57. 26.32 poundals. 58. Tension = weight of 4024.3 gm. 59. 0.1113 ft.-sec. units. 62. g=981.05. 63. g=32.227. 64. It would have to be shortened to 3/19ths of its original length. 66. g=32.191; 39.69 in. 67. The pendulum is 0.2717 in. too long, therefore 8.15 turns are required to correct it. 77. 9.6 x 106 gramme-centimetres; 9.4176 x 109 ergs. 78. 288,000 ft.-lbs. 79. 10:27. 80. 1820 √3 ft.-lbs. 81. 9600 ft.-lbs. 82. 1.44 × 109 ergs. 83. (1) equal; (2) inversely proportional to the masses. 84. 20.16 ft.-lbs. 85. 69.12 ft.-lbs. 86. 6788 ft.-tons. 87. In the ratio of 400 to I. 88. 1634.3 ft.-tons. 89. 706.9 ft.-lbs. 90. Total work 91. 98,175,000 ft.-lbs. = 11,000 ft.-lbs. 99. (1) 336,000 ft.-lbs.; (2) 10,752,000 ft.-poundals. 100. 33 ft. 101. 2500 ft.-lbs. 102. 3.767 × 101o ergs. 103. 9.81 x 107 ergs. 104. 6.25 × 1010 ergs. 105. 4.5 × 1012 ergs. 106. Momentum = 8960 √2; K.E = 143,360 ft.-poundals, or in ft.-lbs. = 143,360/32 = 4480. 107. 1536 ft.-poundals, or 48 ft.-lbs. 108. The initial velocity must have been 96 ft. per sec., and the final energy must = K.E. at starting = 5 × (96)2/2g=720 ft.-lbs. 109. Force = 300 poundals; K.E. = 225,000 ft. poundals. 110. 8 x Io10 ergs. 112. 2.5 × Iolo ergs. 113. 16,940 ft.-lbs. 114. 8.5 × 106 ergs. 116. 18 cm. 117. K.E. = 1210 ft.-tons; a force equal to the weight of 1 ton. 118. 5461 ft.-poundals. 119. As 112:625. 120. 9.375 × 108 dynes. 121. 13,090 ft.-lbs. 123. 32,000 ft.-lbs.; 320 ft. per. sec. pact I 50,000. 125. Force = = 114,688 ft.-poundals. 126. 2/3. 127. As (√3 − 1) : 1. = 124. K.E. before im716.8 poundals; work Mass - 45; velocity= = 130. (1) 4561 kilogramme-metres per min.; (2) 132. 911 7.456 × 109 ergs per sec. 131. 38,016 cub. ft. H.P. 133. 52 miles per hour. 134. 16.8 H.P. 135. 151.2 H.P. 136. 192 H.P. 137. 480 H.P. 138. At the rate of 3.367 H.P. 139. The unit of work would be increased ten-fold; the numerical value of the horse-power would not be changed. EXAMINATION QUESTIONS 141, 143. See Introduction, § 8, and Ch. I., Exs. 16 and 26. 145, 146. See §§ 8 and 9 and Exs. 27 and 87. 150. The proof follows easily when the dimensions [MLT2] of force are known. 156. When describing a circle of 100 ft. radius, his inclination to the ice is 84°. 162. (1) 2000 ft.-lbs.; (2) 1562.5 ft-lbs. 171. See Ex. 97. 174. This example is also inserted (and solved) in the chapter on Thermodynamics (Ch. V., 40). 175. K.E. of tram-car = 431,500 ft.-poundals; work done in a run of 3 miles 22,065, 120 ft.-poundals. CHAPTER II-HYDROSTATICS 3. The dimensions of pressure are the same as those of force, viz. MLT2; the dimensions of intensity of pressure (force per unit area) are ML1T-2. 6. 124.4 lbs. per cub. ft. 7. 0.5787 oz. per cub. in. 8. 1.736 oz. per cub. in. 10. 13.824. 11. 10.98 lbs. 12. 0.7055 gm. per c.c. 13. 300.8 lbs. 16. Cross-section = 0.3676 sq. cm.; diameter 0.683 cm. 17. 111.97 gm. 18. 1.4. 19. 4.5 cm. 20. As 27: 10. 21. 19.712 gm. 23. 4.516. 14. 3.2 cub. ft. = 24. 4.57. 27. 25. 15s/8. 26. Its density is 0.925 that of air. (1) 100 gm. per sq. cm.; (2) 35. 206,991 dynes per sq. cm. lbs. per sq. in. 49. Sufficient to cm. 36. 10.193 metres. 37. 98,100 dynes per sq. 38. 12,750 gm. weight. 39. 1/384 lb. per cub. in. 40. 18,750 lbs. per sq. ft. 41. 17.36 lbs. per sq. in. 42. 138.2 ft. 43. 337,920 lbs. per sq. ft. 44. 28.02 lbs. per sq. in. 45. 10.193 metres. 46. 0.9. 47. 49.08 ft. 48. 21.3 occupy 5 in. of the tube. 50. l'cos 30°(1 − s). 51. 150 gm. 52. 68,360 gm. 53. 99,000 gm. 54. 18,150 gm. 55. 10,000 gm. on upper surface, 11,000 gm. on lower surface, 10,500 gm. on each of the vertical sides. 70. Acceleration = 58. Volume = 20 c.c.; sp. gr. = 3.1. 59. 22.04 gm. 60. 21.08 gm. 61. 34 c.c. 62. 425.9 oz. 63. 20.8 gm. 64. 27.5 gm. 66. Sq=ms/(m25μ − m11+ m1). 67. 20.97 gm. 68. 21.57 gm. 69. A weight equal to that of the copper. 20 ft.-sec. units; time = 1.247 sec. 71. 6.04 lbs. 72. 0.32 gm. 74. 1000 c.c. 75. 300 C.C. 76. 6437.5 cub. yds. 77. 50 c.c. 78. Sp. gr. of both solid and liquid = 0.5. 79. The sphere will rest in equilibrium with 1/7th of its volume immersed in the mercury. 80. 3.3. 81. 1.625. 82. 1.4. 83. 1.204. 84. 0.8271. 85. 1.434. 86. 0.9127. 88. 11.31. 89. 1.948. 90. 0.7351. 91. 1.059. 99. Sp. gr. 0.8. 92. 4.75. 93. 3.219. 94. of pebble = 2.74; of spirit = 0.825. 100. 240 lbs. 101. 10,080 lbs. 102. Ratio of arms = Pressure = 8064 lbs.; distance Mechanical advantage = 1440. = : 8:5. 103. 0.1042 in. 104. 107. Apparent height = 20√3 in. = 34.64 in. 108. 34 ft. 109. 13.6 in. 110. 25.197 ft. 111. (1) 0.0938; (2) 1.276. 112. 11.86 metres. 113. 10,546 kgm. per sq. metre. 115. 1,039,890 dynes per sq. cm. 116. 14.01 lbs. per sq. in. 117. 30,000 oz. per sq. ft. 118. 1006.1 gm. per sq. cm., or 10,061 kgm. per sq. metre. 119. 4.045 × 107 dynes. 120. A change of about half a pound (0.4917 lb.) per sq. in. 124. p: p' = 9 ; 1. 125. p:p'=r'3 ; r3 ̧ per sq. in.; 3.6 in. 126. 71.93 cm. 127. 360 lbs. 128. As I 1.306. 129. 1.395 131. 96.6 cm. gm. 130. Pressure = 2 atmospheres. 132. It will descend 1.2 in. 133. p: pr': r. 135. 54.1 cm. 136. The tube should have been raised until a length of 80 cm. stood out of the mercury. 138. 68.85 cm. 139. About 18 cm. 140.cub. 144. 4.34 ft. in. 142. (1) 34.2 c.c.; (2) 87 c.c. 145. 4.49 ft. 146. It must be lowered until its top is 66 ft. below the surface. 148. 164.3 cub. ft.; 80.3 in. 149. 0.876 in. 150. The jar must be sunk until its mouth is 33.64 ft. below the surface. (3/4)10D=0.05631D. 154. 0.4363 gm. gm. 158. 260 kgm. 152. Dn= 155. 0.741 EXAMINATION QUESTIONS 164. See p. 58; pres166. Whole pressure 159. See Ch. VII. p. 172. sure =2379.3 gm. per sq. cm. =3750 lbs.; resultant pressure=625 lbs. force equal to the weight of 320 lbs. 172-175. 62, 63, and Ex. 171. 189. See Ex. 137. 167. A See pp. CHAPTER III-HEAT 2.0068 metres; 7. 263.16 cm. 10. 22.128 cm. temperature = 14. 0.0144 sq. 18. 1.00649 1. Expansion of Solids. -4. 150°. 5. 153.86 cm. 6. 0.0432 in. 8. 87.2464 cm. 9. 1.00095 yard. 11. Coefficient of expansion = 0.00008; 260°. 12. 100 cm. 13. 0.02864 in. ft. 15. 0.01764 ft. 17. 300.912 sq. cm. metre. 19. 36 cm. 20. When the pendulum keeps correct time (¿.e. at 5°), it swings 86,400 times per day (supposing it to be a seconds pendulum; see p. 30). At 30° its length is increased in the ratio of 1.0003: 1, and it now swings 86,400 √1/1.0003 times per day. Referring to p. 19, it will be seen that √1/1.003 = 1/1.000151 -0.00015 approximately. Therefore the clock loses 86,400 x 0.00015= 12.96 secs. per day. 21. It will gain 20.52 secs. per day. 23. The required temperature is 196 ̊.4, and the common length at this temperature is 251.424 cm. 24. 12°. See § II for this and the next example. 25. Increase in volume= 2.592 cub. in. 27. 10.22. 28. 48.373 c.c. 2. Expansion of Liquids. 29. 0.00002797. 33. 0.000302. 34, 0.000301. 37. 0.0001817. 38. 103°.8. 45. 0.0001558. 46. The coefficient of apparent expansion of the mercury is 0.0001546, and this gives 0.0000274 as the coefficient of expansion of the glass. 47. 1.57 c.c. 48. 132°.3 49. 109°.65. 50. 9.517 gm. 54. The volume of the solid is 12.9752 c.c. at 10°, and 12.9914 c.c. at 95°; ... its coefficient of expansion is 0.00001468. 3. Expansion of Gases.-58. 3.187 litres; 54°.6. 59. (1) 11.16 litres; (2) 12.38 litres. 60. 221.978 c.c. 61. 0.10231 gm. 62. 333°. 63. 69.63 cm. at o°; 95.12 cm. at 100°. 64. 2190.1 C.C. 65. 0.00367. 66. 77°.6. 69. 18°.74. 70. The temperature must rise from 10° to 57°.16. 71. 322.6 c.c. 87. 11.97 77. As I: 0.9269. 78. 998.9 c.c. 79. 10.47 atmospheres. 80. As I: 0.7808. 81. As I 2.256. 82. The temperature must fall to - 2°. 84. 0.599 gm. 85. 924.9 kgm. 86. 11.66 litres. gm. 88. 0.3743 gm. 90. 1991 C.C. 91. 37-98 in.; at 459°.5. 92. As I: 1.02. 93. The temperature must rise to 7°.18. 94. (1): (2)=1:0.891. 95. 299 ̊. 96. 91°.8. 99. 13°.65. EXAMINATION QUESTIONS 100. The lengths must be the coefficients of expansion. inversely proportional to 101. Observe that the |