (1.00 *103 kg/m3)(9.80 m/s2)(71 m) =

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Problem 48 REASONING AND SOLUTION Only the weight of the block compresses the spring. Applying Hooke's law gives W = kx. The spring is stretched by the buoyant force acting on the block minus the weight of the block. Hooke's law again gives FB – W = 2kx. Eliminating kx gives FB = 3W. Now FB = w gV, so that the volume of the block is V = 3M / w = 3(8.00 kg)/(1.00 * 103 kg/m3) = 2.40 * 10–2 m3 Vw = M/ b = (8.00 kg)/(840 kg/m3) = 9.52 * 10–3 m3 V – Vw = 1.45*10-2 m3 100(1.45* 10–2)/(2.40*10–2) =

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**Problem 52 Solution REASONING**

a. According to Equation 11.10, the volume flow rate Q is equal to the product of the cross-sectional area A of the artery and the speed v of the blood, Q = Av. Since Q and A are known, we can determine v . b. Since the volume flow rate Q2 through the constriction is the same as the volume flow rate Q1 in the normal part of the artery, Q2 = Q1. We can use this relation to find the blood speed in the constricted region. Solution a. Since the artery is assumed to have a circular cross-section, its cross-sectional area is A1 = , where r1 is the radius. Thus, the speed of the blood is

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b. The volume flow rate is the same in the normal and constricted parts of the artery, so Q2 = Q1. Since the blood speed is v2 = Q2/A2 = Q1/A2. We are given that the radius of the constricted part of the artery is one-third that of the normal artery, so Thus, the speed of the blood at the constriction is

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**Problem 57 REASONING AND SOLUTION**

a. Using Equation 11.12, the form of Bernoulli's equation with y1=y2 , we have b. The pressure inside the roof is greater than the pressure on the outside. Therefore, there is a net outward force on the roof. If the wind speed is sufficiently high, some roofs are "blown outward."

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Problem 60 Reasoning a. The drawing shows two points, labeled 1 and 2, in the fluid. Point 1 is at the top of the water, and point 2 is where it flows out of the dam at the bottom. Bernoulli’s equation, Equation 11.11, can be used to determine the speed v2 of the water exiting the dam. 1 P1=1 atmosphere v1 = 0 m/s y1 2 y2 P2=1 atmosphere v2

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b. The number of cubic meters per second of water that leaves the dam is the volume flow rate Q. According to Equation 11.10, the volume flow rate is the product of the cross-sectional area A2 of the crack and the speed v2 of the water; Q = A2v2 Solutions a. According to Bernoulli’s equation, as given in Equation 11.11, we have

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**Setting P1 = P2 , v1 = 0 m/s, and solving for v2, we obtain**

b. The volume flow rate of the water leaving the dam is

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Problem 77 REASONING The drawing at the right shows the situation. As discussed in Conceptual Example 6, the job of the pump is to draw air out of the pipe that dips down into the water. The atmospheric pressure in the well then pushes the water upward into the pipe.

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In the drawing, the best the pump can do is to remove all of the air, in which case, the pressure P1 at the top of the water in the pipe is zero. The pressure P2 at the bottom of the pipe at point A is the same as that at the point B, namely, it is equal to atmospheric pressure ( ), because the two points are at the same elevation, and point B is open to the atmosphere. Equation 11.4, can be applied to obtain the maximum depth h of the well. SOLUTION Setting Pa, and solving Equation 11.4 for h, we have

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Problem 87 REASONING AND SOLUTION Let rh represent the inside radius of the hose, and rp the radius of the plug, as suggested by the figure below. r h p hose plug

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**Then, from Equation 11.9, A1v1 = A2v2 , we have**

or According to the problem statement, v2 = 3v1 , or

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