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The Fluid States

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The Fluid States CK-12 Physics Concepts - Intermediate Answer Key Chapter 11: The Fluid States 11.1 Pressure in Fluids Practice Questions 1. Why do the streams of water at the bottom of the bottle go the farthest? 2. Why does water stop flowing out of the top hole even before the water level falls below it? Answers 1. At the bottom of the bottle there is more water pushing down and therefore more pressure coming out of the holes. 2. As the water level falls there isn't as much water pushing down and therefore less pressure to push water out of the holes. Additional Practice Problems: Questions 1. Calculate the pressure produced by a force of 800. N acting on an area of 2.00 m2. 2. A swimming pool of width 9.0 m and length 24.0 m is filled with water to a depth of 3.0 m. Calculate pressure on the bottom of the pool due to the water. 3. What is the pressure on the side wall of the pool at the junction with the bottom of the pool in the previous problem? Answers 퐹 800푁 400푁 1. 푃 = ∶ 푃 = = = 400푃푎 퐴 2푚2 푚2 푘푔 푚 2. Using 퐹 = 푎ℎ푝 ∶ (9.0푚 ∗ 24.0푚) ∗ 3푚 ∗ 1000 ∗ 9.8 = ퟔ. ퟑ × ퟏퟎퟔ 푷풂 푚3 푠2 푘푔 푚 3. Using 푃 = 푝ℎ ∶ 1000 ∗ 9.8 ∗ 3.0푚 = 29400 푃푎 = ퟐ. ퟗퟒ × ퟏퟎퟒ 푷풂 푚3 푠2 1 Review Questions 1. If you push the head of a nail against your skin and then push the point of the same nail against your skin with the same force, the point of the nail may pierce your skin while the head of the nail will not. Considering that the forces are the same, what causes the difference? 2. A brick of gold is 10.0 cm wide, 10.0 cm high, and 20.0 cm long. The density of gold is 19.3 g/cm3. a. What pressure does the brick exert on the table if the brick is resting on its side? b. What pressure does the brick exert on the table if it is resting on its end? 3. What is the total force and the pressure on the bottom of a swimming pool 8.0 m by 15.0 m whose uniform depth is 2.0 m? Answers 1. Because the point of the nail has a smaller area, the pressure exerted by the point on your skin is greater than the pressure exerted by the head with the same force. Therefore, the point may break your skin while the head will not. 2. The brick’s total volume is 10푐푚 ∗ 10푐푚 ∗ 20푐푚 = 2000푐푚3, total mass is 푔 푚 2000푐푚3 ∗ 19.3 = 38,600 = 38.6 kg, and its weight is 38.6푘 ∗ 9.8 = 378 푐푚3 푠2 N. a. The pressure of the brick resting on its side, using 푃 = 푝ℎ is: 푔 푚 푃 = 19.3 ∗ 9.8 ∗ 10.0푐푚 = 1.89 x 104 Pa. 푐푚3 푠2 b. The pressure of the brick resting on its end is: 푔 푚 푃 = 19.3 ∗ 9.8 ∗ 20.0푐푚 = 3.78 x 104 Pa. 푐푚3 푠2 푘푔 푚 3. The total force, using 퐹 = 푎ℎ푝 is 8푚 ∗ 15푚 ∗ 2푚 ∗ 1000 ∗ 9.8 = 2.35 x 106 푚3 푠2 푘푔 푚 N m2, and the pressure, using 푃 = 푝ℎ is: 1000 ∗ 9.8 ∗ 2푚 = 푚3 푠2 1.96 x 104 Pa. 11.2 Pascal's Principle Practice Questions 1. What property of liquids allows for a great mechanical advantage of a hydraulic lift? 2. If the ratio of A2 to A1 was 100, what force would be required to lift a 10000 N car? 2 Answers 1. Pascal's Principle – pressure is the same throughout a fluid system. 2. The output:input ratio is 100:1, therefore an input force of 100N would result in an output force of 100*100N = 10,000N Review Questions 1. In a hydraulic lift whose input line has a cross-sectional area of 1.00 cm2 and whose output line has a cross-sectional area of 20.0 cm2, what is the largest mass (kg) that can be lifted by an input force of 1000. N? 2. In a hydraulic lift whose IMA is 50, how far can the output platform be lifted when the input platform is depressed 100. cm? 3. A 20.0 N force is exerted on the small piston of a hydraulic system. The cross- sectional area of the small piston is 0.0500 m2. What is the magnitude of the weight than can be lifted by the large piston, which has a cross-sectional area of 0.100 m2? Answers 1. The output:input force ratio is 20:1. The input force is 1000N, resulting in an output of 20,000N. Convert 20,000N to kg by dividing by g: 20000푁 = 2041 kg. 9.8 푚/푠2 2. The output:input distance ratio is 1:50. An input distance of 100cm would result 100푐푚 in an output distance of = 2 cm. 50 0.100푚2 3. The IMA of the system is = 2. An input force of 20.0N results in (ideally) 0.0500푚2 an output force of 40N. Converting 40N to kg by dividing by 9.8m/s2 yields 40푁 푚 = 4.08 kg. 9.8 푠2 11.3 Archimedes' Principle and Buoyancy Practice Questions 1. Why were they unable to use the overturned boat as a submarine? 2. Why does the bubble in the tube shrink as it is taken lower in the pool Answers 1. The upward pressure of the wood and air under the boat was greater than their combined weight pulling down. 3 2. There is more water pressure at lower depths. This pressure pushes up against the bubble, shrinking it. Additional Practice Questions: Questions 1. If a ship has a mass of 200,000,000 kg, how much water would it need to displace in order to float? 2. A 40.0 kg cylinder that is 1.2 m tall and has a 0.20 m diameter falls off the pier and gets completely submerged when it hits the water. What is the buoyant force on the cylinder when it is under water? Will it sink or float? The formula for finding the volume of a cylinder is 휋푟2ℎ. 3. The formula for the volume of a sphere is 4휋푟3. A 1.7 × 10-6 kg air bubble has a diameter of 1.00 cm and is submerged in water. What is the buoyant force on the bubble? Answers 200000000kg 1. Using ∶ 200,000 m3 1000 kg/m3 2. Volume of cylinder = π ∗ (.1m)2 ∗ 1.2m = .038m3. Weight of water (buoyant kg force) = 0.038m3 ∗ 1000 = ퟑퟖ퐤퐠. The object will sink, since 40kg > 38kg m3 3. Volume of displaced water = volume of bubble: 4π(0.5cm)3 = 1.57cm3 = 1.57 × 1000kg 10−6m3. Weight of displaced water = 1.57 × 10−6m3 ∗ = ퟏ. ퟓퟕ × ퟏퟎ−ퟑ퐤퐠. m3 m Buoyant force in N = 1.57 × 10−3kg ∗ 9.8 = ퟎ. ퟎퟏퟓ퐍 s2 Review Questions 1. A cylinder with a radius of 11 cm and a height of 3.4 cm has a mass of 10.0 kg. a. What is the weight of this cylinder? b. What is the weight of this cylinder when it is submerged in water? 2. A wooden raft is 2.00 m wide, 3.00 m long, and 0.200 m deep. The raft and its occupants have a mass of 700. kg. How deep will the raft sink below the water when floating? 3. For the raft in problem #2, how many 50. kg people can be added to the raft before it sinks completely under water? 4. The density of gold is 19,320 kg/m3 and the density of mercury is 13,500 kg/m3. If a cube of gold that is 0.100 m on each side is placed in a tank of mercury and weighed while under the surface, what is the apparent weight of the cube? 4 Answers 1. a. 10kg * 9.8 m/s2 = 98 N. 푘푔 푚 b. Buoyant force = (휋푟2ℎ (1000 )) ∗ 9.8 = ((휋 ∗ (. 11푚)2 ∗ 0.034푚) ∗ 푚3 푠2 푘푔 1000 ) ∗ 9.8 푚/푠2 = 12.7N. Weight under water = 98N – 12.7N = 85.3 N. 푚3 700푘푔 2. The raft floats when displaced water = 700 kg: 푥 = 푘푔 = 0.117 m. (2푚)(3푚)(1000 ) 푚3 3. When submerged, the raft displaces 1.2m3*1000 kg/m3 = 1200kg of water. The raft and initial occupants weight 700kg, so 1200푘 − 700푘 = 500푘 ∶ 500푘푔 =10 more people can be added to the raft. 50푘푔 푘푔 4. The weight of the displaced mercury is 13500 ∗ (0.1푚)3 = 13.5푘 = 푚3 푘푔 132.3푁. The gold has an initial weight of 19320 ∗ (0.1푚)3 = 19.32푘 = 189.3푁. 푚3 The submerged gold has an apparent weight of 189.3푁 − 132.3푁 = 57 N. 11.4 Combined Gas Law Practice Questions 1. What must be held constant for the combined gas law to be true? 2. What happens to the combined gas law if temperature, pressure, or volume are held constant? Answers 1. The numbers of moles of gas must remain constant. 2. The combined gas law reduces to Boyle's, Charles', or Gay-Lussac's Law. Review Questions 1. A sample of gas has a volume of 800. mL at -23.0°C and 300. Torr. What would the volume of the gas be at 227.0°C and 600. Torr? 2. 500.0 L of gas are prepared at 0.921 atm pressure and 200.0°C.
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