By Greg Jacobs  Joshua Schulman — McGrawHill Professional
Updated on Feb 11, 2011
Solutions
 A—The fluid is pushed into the mouth by the atmospheric pressure. Because the surface of a drink is open to the atmosphere, the surface is at atmospheric pressure, and the pressure in the mouth must be lower than atmospheric.
 A—Flow rate (volume of flow per second) is the area of a pipe times the speed of flow. So if we increase the volume of flow (choices B, C, and D), we increase the speed. By Bernoulli's equation, choice D also increases the fluid speed. Expanding the vein increases the area of the pipe, so if flow rate is constant, then the velocity must decrease.
 A—When the treasure is floating in the boat, it displaces an amount of water equal to its weight. When the treasure is on the lake bottom, it displaces much less water, because the lake bottom supports most of the weight that the buoyant force was previously supporting. Thus, the lake level drops.
 D—Since Brian will be floating in equilibrium, his weight must be equal to the buoyant force on him. The buoyant force is ρ_{water}V_{submerged}g, and must equal Brian's weight of 800 N. Solving for V_{submerged}, we find he needs to displace 8/100 of a cubic meter. Converting to liters, he needs to displace 80 L of water, or 40 bottles. (We would suggest that he use, say, twice this many bottles—then the raft would float only half submerged, and he would stay drier.)

 If the valve is closed, we have a static column of water, and the problem reduces to one just like the example in the chapter. P = P_{0} + ρgh = 10^{5} N/m^{2} + (1000 kg/m^{3})(10 N/kg)(15 m) = 250,000 N/m^{2}, (atmospheric pressure is given on the constant sheet).
 Use Bernoulli's equation for a flowing fluid. Choose point P and the valve as our two positions. The pressure at both positions is now atmospheric, so the pressure terms go away. Choose the height of the valve to be zero. The speed of the water at the top is just about zero, too. So, four of the six terms in Bernoulli's equation cancel! The equation becomes ρgy_{top} = 1/2 ρv_{bottom}^{2}. Solving, v_{bottom} = 17 m/s.
 Flow rate is defined as Av. The crosssectional area of the pipe is π (0.1 m)^{2} = 0.031 m^{2}. (Don't forget to use meters, not centimeters!) So the volume flow rate is 0.53 m^{3}/s.
 The volume of the spherical portion of the tank can be estimated as (4/3) πr^{3} (this equation is on the equation sheet), where the radius of the tank looks to be somewhere around 2 or 3 meters. Depending what actual radius you choose, this gives a volume of about 100 m^{3}. The tank looks to be something like 3/4 full… So call it 70 m^{3}. (Any correct reasoning that leads to a volume between, say, 30–300 m^{3} should be accepted.)
 The flow rate is 0.53 m^{3}/s; we need to drain 70 m^{3}. So, this will take 70/0.53 = 140 seconds, or about two minutes. (Again, your answer should be counted as correct if the reasoning is correct and the answer is consistent with part (d).)
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From 5 Steps to a 5 AP Physics B & C. Copyright © 2010 by The McGrawHill Companies. All Rights Reserved.
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