Baixe Solucionário Halliday 8ª edição vol. 2 - ch14 e outras Notas de aula em PDF para Física, somente na Docsity! Chapter 14 1. 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. 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 3 3 2 6 2 1 1 1 1 (2.6 g / cm )(0.50 L)(1000 cm / L)(980 cm/s ) 1.27 10 g cm/s 12.7 N. W m g V gρ= = = = × ⋅ = In the last step, we have converted grams to kilograms and centimeters to meters. Similarly, for the second and the third liquids, we have 3 3 2 2 2 2 2 (1.0 g/cm )(0.25 L)(1000 cm L)(980 cm s ) 2.5 NW m g V gρ= = = = and 3 3 2 3 3 3 3 (0.80 g/cm )(0.40 L)(1000 cm / L)(980 cm/s ) 3.1 N.W m g V gρ= = = = 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, 5 4(1.0 atm 0.96 atm)(1.013 10 Pa/atm)(3.4 m)(2.1 m) = 2.9 10 N.F = − × × 4. Knowing the standard air pressure value in several units allows us to set up a variety of conversion factors: (a) ( ) 5 2 2 1.01 10 Pa28 lb/in. 190 kPa 14.7 lb/in P  ×= =    . 619 CHAPTER 14 (b) 5 51.01 10 Pa 1.01 10 Pa (120 mmHg) 15.9 kPa, (80 mmHg) 10.6 kPa. 760 mmHg 760 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 3 3fish fish fish 1.08 g/cm and 1.00 g/cmw a m m V V V ρ ρ= = = = + 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 5 4 4 2 480 N1.0 10 Pa 3.8 10 Pa. 77 10 mi o Fp p 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 = 2πR2 sin θ dθ. Thus the net horizontal component of the force of the air is given by / 2 0 22 2 2 2 0 2 sin cos sin .hF R p d R p R p π π θ θ θ π θ π π = ∆ = ∆ = ∆∫ (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 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. 108 (b) The gauge pressure at the feet of the Argentinosaurus is 3 3 2 feet brain 3 1 torr80 torr (1.06 10 kg/m )(9.8 m/s )(21 m) 133.33 Pa 80 torr 1642 torr 1722 torr 1.7 10 torr. p p ghρ ′= + = + × = + = ≈ × 18. At a depth h without the snorkel tube, the external pressure on the diver is 0p p ghρ= + 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 0p p p ghρ∆ = = = . (a) If 0.20 m,h = then 3 2 5 1atm(998 kg/m )(9.8 m/s )(0.20 m) 0.019 atm 1.01 10 Pa p ghρ∆ = = = × . (b) Similarly, if 4.0 m,h = then 3 2 5 1atm(998 kg/m )(9.8 m/s )(4.0 m) 0.39 atm 1.01 10 Pa p ghρ∆ = = ≈ × . 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 ( ) 2 3 3 2 4 2 21 2 1 1 (1.30 10 kg/m )(9.80 m/s )(4.00 10 m )(1.56 m 0.854 m) 4 4 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 4a g= , we have 3 3 2 brain heart 1 torr120 torr (1.06 10 kg/m )(4 9.8 m/s )(0.30 m) 133 Pa 120 torr 94 torr 26 torr. p p arρ= − = − × × = − = 111 CHAPTER 14 21. Letting pa = pb, we find ρcg(6.0 km + 32 km + D) + ρm(y – D) = ρcg(32 km) + ρmy and obtain ( ) ( ) ( )3 3 3 6.0km 2.9g cm6.0km 44km. 3.3g cm 2.9g cm c m c D ρ ρ ρ = = = − − 22. (a) The force on face A of area AA due to the water pressure alone is ( ) ( ) ( ) 32 3 3 2 6 (2 ) 2 1.0 10 kg m 9.8m s 5.0m 2.5 10 N. A A A w A A wF p A gh A g d dρ ρ= = = = × = × Adding the contribution from the atmospheric pressure, F0= (1.0 × 105 Pa)(5.0 m)2 = 2.5 × 106 N, we have 6 6 6 0' 2.5 10 N 2.5 10 N 5.0 10 N.A AF F F= + = × + × = × (b) The force on face B due to water pressure alone is ( ) ( ) ( ) 32 3 3 3 2avg 6 5 5 5 1.0 10 kg m 9.8m s 5.0m 2 2 2 3.1 10 N. B B B w dF p A g d gdωρ ρ  = = = = ×   = × Adding the contribution from the atmospheric pressure, F0= (1.0 × 105 Pa)(5.0 m)2 = 2.5 × 106 N, we obtain 6 6 6 0' 2.5 10 N 3.1 10 N 5.6 10 N.B BF F F= + = × + × = × 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 112 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 ρgw ( hf 2 – hi 2 ) = (998 kg/m3)(9.80 m/s2)(8.00 m)(4.002 – 2.002)m2 = 4.69 × 105 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 ( ) ( ) ( ) ( ) 22 3 3 2 0 9 1 1 1.00 10 kg m 9.80m s 314m 35.0m 2 2 1.88 10 N. D F gyW dy gWDρ ρ= = = × = × ∫ (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 ( ) ( ) ( ) ( ) ( ) 3 3 3 0 33 3 2 10 1 1 1 2 3 6 1 1.00 10 kg m 9.80m s 314m 35.0m 2.20 10 N m. 6 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, 31 6 21 2 35.0 m 11.7 m. 3 3 gWD Dr F gWD ρτ ρ = = = = = 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: 0p ghρ= . Substituting the values given in the problem statement, we find the atmospheric pressure to be 4 3 2 0 1 torr(1.3608 10 kg/m )(9.7835 m/s )(0.74035 m) 739.26 torr. 133.33 Pa p ghρ= = × = 26. The gauge pressure you can produce is ( ) ( ) ( )3 2 2 3 5 1000kg m 9.8m s 4.0 10 m 3.9 10 atm 1.01 10 Pa atm p ghρ − − × = − = − = − × × 113 CHAPTER 14 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' = 1100 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 3 3 3 3 2 3.56 10 N' 3.30 m . ' (1.10 10 kg/m )(9.80 m/s ) WV gρ ×= = = × 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. 33. The problem intends for the children to be completely above water. The total downward pull of gravity on the system is ( ) wood3 356 N N gVρ+ where N is the (minimum) number of logs needed to keep them afloat and V is the volume of each log: V = π(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 3 356 N 3 356 N N gV gNV N gV ρ ρ ρ ρ + = ⇒ = − 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” = 5.00 kg, and “3” = 3.00 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 ρb = 2ρw/3 = 2(1000 kg/m3)/3 ≈ 6.7 ×102 kg/m3. (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. 116 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) = 4.11 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 ( ) 312 4 3s oV r= π , where ro is the outer radius. Therefore, ( )3 3 32 2 800 kg/m (0.090 m) 1.22 kg. 3 3o m rπ πρ  = = =   (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 ( ) ( )( )3 33 3 4 34 4( ) 0.090 m 0.080 m 9.09 10 m . 3 3o i V r rπ π −= − = − = × The density is 3 3 4 3 1.22 kg 1.3 10 kg/m . 9.09 10 mm ρ −= = ×× 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 m m m′→ = + ∆ , then 2H O ( )b bF F A h h gρ′→ = + ∆ . With 0.010b b bF F F mg′∆ = − = , the alligator sinks by 2 2 3 3 2 H O H O 0.01 0.010(130 kg) 6.5 10 m 6.5 mm (998 kg/m )(0.20 m ) bF mgh Ag Agρ ρ −∆∆ = = = = × = . 117 CHAPTER 14 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 V V V V V − = = − . Since iceberg is floating, Eq. 14-18 applies: .g i f i fF m g m g m m= = ⇒ = Since m Vρ= , the above equation implies f i i i f f i f V V V V ρρ ρ ρ = ⇒ = . Thus, the visible fraction is frac 1 1f i i f V V ρ ρ = − = − (a) If the iceberg ( 3917 kg/miρ = ) floats in saltwater with 31024 kg/mfρ = , then the fraction would be 3 3 917 kg/mfrac 1 1 0.10 10% 1024 kg/m i f ρ ρ = − = − = = . (b) On the other hand, if the iceberg floats in fresh water ( 31000 kg/mfρ = ), then the fraction would be 3 3 917 kg/mfrac 1 1 0.083 8.3% 1000 kg/m i f ρ 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. 118 wood wood wood lead 3 3 4 3 lead 0.900( / ) 1.84 kg 1 / 1 (1.00 10 kg/m /1.13 10 kg/m ) 2.01 kg. w w m m m ρ ρ ρ ρ − = = − − × × = 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 ∆m. Thus, ( )g g bF F F mg− − = ∆ which leads to Fb = ∆mg. Since Fb = ρwgVm (the weight of water displaced by the model) we obtain 4 30.63776kg 6.378 10 m . 1000 kg/mm w mV ρ −∆= = ≈ × (b) The 120 scaling factor is discussed in the problem (and for purposes of significant figures is treated as exact). The actual volume of the dinosaur is 3 3 dino 20 5.102 m .mV V= = (c) Using ρ ≈ ρw = 1000 kg/m3, we find 3 3dino dino dino (1000kg/m ) (5.102 m )m m V ρ = ⇒ = 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 = (6×12) cm3 = 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 × 12) cm3 = 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 ⇒ 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. 121 CHAPTER 14 49. We use the equation of continuity. Let v1 be the speed of the water in the hose and v2 be its speed as it leaves one of the holes. A1 = πR2 is the cross-sectional area of the hose. If there are N holes and A2 is the area of a single hole, then the equation of continuity becomes ( ) 2 1 1 1 2 2 2 1 12 2 A Rv A v NA v v v NA Nr = ⇒ = = where R is the radius of the hose and r is the radius of a hole. Noting that R/r = D/d (the ratio of diameters) we find ( ) ( ) ( ) 22 2 1 22 1.9cm 0.91m s 8.1m s. 24 0.13cm Dv v Nd = = = 50. We use the equation of continuity and denote the depth of the river as h. Then, ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )8.2m 3.4m 2.3m s 6.8m 3.2m 2.6m s 10.5m 2.9m sh+ = which leads to h = 4.0 m. 51. This problem involves use of continuity equation (Eq. 14-23): 1 1 2 2A v A v= . (a) Initially the flow speed is 1.5 m/siv = and the cross-sectional area is iA HD= . At point a, as can be seen from Fig. 14-47, the cross-sectional area is ( ) ( )aA H h D b h d= − − − . Thus, by continuity equation, the speed at point a is (14 m)(55 m)(1.5 m/s) 2.96 m/s ( ) ( ) (14 m 0.80 m)(55 m) (12 m 0.80 m)(30 m) 3.0 m/s. i i i a a Av HDvv A H h D b h d = = = = − − − − − − ≈ (b) Similarly, at point b, the cross-sectional area is bA HD bd= − , and therefore, by continuity equation, the speed at point b is (14 m)(55 m)(1.5 m/s) 2.8 m/s. (14 m)(55 m) (12 m)(30 m) i i i b b Av HDvv A HD bd = = = = − − 52. The left and right sections have a total length of 60.0 m, so (with a speed of 2.50 m/s) it takes 60.0/2.50 = 24.0 seconds to travel through those sections. Thus it takes (88.8 – 24.0) s = 64.8 s to travel through the middle section. This implies that the speed in the 122 middle section is vmid = (110 m)/(64.8 s) = 0.772 m/s. Now Eq. 14-23 (plus that fact that A = πr2) implies rmid = rA where rA = 2.00 cm. Therefore, rmid = 3.60 cm. 53. Suppose that a mass ∆m of water is pumped in time ∆t. The pump increases the potential energy of the water by ∆mgh, where h is the vertical distance through which it is lifted, and increases its kinetic energy by 212 mv∆ , where v is its final speed. The work it does is 212W mgh mv∆ = ∆ + ∆ and its power is 21 . 2 W mP gh v t t ∆ ∆  = = + ∆ ∆   Now the rate of mass flow is ∆m/ ∆t = ρwAv, where ρw is the density of water and A is the area of the hose. The area of the hose is A = πr2 = π(0.010 m)2 = 3.14 × 10–4 m2 and ρwAv = (1000 kg/m3) (3.14 × 10–4 m2) (5.00 m/s) = 1.57 kg/s. Thus, ( ) ( ) ( ) ( ) 2 2 2 5.0m s1 1.57 kg s 9.8m s 3.0m 66 W. 2 2 P Av gh vρ     = + = + =       54. (a) The equation of continuity provides (26 + 19 + 11) L/min = 56 L/min for the flow rate in the main (1.9 cm diameter) pipe. (b) Using v = R/A and A = πd 2/4, we set up ratios: 2 56 2 26 56 / (1.9) / 4 1.0. 26 / (1.3) / 4 v v π π = ≈ 55. (a) We use the equation of continuity: A1v1 = A2v2. Here A1 is the area of the pipe at the top and v1 is the speed of the water there; A2 is the area of the pipe at the bottom and v2 is the speed of the water there. Thus v2 = (A1/A2)v1 = [(4.0 cm2)/(8.0 cm2)] (5.0 m/s) = 2.5m/s. (b) We use the Bernoulli equation: 2 21 1 1 1 1 2 2 22 2p v gh p v ghρ ρ ρ ρ+ + = + + , where ρ is the density of water, h1 is its initial altitude, and h2 is its final altitude. Thus 123 CHAPTER 14 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, 5 3 5(1.0 10 Pa) (1.4 m ) 1.4 10 J.W = × = × 62. (a) The volume of water (during 10 minutes) is ( ) ( ) ( ) ( ) ( ) 2 31 1 15m s 10min 60s min 0.03m 6.4m .4V v t A π = = =   (b) The speed in the left section of pipe is ( ) 2 2 1 1 2 1 1 2 2 3.0cm15m s 5.4m s. 5.0cm A dv v v A d       = = = =           (c) Since 2 21 11 1 1 2 2 2 1 2 1 02 2 and ,p v gh p v gh h h p pρ ρ ρ ρ+ + = + + = = , which is the atmospheric pressure, ( ) ( ) ( ) ( )2 22 2 5 3 32 0 1 2 5 1 11.01 10 Pa 1.0 10 kg m 15m s 5.4m s 2 2 1.99 10 Pa 1.97atm. p p v vρ  = + − = × + × −  = × = Thus, the gauge pressure is (1.97 atm – 1.00 atm) = 0.97 atm = 9.8 × 104 Pa. 63. (a) The friction force is 3 3 2 2(1.0 10 kg/m ) (9.8 m/s ) (6.0m) (0.040 m) 74 N. 4 f A p gdAωρ π = ∆ = = × =   (b) The speed of water flowing out of the hole is v = 2 .gd Thus, the volume of water flowing out of the pipe in t = 3.0 h is 2 2 2 2 3(0.040 m) 2(9.8 m/s ) (6.0 m) (3.0 h) (3600 s/h) 1.5 10 m . 4 V Avt π= = = × 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 126 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: 02 for the projectile motion.v gh v v= ⇒ = 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” 2( ) 2 ( ).H ht H h g g − −= = − − Using this in Eq. 4-21, where x0 = 0 by choice of coordinate origin, we find 0 2( )2 2 ( ) 2 (10 cm)(40 cm 10 cm) 35 cm.H hx v t gh h H h g −= = = − = − = (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: 2 2 2 2 0 4 2 x H H xh Hh h ± −− + = ⇒ = 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 2 2 2 2 1 2 .2 2 H H x H H xh h H+ − − −+ = + = Thus, one root is related to the other (generically labeled h' and h) by h' = H – h. Its numerical value is ' 40cm 10 cm 30 cm.h = − = 127 CHAPTER 14 (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 4 8 0 2 df HH h h dh = − = ⇒ = 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 2 21 1 2 2p v Vρ ρ∆ + = , where ∆p = p1 – p2. The first equation gives V = (A/a)v. We use this to substitute for V in the second equation, and obtain ( ) 22 21 12 2p v A a vρ ρ∆ + = . We solve for v. The result is ( ) ( ) 2 2 2 2 2 2 . ( / ) 1 p a pv A a A aρ ρ ∆ ∆= = − − (b) We substitute values to obtain ( ) 4 2 2 3 3 3 4 2 2 4 2 2 2(32 10 m ) (55 10 Pa 41 10 Pa) 3.06m/s. (1000kg / m ) (64 10 m ) (32 10 m ) v − − − × × − ×= = × − × Consequently, the flow rate is 4 2 2 3(64 10 m ) (3.06 m/s) 2.0 10 m / s.Av − −= × = × 68. We use the result of part (a) in the previous problem. (a) In this case, we have ∆p = p1 = 2.0 atm. Consequently, 5 2 3 2 2 4(1.01 10 Pa) 4.1m/s. (( / ) 1) (1000 kg/m ) [(5 / ) 1] pv A a a aρ ∆ ×= = = − − (b) And the equation of continuity yields V = (A/a)v = (5a/a)v = 5v = 21 m/s. (c) The flow rate is given by 128 V = πr2hw = π(1.50 cm)2(6.41 cm) = 45.3 cm3. 76. Since (using Eq. 5-8) Fg = mg = ρskier g V and (Eq. 14-16) the buoyant force is Fb = ρsnow g V, then their ratio is = = = = 0.094 (or 9.4%). 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 2 21 1 2 2D D D C C C p v gh p v ghρ ρ ρ ρ+ + = + + which leads to 2 2 2( ) 2 ( ) 2 ( )D CC D C D p pv g h h v g d h ρ −= + − + ≈ + where in the last step we set pD = pC = pair and vD/vC ≈ 0. Plugging in the values, we obtain 22(9.8 m/s )(0.40 m 0.12 m) 3.2 m/s.cv = + = (b) We now consider points B and C: 2 21 1 . 2 2B B B C C C p v gh p v ghρ ρ ρ ρ+ + = + + Since vB = vC by equation of continuity, and pC = pair, Bernoulli’s equation becomes air 1 2 5 3 3 2 4 ( ) ( ) 1.0 10 Pa (1.0 10 kg/m )(9.8 m/s )(0.25 m 0.40 m 0.12 m) 9.2 10 Pa. B C C Bp p g h h p g h h dρ ρ= + − = − + + = × − × + + = × (c) Since pB ≥ 0, we must let pair – ρg(h1 + d + h2) ≥ 0, which yields air air 1 1,max 2 10.3 m. p ph h d h ρ ρ ≤ = − − ≤ = 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, b g w wF F gV gVρ ρ= ⇒ = 131 CHAPTER 14 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 w w V f V ρ ρ = = 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 1.6 1.6 w w V f V ρ ρ = = which means that 59% of the body is submerged and thus 41% is above the quicksand surface. (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 = πr3 for the bubble, we then find its mass: mbub = 5.11 × 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 – 1.00)(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 m ) must be equal to the buoyant force upward: 132 3 3( ) (1g/cm ) (1200 cm ) 130 gt wm m g Vg mρ+ = ⇒ = −l l 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 = 11.2 cm. In these units, ρw = 1.00 g/cm3 and ρHg = 13.6 g/cm3 (see Table 14-1). We obtain 3 3 Hg (1.00 g/cm ) (11.2 cm) 0.412 cm. 2 2(13.6 g/cm ) wghx g ρ ρ = = = 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 3 3 3 2 10 N 1.02 10 m (1000 kg/m )(9.8m/s )b w F Vg Vρ −= ⇒ = = × 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 2 3 2 2 3 3 6.0 N 6.0 10 kg/m . (9.8 m/s ) (1.02 10 m ) ρ −= = ×× 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 V V ρ ρ = for the fraction of volume submerged of a floating object. When the liquid is water, as described in this problem, this relation leads to 1 w ρ ρ = 133