Respuesta Del Capitulo 15 Fisica Elemental Paul Tippens

Chapter 15. Fluids

Physics, 6th Edition

Chapter 15. FLUIDS
Density
15.1. What volume does 0.4 kg of alcohol occupy? What is the weight of this volume?

ρ=

m ; V

V=

m 0.4 kg = ; ρ 790 kg/m 2

V = 5.06 x 10-4 m3 W = 3.92 N

W = DV = ρgV = 790 kg/m3)(9.8 m/s2)(5.06 x 10-4 m3);

15-2. An unknown substance has a volume of 20 ft3 and weighs 3370 lb. What are the weight densityand the mass density? D= W 3370 lb = ; D = 168 lb/ft3 V 20 ft 3

ρ=

D 168 lb/ft 3 = ; g 9.8 m/s 2

ρ = 5.27 slugs/ft3

15-3. What volume of water has the same mass of 100 cm3 of lead? What is the weight density of lead? First find mass of lead: m = ρV = (11.3 g/cm3)(100 cm3); m = 1130 g Now water: Vw = m 1130 g = = 1130 cm3 ; Vw = 1130 cm3 ρ w 1 g/cm3 D = ρg

D = (11,300 kg/m3)(9.8m/s2) = 110,740 N/m3;

D = 1.11 x 105 N/m3

15-4. A 200-mL flask (1 L = 1 x 10-3 m3) is filled with an unknown liquid. An electronic balance indicates that the added liquid has a mass of 176 g. What is the specific gravity of the liquid? Can you guess the identity of the liquid? V = 200 mL = 0.200 L = 2 x 10-4 m3; m = 0.176 kg

ρ=

m 0.176 kg = ; V 2 x 10−4 m3

ρ = 880 kg/m3, Benzene

203Chapter 15. Fluids

Physics, 6th Edition

Fluid Pressure
15-5. Find the pressure in kilopascals due to a column of mercury 60 cm high. What is this pressure in lb/in.2 and in atmospheres? P = ρgh = (13,600 kg/m3)(9.8 m/s2)(0.60 m);  0.145 lb/in.2  P = 80 kPa  ;  1 kPa  P= 80 kPa ; 101.3 kPa/atm P = 80 kPa

P = 11.6 lb/in.2

P = 0.790 atm

15-6. A pipe contains water under agauge pressure of 400 kPa. If you patch a 4-mm-diameter hole in the pipe with a piece of tape, what force must the tape be able to withstand? A=

π D 2 π (0.004 m) 2 = = 1.257 x 10-5 m 2 ; 4 4

P=

F ; A

F = PA = (400,000 Pa)(1.257 x 10-5 m2);

P = 5.03 N

15-7. A submarine dives to a depth of 120 ft and levels off. The interior of the submarine is maintained at atmospheric pressure.What are the pressure and the total force applied to a hatch 2 ft wide and 3 ft long? The weight density of sea water is around 64 lb/ft.3 P = Dh = (64 lb/ft3)(120 ft); P = 7680 lb/ft2; P = 53.3 lb/in.2

F = PA = (7680 lb/ft2)(3 ft)(2 ft);

F = 46,100 lb

15-8. If you constructed a barometer using water as the liquid instead of mercury, what height of water would indicate a pressure of oneatmosphere? P = ρ gh; h= P 101,300 Pa = ; ρ g (1000 kg/m3 )(9.8 m/s 2 ) or 34 ft !

h = 10.3 m

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Chapter 15. Fluids

Physics, 6th Edition

15-9. A 20-kg piston rests on a sample of gas in a cylinder 8 cm in diameter. What is the gauge pressure on the gas? What is the absolute pressure?

π D 2 π (0.08 m) 2 A= = = 5.027 x 10-3 m 2 ; 4 4
P= (20 kg)(9.8 m/s 2 ) = 3.90 x 104 kPa; -3 25.027 x 10 m

P=

F mg = A A

P = 39.0 kPa Pabs = 140 kPa

Pabs = 1 atm + Pgauge = 101.3 kPa + 39.0 kPa;

*15-10. An open U-shaped tube such as the one in fig. 15-21 is 1 cm2 in cross-section. What volume of water must be poured into the right tube to cause the mercury in the left tube to rise 1 cm above its original position?

ρ m ghm = ρ 2 gh2 ;

h2 =

ρ m hm (13.6 g/cm3 )(1 cm) == 13.6 cm ρ2 1 g/cm3
V = 13.6 cm3 or 13.6 mL

V = Ah = (1 cm2)(13.6 cm);

15-11. The gauge pressure in an automobile tire is 28 lb/in.2. If the wheel supports 1000 lb, what area of the tire is in contact with the ground? A= F 1000 lb = ; P 28 lb/in.2 A = 35.7 in.2

15-12. Two liquids that do not react chemically are placed in a bent tube like the one in Fig. 1521. Show that the heights ofthe liquids above their surface of separation are inversely

proportional to their densities:

h1 ρ 2 = h2 ρ1

The gauge pressure must be the same for each column: ρ1gh1 = ρ2gh2 So that:

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Chapter 15. Fluids h1 ρ 2 = h2 ρ1

Physics, 6th Edition

15-13. Assume that the two liquids in the U-shaped tube in Fig. 15-21 are water and oil. Compute the density of the oil if the water...