Aplicaciones del teorema de bernouli

15-19. A 100-g cube 2 cm on each side is attached to a string and then totally submerged in water. What is the buoyant force and what is the tension in the rope?
V = (0.02 m)3 = 8 x 10–6 m3; m = 0.100 kg; FB = gV
FB = (1000 kg/m3)(9.8 m/s2)(8 x 10-6 m3); FB = 0.0784 N
Fy = 0; T + FB – mg = 0; T = mg – FB ;
T = (0.100 kg)(9.8m/s2) – 0.0784 N; T = 0.980 N – 0.0784 N; T = 0.902 N

8 N
FB
T
T=6.5 N
*15-20. A solid object weighs 8 N in air. When it is suspended from a spring scale and submerged in water, the apparent weight is only 6.5 N. What is the density of the object?

To find density, we need to know the volume V of the block
which is the same as the volume of the water displaced.FB = 8 N – 6.5 N; FB = 1.5 N; FB = gV;
;  = 5333 kg/m3
*15-21. A cube of wood 5.0 cm on each edge floats in water with three-fourths of its volume submerged. (a) What is the weight of the cube? (b) What is the mass of the cube? (c) What is the specific gravity of the wood? [ V = (0.05 m)3 = 1.25 x 10–4 m3 ]
W
FB
The volume of water displaced is ¾ ofthe volume of the block:
VD = ¾(1.25 x 10-4 m3) = 9.38 x 10-5 m3 FB = W
FB = gVD = (1000 kg/m3)(9.8 m/s2)(9.38 x 10-5 m3) = 0.919 N
The weight is the same as the buoyant force if the block floats: W = 0.919 N
; m = 0.0938 kg or m = 93.8 g
Specific gravity: ; Specific gravity = 0.75

mg
FB
T
T
*15-22. A 20-g piece of metal has a density of4000 kg/m3. It is hung in a jar of oil (1500 kg/m3) by a thin thread until it is completely submerged. What is the tension in the thread?
First find volume of metal from its mass and density:
; FB = gV
FB = (1500 kg/m3)(9.8 m/s2)(5 x 10-6 m3); FB = 0.0735 N
Fy = 0; T + FB – mg = 0; T = mg – FB ; mg = (0.02 kg)(9.8 m/s2)T = 0.196 N – 0.0735 N; T = 0.123 N

*15-23. The mass of a rock is found to be 9.17 g in air. When the rock is submerged in a fluid of density 873 kg/m3, its apparent mass is only 7.26 g. What is the density of this rock?
Apparent weight = true weight – buoyant force; mAg = mg – mfg
mA = m – mf ; mf = m - mA = 9.17 g – 7.26 g = 1.91 g
*15-23.(Cont.) The volume Vf displaced is found from the mass displaced: mf = 1.91 g
; Vf = Vrock = 2.19 x 10-6 m3
; rock = 4191 kg/m3

Wb
WL
FB
WHe
*15-24. A balloon 40 m in diameter is filled with helium. The mass of the balloon and attached basket is 18 kg. What additional mass can be lifted by this balloon?
First find volume of balloon: Vb = (4/3)R3 = (4/3) (20 m)3Vb = 3.351 x 104 m3; FB = Wballoon + WHelium + Wadded
Fb = airgVb = (1.29 kg/m3)(9.8 m/s2)(3.351 x 104 m3) = 4.24 x 105 N
WHe = gV = (0.178 kg/m3)(9.8 m/s2)(3.351 x 104 m3) = 5.845 x 104 N; Wb = mbg
Wadd = FB – Wb – WHelium = 4.24 x 105 N – (18 kg)(9.8 m/s2) – 5.845 x 104 N
Wadd = 3.65 x 105 N; ; madd =37,200 kg
The densities of air and helium were taken from Table 15-1 in the text.

Applications of Bernoulli’s Equation

15-30. Consider the situation described by Problem 15-29. If the centers of each pipe are on the same horizontal line, what is the difference in pressure between the two connecting pipes?
For a horizontal pipe, h1 = h2: P1 + gh1 +½v12 = P2 + gh2 + ½v22
P1 - P2 =½v22 - ½v12 = ½(v22 – v12);  = 1000 kg/m3
P1 – P2 = ½(1000 kg/m3)[(24 m/s)2 – (6 m/s)2] = 2.70 x 105 Pa; P = 270 kPa

15-31. What is the emergent velocity of water from a crack in its container 6 m below the surface? If the area of the crack is 1.2 cm2, at what rate of flow does water leave the container? [ The pressures are the same at top and at crack: P1 = P2 ]
P1 +...