{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# Ch13Word - Chapter 13 CHAPTER 13 Fluids 1 When we use the...

This preview shows pages 1–3. Sign up to view the full content.

Chapter 13 Page 1 CHAPTER 13 – Fluids 1. When we use the density of granite, we have m = ρ 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 = ρ V = ρ 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 = ρ V = ρ 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 = ρ 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 = ρ fluid / ρ 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 = ρ mixture / ρ water . The density of the mixture is ρ mixture = m mixture / V = ( SG antifreeze ρ water V antifreeze + SG water ρ 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 = ρ g ? h ; (80 mm-Hg)(133 N/m 2 · mm-Hg) = (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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Chapter 13 Page 2 12. The pressure from the height of alcohol must balance the atmospheric pressure: P = ρ 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 . 13. The pressure at a depth h is P = P 0 + ρ gh = 1.013 × 10 5 N/m 2 + (1.00 × 10 3 kg/m 3 )(9.80 m/s 2 )(2.0 m) = 1.2 × 10 5 N/m 2 . The force on the bottom is F = PA = (1.2 × 10 5 N/m 2 )(22.0 m)(8.5 m) = 2.3 × 10 7 N (down) . The pressure depends only on depth, so it will be the same: 1.2 × 10 5 N/m 2 . 14. The pressure is produced by a column of air: P = ρ gh ; 1.013 × 10 5 N/m 2 = (1.29 kg/m 3 )(9.80 m/s 2 ) h , which gives h = 8.0 × 10 3 m = 8.0 km .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern