**UFBA**

# Resolução Halliday Volume 2 - Oscilações, Ondas, Termodinâmica

(Parte **1** de 3)

1. The air inside pushes outward with a force given by piA, where pi is the pressure inside the room and A is the area of the window. Similarly, the air on the outside pushes inward with a force given by poA, where po is the pressure outside. The magnitude of the net force is F = (pi – po)A. Since 1 atm = 1.013 × 105 Pa,

2. We note that the container is cylindrical, the important aspect of this being that it has a uniform cross-section (as viewed from above); this allows us to relate the pressure at the bottom simply to the total weight of the liquids. Using the fact that 1L = 1000 cm3, we find the weight of the first liquid to be

33262 | (2.6 g/cm)(0.50 L)(1000 cm/L)(980 cm/s)1.2710gcm/s12.7 N. |

In the last step, we have converted grams to kilograms and centimeters to meters. Similarly, for the second and the third liquids, we have and

The total force on the bottom of the container is therefore F = W1 + W2 + W3 = 18 N.

3. The pressure increase is the applied force divided by the area: ∆p = F/A = F/πr2, where r is the radius of the piston. Thus ∆p = (42 N)/π(0.011 m)2 = 1.1 × 105 Pa. This is equivalent to 1.1 atm.

4. The magnitude F of the force required to pull the lid off is F = (po – pi)A, where po is the pressure outside the box, pi is the pressure inside, and A is the area of the lid. Recalling that 1N/m2 = 1 Pa, we obtain

5. Let the volume of the expanded air sacs be Va and that of the fish with its air sacs collapsed be V. Then

fish1.08 g/cm | and 1.0 g/cmw |

33fish fish amm V V

whereρw is the density of the water. This implies ρfishV = ρw(V + Va) or (V + Va)/V = 1.08/1.0, which gives Va/(V + Va) = 7.4%.

6. Knowing the standard air pressure value in several units allows us to set up a variety of conversion factors:

14.7 lb/in

(b)551.0110Pa1.0110 Pa (120 mmHg)15.9 kPa, | (80 mmHg)10.6 kPa. |

760 mmHg760 mmHg

7. (a) The pressure difference results in forces applied as shown in the figure. We consider a team of horses pulling to the right. To pull the sphere apart, the team must exert a force at least as great as the horizontal component of the total force determined by “summing” (actually, integrating) these force vectors.

We consider a force vector at angle θ. Its leftward component is ∆p cos θdA, where dA is the area element for where the force is applied. We make use of the symmetry of the problem and let dA be that of a ring of constant θ on the surface. The radius of the ring is r = R sin θ, where R is the radius of the sphere. If the angular width of the ring is dθ, in radians, then its width is R dθ and its area is dA = 2πR2 sin θdθ. Thus the net horizontal component of the force of the air is given by

2 sin cos sin.hFRpdRpRpππθθθπθππ =∆ = ∆ = ∆³

(b) We use 1 atm = 1.01 × 105 Pa to show that ∆p = 0.90 atm = 9.09 × 104 Pa. The sphere radius is R = 0.30 m, so

(c) One team of horses could be used if one half of the sphere is attached to a sturdy wall. The force of the wall on the sphere would balance the force of the horses.

8. Note that 0.05 atm equals 5065 N/m2. Application of Eq. 14-7 with the notation in this problem leads to dmax = 5065 ρliquidg with SI units understood. Thus the difference of this quantity between fresh water (998 kg/m3) and Dead Sea water (1500 kg/m3) is

10. Recalling that 1 atm = 1.01 × 105 Pa, Eq. 14-8 leads to

1. The pressure p at the depth d of the hatch cover is p0 + ρgd, where ρ is the density of ocean water and p0 is atmospheric pressure. The downward force of the water on the hatch cover is (p0 + ρgd)A, where A is the area of the cover. If the air in the submarine is at atmospheric pressure then it exerts an upward force of p0A. The minimum force that must be applied by the crew to open the cover has magnitude

F = (p0 + ρgd)A – p0A = ρgdA = (1024 kg/m3)(9.8 m/s2)(100 m)(1.2 m)(0.60 m) = 7.2 × 105 N.

13. With A = 0.000500 m2 and F = pA (with p given by Eq. 14-9), then we have ρghA = 9.80 N. This gives h≈ 2.0 m, which means d + h = 2.80 m.

14. Since the pressure (caused by liquid) at the bottom of the barrel is doubled due to the presence of the narrow tube, so is the hydrostatic force. The ratio is therefore equal to 2.0. The difference between the hydrostatic force and the weight is accounted for by the additional upward force exerted by water on the top of the barrel due to the increased pressure introduced by the water in the tube.

15. When the levels are the same the height of the liquid is h = (h1 + h2)/2, where h1 and h2 are the original heights. Suppose h1 is greater than h2. The final situation can then be achieved by taking liquid with volume A(h1 – h) and mass ρA(h1 – h), in the first vessel, and lowering it a distance h – h2. The work done by the force of gravity is

W = ρA(h1 – h)g(h – h2). We substitute h = (h1 + h2)/2 to obtain

0.635 J

16. Letting pa = pb, we find ρcg(6.0 km + 32 km + D) + ρm(y – D) = ρcg(32 km) + ρmy and obtain

3.3g cm 2.9g cmcmc

17. We can integrate the pressure (which varies linearly with depth according to Eq. 14-7) over the area of the wall to find out the net force on it, and the result turns out fairly intuitive (because of that linear dependence): the force is the “average” water pressure multiplied by the area of the wall (or at least the part of the wall that is exposed to the water), where “average” pressure is taken to mean 12(pressure at surface + pressure at bottom). Assuming the pressure at the surface can be taken to be zero (in the gauge pressure sense explained in section 14-4), then this means the force on the wall is 12ρgh multiplied by the appropriate area. In this problem the area is hw (where w is the 8.0 m width), so the force is 12ρgh2w, and the change in force (as h is changed) is

18. (a) The force on face A of area A due to the water pressure alone is

A A w A A wFp A gh A g d dρρ== = = × =×

Adding the contribution from the atmospheric pressure, F0= (1.0 × 105 Pa)(5.0 m)2 =

Adding the contribution from the atmospheric pressure, F0= (1.0 × 105 Pa)(5.0 m)2 =

19. (a) At depth y the gauge pressure of the water is p = ρgy, where ρ is the density of the water. We consider a horizontal strip of width W at depth y, with (vertical) thickness dy, across the dam. Its area is dA = W dy and the force it exerts on the dam is dF = p dA = ρgyW dy. The total force of the water on the dam is

D Fg yW dy gWDρρ==

(b) Again we consider the strip of water at depth y. Its moment arm for the torque it exerts about O is D – y so the torque it exerts is dτ = dF(D – y) = ρgyW (D – y)dy and the total torque of the water is

D gyW D y dy gW D D gWDτρ ρ ρ§·

(c) We write τ = rF, where r is the effective moment arm. Then, gWD D r

Fg WD

20. The gauge pressure you can produce is

where the minus sign indicates that the pressure inside your lung is less than the outside pressure.

21. (a) We use the expression for the variation of pressure with height in an incompressible fluid: p2 = p1 – ρg(y2 – y1). We take y1 to be at the surface of Earth, where the pressure is p1 = 1.01 × 105 Pa, and y2 to be at the top of the atmosphere, where the pressure is p2 = 0. For this calculation, we take the density to be uniformly 1.3 kg/m3 .

Then,

(b) Let h be the height of the atmosphere. Now, since the density varies with altitude, we integrate

Assuming ρ = ρ0 (1 - y/h), where ρ0 is the density at Earth’s surface and g = 9.8 m/s2 for 0≤y≤h, the integral becomes

2 h ypp g dy p ghh

Since p2 = 0, this implies

(b) We obtain

The ratio of the squares of diameters is equivalent to the ratio of the areas. We also note that the area units cancel.

23. Eq. 14-13 combined with Eq. 5-8 and Eq. 7-21 (in absolute value) gives mg = kxA1 A2 .

(Parte **1** de 3)