Baixe Halliday 8 ediçao e outras Notas de estudo em PDF para Física, somente na Docsity! Chapter 14 1. The pressure increase is the applied force divided by the area: Δp = F/A = F/πr2, where r is the radius of the piston. Thus Δp = (42 N)/π(0.011 m)2 = 1.1 × 105 Pa. 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: (a) . (b) 619 5. Let the volume of the expanded air sacs be Va and that of the fish with its air sacs collapsed be V. Then where ρw is the density of the water. This implies ρfishV = ρw(V + Va) or (V + Va)/V = 1.08/1.00, 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 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 Δp 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 θ dθ. Thus the net horizontal component of the force of the air is given by (b) We use 1 atm = 1.01 × 105 Pa to show that Δp = 0.90 atm = 9.09 × 104 Pa. The sphere radius is R = 0.30 m, so Fh = π(0.30 m)2(9.09 × 104 Pa) = 2.6 × 104 N. (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 620 . 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 , we have 21. Letting pa = pb, we find ρcg(6.0 km + 32 km + D) + ρm(y – D) = ρcg(32 km) + ρmy and obtain 22. (a) The force on face A of area AA due to the water pressure alone is Adding the contribution from the atmospheric pressure, F0= (1.0 × 105 Pa)(5.0 m)2 = 2.5 × 106 N, we have (b) The force on face B due to water pressure alone is Adding the contribution from the atmospheric pressure, F0= (1.0 × 105 Pa)(5.0 m)2 = 2.5 × 106 N, we obtain 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 8.00 m width), so the force is ρgh2w, and the change in force (as h is changed) is 623 CHAPTER 14 ρgw ( hf 2 – hi 2 ) = (998 kg/m3)(9.80 m/s2)(8.00 m)(4.002 – 2.002)m2 = 4.69 F 0B 4 10 5 N. 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 (c) We write τ = rF, where r is the effective moment arm. Then, 25. As shown in Eq. 14-9, the atmospheric pressure bearing down on the barometer’s mercury pool is equal to the pressure at the base of the mercury column: . Substituting the values given in the problem statement, we find the atmospheric pressure to be 26. The gauge pressure you can produce is 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/m 3. 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 624 Since p2 = 0, this implies 28. (a) According to Pascal’s principle F/A = f/a → F = (A/a)f. (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. 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 F top = 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 – F top 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 625 CHAPTER 14 . Since iceberg is floating, Eq. 14-18 applies: Since , the above equation implies . Thus, the visible fraction is (a) If the iceberg () floats in saltwater with , then the fraction would be . (b) On the other hand, if the iceberg floats in fresh water (), then the fraction would be . 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 ρ liquid. 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 F 0B 4 10−3 m3. 41. For our estimate of Vsubmerged we interpret “almost completely submerged” to mean Thus, equilibrium of forces (on the iron sphere) leads to where ri is the inner radius (half the inner diameter). Plugging in our estimate for Vsubmerged as well as the densities of water (1.0 g/cm3) and iron (7.87 g/cm3), we obtain the inner diameter: 42. From the “kink” in the graph it is clear that d = 1.5 cm. Also, the h = 0 point makes it clear that the (true) weight is 0.25 N. We now use Eq. 14-19 at h = d = 1.5 cm to obtain Fb = (0.25 N – 0.10 N ) = 0.15 N. 628 Thus, ρliquid g V = 0.15, where V = (1.5 cm)(5.67 cm2) = 8.5 F 0B 4 10−6 m3. Thus, ρliquid = 1800 kg/m3 = 1.8 g/cm3. 43. The volume Vcav of the cavities is the difference between the volume Vcast of the casting as a whole and the volume Viron contained: Vcav = Vcast – Viron. The volume of the iron is given by Viron = W/gρiron, where W is the weight of the casting and ρiron is the density of iron. The effective weight in water (of density ρw) is Weff = W – gρw Vcast. Thus, Vcast = (W – Weff)/gρw and 44. Due to the buoyant force, the ball accelerates upward (while in the water) at rate a given by Newton’s second law: ρwaterVg – ρballVg = ρballVa F 0D E ρball = ρwater (1 + “a”) where – for simplicity – we are using in that last expression an acceleration “a” measured in “gees” (so that “a” = 2, for example, means that a = 2(9.80 m/s2) = 19.6 m/s2). In this problem, with ρball = 0.300 ρwater, we find therefore that “a” = 7/3. Using Eq. 2-16, then the speed of the ball as it emerges from the water is v = , were a = (7/3)g and Δy = 0.600 m. This causes the ball to reach a maximum height hmax (measured above the water surface) given by hmax = v2/2g (see Eq. 2-16 again). Thus, hmax = (7/3)Δy = 1.40 m. 45. (a) If the volume of the car below water is V1 then Fb = ρwV1g = Wcar, which leads to (b) We denote the total volume of the car as V and that of the water in it as V2. Then which gives 46. (a) Since the lead is not displacing any water (of density ρw), the lead’s volume is not contributing to the buoyant force Fb. If the immersed volume of wood is Vi, then which, when floating, equals the weights of the wood and lead: Thus, 629 CHAPTER 14 (b) In this case, the volume V lead = mlead/ρlead also contributes to Fb. Consequently, which leads to 47. (a) When the model is suspended (in air) the reading is Fg (its true weight, neglecting any buoyant effects caused by the air). When the model is submerged in water, the reading is lessened because of the buoyant force: Fg – Fb. We denote the difference in readings as Δm. Thus, which leads to Fb = Δmg. Since Fb = ρwgVm (the weight of water displaced by the model) we obtain (b) The scaling factor is discussed in the problem (and for purposes of significant figures is treated as exact). The actual volume of the dinosaur is (c) Using ρ ≈ ρw = 1000 kg/m3, we find which yields 5.102 × 103 kg for the T. rex mass. 48. Let ρ be the density of the cylinder (0.30 g/cm3 or 300 kg/m3) and ρFe be the density of the iron (7.9 g/cm3 or 7900 kg/m3). The volume of the cylinder is Vc = (6F 0B 412) cm 3 = 72 cm3 = 0.000072 m3, and that of the ball is denoted Vb . The part of the cylinder that is submerged has volume Vs = (4 F 0B 4 12) cm 3 = 48 cm3 = 0.000048 m3. Using the ideas of section 14-7, we write the equilibrium of forces as ρgVc + ρFe gVb = ρw gVs + ρw gVb F 0D E Vb = 3.8 cm3 where we have used ρw = 998 kg/m3 (for water, see Table 14-1). Using Vb = πr3 we find r = 9.7 mm. 630 (b) Since the pressure difference is 59. (a) We use the Bernoulli equation: , where h1 is the height of the water in the tank, p1 is the pressure there, and v1 is the speed of the water there; h2 is the altitude of the hole, p2 is the pressure there, and v2 is the speed of the water there. ρ is the density of water. The pressure at the top of the tank and at the hole is atmospheric, so p1 = p2. Since the tank is large we may neglect the water speed at the top; it is much smaller than the speed at the hole. The Bernoulli equation then becomes and The flow rate is A2v2 = (6.5 × 10–4 m2)(2.42 m/s) = 1.6 × 10–3 m3/s. (b) We use the equation of continuity: A2v2 = A3v3, where and v3 is the water speed where the area of the stream is half its area at the hole. Thus v3 = (A2/A3)v2 = 2v2 = 4.84 m/s. The water is in free fall and we wish to know how far it has fallen when its speed is doubled to 4.84 m/s. Since the pressure is the same throughout the fall, . Thus 60. (a) The speed v of the fluid flowing out of the hole satisfies . Thus, ρ1v1A1 = ρ2v2A2, which leads to (b) The ratio of volume flow is (c) Letting R1/R2 = 1, we obtain Thus . 61. We rewrite the formula for work W (when the force is constant in a direction parallel to the displacement d) in terms of pressure: where V is the volume of the water being forced through, and p is to be interpreted as the pressure difference between the two ends of the pipe. Thus, 633 CHAPTER 14 62. (a) The volume of water (during 10 minutes) is (b) The speed in the left section of pipe is (c) Since , which is the atmospheric pressure, Thus, the gauge pressure is (1.97 atm – 1.00 atm) = 0.97 atm = 9.8 × 104 Pa. 63. (a) The friction force is (b) The speed of water flowing out of the hole is v = Thus, the volume of water flowing out of the pipe in t = 3.0 h is 64. (a) We note (from the graph) that the pressures are equal when the value of inverse- area-squared is 16 (in SI units). This is the point at which the areas of the two pipe sections are equal. Thus, if A1 = 1/ when the pressure difference is zero, then A2 is 0.25 m2. (b) Using Bernoulli’s equation (in the form Eq. 14-30) we find the pressure difference may be written in the form a straight line: mx + b where x is inverse-area-squared (the horizontal axis in the graph), m is the slope, and b is the intercept (seen to be –300 kN/ m2). Specifically, Eq. 14-30 predicts that b should be – ρ v22. Thus, with ρ = 1000 kg/ m3 we obtain v2 = m/s. Then the volume flow rate (see Eq. 14-24) is R = A2 v2 = (0.25 m2)( m/s) = 6.12 m3/s. If the more accurate value (see Table 14-1) ρ = 998 kg/m3 is used, then the answer is 6.13 m3/s. 65. (a) Since Sample Problem 14-8 deals with a similar situation, we use the final equation (labeled “Answer”) from it: 634 The stream of water emerges horizontally (θ0 = 0° in the notation of Chapter 4), and setting y – y0 = –(H – h) in Eq. 4-22, we obtain the “time-of-flight” Using this in Eq. 4-21, where x0 = 0 by choice of coordinate origin, we find (b) The result of part (a) (which, when squared, reads x2 = 4h(H – h)) is a quadratic equation for h once x and H are specified. Two solutions for h are therefore mathematically possible, but are they both physically possible? For instance, are both solutions positive and less than H? We employ the quadratic formula: which permits us to see that both roots are physically possible, so long as x < H. Labeling the larger root h1 (where the plus sign is chosen) and the smaller root as h2 (where the minus sign is chosen), then we note that their sum is simply Thus, one root is related to the other (generically labeled h' and h) by h' = H – h. Its numerical value is (c) We wish to maximize the function f = x2 = 4h(H – h). We differentiate with respect to h and set equal to zero to obtain or h = (40 cm)/2 = 20 cm, as the depth from which an emerging stream of water will travel the maximum horizontal distance. 66. By Eq. 14-23, we note that the speeds in the left and right sections are vmid and vmid, respectively, where vmid = 0.500 m/s. We also note that 0.400 m3 of water has a mass of 399 kg (see Table 14-1). Then Eq. 14-31 (and the equation below it) gives W = m vmid2 = –2.50 J . 67. (a) The continuity equation yields Av = aV, and Bernoulli’s equation yields , where Δp = p1 – p2. The first equation gives V = (A/a)v. We use this to substitute for V in the second equation, and obtain . We solve for v. The result is (b) We substitute values to obtain 635 CHAPTER 14 77. (a) We consider a point D on the surface of the liquid in the container, in the same tube of flow with points A, B and C. Applying Bernoulli’s equation to points D and C, we obtain which leads to where in the last step we set pD = pC = pair and vD/vC ≈ 0. Plugging in the values, we obtain (b) We now consider points B and C: Since vB = vC by equation of continuity, and pC = pair, Bernoulli’s equation becomes (c) Since pB ≥ 0, we must let pair – ρg(h1 + d + h2) ≥ 0, which yields 78. To be as general as possible, we denote the ratio of body density to water density as f (so that f = ρ/ρw = 0.95 in this problem). Floating involves equilibrium of vertical forces acting on the body (Earth’s gravity pulls down and the buoyant force pushes up). Thus, where V is the total volume of the body and Vw is the portion of it which is submerged. (a) We rearrange the above equation to yield which means that 95% of the body is submerged and therefore 5.0% is above the water surface. (b) We replace ρw with 1.6ρw in the above equilibrium of forces relationship, and find which means that 59% of the body is submerged and thus 41% is above the quicksand surface. 638 (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 F 0D E 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 = πr3 for the bubble, we then find its mass: mbub = 5.11 F 0B 4 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 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 – 1.00)(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 ) must be equal to the buoyant force upward: which yields 1.07F 0B 4103 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 Δp = 2ρHggx, which should be compensated for by the water pressure pw = ρwgh, where h = 11.2 cm. In these units, ρw = 1.00 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 639 CHAPTER 14 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: for the fraction of volume submerged of a floating object. When the liquid is water, as described in this problem, this relation leads to 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, 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, so that ρB/ρw = 3. (c) For liquid C, noting that one-fourth above means three-fourths below, so that ρC/ρw = 4/3. 85. Equilibrium of forces (on the floating body) is expressed as which leads to 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). 640