 # Physics exercises_solution: Chapter 14

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Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 14

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## Nội dung Text: Physics exercises_solution: Chapter 14

1. 14.1: w  mg  ρVg     7.8  103 kg m 3 0.858 m π 1.43  102 m 9.80 m s 2  41.8 N  2  or 42 N to two places. A cart is not necessary. 14.2: ρ m m  4 3  7.35  1022 kg   3.33  103 kg m3 . V 3 πr 4 3 π 1.74  10 m 6 3  0.0158 kg  3 14.3: ρV  m 5.015.030.0  mm 3  7.02  103 kg m . You were cheated. 14.4: The length L of a side of the cube is 1 1  m 3  1 40.0 kg 3 L V       3  21.4  10 3 kg m 3   12.3 cm.      14.5: m  ρV  4 πr 3 ρ 3 Same mass means ra3 ρa  r13 ρ1 a  aluminum, l  lead  13 13 ra  ρ1   11.3  103      2.7  103   1.6 r1  ρa      M sun 1.99  1030 kg 1.99  1030 kg 14.6: a) D   Vsun 4 3  π 6.96  108 m 3  1.412  1027 m3  1.409  10 kg m 3 3 1.99  1030 kg 1.99  1030 kg b) D   0.594  1017 kg m3 4 3  π 2.00  10 m 4 3 3.351  10 m 13  3  5.94  1016 kg m 3 14.7: p  p0  ρgh p  p0 1.00  105 Pa h   9.91m ρg (1030 kg m 3 ) (9.80 m s 2 )
2. 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: ρgh  5980 Pa 5980 Pa 5980 N m 2 h   0.581m gh (1050 kg m3 ) (9.80 m s 2 ) 14.9:    a) ρgh  600 kg m3 9.80 m s 2 0.12 m   706 Pa.    b) 706 Pa  1000 kg m 3 9.80 m s 2 0.250 m   3.16  103 Pa. 14.10: a) The pressure used to find the area is the gauge pressure, and so the total area is (16.5  10 3 N)  805 cm 2  (205  10 Pa ) 3 b) With the extra weight, repeating the above calculation gives 1250 cm 2 . 2 14.11: a) ρgh  (1.03  103 kg m3 )(9.80 m s )(250 m)  2.52  106 Pa. b) The pressure difference is the gauge pressure, and the net force due to the water and the air is (2.52  106 Pa )(π (0.15 m) 2 )  1.78  105 N. 14.12: p  ρgh  (1.00  103 kg m 3 )(9.80 m s 2 )(640 m)  6.27  106 Pa  61.9 atm. 14.13: a) pa  ρgy2  980  10 2 Pa  (13.6  103 kg m 3 )(9.80 m s 2 )(7.00  102 m)  1.07  10 5 Pa. b) Repeating the calcultion with y  y 2  y1  4.00 cm instead of y 2 gives 1.03  10 5 Pa. c) The absolute pressure is that found in part (b), 1.03  10 5 Pa. d) ( y2  y1 ) ρg  5.33  103 Pa (this is not the same as the difference between the results of parts (a) and (b) due to roundoff error). 14.14: ρgh  (1.00  103kg m 3 )(9.80 m s 2 )(6.1 m)  6.0  104 Pa.
3. 14.15: With just the mercury, the gauge pressure at the bottom of the cylinder is p  p 0  p m ghm With the water to a depth hw , the gauge pressure at the bottom of the cylinder is p  p0  ρm ghm  pw ghw . If this is to be double the first value, then ρw ghw  ρm ghm. hw  hm ( ρm ρw )  (0.0500 m)(13.6  103 1.00  103 )  0.680 m The volume of water is V  hA  (0.680 m)(12.0  104 m 2 )  8.16  104 m 3  816 cm3 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 2500 N m 2 h  0.255 m (1000 kg m3 ) (9.80 m s 2 ) 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: 1000 N m 2 h  0.102 m (1000 kg m 3 )(9.80 m s 2 ) Thus the water must be lowered by 0.255 m  0.102 m  0.153 m 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 F  ( ρgh) A  w  (1.03  103 ) (9.80 m s 2 ) (30 m) (0.75 m 2 )  300 N  2.27  105 N. 14.18: 130  10 3  Pa  (1.00  103 kg m 3 )(3.71 m s 2 )(14.2 m)  93  103 Pa (2.00 m 2 )  1.79  10 5 N.
4. 14.19: The depth of the kerosene is the difference in pressure, divided by the product ρg  mg , V (16.4  103 N) (0.0700 m 2 )  2.01  105 Pa h  4.14 m. (205 kg)(9.80 m s 2 ) (0.250 m3 ) F mg (1200 kg)(9.80 m s 2 ) 14.20: p  2  2  1.66  105 Pa  1.64 atm. A π (d 2) π (0.15 m) 14.21: The buoyant force must be equal to the total weight; ρwaterVg  ρiceVg  mg , so m 45.0 kg V   0.563 m 3 , ρwater  ρice 1000 kg m  920 kg m 3 3 or 0.56 m 3 to two figures. 14.22: The buoyant force is B  17.50 N  11.20 N  6.30 N, and B (6.30 N) V   6.43  10 4 m3 . ρwater g (1.00  10 kg m )(9.80 m s ) 3 3 2 The density is m w g w  17.50  ρ   ρwater  (1.00  103 kg m 3 )    2.78  10 kg m . 3 3 V B ρwater g B  6.30 
5. 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 V0 . Then, Vo  ρfluid V , so V0  ρfluid  The fraction V ρ above the fluid is then 1  P Pfluid . If p  0, the entire object floats, and if    fluid , none of the object is above the surface. d) Using the result of part (c),  (0.042 kg) (5.0  4.0  3.0  10 -6 m 3 ) 1  1  0.32  32%.  fluid 1030 kg m 3 14.24: a) B  ρwater gV  1.00  103 kg m3 9.80 m s 2 0.650 m 3   6370 N. b) m  w  B T  6370 Nm900 N  558 kg. g g - 9.80 s 2 c) (See Exercise 14.23.) If the submerged volume is V , w V w 5470 N V  and    0.859  85.9%. ρwater g V ρwater gV 6370 N 14.25: a) ρoil ghoil  116 Pa.     b) 790 kg m 3 0.100 m   1000 kg m3 0.0150 m  9.80 m s 2  921 Pa.   w  pbottom  ptop A 805 Pa 0.100 m  2 c) m    0.822 kg. g g 9.80 m s 2   The density of the block is p  0.822 kg 0.10 m 3  822 m 3 . Note that is the same as the average kg  density of the fluid displaced, 0.85 790 kg m3  0.15 (1000 kg m 3 ) .
6. 14.26: a) Neglecting the density of the air, V m w g w    89 N   3.36  103 m3 , ρ ρ   gρ 9.80 m s 2.7  103 kg m3 2  or 3.4  103 m3 to two figures.  ρ   1.00  b) T  w B  w  gρwaterV  ω1  water    89 N 1     56.0 N.  ρaluminum   2.7  14.27: a) The pressure at the top of the block is p  p 0  gh, 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) B   flVobj g . The blocks have the same volume Vobj so experience the same buoyant force. c) T  w  B  0 so T  w B. w  ρVg. 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 39.2 N - 28.4 N  10.8 N. The mass of this much water is thus 10.8 N 9.80 m s 2  1.102 kg and its volume, equal to the rock’s volume, is 1.102 kg  1.102  10 3 m3 1.00  10 kg m 3 3 The weight of unknown liquid displaced is 39.2 N  18.6 N  20.6 N, and its mass is 20.6 N 9.80 m s 2  2.102 kg. The liquid’s density is thus 2.102 kg 1.102  103 m 3  1.91  103 kg m 3 , or roughly twice the density of water. 14.29: v1 Α1  v2 Α2 , v2  v1 ( Α1 Α2 ) Α1  π (0.80 cm)2 , Α2  20π (0.10 cm)2 π (0.80)2 v2  (3.0 m s)  9.6 m s 20π (0.10) 2
7. Α1 (3.50 m s)(0.0700 m 2 ) 0.245 m3 s 14.30: v2  v1    Α2 Α2 Α2 a) (i) Α2  0.1050 m 2 , v2  2.33 m s. (ii) Α2  0.047 m 2 , v2  5.21 m s. b) v1 Α1t  υ2 Α2t  (0.245 m3 s) (3600 s)  882 m3 . dV dt (1.20 m 3 s) 14.31: a) v    16.98. A π (0.150 m) 2 b) r2  r1 v1 v2  (dV dt ) πv2  0.317 m. 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 v  2( gy  ( p ρ)) 3 = 2((9.80 m s 2 )(11.0 m)  (3.00)(1.013  10 5 Pa) (1.03  10 3 kg m ))  28.4 m s. Note that y = 0 and p  pa 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), v  2 gh  2(9.80 m s 2 )(14.0 m)  16.6 m s. b) vΑ  (16.57 m s)(π (0.30  102 m)2 )  4.69  104 m3 s. Note that an extra figure was kept in the intermediate calculation. 14.35: The assumption may be taken to mean that v1  0 in Eq. (14.17). At the maximum height, v2  0, and using gauge pressure for p1 and p2 , p2  0 (the water is open to the atmosphere), p1  ρgy2  1.47  105 Pa.
8. 14.36: Using v2  1 v1 in Eq. (14.17), 4 1  15   p2  p1  ρ(v12  v2 )  ρg ( y1  y2 )  p1  ρ  υ12  g ( y1  y2 ) 2 2  32    15   5.00  104 Pa  (1.00  103 kg m3 )  (3.00 m s) 2  (9.80 m s 2 )(11.0 m)   32   1.62  10 Pa. 5 14.37: Neglecting the thickness of the wing (so that y1  y2 in Eq. (14.17)), the pressure difference is p  (1 2) ρ(v2  v12 )  780 Pa. The net upward force is then 2 (780 Pa)  (16.2 m 2 )  (1340 kg)(9.80 m s 2 )  496 N. 14.38: a) 220 0..0 s kg   1.30 kg s. b) The density of the liquid is 60 355 0.355 kg 0.35510 3 m 3  1000 kg m3 , and so the volume flow rate is 1.30 kg s 1000 kg m 3  1.30  103 m3 s  1.30 L s. This result may also be obtained 220 0.355 L  1.3010 m s 3 3 from 60.0 s  1.30 L s. c) v1  2.0010 4 m 2  6.50 m s, v2  v1 4  1.63 m s. 1  d) p1  p2  ρ v2  v12  ρg  y2  y1  2 2    152 kPa  1 2 1000 kg m3  1.63 m s   6.50 m s  2 2     1000 kg m3 9.80 m s 2  1.35 m    119 kPa 4.6510 4 m 3 s 14.39: The water is discharged at a rate of v1  1.3210 3 m 2  0.352 m s. The pipe is given as horizonatal, so the speed at the constriction is v2  v12  2 p ρ  8.95 m s, keeping an extra figure, so the cross-section are at the constriction is 4.6510 4 m 3 s 8.95 m s  5.19  105 m 2 , and the radius is r  A   0.41 cm.
9. 14.40: From Eq. (14.17), with y1  y 2 , 1  v2  p2  p1  1 2  2   2  4 3 ρ v12  v2  p1  ρ v12  1   p1  ρv12 8  1.80  10 4 Pa  3 8   1.00  10 3 kg m 3 2.50 m s   2.03  10 4 Pa, 2 v where the continutity relation v 2  1 has been used. 2 14.41: Let point 1 be where r1  4.00 cm and point 2 be where r2  2.00 cm. The volume flow rate has the value 7200 cm 3 s at all points in the pipe. v1 A1  v1πr12  7200 cm3 , so v1  1.43 m s v2 A2  v2πr22  7200 cm3 , so v2  5.73 m s 1 1 2 p1  ρgy1  ρv12  p2  ρgy2  ρv2 2 2 y1  y2 and p2  2.40  105 Pa, so p2  p1  ρv12  v2   2.25  105 Pa 1 2 2 2 14.42: a) The cross-sectional area presented by a sphere is π D4 , therefore F   p0  p π D4 . b) The force on each hemisphere due to the atmosphere is 2  π 5.00  10 2 m  1.013  10 2 5  Pa 0.975  776.     14.43: a) ρgh  1.03  103 kg m3 9.80  m s 2 10.92  103 m  1.10  108 Pa. b) The fractional change in volume is the negative of the fractional change in density. The density at that depth is then     ρ  ρ0 1  kp   1.03  103 kg m3 1  1.16  108 Pa 45.8  1011 Pa 1   1.08  103 kg m3 , A fractional increase of 5.0%. Note that to three figures, the gauge pressure and absolute pressure are the same.
10. 14.44: a) The weight of the water is   ρgV  1.00  103 kg m 3 9.80 m s 2  5.00 m4.0 m3.0 m   5.88  10 5 N, or 5.9  10 5 N 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; d F  gA 2    1.00  103 kg m 3 9.80 m s 2  4.0 m 3.0 m 1.50 m   1.76  10 5 N, or 1.8  10 5 N 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 dF  ρghwdh  and the torque about the hinge is dτ  ρgwhh  H 2dh; integrating from h  0 to h  H gives   gH 3 12  2.61  10 4 N  m. 14.46: a) See problem 14.45; the net force is  dF from h  0 to h  H , F  gH 2 2  gAH 2, where A  H . b) The torque on a strip of vertical thickness dh about the bottom is dτ  dF H  h   gwhH  h dh, and integrating from h  0 to h  H gives τ  ρgwH 3 6  ρgAH 2 6. 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 p p g  m ρd V d and so the planet’s mass is gR 2 pVR 2 M   G mGd
11. 14.48: The cylindrical rod has mass M , radius R, and length L with a density that is proportional to the square of the distance from one end,   Cx 2 . a) M   dV   Cx 2 dV . The volume element dV  πR 2 dx. Then the integral L3 becomes M   0 Cx 2R 2 dx. Integrating gives M  CR 2  0 x 2 dx  CR 2 L L . Solving 3 for C , C  3M R 2 L3 . b) The density at the x  L end is ρ  Cx 2   L    . The denominator is 3M πR 2 L3 2 3M πR 2 L just the total volume V , so   3M V , or three times the average density, M V . So the average density is one-third the density at the x  L end of the rod.
12. 14.49: a) At r  0, the model predicts   A  12,700 kg m 3 and at r  R, the model predicts   A  BR  12,700 kg m 3  (1.50  10 3 kg m 4 )(6.37  10 6 m)  3.15  10 3 kg m 3 . b), c) R  AR 3 BR 4   4πR 3   3BR  M   dm  4π  [ A  Br ]r 2 dr  4π      3  A  4    0  3 4      4π (6.37  106 m)3   3(1.50  103 kg m 4 )(6.37  106 m)    12,700 kg m3     3  4   5.99  10 kg, 24 which is within 0.36% of the earth’s mass. d) If m (r ) is used to denote the mass contained in a sphere of radius r , then g  Gm ( r ) r 2 . Using the same integration as that in part (b), with an upper limit of r instead of R gives the result. e) g  0 at r  0, and g at r  R, g  Gm( R) R 2  (6.673  1011 N  m 2 kg 2 ) (5.99  1024 kg) (6.37  106 m)2  9.85 m s 2 . dg  4πG  d  3Br 2   4πG   3Br  f)    Ar     3  A  2  ; dr  3  dr  4     setting ths equal to zero gives r  2 A 3B  5.64  10 6 m , and at this radius  4πG  2 A    3   2 A  g     A    B   3  3B    4   3B   4πGA2  9B 4π (6.673  1011 N  m 2 kg 2 ) (12,700 kg m3 ) 2   10.02 m s 2 . 9(1.50  10 3 kg m 4 )
13. 14.50: a) Equation (14.4), with the radius r instead of height y, becomes dp   ρg dr   ρgs (r R)dr. This form shows that the pressure decreases with increasing radius. Integrating, with p  0 at r  R, ρg s r ρg s R ρg s 2 p R  r dr  R R  r r dr  2R ( R  r 2 ). b) Using the above expression with r  0 and   M V  3M 4R 3 , 3(5.97  1024 kg)(9.80 m s 2 ) p(0)   1.71  1011 Pa. 8π (6.38  106 m)2 c) While the same order of magnitude, this is not in very good agreement with the estimated value. In more realistic density 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 the hint, h F   ( ρgy )(2πR )dy  ρgπRh 2 o where R and h are the radius and height of the tank (the fact that 2 R  h is more or less coincidental). Using the given numerical values gives F  5.07  10 8 N. 14.53: For the barge to be completely submerged, the mass of water displaced would need to be ρwaterV  (1.00  103 kg m3 )(22  40  12 m3 )  1.056  107 kg. The mass of the barge itself is (7.8  103 kg m3 ) ((2(22  40)  12  22  40)  4.0  102 m 3 )  7.39  105 kg, so the barge can hold 9.82  10 6 kg of coal. This mass of coal occupies a solid volume of 6.55  10 3 m 3 , which is 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. 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 (5800 N) ρave  1.23 kg m 3  2 3  0.96 kg m3 . (9.80 m s )(2200 m ) 14.55: a) The submerged volume V  is w ρ water g , so V  w ρwater g m (900 kg)     0.30  30%  V V ρwaterV (1.00  103 kg m3 ) (3.0 m3 ) b) As the car is about to sink, the weight of the water displaced is equal to the weight of the car plus the weight of the water inside the car. If the volume of water inside the car is V  , V  w Vρwater g  w  V pwater g , or 1  1  0.30  0.70  70%  V Vpwater g 14.56: a) The volume displaced must be that which has the same weight and mass as the ice, 1.00.70 gm 3  9.70 cm3 (note that the choice of the form for the density of water 9 gm cm avoids conversion of units). b) No; when melted, it is as if the volume displaced by the 9.70 gm of melted ice displaces the same volume, and the water level does not change. 9.70 gm c) 1.05 gm cm 3  9.24 cm3  d) The melted water takes up more volume than the salt water displaced, and so 0.46 cm3 flows over. A way of considering this situation (as a thought experiment only) is that the less dense water “floats” on the salt water, and as there is insufficient volume to contain the melted ice, some spills over.
15. 14.57: The total mass of the lead and wood must be the mass of the water displaced, or VPb ρ Pb  Vwood ρ wood  (VPb  Vwood ) ρ water ; solving for the volume VPb ,    wood VPb  Vwood water  Pb   water 1.00  103 kg m3  600 kg m 3  (1.2  10 2 m3 ) 11.3  103 kg m 3  1.00  103 kg m3  4.66  104 m3 , which has a mass of 5.27 kg.  14.58: The fraction f of the volume that floats above the fluid is f  1  ,  fluid where ρ is the average density of the hydrometer (see Problem 14.23 or Problem 14.55), 1 which can be expressed as ρ fluid  ρ . Thus, if two fluids are observed to have 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 2 2 f1  (8.00 (cm)(0.400)cm )  0.242, f 2  (3.20 (cm)(0.400) cm )  0.097 gives 3 13.2 cm 3 13.2 cm ρalcohol  (0.839) ρwater  839 kg m3 . 14.59: a) The “lift” is V ( ρair  ρH 2 ) g , from which 120,000 N V  11.0  10 3 m 3 . (1.20 kg m  0.0899 kg m )(9.80 m s ) 3 3 2 b) For the same volume, the “lift” would be different by the ratio of the density differences,  ρ  ρHe  (120,000 N) air   11.2  104 N.  ρair  ρH   2  This increase in lift is not worth the hazards associated with use of hydrogen.
16. M 14.60: a) Archimedes’ principle states gLA  Mg , so L  . A b) The buoyant force is gA( L  x)  Mg  F , and using the result of part (a) and solving for x gives x  ρgA . F 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 T  2π  2π . k ρgA 14.61: a) x  w  mg  m  70.0 kg   0.107 m.  ρgA ρgA ρA 1.03  10 kg m3 π 0.450 m 2 3  b) Note that in part (c) of Problem 14.60, M is the mass of the buoy, not the mass of the man, and A is the cross-section area of the buoy, not the amplitude. The period is then T  2π 950 kg   2.42 s. 1.03  10 3 kg m 3 9.80 m s π 0.450 m  2 2
17. 14.62: To save some intermediate calculation, let the density, mass and volume of the life preserver be  0 , m and v, and the same quantities for the person be ρ1 , M and V . Then, equating the buoyant force and the weight, and dividing out the common factor of g, ρwater 0.80V  v   ρ0v  ρ1V , Eliminating V in favor of ρ1 and M , and eliminating m in favor of ρ0 and v,  M  ρ0v  M  ρwater  0.80   v .    ρ1  Solving for  0 , 1  M   ρ0   ρwater  0.80  v   M      v  ρ1   M ρ   ρwater  1  0.80  water   v  ρ1  75.0 kg  1.03  103 kg m 3   1.03  103 kg m 3  1  (0.80)   0.400 m3  980 kg m 3    732 kg m3  14.63: To the given precision, the density of air is negligible compared to that of brass, but not compared to that of the wood. The fact that the density of brass may not be known the three-figure precision does not matter; the mass of the brass is given to three figures. The weight of the brass is the difference between the weight of the wood and the buoyant force of the air on the wood, and canceling a common factor of g , Vwood ( ρwood  ρair )  M brass, and 1 ρwood  ρ  M wood  ρwoodVwood  M brass  M brass 1  air   ρwood  ρair  ρwood   1  1.20 kg m3   (0.0950 kg)1   150 kg m3   0.0958 kg.   
18. 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 is 4.70 kg 3.8010 - 3 m 3  1.24  10 3 kg m 3 . b) Scale D will read the mass of the block, 8.20 kg, as found above. Scale E will read the sum of the masses of the beaker and liquid, 2.80 kg. 14.65: Neglecting the buoyancy of the air, the weight in air is g ( ρAuVAu  ρA1VA1 )  45.0 N. and the buoyant force when suspended in water is ρwater (VAu  VA1 ) g  45.0 N  39.0 N  6.0 N. These are two equations in the two unknowns VAu and VA1. Multiplying the second by  A1 and the first by  water and subtracting to eliminate the VA1 term gives ρwaterVAu g ( ρAu  ρA1 )  ρwater (45.0 N)  ρA1 (6.0 N) ρAu wAu  ρAu gVAu  ( ρwater (45.0 N)  ρAu (6.0)) ρwater ( ρAu  ρA1 ) (19.3)  ((1.00)(45.0 N)  (2.7)(6.0 N)) (1.00)(19.3  2.7)  33.5 N. Note that in the numerical determination of wAu , specific gravities were used instead of densities.
19. 14.66: The ball’s volume is 4 4 V  πr 3  π (12.0 cm)3  7238 cm3 3 3 As it floats, it displaces a weight of water equal to its weight. a) By pushing the ball under water, you displace an additional amount of water equal to 84% of the ball’s volume or (0.84)(7238 cm 3 )  6080 cm 3 . This much water has a mass of 6080 g  6.080 kg and weighs (6.080 kg)(9.80 m s 2 )  59.6 N, which is how hard you’ll have to push to submerge the ball. b) The upward force on the ball in excess of its own weight was found in part (a): 59.6 N. The ball’s mass is equal to the mass of water displaced when the ball is floating: (0.16)(7238 cm3 )(1.00 g cm3 )  1158 g  1.158 kg, and its acceleration upon release is thus Fnet 59.6 N a   51.5 m s 2 m 1.158 kg 14.67: a) The weight of the crown of its volume V is w  ρcrown gV , and when suspended the apparent weight is the difference between the weight and the buoyant force, fw  fρcrown gV  ( ρcrown  ρwater ) gV . Dividing by the common factors leads to ρcrown 1  ρwater  ρcrown  fρcrown or  . ρwater 1  f As f  0, the apparent weight approaches zero, which means the crown tends to float; from the above result, the specific gravity of the crown tends to 1. As f  1, the apparent weight is the same as the weight, which means that the buoyant force is negligble compared to the weight, and the specific gravity of the crown is very large, as reflected in the above expression. b) Solving the above equations for f in terms of the ρ water specific gravity, f  1  ρcrown , and so the weight of the 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.
20. 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 - wfluid  . ρwater w - wwater b) When wfluid is greater than wwater, the term on the right in the above expression is less than one, indicating that the fluids is less dense than water, and this is consistent with the buoyant force when suspended in liquid being less than that when suspended in water. If the density of the fluid is the same as that of water wfluid  wwater , as expected. Similarly, if wfluid is less than wwater , the term on the right in the above expression is greater than one, indicating the the fluid is denser than water. c) Writing the result of part (a) as ρfluid 1  f fluid  . ρwater 1  f water and solving for f fluid , ρfluid f fluid  1  1  f water   1  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 )  B  ρwater gV  ρwater g   ρ  V0 ,   m  or B w V0    ρwater g w g 4 b) ρwater g  B w ρ Cu g  2.52  10 m . Since the total volume of the casting is 3 B ρ water g , the cavities are 12.4% of the total volume. CÓ THỂ BẠN MUỐN DOWNLOAD