Chapter 15: FluidsAnswers to Even-Numbered Conceptual Questions2. No, because the Moon has no atmosphere to press down on the surface of the liquid.4. A suction cup is held in place by atmospheric pressure. When the cup is applied, you push it flat against thesurface for which you want it to stick. This expels most of the air in the cup, and leads to a larger pressure onthe outside of the cup. Thus, atmospheric pressure pushes the outside of the cup against the surface.6. Mercury is more practical in a barometer than water because of its greater density. With such a large density,the height of the mercury column is only about 0.760 m. The density of water is less than that of mercury byroughly a factor of 14. Therefore, the height of a water column in a barometer would be about 10 m—the height of a three-story building.8. In a hot-air balloon, vertical motion is controlled by adding heat to the air in the balloon, or by letting it cooloff. As the temperature of the air in the balloon changes, so too does its density. By controlling the overalldensity of the balloon, one can control whether it rises, falls, or is neutrally buoyant.10. The physics in this case is pretty ugly. Ice floats in water, whether it is a house-sized iceberg, a car-sizedchunk, or a thimble-sized ice cube. If the Earth is warming and icebergs are breaking up into smaller pieces,each of the smaller pieces will be just as buoyant as the original berg.12. A metal boat can float if it displaces a volume of water whose weight is equal to the weight of the boat. Thiscan be accomplished by giving the boat a bowl-like shape, as illustrated in Figure 15-11.14. As wind blows across the top of the chimney, a pressure difference is established between the top andbottom of the chimney, with the top having the lower pressure. This will cause smoke to rise more rapidly.16. If a ball is placed in the stream of air such that the speed of air over its upper surface is greater than thespeed across its lower surface, the result will be a lower pressure at the top of the ball. This results in anupward force that can equal the weight of the ball.Solutions to Problems and Conceptual Exercises1. Picture the Problem: Air, although light in comparison to water or solid objects, has mass. We wish to calculate the mass of air in a typical classroom.Strategy: The volume of air in a typical classroom is on the order of 102 m3. Multiply the mass by the acceleration of gravity to calculate the weight. Solve equation 15-1 for the mass in terms of density and volume. The density of air is given in Table 15-1.Solution: Use equation 15-1 to write the weight in terms of density and volume: W =mg =ρVg = 1.29 kg/m3( )102 m3( )9.81 m/s2( )= 103 NInsight: The air in a physics classroom weighs about the same as two students.2. Picture the Problem: A 25-gallon aquarium is filled with water. We wish to calculate the weight of the water.Strategy: Multiply the mass times the acceleration of gravity to calculate the weight. Solve equation 15-1 for the mass in terms of density and volume. The density of water is given in Table 15-1. Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.15 – 1Chapter 15: Fluids James S. Walker, Physics, 4th EditionSolution: Use equation 15-1 to write the weight in terms of density and volume: W =mg =ρVg= 1000 kg/m3( )25 gal( )3.79×10−3m3/gal( )9.81 m/s2( )= 0.93 kNInsight: The 25-gallon tank weighs about 200 lbs when filled.Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.15 – 2Chapter 15: Fluids James S. Walker, Physics, 4th Edition3. Picture the Problem: A gold ring has a density equal to its mass divided by its volume.Strategy: If the ring is pure gold, its density will be equal to the density of gold. Since the mass and volume of the ring are known, use equation 15-1 to calculate the density. Compare the result with the density of gold given in Table 15-1.Solution: 1. Divide the volume by the mass: ρ =mV=0.014 g0.0022 cm3=6.4 g cm32. Compare with the density of gold from Table 15-1: ρgold=19.3 g cm3. No, the ring is not solid gold.Insight: If the ring were pure gold of the same volume given in the problem, its mass would be 42.5 g. 4. Picture the Problem: A treasure chest is filled with gold doubloons. A typical chest might measure 8 inches by 8 inches by 12 inches and have a volume on the order of 10-1 m3Strategy: Solve equation 15-1 for the mass of the chest in terms of the density of gold and the volume of the chest. Calculate the weight of the chest by multiplying its mass times the acceleration of gravity. The density of gold is given in Table 15-1.Solution: Use equation 15-1 to write the weight in terms of density and volume: W =mg =ρVg = 1.93×104kg/m3( )10–1 m3( )9.81 m/s2( )= 104NInsight: Even though this chest seems fairly small, it weighs a ton!5. Picture the Problem: A cube has a mass of 0.347 kg and sides of 3.21 cm each.Strategy: Use equation 15-1 to calculate the density of the cube. Compare the resulting density with the densities given in Table 15-1 to determine the likely composition.Solution: 1. Calculate the density of the cube: ρ =mV=0.347 kg3.21 cm1 m100 cm( )⎡⎣⎤⎦3= 1.05×104kg m32. Compare with the densities in Table 15-1:The cube has the density of silver.Insight: Cubes made of different materials could have considerably different masses. For example, a cube of gold (with the same volume as the silver cube) would have a mass 0.638 kg, while a cube of aluminum would have a mass of0.089 kg.6. Picture the Problem: The football field has an area given by its length multiplied by width. Atmospheric pressure produces a downward force on the field.Strategy: Solve equation 15-2 for the force exerted over the area of the football field. Set the pressure equal to atmospheric pressure, 1.01×105 N/m2.Solution: Solve equation 15-2 for F: F =PA = 1.01×105N/m2( )360 ft( )160 ft( )m3.281 ft⎛⎝⎜⎞⎠⎟2= 5.40×108 NInsight: This weight is over 50,000 tons! 7. Picture