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Unformatted text preview: Chapter 13 Page 1 CHAPTER 13 Fluids 1. When we use the density of granite, we have m = V = (2.7 10 3 kg/m 3 )(1 10 8 m 3 ) = 3 10 11 kg . 2. When we use the density of air, we have m = V = LWH = (1.29 kg/m 3 )(4.8 m)(3.8 m)(2.8 m) = 66 kg . 3. When we use the density of gold, we have m = V = LWH = (19.3 10 3 kg/m 3 )(0.60 m)(0.25 m)(0.15 m) = 4.3 10 2 kg ( 950 lb!). 4. If we assume a mass of 65 kg with the density of water, we have m = V ; 65 kg = (1.0 10 3 kg/m 3 ) V , which gives V = 6.5 10 2 m 3 ( 65 L). 5. From the masses we have m water = 98.44 g 35.00 g = 63.44 g; m fluid = 88.78 g 35.00 g = 53.78 g. Because the water and the fluid occupy the same volume, we have SG fluid = fluid / water = m fluid / m water = (53.78 g)/(63.44 g) = 0.8477 . 6. The definition of the specific gravity of the mixture is SG mixture = mixture / water . The density of the mixture is mixture = m mixture / V = (SG antifreeze water V antifreeze + SG water water V water )/ V , so we get SG mixture = (SG antifreeze V antifreeze + SG water V water )/ V = [(0.80)(5.0 L) + (1.0)(4.0 L)]/(9.0 L) = 0.89 . 7. ( a ) The normal force on the four legs must equal the weight. The pressure of the reaction to the normal force, which is exerted on the floor, is P = F N / A = mg / A = (60 kg)(9.80 m/s 2 )/4(0.05 10 4 m 2 ) = 3 10 7 N/m 2 . ( b ) For the elephant standing on one foot, we have P = F N / A = mg / A = (1500 kg)(9.80 m/s 2 )/(800 10 4 m 2 ) = 2 10 5 N/m 2 . Note that this is a factor of 100 less than that of the loudspeaker! 8. ( a ) The force of the air on the table top is F = PA = (1.013 10 5 N/m 2 )(1.6 m)(2.9 m) = 4.7 10 5 N (down) . ( b ) Because the pressure is the same on the underside of the table, the upward force has the same magnitude: 4.7 10 5 N . This is why the table does not move! 9. The pressure difference on the lungs is the pressure change from the depth of water: ? P = g ? h ; (80 mmHg)(133 N/m 2 mmHg) = (1.00 10 3 kg/m 3 )(9.80 m/s 2 )? h , which gives ? h = 1.1 m . 10. There is atmospheric pressure outside the tire, so we find the net force from the gauge pressure. Because the reaction to the force from the pressure on the four footprints of the tires supports the automobile, we have 4 PA = mg ; 4(240 10 3 N/m 2 )(200 10 4 m 2 ) = m (9.80 m/s 2 ), which gives m = 2.0 10 3 kg . 11. Because the force from the pressure on the cylinder supports the automobile, we have PA = mg ; (17.0 atm)(1.013 10 5 N/m 2 atm) # p(24.5 10 2 m) 2 = m (9.80 m/s 2 ), which gives m = 8.28 10 3 kg . Note that we use gauge pressure because there is atmospheric pressure on the outside of the cylinder. Chapter 13 Page 2 12. The pressure from the height of alcohol must balance the atmospheric pressure: P = gh ; 1.013 10 5 N/m 2 = (0.79 10 3 kg/m 3 )(9.80 m/s 2 ) h , which gives h = 13 m ....
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This note was uploaded on 02/05/2010 for the course PHYSICS 37,38,39 taught by Professor Fielding during the Spring '09 term at Los Angeles Valley College.
 Spring '09
 Fielding
 Physics

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