**UFBA**

# Respostas e resolução Física Halliday 1 2 3 4 8 edição - ch14

(Parte **3** de 3)

(c) The answer to part (b) suggests that a person in that situation is able to breathe.

79. We note that in “gees” (where acceleration is expressed as a multiple of g) the given acceleration is 0.225/9.8 = 0.02296. Using m = ρV, Newton’s second law becomes

ρwatVg – ρbubVg = ρbubVa | ⇒ ρbub = ρwat (1 + “a”) |

where in the final expression “a” is to be understood to be in “gees.” Using ρwat = 998 kg/m3 (see Table 14-1) we find ρbub = 975.6 kg/m3. Using volume V = pir3 for the bubble, we then find its mass: mbub = 5.1 × 10−7 kg.

80. The downward force on the balloon is mg and the upward force is Fb = ρoutVg. Newton’s second law (with m = ρinV) leads to

out out in in in 1 .Vg Vg Va g aρρ ρ ρ ρ

The problem specifies ρout / ρin = 1.39 (the outside air is cooler and thus more dense than the hot air inside the balloon). Thus, the upward acceleration is (1.39 – 1.0)(9.80 m/s2) = 3.82 m/s2.

81. We consider the can with nearly its total volume submerged, and just the rim above water. For calculation purposes, we take its submerged volume to be V = 1200 cm3. To float, the total downward force of gravity (acting on the tin mass mt and the lead mass m ) must be equal to the buoyant force upward:

which yields 1.07×103 g for the (maximum) mass of the lead (for which the can still floats). The given density of lead is not used in the solution.

82. If the mercury level in one arm of the tube is lowered by an amount x, it will rise by x in the other arm. Thus, the net difference in mercury level between the two arms is 2x, causing a pressure difference of ∆p = 2ρHggx, which should be compensated for by the water pressure pw = ρwgh, where h = 1.2 cm. In these units, ρw = 1.0 g/cm3 and ρHg = 13.6 g/cm3 (see Table 14-1). We obtain

83. Neglecting the buoyant force caused by air, then the 30 N value is interpreted as the true weight W of the object. The buoyant force of the water on the object is therefore (30 – 20) N = 10 N, which means

is the volume of the object. When the object is in the second liquid, the buoyant force is (30 – 24) N = 6.0 N, which implies

84. An object of mass m = ρV floating in a liquid of density ρliquid is able to float if the downward pull of gravity mg is equal to the upward buoyant force Fb = ρliquidgVsub where Vsub is the portion of the object which is submerged. This readily leads to the relation:

sub iquidl Vρ ρ

CHAPTER 14 for the fraction of volume submerged of a floating object. When the liquid is water, as described in this problem, this relation leads to

1wρ ρ since the object “floats fully submerged” in water (thus, the object has the same density as water). We assume the block maintains an “upright” orientation in each case (which is not necessarily realistic).

(a) For liquid A, 1

2Aρ ρ so that, in view of the fact that ρ = ρw, we obtain ρA/ρw = 2. (b) For liquid B, noting that two-thirds above means one-third below,

3Bρ ρ so that ρB/ρw = 3. (c) For liquid C, noting that one-fourth above means three-fourths below,

4Cρ ρ so that ρC/ρw = 4/3. 85. Equilibrium of forces (on the floating body) is expressed as body liqui d submerged body totalbF m g gV gVρ ρ= ⇒ = which leads to submerged body total liquid

We are told (indirectly) that two-thirds of the body is below the surface, so the fraction above is 2/3. Thus, with ρbody = 0.98 g/cm3, we find ρliquid ≈ 1.5 g/cm3 — certainly much more dense than normal seawater (the Dead Sea is about seven times saltier than the ocean due to the high evaporation rate and low rainfall in that region).

(Parte **3** de 3)