Solucionario serway capitulo 15

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Solucionario serway capitulo 15

  1. 1. Chapter 15 Solutions 15.1 M = pimnV = (7860 kg/ m3) 7: (0.0150 m)3:| M= 15.2 The density of the nucleus is of the same order of magnitude as that of one proton, according to the assumption of close packing: _ 1.67 x 10-27 k _ 10,, kg/ m, p: ¿»(1045 m)3 <| E _ F __ 50.0(9.80) _ 15.3 P-A — K (0.500 X104? — 6.24 x106 N/ mz . . . F EL "15.4 Let FS be Its weight. Then each tire supports Fg/4, so P = í = 4A yielding FS = 4AP = 4(0.0240 m2)(2O0 x 103 N/ mz) = 1.92 x 104 N 15.5 The Earth's surface area is 471K? The force pushing inward over this area amounts to F = POA = P04JtR2 This force is the weight of the air: FS = mg = PO4IIRZ so the mass of the air is P042118 (1.013 x 105 N/ mz) 47116.37 x 106 m)2 g 7 9.80 m/ sl = 5'27 x 1°” k3 15.6 (a) P = Po + pgh = 1.013 X 105 Pa + (1024 kg/ m3)(9.80 m/ s2)(1000 m) P: 1.01 X 107 Pa (b) The gauge pressure is the difference in pressure between the water outside and the air inside the submarine, which we suppose is at 1.00 atmosphere. 195m8, = P — Po = pgh =1.00 x 107 Pa The resultant inward force on the porthole is then F = PKWFA = (1.00 x107 Pa) [M0150 m)2] = 7.09 x 105 N © 2000 by l-larcourt College Publishers. All rights reserved.
  2. 2. 2 Chapter 15 Solutions 15.7 F31 = Fm, “ or kx = pghA A kx and li = Ñ 2 —. 'I 1, = :1_62m (103 kg/ m3)(9.80 m/ s2)7t(l.00 x10“: m)Z F F 15.8 Since the pressure is the same on both sides, f = zi l 2 _ 15 000 F2 In this case, 200 — 300 F 15.9 Fx = (80.0 kg)(9.8O m/ sz) = 784 N = PA = (1.013 x 105 Pa)(A) F FV 784 f‘ "i? = 7-74“°"m2 a Goal Solution G: The suction cups used by burglars seen in movies are about 10 cm in diameter, and it seems reasonable that one of these might be able to support the weight of an 80-kg student. The area of a 10-cm cup is approximately: 7r(0.05 m)2 z 8 x 104 m2 : "Suction" ¡s not a new kind of force. Familiar forces hold the cup in equilibrium, one of which is the atmospheric pressure acting over the area of the cup. This problem is simply one situation where Newton's 2nd law can be applied. : The vacuum between cup and ceiling exerts no force on either. The atmospheric pressure of the air below the cup pushes up on it with a force. If the cup barely supports the student's weight, then the normal force of the ceiling is approximately zero, and +PDA — mg = 0 -2 Therefore, A = 3% = ïqgmíïfïgíïa) = 7.74 x 10-3 m2 <> This calculated area agrees with our prediction and corresponds to a suction cup that is 9.93 cm in diameter (Our 10 cm estimate was right on-a lucky guess considering that a burglar would probably use at least two suction cups, not one. ) Also, the suction cup in the drawing appears to be about 30 cm in diameter, plenty big enough to support the weight of the student. © 2000 by Harcourt College Publishers. All rights reserved.
  3. 3. Chapter 15 Solutions 3 ‘15.10 (a) Suppose the “vacuum Cleaner” functions as a high-vacuum pump. The air below the brick will exert on ita lifting force r = PA = (1.013 x 103 Pa)[n(1.43 x 102 n02} = 65.1 N (b) The octopus can pull the bottom away from the top shell with a force that could be no larger than F = PA = (P0 + pgh)A = [1.013 x 103 Pa + (1030 kg/ m3)(9.80 m/ s2)(32.3 m)][n(1.43 x104 m)2] F= "15.11 The excess water pressure (over air pressure) halfway down is Pgaug, = pgh = (1000 kg/ m3)(9.80 m/ s2)(1.20 m) = 1.18 x 104 Pa The force on the wall due to the water is F = PgauzeA = (1.18 x 104 Pa)(2.4O m)(9.60 m) = 2.71 x 105 N horizontally toward the back of the hole. 15.12 The pressure on the bottom due to the water is P1, = pgz = 1.96 x104 Pa so, I= ,,= P,, A= 5.88x10°N On each end, F = P A = (9.80 x 103 Pa)(20.0 m2) = 196 kN On the side F = P A = (9.80 x103 Pa)(60.0 m2) = 588 kN *15.13 In the reference frame of the fluid, the cart's acceleration causes a fictitious force to act backward, as if the acceleration of gravity were Vgz + a2 directed downward and backward at 9 = tan" (a / g) from the vertical. The center of the spherical shell is at depth d/ 2 below the air bubble and the pressure there is P= P0+pgeffh= Po+%pd Vg2+ a2 © 2000 by Harcourt College Publishers. All rights reserved.
  4. 4. 4 Chapter 15 Solutions 15.14 The air outside and water inside both exert atmospheric pressure, so only the excess water pressure pgh counts for the net force. Take a strip of hatch between depth II and h + dh. It feels force dF = PdA = pgh(2.00 m)dh (a) The total force is T F= Íd1= = "Í: nompgh (2.00 m)dh 200m " mom _I_ T 2 2.00 m . F= pg(2.00m) h? hmm 1.00 m 2.00 = (1000 kg/ m3)(9.80 m/ sz) ( m) [(2.00 m)2 — (1.00 m)2] 2 F: 29.4 kN (to the right) (b) The lever arm of tiF is the distance (h — 1.00 m) from hinge to strip: rzfar 41'00"" pgh 12001100: — 1.00m)dh II = l.00m h 3 h; 2.00 m t: pg(2.00 m) [ï- (1.00 mhz-j 1.00 m 3 2 r: 100Oí53-(980 m/ s2)(2.00 m) (m0 m3 3'00 ma) r: 16.3 kN - m counterclockwise 15.15 The pressure on the ball is given by: P = Pam, + pugh so the change in pressure on the ball from when it is on the surface of the ocean to when it is at the bottom of the ocean is AP = pngh. In addition: —V AP impgltV _ ‘hrpishr’ AV = B = d-E- — -—3-B—' , where B is the Bulk Modulus. = ' 4,31030 k hÏÏiÏÍÏÏoÏ/ ÏÏÏÍJ mmm mi‘ = — 0.0102 m3 Therefore, the volume of the ball at the bottom of the ocean is V — AV = g 7r(1.50 m)3 - 0.0102 m3 =14.137 m3 — 0.0102 m3 =14.127 m3 This gives a radius of 1.49964 m and a new diameter of 2.9993 m. Therefore the diameter decreases by . (i) 2000 by Harcourt College Publishers. All rights reserved.
  5. 5. 15.16 15.17 15.18 Chapter 15 Solutions 5 APO = pgAh = -2.66 x 103 Pa P = P0 + APO = (1.013 — 0.0266) x 105 Pa = 0.986 x 105 Pa P0 = 98h _: _ _ _ pg T (0.984 x 103 kg/ m3)(9.80 m/ sz) _ Po 1.013 x 105 Pa Some alcohol and water will evaporate. (a) Using the definition of density, we have m water 100 "0 = A, pm, = (5.00 cm2)(1.o0 g/ cm3) = The sketch at the right represents the situation after the water is added. A volume (A¡h¡) of mercury has been displaced by water in the right tube. The additional volume of mercury now in the left tube is Alil. Since the total volume of mercury has not changed, Azhz = Alh or i157; h (1) (b) At the level of the mercury-water interface in the right tube, we may write the absoluta pressure as: P = P0 + pwaterghw The pressure at this same level in the left tube is given by P = P0 + Pligg“! + k2) = P0 + Pwaterghw which, using equation (l) above, reduces to A PHK h [1 + = Pwater hw h : pwaler hw pHg (1 + 01' © 2000 by Harcourt College Publishers. All rights reserved.
  6. 6. 6 15.19 ‘15.20 Chapter 15 Solutions Thus, the level of mercury has risen a distance of _ . g 3 2 . _ , _ h — (1 00 km x o 0 Cm) — 0.490 cm above the original level. 10.0 (13.6 g/ cm3)[1+ Let h be the height of the water column added to the right side of the U-tube. Then when equilibrium is reached, the situation is as shown in the sketch at right. Now consider two points, A and B shown in the sketch, at the level of the water-mercury interface. By Pascal's Principle, the absolute pressure at B is the same as that at A. But, P¿ = P0 + pwgh + pHgghz and PB = P0 + p, ,,g(h¡ + h + hz) Thus: From PA = PBI Pichi ‘l’ pu! ’ "l" P111112 = pri-h ‘l’ PHghZI or (a) (b) 111 = [%ï-1j h, = (13.6 —1)(1.00 cm) = The balloon is nearly in equilibrium: ÏFy = "my = B - (Fglheiium - (Fglpayload = 0 0T P6118 V - pheliunvgv - mpayloadg = 0 This reduces to mpayload = (pair " Pheuumll/ = (1.29 kg/ m3 - 0.179 kg/ m3)(400 m3) mpaykmd I 444 kg Similarly, "lpayload = (pair ’ PhydmgenIV = (1.29 kg/ m3 - 0.0899 kg/ m3)(400 m3) mpaym = 480 kg The air does the lifting, nearly the same for the two balloons, © 2000 by Harcourt College Publishers. All rights reserved. wa ter Mercury
  7. 7. Chapter 15 Solutions 7 15. 21 The total weight supported is: Fx = (m + psV)g FX = [750 kg + (300 (0.100 mm] (9.80 FS = 735 N + (294 N/ m7-)A The buoyancy force is: n, = pmgV = (103 ‘¿É-jfiiso [(0.100 m)A] Ft = (980 N/ m2)A Since F1, = Fx, 1 (980 N/ mzM = 735 N + (294 N/ mZM 735 N °’ A = (980 — 294) N/ mz = 15.22 FS = (m + p, .V)g = P], = pwVg (see the figure with 15.21) Since V = Ah, m + psAIz = pwAh, nl and A : (Pm ' 15.23 (a) Before the metal is immersed: ZEV= T¡-Mg=0 or T1 = Mg = (1.00 kg) (9.80 = 9.80 N © 2000 by Harcourt College Publishers. All rights reserved.
  8. 8. 8 15.24 Chapter 15 Solutions (b) (a) (b) (C) After the metal is immersed: ZFy= T¡+B—Mg=0 or T2 = M8“ B = Mg- (puvwg V _ M _ 1.00 k ' p _ 2700 kg/ m3 Thus, 3 1.00 kg m r2 = Mg - B = 9.80 N — (1000 kg/ m) 2700 ¡(g/ ma 9.80 S7 = P = 130+ pgh Taking P0 = 1.0130 x103 N/ mz and h = 5.00 cm we find Pmp = 1.0179 x 105 N/ mz For h = 17.0 cm, we get Pm = 1.0297 x 105 N/ mz Since the areas of the top and bottom are A = (0.100 m x 0.100 m) = 10‘? m2 we find 5°, , = PmPA = .0179 >< 103 N Pm = 1.0297 x 103 N T+B—Mg=0 where and B = pwVg = (103 kg/ m3)(1.20 x104‘ m3)(9.80 m/ sz) = 11.8 N and Mg = 10.0(9.80) = 98.0 N Therefore, T = Mg- B = 98.0 -11.8 = 86.2 N Fm - Fw = (1.0297 —1.0179) x103 N = 11.8 N which is equal to B found in part (b). © 2000 by Harcourt College Publishers. All rights reserved. @
  9. 9. Chapter 15 Solutions 9 15.25 (a) According to Archimedes, B = pwm, Vg = (1.00 g/ cm3)[2O.0 x 20.0 x (20.0 - h)]g But B = Weight of Block = Mg = pwmvwmg = (0.650 g/ cm3)(20.0 cm)3g (0.650)(20.0)3g = (1.00)(20.0)(20.0)(20.0 —h)g 20.0 - h = 20.0(0.650) h = 2o. o(1— 0.650) = (b) B = Pg + Mg where M = mass of lead (1.00)(20.0)3g = (0.650)(20.O)3g + Mg M = (20.0)3(1.00 — 0.650) = (20.0)3(O.350) = 2800 g = 2.80 kg ”15.26 Consider spherical balloons of radius 12.5 cm containing helium at STI’ and immersed in air at 0°C and 1 atm. If the rubber envelope has mass 5.00 g, the upward force on each is B ‘Fgkle’ Fggnv = pair Vg"PHe V8’ menvg 4 Fup = (pair’pHe) '3- "738 " "¡m8 Jai? (¿Haz Fup = (1.29 - 0.179) 7r(0.125 m)3(9.80 m/ sz) — 5.00 x 104 kg(9.80 m/ sz) m. Fup = 0.0401 N If your weight (including hamess, Strings, and submarine sandwich) is (70.0 kg)(9.80 m / s2) = ese N you need this many balloons: © 2000 by Harcourt College Publishers. All rights reserved.
  10. 10. 10 Chapter 15 Solutions V 15'27 Pifiogï = pspheregv psphere = % PHQO : 4 poilg 1T) V ' pspheregv 7' 0 0 pm] =1T(500 kg/ m3) = 1250 kg/ m3 15.28 Since the frog floats, the buoyant force = the weight of the frog. Also, the weight of the displaced water = weight of the frog, so pooze V8 = "¡no38 1 or mm. = pum V = pum - í rr r3 = (1.35 x103 k / m3)E(6.O0 x 10‘? m) 3 b 2 3 5 3 l-Ience, mm, “ = 0.611 kg 15.29 The balloon stops rising when (pa-u - pHggV = Mg or, when (Pair - PHJV = M M 400 3 Therefore, V — (pair _ pHe) — (12564 _ 0.180) — 1430 m 15.30 Let I’ represent the length below water at equilibrium and M the tubes mass: ZF’, ,=0=>-Mg+pm¿flg= O Now with any excursion x from equilibrium —Mg + p7rr¡(/ ' — x)g = Ma Subtracting the equilibrium equation gives —-p7rr2gx = Ma n = -(p7rr2g/ M)x = —a>2x The opposite direction and direct proportionality of a to x imply SHM with angular frequency co: Vprrrzg/ M © 2000 by Harcourt College Publishers. All rights reserved.
  11. 11. *15.31 *15.32 "15.33 15.34 Chapter 15 Solutions Constant Velocity implies zero acceleration, which means that the submersible is in equilibrium under the gravitational force, the upward buoyant force, and the upward resistance force: ZFy = muy = 0 -(1.20 x 104 kg + m)g + pwgV +1100 N = O where m is the mass of the added water and V is the sphere’s volume. 47!’ 1100 N (1.20 x 104 kg + m) = (1.03 x 103) (1.50) 3 + so m = 2.67 x 103 kg By Archimedes’s principle, the weight of the fifty planes is equal to the weight of a horizontal slice of water 11.0 cm thick and circumscribed by the water line: AB = pwaterg(AV) 50(2.90 x 10" kg)g = (1030 kg/ m3)g(0.110 m)A giving A = 1.28 x 10"‘ m2 The acceleration of gravity does not affect the answer. Volume flow rate = A¡v¡ = A2122 20.0 L 1000 cm3 = 711.00 Cmyvhose = M0500 cmyvmuh 333 cm3/s (a) vhose = ‘¿MT = 333 3/ (b) vnozzle = = By Bernoulli's equation, N 1 N 1 8.00 X 10" Ñ +5 1000122 = 6.00 x 10" r-n-ï +5 1000(16v2) : >__ í 2.00 x 10* 'N— = 100005122) m2 l 2 v: 1.63 m/ s % = Av =1000It(5.00 x 10'¡)¡(1.63 m/ s) = 12.8 kg/ s © 2000 by Harcourt College Publishers. All rights reserved. 11 l: >4v
  12. 12. 12 Chapter 15 Solutions 15.35 Assuming the top is open to the atrnosphere, then P1 = P0 Note l’; = P0 Flow rate = 2.50 x 10*‘ m3/min = 4.17 x 10-5 m3/s (a) A1 >> A2 so v, << v2 Assuming v1 = 0, I, pvÏ pvÉ ¡+-2— +pgy¡= P2+T +pgyz v2 = tzgyol“ = l2(9.80)(16-0)l"2 = ¡n12 (b) Flow rate = A; v2 = (T)(17.7) = 4.17 x10‘5 m3/s 15.36 (a) Suppose the flow is very slow (¡Hi-pvlwvgy) _ = (P+%pv3+psyj . HVET I'll“ P+0+pg(564m)=1atrn+0+pg(2096m) P = 1 atm + 1000 kg/ m3(9.80 m/ s2)1532 m P: 1 atm + 15.0 MPa (b) The volume flow rate is 4500m3 _A _Il. 'd2U a ‘ ”‘ 4 v__——-¿——¿—: u¿í __45oo m3 1d 4 d 864005 7r(0.150 m)2 v= © 2000 by Harcourt College Publishers. All rights reserved.
  13. 13. Chapter 15 Solutions 13 (c) Imagine the pressure as applied to stationary water at the bottom of the pipe: 1 1 (P + ¡pvz + pgy) {P + ¡pvz + pgs/ J bottom top l P+0=1atm+2 (1000 kg/ m3)(2.98 m/ s) 2 + (1000 kg)9.80 m/ s2(1532 m) P = 1 atm + 15.0 MPa + 4.45 kPa The additional pressure is 4.45 kPa 15.37 Flow rate Q = 0.0120 m3/s = v¡A¡ = vzAz 0.0120 *15.38 (a) For upward flight of a water—drop projectile from geyser vent to fountain-top: v? = v; + Zag/ sy) o = v, ’ + 2(—9.so m/ s2)(+40.0 m) (b) Between geyser vent and fountain-top: 1 1 Pwïnvi +pgy1=P2+5 2 +pgy2 Air is so low in density that very nearly P1 = P2 = 1 atm. Then, 1 ï of + o = o + (9.30 m/ s2)(4o. o m) v1 = 23.0 m/ s (c) Between the Chamber and the fountain-top: 1 2 1 2 PH‘; avi +Pgyl= P2+ï P02 +pgyz P, + 0 + (1000 kg/ m3)(9.80 m/ s2)(-175 m) = P0 + 0 + (1000 kg/ m3)(9.80 m/ s2)(+40.0 m) P1 — P0 = (1000 kg/ m3)(9.80 m/ s2)(215 m) = 2.11 MPa © 2000 by Harcourt College Publishers. All rights reserved.
  14. 14. 14 Chapter 15 Solutions 15.39 Mg = (P1 - P¿)A for a balanced condition (16000)(9.80) A = 7.00x 104-13, where A = 80.0 m2, P2 = 7.00 x104 -0.196 x 104 = 6.80 x 104 Pa 2 2 15.40 P1 +2? = P2 +3202- (Bernoulli equation) Ai L7¡A1=U¡A2 where A—=4 2 A2 2 Pvi AP = T(15) = 21000 Pa v1: 2.00 m/ s v¿=4v¡ =8.00m/ s and Q = v, A¡ = 2.51 x104 m3/s 15.41 p“? = AP = pHgg Ah "15.42 The assumption of incompressibility is surely unrealistic, but allows an estimate of the speed: 1 z 1 z Pi+psyi+5pvi = P2+P8y2+ïPU2 1.00 atm + o + o = 0.237 atm + o «¿(1.20 kg/ mï‘) vi _ 2(1.oo — o.2s7)(1.o3 x 105 N/ mz) _ v2 “ V 1.20 kg/ m3 ‘ © 2000 by Harcourt College Publishers. All rights reserved.
  15. 15. 1 15.43 (a) Pg+pgh+0=P°+0+ïpzÉ v3=v2gh lfh=1.00m, v3: 4.43m/ s 1 2 1 2 (b) P+pgy+ï pv2 = P0+0+ï pv3 Since v2 = v3, P= PO'PSV SinceP20, Po (1.013 x105 Pa) P’ 5 z = (103 kg/ m3)(9.80 m/ sz) *15.44 ln the reservoir, the gauge pressure is 2.00 N =8.O0x104 Pa AP: From the equation of continuity: A101 = A2112 (2.50 x 10’5 m2)v¡ = (1.00 x 10*‘ mzlv2 v1 = (4.00 x 104)v2 = 10.3m Chapter 15 Solutions -‘. l: Ïl Í. 4 g xk l y l l] il l ÏIf; ï:v Thus, vÏ is negligible in comparison to v5. Then, from Bemoullïs equa tien: 1 1 (Pl"P2)*'P8Ï/1+ïPvf = PSH2+ÏPDÉ 8.00 x 104 Pa + 0+0 = o «¿(1000 kg/ m3) 022 2 .0 10‘? ! v2: (80 x a = 12.6m/ s 1000 kg/ m3 © 2000 by Harcourt College Publishers. All rights reserved. 15
  16. 16. 16 Chapter 15 Solutions 15.45 Apply Bemoulli’s equation between the top surface and the exiting stream. 1 Pg+O+pghg= P0+ïpvÏ +pgh 12,2, = 2g(h0 — h) 1), : /2g(h0 — h) x = v‘! y = h = á- gtz 2h 2= — 8 1 15.46 x = v, ,t h: ï gtz v, = 2g(hg - h) from Problem 45 and t= ‘f2—h 8 x= /2g(h0 — h) il = /4h(h0 — h) (a) Maximize x with respect to h: 1 e = HÏÏ :0 when dh , f4h(ho_¡1) 2 h h (b) Forh= ï0, vpvghg, t: ía, then x= vxf= © 2000 by Harcourt College Publishers. All rights reserved.
  17. 17. Chapter 15 Solutions 17 15.47 At equilibrium XF = 0 Filpp or FW, + mg = B where B is the buoyant force The applied force, w Fam, = B — mg where B E (Vol)(pw, ,,, .,)g and m = (Vohpban 4 so: Fapp : (V0l)g(pwa(er _ pball) = 5 7tr3g(pv'nler_ pbnll) 4 Pa”, = 3- 7r(1.90 x 10‘? m)3(9.80 m/ s2)(103 kg/ m3 — 84.0 kg/ m3) F ¿W = 0.258 N Goal Solution G: According to Archimedes’s Principle, the buoyant force acting on the submerged the ball will be equal to the weight of the water the ball will displace. The ball has a volume of about 30 cmv‘, so the weight of this water is approximately: B = FX = pVg = (1 g/ cm3)(3O cm3)(10 m/ sz) = 0.3 N Since the ball is much less dense than the water, the applied force will approximately equal this buoyant force. : Apply Newton's 2nd law to find the applied force. : At equilibrium, XF = 0 or Pa”, + mg — B = 0 Where the buoyant force is B = pwVg and p“, = 1000 kg/ m3 The applied force is then, Fam, = p, ,.Vg - mg Using m = pbauV to eliminate the unknown mass of the ball, this becomes 4 Fapp = Vg(Pru ' pball) = 3 W38(Pm - pball) 4 Fam, = 3 7t(1.90 x 104 m)3(9.80 m/ sZ)(103 kg/ m3 — 84 kg/ m3) Pa”, = 0.258 N The force is approximately what we expected, so our result is reasonable. lf the applied force would be greater than 0.258 N, the ball would sink until it hit the bottom (which would then provide a normal force directed upwards). © 2000 by Harcourt College Publishers. All rights reserved.
  18. 18. 18 Chapter 15 Solutions 15.48 Consider the diagram and apply Bemoulli’s equation to points A and B, Á . _ _. . . ._ taking y = 0 at the level of point B, and recognizing that UA is approximately , zero. x l . Í gr-. .“ h / B This gives: F L Vai/ Ed p p «_o 1 . PA + 5 p, ¿,(0)2 + p, ,,g(h - L sin 6) 1 : PB ‘l’ ï pwvlzï + Pu8(Ü) Now, recognize that PA = P5 = Pamosphem since both points are open to the atmosphere (neglecting variation of atmospheric pressure with altitude). Thus, we obtain v5 = V2g(h — L sin 9) = 2(9.80 m/ s2)[10.O m — (2.00 m) sin 30.0°] v3 = 13.3 m/ s Now the problem reduces to one of projectile motion with uy, - = v5 sin 30.0° = 6.64 m/ s.. Then, v5 = v5, + 2a(Ay) gives at the top of the arc (where y = ym“ and v_, , = 0) O = (6.64 m/ s)2 + 2(—-9.80 m/ szxym, — 0) or ymax = Error! 15.49 When the balloon comes into equilibrium, we must have zFy = B ' Fx, balloon ' FgHe ' Fgstring = 0 Fx, m“ is the weight of the string above the ground, and B is the buoyant force. Now F3, balloon = mballoon 8 Fg, He = pHe B = pair h and Fg, string = mstringï g © 2000 by Harcourt College Publishers. All rights reserved.
  19. 19. Chapter 15 Solutions 19 Therefore, we have I pair V8 " "¡balloon 8 ' pHe V8 - "lstrlngïl S’ :0 _ (pair ' pHe)V T "¡balloon L or h "¡string giving, 3 (1.29 — 0.179) kg/ m3 l- 0.250 kg h = 0.0500 kg (2.00 m) = 1.91 m 15.50 Assume vmme=0 P + 0 + 0 = 1 atm + %(1000)(30.0 m/ s) 2 + 1000(9.80)(0.500) Pgauge = P — 1 atm = 4.50 x 105 + 4.90 x103 = 455 kPa 15.51 The "balanced" condition is one in which the apparent weight of the body equals the apparent weight of the weights. This condition can be written as: FX—B= F¿-B' where B and B’ are the buoyant forces on the body and weights respectively. The buoyant force experienced by an object of volume V in air equals: Buoyant force = (Volume of objecflpmg P. so we have B = Vpmg and B‘ = pmg L F. Therefore, Fx n F; + [V - pmg Goal Solution The "balanced" condition is one in which the net torque on the balance is zero. Since the balance has lever arms of equal length, the tota] force on each pan is equal. Applying X1 = 0 around the pívot leads to I-‘¿—B= F_¿ —B' where B and B‘ are the buoyant forces on the body and weights respectively. The buoyant force experienced by an object of volume V in air is B = Vpaug and B‘: Vpmg © 2000 by Harcourt College Publishers. All rights reserved.
  20. 20. 20 Chapter 15 Solutions Since the volume of the weights is not given explicitly, we must use the density equation to eliminate it: With this substitution, the buoyant force on the weights becomes si o " pg Pau-S F1 Therefore, FA, = +(V - pmg Side note: We can now answer the popular riddle: Which weighs more, a pound of feathers or a pound of bricks? Answer: Like in the problem above, the feathers have a greater buoyant force than the bricks , so if they "weigh” the same on a scale as a pound of bricks, then the feathers must have more mass and therefore a greater "true weight. " 15.52 P = pgh 15.53 1.013 x105 = (l.29)(9.80)h n= For Mt. Everest, 29 30o ft = 8.88 km The cross-sectional area above water is 1 0.400. cm 2.46 rad T Jr (0.600 cm): - (0.200 cm)(0.566 cm) = 0.330 cmz (1.80 Cm "TTT- Cwj AM1 = .*r(O.600)2 = 1.13 cmz pwaler ¿’Aunder = pwoud Aall 8 = 0.709 g/ cm3 = 1.13 — 0.330 pwood : 1 .1 3 © 2000 by Harcourt College Publishers. All rights reserved.
  21. 21. Chapter 15 Solutions 21 15.54 At equilibrium, XFy = 0 0T B - Fspring ‘ Fx. He "' Pg, balloun = 0 giving Fspnng = kL = B - (Him. + mbauoon)g But B = weight of displaced air = pm Vg and In“? = pHe V Therefore, we have: k’- = pair V8 ' pHe V8 - mballoong _ (Pair " pHe)V ‘ "¡balloon or L k From the given data, this gives _ 3_ _ 3 _ 3 _ _ 3 L= (129kg/ m 0180k9gág1I)J(/ E3I:0m) 200 x10’ kg(9.80) z 15.55 Looking first at the top scale and the iron block, we have: T1+ B = Fgjron where T¡ is the tension in the spring scale, B is the buoyant force, and FS, ¡mn is the weight of the iron block. Now if mmm is the mass of the iron block, we have _ ¡mm piron "¡iron = Puan V 5° V = ‘¡displaced oil Then: B = Poil Virong miran P iron or T¡= [ —%‘-) m¡, °ng= ( —79;T60)(2.00)(9.80) = 17.3N U0“ Therefoïe/ T] = F3, ¡mn ' P011 Virung = mirong " poil 8 Next, we look at the bottom scale which reads T2 (i. e., exerts an upward force T1 on the system). Consider the extemal vertical forces acting on the beaker-oil-iron combination. Éfy=0 gives T1+T2-Fg, beaker'Fg, oil"Fg, iron:0 or Tz = (mmke, + mo“ + nz¡, o,, )g - T1 = (5.00 kg)(9.80 m/ sZ) - 17.3 N Thus, T2 = 31.7 N is the lower scale reading. © 2000 by Harcourt College Publishers. All rights reserved.
  22. 22. 22 Chapter 15 Solutions 15.56 Looking at the top scale and the iron block: "¡pe T, + B = Fgre where B = pOVRg = g is the buoyant force exerted on the iron block by the oil. Thus, mFe T1 = Fgre - B = "1Fe8 ‘ Po 8 or T¡ = (1 — -: %)mpeg is the reading on the top scale. Now, consider the bottom scale which exerts an upward force of T2 on the beaker-oil-iron combination. zFyzoáTl+T2"Fg, beaker-Fg, oiJ-Fg, Fe=0 T2 = Fxlbeake, + F810“ + Fxïpe - T1 = (mb + m0 + nzpgg -(1 - 152:) Inpfig or T2 = [mb + m0 + me] g is the reading on the bottom scale. PFe 15.57 The torque is r = I dr = I rdF From the figure, H 1 mi ylpsflí - yhvdy] = 0 The total force is given as á pgwI-Iz If this were applied at a height ye“ such that the torque remains unchanged, we have É 9310”‘ = yetf pgwHz] 1 ¿nd Her: = © 2000 by Harcourt College Publishers. All rights reserved.
  23. 23. Chapter 15 Solutions 23 15.58 (a) The pressure on the surface of the two hemispheres is constant at all points, and the force on each element of surface area is directed along the radius of the hemispheres. The applied force along the axis must balance the force on the "effective" area which is the projection of the actual surface onto a plane perpendicular to the x axis, A = ¡{R2 Therefore, F: (P0 — ¡muii (b) For the values given F = (Pg — Ü.100P0)II. '(0.300 m)2 = 0.254190 = 2.58 x 10‘ N 15.59 pc“ v = 3.083 g p¿, ,(xV) + po, (1 -x)V = 2.517 g pz“ (3083) x + 3.083(1- x) = 2.517 pCu 1 7.133 _1 2.517 _8.960 x‘ ‘sosa x=0.9004 15.60 (a) From XF = ma B - mshellg " "¡m8 = mtotala = (mshell + ml-Ie)" (1) Where B = pwm, Vg and 111,19 = p”? V 4 3 ¡n13 Also, V_3 ¡rr — 6 Putting these into equation (l) above, ¡n13 Ird3 M3 "¡shell + pHe T a : pwater? ’ mshell ' PHET 8 © 200D by Harcourt College Publishers. All rights reserved.
  24. 24. 24 Chapter 15 Solutions which gives ¡n13 (pwater ‘ PHe) T" mshell ¡n13 mshell + PHeT u: g (1000 — 0.130) (Ïjï- 4.00 kg 3 or n = m 9.80 m/ sz = 0.461 m/ sz 3 4.00 kg + (0.180 ‘¿f3 “——°'2%° m’ _ EA; _ 2(h-d) _ 2(4.00m—0.200m) _ (b) ’_/ a- - V a _ d 0.461m/ s2 ' 15.61 Energy is conserved (K+U), -+AE= (K+U)¡ OH-"ÉL +0=% mv2+0 v = x/ ¿í = v2.00 m(9.80 m/ sz) = 1 *15.62 lnertia of the disk: l = ï MRZ = ;-(10.0 kg)(0.250 m) 2 = 0.312 kg - m3 Angular acceleration: a), = co, » + at a = ( )(2" "df mi“) = —o.524 rad/ sz 60.0 s 1 rev 60.0 s Braking torque: 21': la: —fl1 = Ia, sof= -‘g—a , _ . _0.12kg-1.24 2_ Friction force. f—( 3 rad/ S) —0.744N _ _ _ L _ 0.744N _ Normal force. f- ¡(kn = n — M — 0.500 — 1.49 N n 1.49 N 5a“? PWSW": P = X = ¡r(2.50 >< 10-2 m)2 = © 2000 by Harcourt College Publishers. All rights reserved.
  25. 25. Chapter 15 Solutions 25 "15.63 (a) We imagine the superhero to produce a perfect vacuum in the straw. Take point 1 at the water surface in the basin and point 2 at the water surface in the straw: 1 2 1 2 P¡+ïpv¡ +pgy1=Pz+ïpvz +pgyz 1.013 x105 N/ mz + 0 + 0 = 0 + 0 + (1000 kg/ mÏ‘)(9.8O m/ s2)y¡ v2= (b) No atmosphere can lift the water in the straw through height difference. 15.64 Differentiating P = pgy we have É _ dy ‘flog Also at any given height the density of air is proportional to pressure, or P Po E T Combining these two equations we have p d]? po h . _ = _ _. d fpu p SP0 f0 y and integrating gives P: P0e_aJI 15.65 Let s stand for the edge of the cube, h for the depth of immersion, pm stand for the density of the ice, p“, stand for density of water, and p, , stand for density of the alcohol. (a) According to Archimedes’s principle, at equilibrium we have p. pm, gs" = pmghsï = > h = s fl pu. With p, “ = 0.917 x 103 kg/ m3 pu, = 1.00 x103 kg/ m3 and s = 20.0 mm we get h = 2o. o(o.917) = 18.34 mm = © 2000 by Harcourt College Publishers. All rights reserved.
  26. 26. 26 Chapter 15 Solutions (b) We assume that the top of the cube is still above the alcohol surface. Letting h, , stand for the thickness of the alcohol layer, we have P pngszha T pzugszhxlv = picegss = ° hnv = 5 “(la ha l} lll With p, = 0.806 x103 kg/ m3 and h, = 5.00 mm we obtain 11,, = 13.34 — (0.806)(5.00) = 14.31 mm = (c) Here h, ',, = s - h}, , so Archimedess principle gives . . . P" gs‘ h; + pu-ss’ (s — h}, ) = press’ = > p" h, ‘ + pm (s — hi )= mas :9 h; = = 2o. = s.557= 15.66 (a) A study of the forces on the balloon shows that the tangential restoring force is given as: Fx = —B sin 6+ mg sin 6 = -(B - mg) sin 9 But B = pa” Vg and m: pHeV Also, sin 0= 9(for small 6) 50 Fr = ‘(pair V8 ' pHe VSW = "(pair ' PHe) V89 / /_ 5 L and P, = ‘(Pair — PHe)Vg% = —ks , V with k = (pair " pHe) —L& But 0: (b) Then T = 2¡r fi- = 2¡r giving _ f (0.180)(3.00 m) _ T’ 2” (1.29 — 0.180)(9.80 m/ sz) ’ © 2000 by Harcourt College Publishers. All rights reserved.
  27. 27. 15.67 15.68 (b) Chapter 15 Solutions The flow rate, Av, as given may be expressed as follows: 25.0 liters/ 30.0 s = 0.833 liters/ s = 833 crn3/ s The area of the faucet tap is ¡r cmz, so we can find the Velocity as flow rate 833cm3/s v= -T- :7?- =265cm/ s= 2.65m/ s We choose point 1 to be in the entrance pipe and point 2 to be at the faucet tap. A11), = Azvz gives v, = 0.295 m/ s. Bemoulli’s equation is: 1 . P] - P2 = 5 mu; — vÏ) + pg(y2 - y¡), and gives P, — p, = %(103 kg/ m3) [(2.65 m/ s)2 — (0.295 m/ s)2] + (103 kg/ mÏ‘)(9.80 m/ s3)(2.00 m) 0|’ Pgauge: P] - P2 = X 104 Pa Since the upward buoyant force is balanced by the weight of the sphere, vnng = pVg = pGrrR’) s ln this problem, p = 0.789 45 g/ cm3 at 20.0°C, and R = 1.00 cm, so we find: m1: pGnRB] = (O.78945g/ c1'n3)(%¡rR3)1.00cm)3: 3.307g 27 Following the same procedure as in part (a), with p’ = 0.78097 g/ cm’ at 30.0°C, we find: , 4 4 m2 = p NIF) = (0.780 97 g/ cm3 )(í ¡rR3 }1.00 cm) 3 or m2 = 3.271 g When the first sphere is resting on the bottom of the tube, n + B = I-‘¿l = m¡g, where n is the normal force. Since B = p'Vg, n = ¡mg — p'Vg = [3.307 g — (0.780 97 (1.00 cm3)] (930 n =34.8g-cm/ s2= 3.48 x 10" N © 2000 by Harcourt College Publishers. All rights reserved.
  28. 28. 28 Chapter 15 Solutions 15.69 Note: Variation of atmospheric pressure with altitude is included in this solution. Because of the small distances involved, this effect is unimportant in the final answers. shield b c (a) Consider the pressure at points A and B in part (b) of the figure: Using the left tube: PA = Pam, + pugh + p, ,.g(L — h), where the second term is due to the variation of air pressure with altitude. Using the right tube: PB = PM", + pggL But Pascal's principle says that PA = PB. Therefore, Pam, + QgL = Pam + pagh + p, ,g(L - h) or (pi- - nah = (Pm - Poll-i giving h = (52:53) «w = (b) Consider part (c) of the diagram showing the situation when the air flow over the left tube equalizes the fluid levels in the two tubes. First, apply Bemoulli’s equation to points A and B (yA = yg, UA = v, and v5 = 0). , , l , 1 This gives: PA + 5 pm + pan = PB + 5 Pii(0)2 + Piigys 1 and since yA = ya, this reduces to: PB — PA = í pazvz (1) Now consider points C and D, both at the level of the oil-water interface in the right tube. Using the variation of pressure with depth in static fluids, we have: Pc = PA + 23H + push and Po = PB + nigH + P03’- © 2000 by Harcourt College Publishers. All rights reserved.
  29. 29. Chapter 15 Solutions 29 But Pascal's principle says that Pc = PD. Equating these two gives: PB + ggH + pugL = PA + p, gH + pugL or ¡’a - PA = (Pm - Po)8L (2) Substitute equation (1) for PB - PA into (2) to obtain 1 7 ï pav- = (pin ' Po)8L Or v: x Fïi“) = 2(9.3o m/ s2)(0.0500 m) ¿mi p” 1.29 © 2000 by Harcourt College Publishers. All rights reserved.

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