Tài liệu Physics exercises_solution: Chapter 14 pdf

14.1: ρVgmgw          N8.41sm80.9m1043.1m858.0mkg108.7 2 2 233   π or 42 N to two places. A cart is not necessary. 14.2:     .mkg1033.3 m1074.1 kg1035.7 33 3 6 3 4 22 3 3 4     π πr m V m ρ 14.3:     .mkg1002.7 3 3 mm0.300.150.5 kg0158.0 3   V m ρ You were cheated. 14.4: The length L of a side of the cube is cm.3.12 mkg104.21 kg0.40 3 1 3 1 3 1 33                      m VL 14.5: ρπrρVm 3 3 4  Same mass means   leadlaluminum,a 1 3 1a 3 a  ρrρr 6.1 107.2 103.11 31 3 3 31 a 1 1 a                      ρ ρ r r 14.6: a)   327 30 3 8 3 4 30 sun sun m10412.1 kg1099.1 m1096.6 kg1099.1       π V M D 33 mkg10409.1  b)   317 313 30 3 4 3 4 30 mkg10594.0 m10351.3 kg1099.1 m1000.2 kg1099.1        π D 316 mkg1094.5  14.7: ρghpp  0 m91.9 )sm80.9()mkg1030( Pa1000.1 23 5 0      ρg pp h 14.8: The pressure difference between the top and bottom of the tube must be at least 5980 Pa in order to force fluid into the vein: Pa5980ρgh m581.0 )sm80.9()mkg1050( mN5980Pa5980 23 2  gh h 14.9: a)      Pa.706m12.0sm80.9mkg600 23 ρgh b)      Pa.1016.3m250.0sm80.9mkg1000Pa706 323  14.10: a) The pressure used to find the area is the gauge pressure, and so the total area is    2 3 3 cm805 )Pa10205( )N105.16( b) With the extra weight, repeating the above calculation gives 2 cm1250 . 14.11: a) Pa.1052.2)m250)(sm80.9)(mkg1003.1( 6 2 33 ρgh b) The pressure difference is the gauge pressure, and the net force due to the water and the air is N.1078.1))m15.0()(Pa1052.2( 526  π 14.12: atm.61.9Pa1027.6)m640)(sm80.9)(mkg1000.1( 6233  ρghp 14.13: a)   )m1000.7)(sm80.9)(mkg106.13(Pa10980 22332 2a ρgyp Pa.1007.1 5  b) Repeating the calcultion with 00.4 12  yyy cm instead of Pa.101.03gives 5 2 y c) The absolute pressure is that found in part (b), 1.03 Pa.10 5  d) Pa1033.5)( 3 12  ρgyy (this is not the same as the difference between the results of parts (a) and (b) due to roundoff error). 14.14: Pa.100.6)m1.6)(sm80.9)(mkg1000.1( 4233 ρgh 14.15: With just the mercury, the gauge pressure at the bottom of the cylinder is   mm0 ghppp With the water to a depth w h , the gauge pressure at the bottom of the cylinder is . wwmm0 ghpghρpp  If this is to be double the first value, then m.mww ghρghρ  m680.0)1000.1106.13)(m0500.0()( 33 wmmw  ρρhh The volume of water is 33424 cm816m108.16)m10m)(12.0(0.680A   hV 14.16: a) Gauge pressure is the excess pressure above atmospheric pressure. The pressure difference between the surface of the water and the bottom is due to the weight of the water and is still 2500 Pa after the pressure increase above the surface. But the surface pressure increase is also transmitted to the fluid, making the total difference from atmospheric 2500 Pa+1500 Pa = 4000 Pa. b) The pressure due to the water alone is 2500 Pa .ρgh Thus m255.0 )sm80.9()mkg1000( mN2500 23 2 h To keep the bottom gauge pressure at 2500 Pa after the 1500 Pa increase at the surface, the pressure due to the water’s weight must be reduced to 1000 Pa: m102.0 )sm80.9)(mkg1000( mN1000 23 2 h Thus the water must be lowered by m0.153m102.0m255.0  14.17: The force is the difference between the upward force of the water and the downward forces of the air and the weight. The difference between the pressure inside and out is the gauge pressure, so N.1027.2N300)m75.0(m)30()sm80.9()1003.1()( 5223  wAρghF 14.18:   )m(2.00Pa1093m)2.14)(sm71.3)(mkg1000.1(Pa10130 232333  N.1079.1 5  14.19: The depth of the kerosene is the difference in pressure, divided by the product , V mg ρg  m.14.4 )m250.0()smkg)(9.80205( Pa1001.2)m(0.0700N)104.16( 32 523    h 14.20: atm.64.1Pa1066.1 m)15.0( )smkg)(9.801200( )2( 5 2 2 2  πdπ mg A F p 14.21: The buoyant force must be equal to the total weight; so,gg icewater mgVρVρ  ,m563.0 mkg920mkg1000 kg0.45 3 33 water      ice ρρ m V or 3 m56.0 to two figures. 14.22: The buoyant force is and N,30.6N20.11N17.50 B .m1043.6 )sm80.9)(mkg1000.1( N)30.6( 34 233 water     gρ B V The density is .mkg1078.2 30.6 50.17 )mkg1000.1( 3333 water water         B w ρ gρB gw V m ρ 14.23: a) The displaced fluid must weigh more than the object, so . fluid   b) If the ship does not leak, much of the water will be displaced by air or cargo, and the average density of the floating ship is less than that of water. c) Let the portion submerged have volume V, and the total volume be  fluid0 so,Then,. fluido0 ρ ρ V V VρVV  The fraction above the fluid is then 0,If .1 fluid  p P P the entire object floats, and if fluid   , none of the object is above the surface. d) Using the result of part (c), %.3232.0 mkg1030 )m103.04.0(5.0kg)042.0( 11 3 3-6 fluid      14.24: a)     N.6370m650.0sm80.9mkg1000.1 3233 water  gVρB b) kg.558 2 sm9.80 N900-N6370   g TB g w m c) (See Exercise 14.23.) If the submerged volume is ,V  %.9.858590. N6370 N5470 and waterwater     gVρ w V V g ρ w V 14.25: a) Pa.116 oiloil ghρ b)            Pa.921sm80.9m0150.0mkg1000m100.0mkg790 233  c)        kg.822.0 sm80.9 m100.0Pa805 2 2 topbottom    g App g w m The density of the block is   .822 33 m kg m10.0 kg822.0 p Note that is the same as the average density of the fluid displaced,       .)mkg1000(15.0mkg79085.0 33  14.26: a) Neglecting the density of the air,      ,m1036.3 mkg107.2sm80.9 N89 33 332     gρ w ρ gw ρ m V 33 m104.3or   to two figures. b)   N.0.56 7.2 00.1 1N891 aluminum water water                  ρ ρ ω VgρwBwT 14.27: a) The pressure at the top of the block is , 0 ghpp   where h is the depth of the top of the block below the surface. h is greater for block  , so the pressure is greater at the top of block  . b) . objfl gVB   The blocks have the same volume obj V so experience the same buoyant force. c) . so 0 BwTBwT  .ρVgw  The object have the same V but ρ is larger for brass than for aluminum so w is larger for the brass block. B is the same for both, so T is larger for the brass block, block B. 14.28: The rock displaces a volume of water whose weight is N.10.8N28.4-N2.39  The mass of this much water is thus kg102.1sm9.80N8.10 2  and its volume, equal to the rock’s volume, is 33 33 m10102.1 mkg101.00 kg102.1    The weight of unknown liquid displaced is N,20.6N18.6N2.39  and its mass is kg.102.2sm9.80N6.20 2  The liquid’s density is thus 33 m101.102kg102.2   ,mkg1091.1 33  or roughly twice the density of water. 14.29: )(, 21122211 ΑΑvvΑvΑv  2 2 2 1 cm)10.0(20,cm)80.0( πΑπΑ  sm6.9 )10.0(20 (0.80) s)m0.3( 2 2 2  π π v 14.30:  2 3 2 2 2 1 12 sm245.0)m0700.0s)(m50.3( ΑΑΑ Α vv a) (i) s.m21.5,m047.0(ii) s.m33.2,m1050.0 2 2 22 2 2  vΑvΑ b) .m882s)3600()sm(0.245 33 2211  tΑυtΑv 14.31: a) .98.16 )m150.0( )sm20.1( 2 3  πA dtdV v b) .m317.0)( 22112  πvdtdVvvrr 14.32: a) From the equation preceding Eq. (14.10), dividing by the time interval dt gives Eq. (14.12). b) The volume flow rate decreases by 1.50% (to two figures). 14.33: The hole is given as being “small,”and this may be taken to mean that the velocity of the seawater at the top of the tank is zero, and Eq. (14.18) gives ))((2 ρpgyv  = ))mkg10(1.03Pa)1013(3.00)(1.0m)0.11)(sm80.9((2 3 352  s.m4.28 Note that y = 0 and a pp  were used at the bottom of the tank, so that p was the given gauge pressure at the top of the tank. 14.34: a) From Eq. (14.18), s.m6.16m)0.14)(sm80.9(22 2  ghv b) s.m1069.4)m)10(0.30s)(m57.16( 3422   πvΑ Note that an extra figure was kept in the intermediate calculation. 14.35: The assumption may be taken to mean that 0 1 v in Eq. (14.17). At the maximum height, ,0 2 v and using gauge pressure for 0,and 221 ppp (the water is open to the atmosphere), Pa.1047.1 5 21  ρgyp 14.36: Using 1 4 1 2 vv  in Eq. (14.17),               )( 32 15 )()( 2 1 21 2 1121 2 2 2 112 yygυρpyyρgvvρpp        m)0.11)(sm80.9()sm00.3( 32 15 )mkg1000.1(Pa1000.5 22334 Pa.1062.1 5  14.37: Neglecting the thickness of the wing (so that 21 yy  in Eq. (14.17)), the pressure difference is Pa.078)(2)1( 2 1 2 2  vvρp The net upward force is then N.496)smkg)(9.801340()m(16.2Pa)780( 22  14.38: a)    s.kg30.1 s0.60 kg355.0220  b) The density of the liquid is ,mkg1000 3 m100.355 kg355.0 33    and so the volume flow rate is s.L1.30sm1030.1 33 mkg1000 skg30.1 3   This result may also be obtained from    s.L30.1 s0.60 L355.0220  c) 24 33 m1000.2 sm1030.1 1     v s.m63.14s,m50.6 12  vv d)     12 2 1 2 221 2 1 yyρgvvρpp                 m35.1sm80.9mkg1000 sm50.6sm63.1mkg100021kPa152 23 22 3   kPa119 14.39: The water is discharged at a rate of s.m352.0 23 34 m1032.1 sm1065.4 1      v The pipe is given as horizonatal, so the speed at the constriction is s,m95.82 2 12  ρpvv keeping an extra figure, so the cross-section are at the constriction is ,m1019.5 25 sm95.8 sm1065.4 34     and the radius is cm.41.0  Ar 14.40: From Eq. (14.17), with , 21 yy    2 11 2 1 2 11 2 2 2 112 8 3 42 1 2 1 ρvp v v ρpvvρpp               Pa,1003.2sm50.2mkg1000.1 8 3 Pa1080.1 4 2 334  where the continutity relation 2 1 2 v v  has been used. 14.41: Let point 1 be where cm00.4 1 r and point 2 be where cm.00.2 2 r The volume flow rate has the value scm7200 3 at all points in the pipe. sm43.1so,cm7200 1 32 1111  vπrvAv sm73.5so,cm7200 2 32 2222  vπrvAv 2 222 2 111 2 1 2 1 ρvρgypρvρgyp    Pa1025.2 2 1 soPa,1040.2and 52 2 2 112 5 221  vvρpppyy 14.42: a) The cross-sectional area presented by a sphere is , 4 2 D π therefore   . 4 0 2 D πppF  b) The force on each hemisphere due to the atmosphere is   2 2 m1000.5  π     .776975.0Pa10013.1 5  14.43: a)     Pa.1010.1m1092.10sm80.9mkg1003.1 83233 ρgh b) The fractional change in volume is the negative of the fractional change in density. The density at that depth is then         111833 0 Pa108.45Pa1016.11mkg1003.11   pkρρ ,mkg1008.1 33  A fractional increase of %.0.5 Note that to three figures, the gauge pressure and absolute pressure are the same. 14.44: a) The weight of the water is         N,1088.5m0.3m0.4m00.5sm80.9mkg1000.1 5233 ρgV or N109.5 5  to two figures. b) Integration gives the expected result the force is what it would be if the pressure were uniform and equal to the pressure at the midpoint; 2 d gAF           N,1076.1m50.1m0.3m0.4sm80.9mkg1000.1 5233  or N108.1 5  to two figures. 14.45: Let the width be w and the depth at the bottom of the gate be .H The force on a strip of vertical thickness dh at a depth h is then   wdhρghdF  and the torque about the hinge is   ;2 dhHhρgwhdτ  integrating from Hhh  to0 gives m.N1061.212 43  Hg  14.46: a) See problem 14.45; the net force is dF from ,22, to0 2 gAHHgFHhh   where .HA   b) The torque on a strip of vertical thickness dh about the bottom is     ,dhhHgwhhHdFdτ  and integrating from Hhh  to0 gives .66 23 ρgAHρgwHτ  c) The force depends on the width and the square of the depth, and the torque about the bottom depends on the width and the cube of the depth; the surface area of the lake does not affect either result (for a given width). 14.47: The acceleration due to gravity on the planet is d p ρd p g V m     and so the planet’s mass is mGd pVR G gR M 22   [...]... models (see Problem 14. 49 or Problem 9.99), the concentration of mass at lower radii leads to a higher pressure 14. 51: a) ρwater ghwater  (1.00  103 kg m3 )(9.80 m s 2 )(15.0  102 m)  1.47  103 Pa b) The gauge pressure at a depth of 15.0 cm  h below the top of the mercury column must be that found in part (a); ρHg g (15.0 cm  h)  ρwater g (15.0 cm), which is solved for h  13.9 cm 14. 52: Following... less than the volume of the interior of the barge (1.06  10 4 m 3 ), but the coal must not be too loosely packed 14. 54: The difference between the densities must provide the “lift” of 5800 N (see Problem 14. 59) The average density of the gases in the balloon is then ρave  1.23 kg m 3  14. 55: (5800 N)  0.96 kg m3 2 3 (9.80 m s )(2200 m ) a) The submerged volume V  is w ρ water g , so (900 kg) V... (see Problem 14. 23 or Problem 14. 55), 1 Thus, if two fluids are observed to have which can be expressed as ρ fluid  ρ 1 f 1  f1 floating fraction f1 and f 2 , ρ2  ρ1 In this form, it’s clear that a larger f 2 1  f2 corresponds to a larger density; more of the stem is above the fluid Using f1  (8.00 (cm)(0.400)cm )  0.242, f 2  (3.20 (cm)(0.400) cm )  0.097 gives 13.2 cm 13.2 cm 14. 58: 2 2... for x gives x  ρgA 14. 60: a) Archimedes’ principle states gLA  Mg , so L  c) The “spring constant,” that is, the proportionality between the displacement x and the applied force F, is k  ρgA, and the period of oscillation is M M  2π k ρgA T  2π 14. 61: a) x  70.0 kg  w mg m     0.107 m 3 ρgA ρgA ρA 1.03  10 kg m3 π 0.450 m 2   b) Note that in part (c) of Problem 14. 60, M is the mass... gauge pressure at E The 14. 85: height of the fluid in the column is 3h1 14. 86: a) v  dV dt A , so the speeds are 6.00  103 m 3 s 6.00  103 m 3 s  6.00 m s and  1.50 m s 10.0  10 4 m 2 40.0  10 4 m 2 2 b) p  1  (v12  v 2 )  1.688  10 4 Pa, or 1.69  10 4 Pa to three figures 2 c) h  p H g g  (1.68810 4 Pa) (13.610 3 kg m 3 )( 9.80 m s 2 )  12.7 cm 14. 87: a) The speed of the... ρwoodVwood  M brass  ρwood ρ   M brass 1  air   ρwood  ρair ρwood    1  1.20 kg m3   (0.0950 kg)1   150 kg m3   0.0958 kg    1 14. 64: The buoyant force on the mass A, divided by g , must be 7.50 kg  1.00 kg  1.80 kg  4.70 kg (see Example 14. 6), so the mass block is 4.70 kg  3.50 kg  8.20 kg a) The mass of the liquid displaced by the block is 4.70 kg, so the density of the liquid... crown would be 1  1 19.312.9 N   12.2 N c) Approximating the average density by that of lead for a “thin” gold plate, the apparent weight would be 1  1 11.312.9 N   11.8 N 14. 68: a) See problem 14. 67 Replacing f with, respectively, wwater w and wfluid w gives ρsteel w ρ w  , steel  , ρfluid w - wfluid ρfluid w - wwater and dividing the second of these by the first gives ρfluid w... 1.2200.128  0.844  84.4% ρwater 14. 69: a) Let the total volume be V; neglecting the density of the air, the buoyant force in terms of the weight is  (w g )   V0 , B  ρwater gV  ρwater g   ρ   m  or V0  b) ρwater g  B w ρ Cu g 4 B ρwater g  w  w g  2.52  10 m Since the total volume of the casting is 3 cavities are 12.4% of the total volume B ρ water g , the 14. 70: a) Let d be the depth... h  or about 0.56 mm ma  steel A  35.0 kg  7860 kg  m 3 8.00 m 2   5.57  10 4 m, 14. 72: a) The average density of a filled barrel is 15.0 m  oil  V  750 kg m3  0.120 kg3  875 kg m3 , which is less than the density of seawater, m so the barrel floats b) The fraction that floats (see Problem 14. 23) is 1  ave 875 kg m 3  1  0.150  15.0%  water 1030 kg m 3 32.0 kg kg c) The average... m3  0.120 kg3  1172 m3 which means the barrel sinks m In order to lift it, a tension kg kg T  (1177 m  )(0.120 m3 )(9.80 sm )  (1030 m 3 )(0.120 m3 )(9.80 sm )  173 N is required 2 2 14. 73: a) See Exercise 14. 23; the fraction of the volume that remains unsubmerged is 1  ρB b) Let the depth of the liquid be x and the depth of the water be y Then ρL Lgx  wgy   B gL and x  y  L Therefore . results of parts (a) and (b) due to roundoff error). 14. 14: Pa.100.6)m1.6)(sm80.9)(mkg1000.1( 4233 ρgh 14. 15: With just the mercury, the gauge pressure. N.1027.2N300)m75.0(m)30()sm80.9()1003.1()( 5223  wAρghF 14. 18:   )m(2.00Pa1093m)2 .14) (sm71.3)(mkg1000.1(Pa10130 232333  N.1079.1 5  14. 19: The depth of the kerosene