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

(Parte 1 de 3)

Chapter 14

1. The pressure increase is the applied force divided by the area: ∆p = F/A = F/pir2, where r is the radius of the piston. Thus

This is equivalent to 1.1 atm.

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

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 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,

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

14.7 lb/in

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

(b) 760 mmHg760 mmHg

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

3 3fish fish a m m where ρw 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) = 0.074 = 7.4%.

6. 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

7710 mioFpp A −

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 = 2piR2 sin θ dθ. Thus the net horizontal component of the force of the air is given by

2 sin cos sin.hFRpdRpRppipiθθθpiθpipi = ∆ = ∆ = ∆∫

(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. We estimate the pressure difference (specifically due to hydrostatic effects) as follows:

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

10. Note that 0.05 atm equals 5065 Pa. Application of Eq. 14-7 with the notation in this problem leads to

max liquid liquid liquid

0.05 atm5065 Papd

Thus the difference of this quantity between fresh water (998 kg/m3) and Dead Sea water (1500 kg/m3) is

max 2 3 3 fw sw

9.8 m/s 998 kg/m 1500 kg/md g ρ ρ

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

 9.80 N This gives h ≈ 2.0 m, which means d + h = 2.80 m.

12. With A = 0.000500 m2 and F = pA (with p given by Eq. 14-9), then we have ρghA = 13. In this case, Bernoulli’s equation reduces to Eq. 14-10. Thus,

14. Using Eq. 14-7, we find the gauge pressure to be gaugepghρ=, where ρ is the density of the fluid medium, and h is the vertical distance to the point where the pressure is equal to the atmospheric pressure.

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The gauge pressure at a depth of 20 m in seawater is

On the other hand, the gauge pressure at an altitude of 7.6 km is

Therefore, the change in pressure is

15. The hydrostatic blood pressure is the gauge pressure in the column of blood between feet and brain. We calculate the gauge pressure using Eq. 14-7.

(a) The gauge pressure at the brain of the giraffe is

(b) The gauge pressure at the feet of the giraffe is

3 3 2 feet heart

(c) The increase in the blood pressure at the brain as the giraffe lower is head to the level of its feet is

16. 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.

17. The hydrostatic blood pressure is the gauge pressure in the column of blood between feet and brain. We calculate the gauge pressure using Eq. 14-7.

(a) The gauge pressure at the heart of the Argentinosaurus is

3 3 2 heart brain

(b) The gauge pressure at the feet of the Argentinosaurus is

3 3 2 feet brain

18. At a depth h without the snorkel tube, the external pressure on the diver is where 0p is the atmospheric pressure. Thus, with a snorkel tube of length h, the pressure difference between the internal air pressure and the water pressure against the body is

(b) Similarly, if 4.0 m,h= then

19. 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

W gA h hρ −

20. To find the pressure at the brain of the pilot, we note that the inward acceleration can be treated from the pilot’s reference frame as though it is an outward gravitational acceleration against which the heart must push the blood. Thus, with 4ag=, we have

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133 Pa

21. 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 cmc m c

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

A A A w A A wF p A gh A g d dρ ρ= = = = × = ×

Adding the contribution from the atmospheric pressure,

Adding the contribution from the atmospheric pressure,

23. 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 (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 ρgh multiplied by the appropriate area. In this problem the area is hw (where w is the 8.0 m width), so the force is ρgh2w, and the change in force (as h is changed) is

24. (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

(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 gWDτ ρ ρ ρ

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

25. As shown in Eq. 14-9, the atmospheric pressure 0p bearing down on the barometer’s mercury pool is equal to the pressure ghρ at the base of the mercury column: 0pghρ=.

Substituting the values given in the problem statement, we find the atmospheric pressure to be

26. The gauge pressure you can produce is

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

27. (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 yp p g dy p ghh

Since p2 = 0, this implies

28. (a) According to Pascal’s principle F/A = f/a → F = (A/a)f.

(b) We obtain 2

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

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

With A2 = 18A1 (and the other values given in the problem) we find m = 8.50 kg.

30. (a) The pressure (including the contribution from the atmosphere) at a depth of htop = L/2 (corresponding to the top of the block) is where the unit Pa (Pascal) is equivalent to N/m2. The force on the top surface (of area A = L2 = 0.36 m2) is

Ftop = ptop A = 3.75 × 104 N.

(b) The pressure at a depth of hbot = 3L/2 (that of the bottom of the block) is where we recall that the unit Pa (Pascal) is equivalent to N/m2. The force on the bottom surface is

Fbot = pbot A = 3.96 × 104 N.

(c) Taking the difference Fbot – Ftop cancels the contribution from the atmosphere (including any numerical uncertainties associated with that value) and leads to which is to be expected on the basis of Archimedes’ principle. Two other forces act on the block: an upward tension T and a downward pull of gravity mg. To remain stationary, the tension must be

(d) This has already been noted in the previous part: 32.1810NbF=×, and T + Fb = mg.

31. (a) The anchor is completely submerged in water of density ρw. Its effective weight is Weff = W – ρw gV, where W is its actual weight (mg). Thus,

W WV gρ

(b) The mass of the anchor is m = ρV, where ρ is the density of iron (found in Table 14-1). Its weight in air is

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32. (a) Archimedes’ principle makes it clear that a body, in order to float, displaces an amount of the liquid which corresponds to the weight of the body. The problem (indirectly) tells us that the weight of the boat is W = 35.6 kN. In salt water of density ρ' = 10 kg/m3, it must displace an amount of liquid having weight equal to 35.6 kN.

(b) The displaced volume of salt water is equal to

In freshwater, it displaces a volume of V = W/ρg = 3.63 m3, where ρ = 1000 kg/m3. The difference is V – V ' = 0.330 m3.

3. The problem intends for the children to be completely above water. The total downward pull of gravity on the system is

where N is the (minimum) number of logs needed to keep them afloat and V is the volume of each log: V = pi(0.15 m)2 (1.80 m) = 0.13 m3. The buoyant force is Fb = ρwatergVsubmerged where we require Vsubmerged ≤ NV. The density of water is 1000 kg/m3. To obtain the minimum value of N we set Vsubmerged = NV and then round our “answer” for N up to the nearest integer:

( )wood water water wood which yields N = 4.28 → 5 logs.

34. Taking “down” as the positive direction, then using Eq. 14-16 in Newton’s second law, we have 5g – 3g = 5a (where “5” = 5.0 kg, and “3” = 3.0 kg and g = 9.8 m/s2). This gives a = g. Then (see Eq. 2-15) at2 = 0.0784 m (in the downward direction).

35. (a) Let V be the volume of the block. Then, the submerged volume is Vs = 2V/3. Since the block is floating, the weight of the displaced water is equal to the weight of the block, so ρw Vs = ρb V, where ρw is the density of water, and ρb is the density of the block. We substitute Vs = 2V/3 to obtain

(b) If ρo is the density of the oil, then Archimedes’ principle yields ρo Vs = ρbV. We substitute Vs = 0.90V to obtain ρo = ρb/0.90 = 7.4 ×102 kg/m3.

36. Work is the integral of the force (over distance – see Eq. 7-32), and referring to the equation immediately preceding Eq. 14-7, we see the work can be written as

W =waterρ∫gA(–y) dy where we are using y = 0 to refer to the water surface (and the +y direction is upward). Let h = 0.500 m. Then, the integral has a lower limit of –h and an upper limit of yf , with yf /h = − ρcylinder /ρwater = – 0.400. The integral leads to

W = ρwatergAh2(1 – 0.42) = 4.1 kJ .

37. (a) The downward force of gravity mg is balanced by the upward buoyant force of the liquid: mg = ρg Vs. Here m is the mass of the sphere, ρ is the density of the liquid, and Vs is the submerged volume. Thus m = ρVs. The submerged volume is half the total volume of the sphere, so ()31243soVr=pi, where ro is the outer radius. Therefore,

(b) The density ρm of the material, assumed to be uniform, is given by ρm = m/V, where m is the mass of the sphere and V is its volume. If ri is the inner radius, the volume is

The density is

38. If the alligator floats, by Archimedes’ principle the buoyancy force is equal to the alligator’s weight (see Eq. 14-17). Therefore,

2 2H O H O( )b gF F m g Ah gρ= = = . If the mass is to increase by a small amount m′→=+∆, then

2H O ( )b bF F A h h gρ′→ = +∆ . With 0.010bbbFFFmg′∆=−=, the alligator sinks by

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39. Let iV be the total volume of the iceberg. The non-visible portion is below water, and thus the volume of this portion is equal to the volume fV of the fluid displaced by the iceberg. The fraction of the iceberg that is visible is frac 1i f f i i

Since iceberg is floating, Eq. 14-18 applies:

.g i f i fF m g m g m m= = ⇒ = Since mVρ=, the above equation implies f i i i f f i f

Thus, the visible fraction is frac 1 1f i i fVV ρ ρ

(a) If the iceberg (3917kg/miρ=) floats in saltwater with 31024kg/mfρ=, then the fraction would be

1024 kg/mifρ ρ

(b) On the other hand, if the iceberg floats in fresh water (31000kg/mfρ=), then the fraction would be

1000 kg/mifρ ρ

40. (a) An object of the same density as the surrounding liquid (in which case the “object” could just be a packet of the liquid itself) is not going to accelerate up or down (and thus won’t gain any kinetic energy). Thus, the point corresponding to zero K in the graph must correspond to the case where the density of the object equals ρliquid. Therefore, ρball = 1.5 g/cm3 (or 1500 kg/m3).

(b) Consider the ρliquid = 0 point (where Kgained = 1.6 J). In this case, the ball is falling through perfect vacuum, so that v2 = 2gh (see Eq. 2-16) which means that K = mv2 = 1.6 J can be used to solve for the mass. We obtain mball = 4.082 kg. The volume of the ball is then given by mball/ρball = 2.72 × 10−3 m3.

(Parte 1 de 3)