Physics exercises_solution: Chapter 01

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Physics exercises_solution: Chapter 01

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Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh chapter 1

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  1.   1.1: 1 mi  5280 ft mi   12 in. ft   2.54 cm in .  1 km 105 cm  1.61 km Although rounded to three figures, this conversion is exact because the given conversion from inches to centimeters defines the inch. 3  1000 cm 3   1in  1.2: 0.473 L    1 L    2.54 cm   28.9 in .    3     1.3: The time required for light to travel any distance in a vacuum is the distance divided by the speed of light; 103 m  3.33  10 6 s  3.33  103 ns. 3.00  10 m s 8 3 g  1 kg   100 cm  4 kg 1.4: 11.3 3   1000 g    1 m  1.13  10 m 3 .    cm     1.5: 327 in  2.54 cm in   1 L 1000 cm   5.36 L. 3 3 3  1000 L   1 gal   128 oz.   1 bottle  1.6: 1 m3    1 m 3    3.788 L    1 gal    16 oz. .                 2111.9 bottles  2112 bottles The daily consumption must then be bottles  1 yr  bottles 2.11  10 3   365.24 da   5.78 da .  yr   1.7: 1450 mi hr   1.61 km mi  2330 km hr . 2330 km hr  103 m km  1 hr 3600 s   648 m s. furlongs  1 mile   1 fortnight   1 day  mi 1.8: 180,000   8 furlongs    14 day    24 h   67 h .      fortnight       km  1 mi   3.788 L  mi 1.9: 15.0    1 gal   35.3 gal .  L  1.609 km   
  2.  mi   1h   5280 ft  ft 1.10: a)  60    3600 s     1 mi   88 s   hr       ft   30.48 cm   1 m  m b)  32 2    1 ft   100 cm   9.8 s 2   s    3  g   100 cm   1 kg  3 kg c) 1.0     1000 g   10 m 3   cm 3   1 m    1.11: The density is mass per unit volume, so the volume is mass divided by density.    V  60  103 g 19.5 g cm3  3077 cm3 4 Use the formula for the volume of a sphere, V  r 3 , 3 to calculate r : r  3V 4   9.0 cm 1/ 3 1.12: (3.16  10 7 s  π  10 7 s) (3.16  10 7 s)  100  0.58% 10 m 1.13: a)  1.1  10 3 %. 890  10 m 3 b) Since the distance was given as 890 km, the total distance should be 890,000 meters. To report the total distance as 890,010 meters, the distance should be given as 890.01 km. 1.14: a) 12 mm   5.98 mm  72 mm2 (two significant figures). b) 512 mm = 0.50 (also two significant figures). .98 mm c) 36 mm (to the nearest millimeter). d) 6 mm. e) 2.0.
  3. 1.15: a) If a meter stick can measure to the nearest millimeter, the error will be about 0.13%. b) If the chemical balance can measure to the nearest milligram, the error will be about 8.3  10 3%. c) If a handheld stopwatch (as opposed to electric timing devices) can measure to the nearest tenth of a second, the error will be about 2.8  10 2 %. 1.16: The area is 9.69  0.07 cm2, where the extreme values in the piece’s length and width are used to find the uncertainty in the area. The fractional uncertainty in the 0 07 cm 2 area is 9..69 cm 2 = 0.72%, and the fractional uncertainties in the length and width are 0.01 cm 5.10 cm = 0.20% and 0.01 cm 1.9 cm = 0.53%. 1.17: a) The average volume is π 8.50 cm2 0.050 cm  2.8 cm 3 4 (two significant figures) and the uncertainty in the volume, found from the extreme values of the diameter and thickness, is about 0.3 cm3 , and so the volume of a cookie is 2.8  0.3 cm3 . (This method does not use the usual form for progation of errors, which is not addressed in the text. The fractional uncertainty in the thickness is so much greater than the fractional uncertainty in the diameter that the fractional uncertainty in the volume is 10% , reflected in the above answer.) b) 8.50 .05  170  20. 1.18: (Number of cars  miles/car.day)/mi/gal = gallons/day (2  108 cars  10000 mi/yr/car  1 yr/365 days)/(20 mi/gal) = 2.75  108 gal/day 1.19: Ten thousand; if it were to contain ten million, each sheet would be on the order of a millionth of an inch thick. 1.20: If it takes about four kernels to fill 1 cm3, a 2-L bottle will hold about 8000 kernels.
  4. 1.21: Assuming the two-volume edition, there are approximately a thousand pages, and each page has between 500 and a thousand words (counting captions and the smaller print, such as the end-of-chapter exercise and problems), so an estimate for the number of words is about 106 . 1.22: Assuming about 10 breaths per minutes, 24  60 minutes per day, 365 days per year, and a lifespan of fourscore (80) years, the total volume of air breathed in a lifetime is about 2 10 5 m 3 . This is the volume of a room 100 m 100 m  20 m , which is kind of tight for a major-league baseball game, but it’s the same order of magnitude as the volume of the Astrodome. 1.23: This will vary from person to person, but should be of the order of 1 105 . 1.24: With a pulse rate of a bit more than one beat per second, a heart will beat 105 times per day. With 365 days in a year and the above lifespan of 80 years, the number of beats in a lifetime is about 3  10 9 . With 1 L (50 cm3) per beat, and about 1 gallon per 20 4 liter, this comes to about 4  10 7 gallons. 1.25: The shape of the pile is not given, but gold coins stacked in a pile might well be in the shape of a pyramid, say with a height of 2 m and a base 3 m  3 m . The volume of such a pile is 6 m 3 , and the calculations of Example 1-4 indicate that the value of this volume is $6  108. 1.26: The surface area of the earth is about 4R 2  5  1014 m 2 , where R is the radius of the earth, about 6 10 6 m , so the surface area of all the oceans is about 41014 m 2 . An average depth of about 10 km gives a volume of 4  1018 m3  4  1024 cm3 . Characterizing the size of a “drop” is a personal matter, but 25 drops cm3 is reasonable, giving a total of 1026 drops of water in the oceans. 1.27: This will of course depend on the size of the school and who is considered a "student''. A school of thousand students, each of whom averages ten pizzas a year (perhaps an underestimate) will total 104 pizzas, as will a school of 250 students averaging 40 pizzas a year each.
  5. 1.28: The moon is about 4 108 m  4 1011 mm away. Depending on age, dollar bills can be stacked with about 2-3 per millimeter, so the number of bills in a stack to the moon would be about 1012. The value of these bills would be $1 trillion (1 terabuck). 1.29: Area of USA  Area bill   number of bills. 9,372,571 km 2    106 m 2 km 2 15.6 cm  6.7 cm  1 m 2 10 4 cm 2  9  1014 bills 9  1014 bills 2.5 108 inhabitants  $3.6 million inhabitant. 1.30: 1.31: 7.8 km, 38  north of east
  6. 1.32: a) 11.1 m @ 77.6 o b) 28.5 m @ 202 o c) 11.1 m @ 258o d) 28.5 m @ 22 o 1.33: 144 m, 41 south of west.
  7. 1.34:  1.35: A; Ax  12.0 m  sin 37.0   7.2 m, Ay  12.0 m  cos 37.0   9.6 m.  B; B x  15.0 m  cos 40.0   11.5 m, B y  15.0 m  sin 40.0   9.6 m.  C ; C x  6.0 m  cos 60.0   3.0 m, C y  6.0 m  sin 60.0   5.2 m. Ay  1.00 m 1.36: (a) tan θ    0.500 AX 2.00 m θ  tan 1  0.500  360   26.6   333 Ay 1.00 m (b) tan θ    0.500 Ax 2.00 m θ  tan 1 0.500  26.6  Ay 1.00 m (c ) tan θ    0.500 Ax  2.00 m θ  tan 1  0.500  180   26.6   153 Ay  1.00 m (d) tan θ    0.500 Ax  2.00 m θ  tan 1 0.500  180   26.6   207 
  8. 1.37: Take the +x-direction to be forward and the +y-direction to be upward. Then the second force has components F2 x  F2 cos 32.4  433 N and F2 y  F2 sin 32.4  275 N. The first force has components F1 x  725 N and F1 y  0. Fx  F1x  F2 x  1158 N and Fy  F1 y  F2 y  275 N The resultant force is 1190 N in the direction 13.4 above the forward direction. 1.38: (The figure is given with the solution to Exercise 1.31). The net northward displacement is (2.6 km) + (3.1 km) sin 45o = 4.8 km, and the net eastward displacement is (4.0 km) + (3.1 km) cos 45o = 6.2 km. The magnitude of the resultant displacement is (4.8 km)2  (6.2 km) 2 = 7.8 km, and the direction is arctan  6..2  = 38o north of east. 48
  9.  1.39: Using components as a check for any graphical method, the components of B are  Bx  14.4 m and B y  10.8 m, A has one component, Ax  12 m . a) The x - and y - components of the sum are 2.4 m and 10.8 m, for a magnitude  10.8  of 2.4 m   10.8 m   11.1 m, , and an angle of   2 2   77.6 .  2.4  b) The magnitude and direction of A + B are the same as B + A. c) The x- and y-components of the vector difference are – 26.4 m and  10.8 m, for a magnitude of 28.5 m and a direction arctan   10..8   202. Note that  26 4 180 must be added to arctan  10..8   arctan 10..8   22 in order to give an angle in the  26 4 26 4 third quadrant.   d) B  A  14.4 mi  10.8 mˆ  12.0 mi  26.4 mi  10.8 mˆ. ˆ j ˆ ˆ j  10.8  Magnitude  26.4 m 2  10.8 m 2  28.5 m at and angle of arctan    22.2 .  26.4  1.40: Using Equations (1.8) and (1.9), the magnitude and direction of each of the given vectors is: a) (8.6 cm) 2  (5.20 cm) 2 = 10.0 cm, arctan  58.20  = 148.8o (which is .60 180 – 31.2o). o b) (9.7 m) 2  (2.45 m) 2 = 10.0 m, arctan  29.45  = 14o + 180o = 194o. .7 c) (7.75 km) 2  (2.70 km) 2 = 8.21 km, arctan  .275  = 340.8o (which is 7 .7 360 – 19.2o). o
  10. 1.41: The total northward displacement is 3.25 km  1.50 km  1.75 km, , and the total westward displacement is 4.75 km . The magnitude of the net displacement is 1.75 km2  4.75 km 2  5.06 km. The south and west displacements are the same, so The direction of the net displacement is 69.80 West of North. 1.42: a) The x- and y-components of the sum are 1.30 cm + 4.10 cm = 5.40 cm, 2.25 cm + (–3.75 cm) = –1.50 cm. b) Using Equations (1-8) and (1-9), (5.4 0 cm) 2 (1.50 cm) 2 = 5.60 cm, arctan  15..50  = 344.5o ccw.  40 c) Similarly, 4.10 cm – (1.30 cm) = 2.80 cm, –3.75 cm – (2.25 cm) = –6.00 cm. d) (2.80 cm) 2  (6.0 cm) 2 = 6.62 cm, arctan  26.80  = 295o (which is 360o – 65 .00
  11.   1.43: a) The magnitude of A  B is   2.80 cm  cos 60.0  1.90 cm  cos 60.0 2      2.48 cm    2.80 cm  sin 60.0  1.90 cm  sin 60.0 2     and the angle is  2.80 cm  sin 60.0  1.90 cm sin 60.0   arctan   2.80 cm  cos 60.0  1.90 cm cos 60.0   18      b) The magnitude of A  B is   2.80 cm  cos 60.0  1.90 cm  cos 60.0 2      4.10 cm    2.80 cm  sin 60.0  1.90 cm  sin 60.0 2     and the angle is  2.80 cm  sin 60.0  1.90 cm sin 60.0   arctan   2.80 cm  cos 60.0  1.90 cm cos 60.0   84          c) B  A   A  B ; the magnitude is 4.10 cm and the angle is 84   180   264 .  1.44: ˆ A = (–12.0 m) i . More precisely,        A  12.0 m  cos 180  i  12.0 m  sin 180  j .      B  18.0 m  cos 37  i  18.0 m  sin 37  ˆ  14.4 m i  10.8 m  ˆ ˆ j ˆ j  1.45: A  12.0 m sin 37.0 i  12.0 m  cos 37.0 ˆ  7.2 m  i  9.6 m  ˆ ˆ j ˆ j  B  15.0 m cos 40.0 i  15.0 m  sin 40.0 j  11.5 m  i  9.6 m  ˆ ˆ ˆ ˆ j  C  6.0 m cos 60.0i  6.0 m  sin 60.0 ˆ  3.0 m  i  5.2 m  ˆ ˆ j ˆ j
  12.  1.46: a) A  3.60 m cos 70.0 i  3.60 m sin 70.0 ˆ  1.23 m i  3.38 m  ˆ ˆ j ˆ j  B  2.40 m  cos 30.0 i  2.40 m  sin 30.0 j   2.08 m i   1.20 m  ˆ ˆ ˆ ˆ j b)    C  3.00  A  4.00 B  3.001.23 m i  3.003.38 m  ˆ  4.00 2.08 m i  4.00 1.20 m  ˆ ˆ j ˆ j  12.01 m i  14.94 ˆ ˆ j (Note that in adding components, the fourth figure becomes significant.) c) From Equations (1.8) and (1.9),  14.94 m  C  12.01 m   14.94 m   19.17 m, arctan  2 2    51.2  12.01 m  1.47: a) A 4.002  3.002  5.00, B  5.002  2.002  5.39   b) A  B  4.00  3.00 i  5.00   2.00  ˆ   1.00 i  5.00  ˆ ˆ j ˆ j c) 1.002  5.002  5.10, arctan 5.00   101.3    - 1.00  d)
  13. 1.48: a) i  ˆ  k  12  12  12  3  1 so it is not a unit vector ˆ j ˆ  b) A Ax2  Ay  Az2 2  If any component is greater than + 1 or less than –1, A  1 , so it cannot be a unit  vector. A can have negative components since the minus sign goes away when the component is squared. c)  A 1 a 2 3.0   a 2 4.0   1 2 2 a 2 25  1 1 a  0.20 5 .0   1.49: a) Let A  Ax i  Ay ˆ, ˆ j B  Bx iˆ  By ˆ. j   A  B   Ax  Bx i  Ay  By  ˆ ˆ j   B  A  Bx  Ax i  By  Ay  ˆ ˆ j     Scalar addition is commutative, so A  B  B  A   A  B  Ax Bx  Ay By   B  A  Bx Ax  By Ay     Scalar multiplication is commutative, so A  B  B  A   b) A  B  Ay Bz  Az B y i   Az Bx  Ax Bz  ˆ  Ax B y  Ay Bx k ˆ j ˆ   B  A  B y Az  Bz Ay i  Bz Ax  Bx Az  ˆ  Bx Ay  B y Ax k ˆ j ˆ Comparison of each component in each vector product shows that one is the negative of the other.
  14. 1.50: Method 1: Pr oduct of magnitudes  cos θ  AB cos θ  12 m  15 m  cos 93  9.4 m 2 BC cos θ  15 m  6 m  cos 80   15.6 m 2 AC cos θ  12 m  6 m  cos 187   71.5 m 2 Method 2: (Sum of products of components) A  B  7.22 (11.49)  (9.58)(9.64)  9.4 m 2 B  C  (11.49)(3.0)  (9.64)(5.20)  15.6 m 2 A  C  (7.22)(3.0)  (9.58)(5.20)  71.5 m 2 1.51: a) From Eq.(1.21),   A  B  4.00 5.00   3.00  2.00   14.00. b) A  B  AB cos θ , so θ  arccos 14.00 5.00  5.39  arccos.5195  58.7 . 1.52: For all of these pairs of vectors, the angle is found from combining Equations (1.18) and (1.21), to give the angle  as    A  B  Ax Bx  Ay B y    arccos    AB   arccos  .     AB  In the intermediate calculations given here, the significant figures in the dot products and in the magnitudes of the vectors are suppressed.   a) A  B   22, A  40, B  13, and so   22    arccos     40 13   165 .      60  b) A  B  60, A  34, B  136,   arccos   28 .  34 136    c) A  B  0,   90.
  15. 1.53: Use of the right-hand rule to find cross products gives (a) out of the page and b) into the page. 1.54: a) From Eq. (1.22), the magnitude of the cross product is 12.0 m 18.0 m sin 180  37   130 m 2 The right-hand rule gives the direction as being into the page, or the – z-direction. Using Eq. (1.27), the only non-vanishing component of the cross product is   C z  Ax B y   12 m  18.0 m  sin 37  130 m 2 b) The same method used in part (a) can be used, but the relation given in Eq. (1.23) gives the result directly: same magnitude (130 m2), but the opposite direction (+z-direction). 1.55: In Eq. (1.27), the only non-vanishing component of the cross product is C z  Ax By  Ay Bx  4.00  2.00  3.00 5.00  23.00,   so A  B  23.00 k , and the magnitude of the vector product is 23.00. ˆ
  16.   1.56: a) From the right-hand rule, the direction of A B is into the page (the – z-direction). The magnitude of the vector product is, from Eq. (1.22), AB sin   2.80 cm 1.90 cm  sin 120  4.61 cm 2 . Or, using Eq. (1.27) and noting that the only non-vanishing component is C z  Ax B y  Ay B x  2.80 cm  cos 60.0   1.90 cm  sin 60   2.80 cm  sin 60.0  1.90 cm  cos 60.0   4.61 cm 2 gives the same result. b) Rather than repeat the calculations, Eq. (1-23) may be used to see that   B  A has magnitude 4.61 cm2 and is in the +z-direction (out of the page). 1.57: a) The area of one acre is 1 8 mi  80 mi  1 1 640 mi2 , so there are 640 acres to a square mile. 2  2  b) 1 acre  1 mi    5280 ft   43,560 ft 2  640 acre   1 mi      (all of the above conversions are exact). c)    7.477 gal  (1 acre-foot)  43,560 ft 3   3   3.26  10 gal, 5  1 ft  which is rounded to three significant figures. 1.58: a)   ($4,950,000 102 acres)  (1acre 43560 ft 2 )  10.77 ft 2 m 2  $12 m 2 . b) ($12 m 2 )  (2.54 cm in) 2  (1 m 100 cm) 2  $.008 in 2 . c) $.008 in 2  (1in  7 8 in )  $.007 for postage stamp sized parcel.
  17. 1.59: a) To three significant figures, the time for one cycle is 1  7.04 10 10 s. 1.420 10 9 Hz  9 cycles   3600 s  12 cycles b) 1.420 10    5.1110  s   1h  h c) Using the conversion from years to seconds given in Appendix F, 1.42 109 Hz   3.156 10 s   4.600 109 y   2.06 10 26. 7      1y     d) 4.600  10 9 y  4.60  10 4 1.00  10 5 y , so the clock would be off by 4.60  10 4 s. 1.60: Assume a 70-kg person, and the human body is mostly water. Use Appendix D to find the mass of one H2O molecule: 18.015 u  1.661  10–27 kg/u = 2.992  10–26 kg/molecule. (70 kg/2.992  10–26 kg/molecule) = 2.34  1027 molecules. (Assuming carbon to be the most common atom gives 3  1027 molecules. 1.61: a) Estimate the volume as that of a sphere of diameter 10 cm: 4 V  r 3  5.2  10  4 m 3 3 Mass is density times volume, and the density of water is 1000 kg m3 , so    m  0.98 1000 kg m3 5.2  10 4 m3  0.5 kg b) Approximate as a sphere of radius r  0.25 μm (probably an over estimate) 4 V  r 3  6.5  10 20 m 3 3    m  0.98 1000 kg m3 6.5  1020 m3  6  1017 kg  6  10-14 g c) Estimate the volume as that of a cylinder of length 1 cm and radius 3 mm: V  r 2l  2.8  10 7 m 3    m  0.98 1000 kg m 3 2.8  10 7 m 3  3  10 4 kg  0.3 g
  18. M M 1.62: a) ρ , so V  V ρ 0.200 kg x3   2.54  10 5 m 3 7.86  10 kg/m 3 3 x  2.94  10  2 m  2.94 cm 4 3 R  2.54  10 5 m 3 b) 3 R  1.82  10 2 m  1.82 cm 1.63: Assume each person sees the dentist twice a year for checkups, for 2 hours. Assume 2 more hours for restorative work. Assuming most dentists work less than 2000 hours per year, this gives 2000 hours 4 hours per patient  500 patients per dentist. Assuming only half of the people who should go to a dentist do, there should be about 1 dentist per 1000 inhabitants. Note: A dental assistant in an office with more than one treatment room could increase the number of patients seen in a single dental office.  6.0  1023 atoms  1.64: a) (6.0  10 kg)   24 mole   14  103 kg   2.6  10 atoms. 50  mole  b) The number of neutrons is the mass of the neutron star divided by the mass of a neutron: (2) (2.0  10 30 kg)  2.4  10 57 neutrons. (1.7  10 27 kg neutron ) c) The average mass of a particle is essentially 2 the mass of either the proton or 3 the neutron, 1.7  10 27 kg. The total number of particles is the total mass divided by this average, and the total mass is the volume times the average density. Denoting the density by  (the notation introduced in Chapter 14). 4  R 3 M (2) (1.5  1011 m) 3 (1018 kg m 3 )  3   27  1.2  10 79. mave 2 (1.7  10 kg ) mp 3 Note the conversion from g/cm3 to kg/m3.
  19.  1.65: Let D be the fourth force.          A  B  C  D  0, so D   A  B  C  Ax   A cos 30.0   86.6 N, Ay   A cos 30.0   50.00 N B x   B sin 30.0   40.00 N, B y   B cos 30.0   69.28 N C x  C cos 53.0   24.07 N, C y  C sin 53.0   31.90 N Then D x  22.53 N, D y  87.34 N D  D x2  D y  90.2 N; 2 tan α  Dy / Dx  87.34 / 22.53 α  75.54 φ  180  α  256  , counterclockwise from  x - axis 1.66: R x  Ax  B x  (170 km) sin 68  (230 km) cos 48   311.5 km R y  Ay  B y  (170 km) cos 68   (230 km) sin 48   107.2 km R  R x2  R y  2 311.5 km 2   107.2 km 2  330 km Ry 107.2 km tanθ R    0.344 Rx 311.5 km θ R  19  south of east
  20. 1.67: a)     b) Algebraically, A  C  B, and so the components of A are Ax  C x  Bx  6.40 cm cos 22.0  6.40 cm  cos 63.0  3.03 cm Ay  C y  B y  6.40 cm  sin 22.0   6.40 cm  sin 63.0  8.10 cm.  8.10 cm  c) A 3.03 cm 2  8.10 cm 2  8.65 cm, arctan   69.5   3.03 cm 
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