Fluid Mechanics with teacher's notes

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$Holt Math Minnesota Test Prep Workbook for Grade 9$

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the water pressure pushing up on the bottom of the box is P o + rgh 2. The net pressure on the box is the sum of these two pressures. P net = P bottom + P top = (P 0 + rgh 2) – (P 0 + rgh 1) = rg(h 2 − h 1) = rgL From this result, we can find the net vertical force due to the pressure on the box as follows: F net = P netA = rgLA = rgV = m fg Note that this is an expression of Archimedes’ principle. In general, we can say that buoyant forces arise from the differences in fluid pressure between the top and the bottom of an immersed object. Atmospheric pressure is pressure from above The weight of the air in the upper portion of Earth’s atmosphere exerts pressure on the layers of air below. This pressure is called atmospheric pressure; the force it exerts on our bodies (assuming a body area of 2 m 2 ) is extremely large, on the order of 200 000 N (40 000 lb). How can we exist under such tremendous forces without our bodies collapsing? The answer is that our body cavities and tissues are permeated with fluids and gases that are pushing outward with a pressure equal to that of the atmosphere. Consequently, our bodies are in equilibrium—the force of the atmosphere pushing in equals the internal force pushing out. An instrument that is commonly used to measure atmospheric pressure is the mercury barometer. Figure 9-9 shows a very simple mercury barometer. A long tube that is open at one end and closed at the other is filled with mercury and then inverted into a dish of mercury. Once inverted, the mercury does not empty into the bowl; rather, the atmosphere exerts a pressure on the mercury in the bowl and pushes the mercury in the tube to some height above the bowl. In this way, the force exerted on the bowl of mercury by the atmosphere is equal to the weight of the column of mercury in the tube. Any change in the height of the column of mercury means that the atmosphere’s pressure has changed. Kinetic theory of gases can describe the origin of gas pressure Many models of a gas have been developed over the years. Almost all of these models attempt to explain the macroscopic properties of a gas, such as pressure, in terms of events occurring in the gas on a microscopic scale. The most successful model by far is the kinetic theory of gases. In kinetic theory, gas particles are likened to a collection of billiard balls that constantly collide with one another. This simple model is successful in explaining many of the macroscopic properties of a gas. For instance, as these particles strike a wall of a container, they transfer some of their momentum during the collision. The rate of transfer of momentum to the container wall is equal to the force exerted by the gas on the container wall (see Chapter 6). This force per unit area is the gas pressure. Copyright © by Holt, Rinehart and Winston. All rights reserved. NSTA TOPIC: Atmospheric pressure GO TO: www.scilinks.org sciLINKS CODE: HF2093 Mercury Fluid Mechanics Empty Figure 9-9 The height of the mercury in the tube of a barometer indicates the atmospheric pressure. 329 SECTION 9-2 Demonstration 6 Hydrostatic pressure Purpose Demonstrate that pressure increases with depth. Materials 32 oz plastic soda bottle, tape, water, bucket; optional: plastic straw or small pieces of glass or metal tubing and modeling clay or silicon Procedure Drill three holes about 10 cm apart along the side of the bottle, with the lowest hole close to the bottom, and tape them. The holes should not be aligned vertically; they should be displaced horizontally about 1 cm so that the water streams do not collide. Put small pieces of tubing into the holes to improve the flow. Cover the holes with tape, fill the bottle with water, and place it at the edge of a table. Place it high enough above the bucket so that the effects of depth on the range of each stream will be easily observed. Ask students to predict how the water streams will compare. Quickly pull the tape away from all three holes, and have students observe the shape of each stream. Water shooting out from near the bottom of the bottle exits the bottle at a higher speed than water shooting out from near the top does. The reason is that the pressure in the bottle increases with depth. 329
SECTION 9-2 Classroom Practice The following may be used as a teamwork exercise or for demonstration at the chalkboard or on an overhead projector. PROBLEM Pressure as a function of depth Find the atmospheric pressure at an altitude of 1.0 × 10 3 m if the air density is constant. Assume that the air density is uniformly 1.29 kg/m 3 and P 0 = 1.01 × 10 5 Pa. Answer 8.8 × 10 4 Pa PRACTICE GUIDE 9C Solving for: P PE Sample, 1, 2; Ch. Rvw. 19a, 34*, 35*, 46*, 57b* PW 5–6 PB 5–7 h PE 3, 4; Ch. Rvw. 57a* PW 3–4 PB Sample, 1–4 r PW Sample, 1–2 PB 8–10 ANSWERS TO Practice 9C Pressure as a function of depth 1. 1.11 × 10 8 Pa 2. a. 1.03 × 10 5 Pa b. 1.05 × 10 5 Pa 3. 0.20 m 4. 20.1 m 330 330 SAMPLE PROBLEM 9C PROBLEM SOLUTION PRACTICE 9C Chapter 9 Pressure as a function of depth Calculate the absolute pressure at an ocean depth of 1.00 × 10 3 m. Assume that the density of the water is 1.025 × 10 3 kg/m 3 and that P 0 = 1.01 × 10 5 Pa. Given: h = 1.00 × 10 3 m P 0 = 1.01 × 10 5 Pa r = 1.025 × 10 3 kg/m 3 g = 9.81 m/s 2 Unknown: P = ? Use the equation for fluid pressure as a function of depth from page 328. P = P 0 + rgh P = P 0 + (1.025 × 10 3 kg/m 3 )( 9.81 m/s 2 )(1.00 × 10 3 m) P = 1.01 × 10 5 Pa + 1.01 × 10 7 Pa P = 1.02 × 10 7 Pa Pressure as a function of depth 1. The Mariana Trench, in the Pacific Ocean, is about 11.0 km deep. If atmospheric pressure at sea level is 1.01 × 10 5 Pa, how much pressure would a submarine need to be able to withstand to reach this depth? (Use the value for the density of sea water given in Table 9-1.) 2. A container is filled with water to a depth of 20.0 cm. On top of the water floats a 30.0 cm thick layer of oil with a density of 0.70 × 10 3 kg/m 3 . a. What is the pressure at the surface of the water? b. What is the absolute pressure at the bottom of the container? 3. A beaker containing mercury is placed inside a vacuum chamber in a laboratory. The pressure at the bottom of the beaker is 2.7 × 10 4 Pa. What is the height of the mercury in the beaker? (See Table 9-1 for the density of mercury. Hint: Think carefully about what value to use for atmospheric pressure.) 4. Calculate the depth in the ocean at which the pressure is three times atmospheric pressure. (Use the value for the density of sea water given in Table 9-1.) Copyright © by Holt, Rinehart and Winston. All rights reserved.
Page 1 and 2: Chapter 9 Overview Section 9-1 defi
Page 3 and 4: Section 9-1 Demonstration 1 Volume
Page 5 and 6: SECTION 9-1 The Language of Physics
Page 7 and 8: SECTION 9-1 Demonstration 4 Lifting
Page 9 and 10: SECTION 9-1 PRACTICE GUIDE 9A Solvi
Page 11 and 12: SECTION 9-2 STOP 326 Misconception
Page 13: SECTION 9-2 Visual Strategy Figure
Page 17 and 18: Section 9-3 Demonstration 7 Fluid f
Page 19 and 20: SECTION 9-3 Misconception STOP Aler
Page 21 and 22: SECTION 9-3 Classroom Practice The
Page 23 and 24: Section 9-4 Demonstration 10 Temper
Page 25 and 26: SECTION 9-4 Classroom Practice The
Page 27 and 28: Chapter 9 Summary Teaching Tip Ask
Page 29 and 30: 9 REVIEW & ASSESS 12. because the p
Page 31 and 32: 9 REVIEW & ASSESS 42. 7.1 m 43. 17
Page 33 and 34: 9 REVIEW & ASSESS 63. 60.9 m/s 2 64
Page 35 and 36: Chapter 9 Laboratory Exercise NOTE
Page 37 and 38: CHAPTER 9 LAB Boyle’s Law Apparat

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