57. Define one foot-pound, one erg, one horse power. Express one horse power in ergs per second, given that it is equal to 550 foot-pounds per second. [One horse power = 550 ft.-lbs. per sec. =550 × 453.6 ft.-gr. per sec. =550 × 453.6 × 30·48 cm.-gr. per sec. =550 × 453.6 × 30.48 × 981-4 cent. dynes or ergs =7.464 × 109 ergs per second.] 58. The mass of an iron fly-wheel is ton and its radius of gyration 2 feet. Find what horse power applied for one minute would get it up to a speed of 2 turns per second. To how many degrees Centigrade would this amount of work raise 18 lbs. of water, supposing the whole work spent in heating the water. 25 [I=Mk2=560× 4 This amount of work would raise 18 lbs. of water to 280 × 625 × 9.87 128.8 × 18 × 1390 =0°·5359 Centigrade.] 59. How much kinetic energy does a bullet weighing 20 lbs. possess while moving with a velocity of 1600 feet per second. Find what average horse power expended in the gun duringth of a second would give this amount of energy to the bullet. If the bullet is stopped by a target, find how much heat is generated, the unit being the quantity required to raise one pound of water through 1o C. COMPRESSIBILITY OF GASES AND LIQUIDS. 60. State Boyle's law. In an experiment with the Mariotte tube, commencing with the mercury in both tubes at the same level (A, B), mercury is poured into the open branch till it stands at the height of 20 inches above the level (A, B). The length of the closed branch is 12 inches, and the barometric height at the time of the experiment is 29.5 inches. the height of the mercury in the closed branch. Find [Let x height of mercury in closed branch. Then 291 × 12=(291+20−x)(12−x) =291 × 12+240−611⁄2x+x2, x2-61x240=0, x=2.8 inches.] 61. Taking the modulus of compression of water as 21 x 106 grammes weight per square centimetre, find how much 500 centimetres of water would be diminished in volume by an increase of pressure of 2000 grammes per square centimetre, [Suppose we take a column of water, 500 cms. long and 1 square centimetre in cross sectional area; then if e be the actual contraction in cms. in 500 cms. length, when 2000 grammes pressure per unit area is applied, we have 62. 105 grammes of alcohol, density 8228, is mixed with 90 grammes of water and the density of the mixture is 909; find by what per cent. the volume has contracted. 63. A tube 48 inches long is filled half with water and half with alcohol. The liquids are then mixed and the mixture is found to occupy only 47 inches of the tube. Find the specific gravity of the mixture. (Specific gravity of alcohol=0·645.) [24+24 x 645=47p, if p is density of mixture, p=84.] ELASTICITY. 64. Define the following terms: Stress, Strain, a Modulus of Elasticity, Rigidity Modulus, Young's Modulus. What is meant by the length of a modulus? From the following data for a steel wire, calculate its modulus for longitudinal strain, and give your result in grammes weight per square centimetre. Length of wire 503 centimetres; stress applied 12700 grammes. Elongation produced 7 cm. Diameter of wire '075 cm. =2.065 × 109 grammes weight per sq. cm. 65. To find the Young's modulus of a certain copper wire the following data were determined Length of wire, 80 feet; Weight of 30.9 inches, 676 grains; Elongation produced by 20 lbs. weight, 1·09 inch; Density of wire, 8.9. Calculate the Young's modulus in pounds weight per square inch from these data. [Let r be the radius of the wire. Then 66. Define Torsional Rigidity and Flexural Rigidity. Taking the rigidity modulus of copper as 450 × 106 grammes weight per square centimetre, calculate the torsional rigidity of a rod of circular section of two centimetres radius. [Torsional rigidity-rigidity modulus × square of radius of gyration × cross sectional area 67. Taking the Young's Modulus of Copper as 1200 x 106 grammes weight per square centimetre, calculate the flexural rigidity of a rod of circular section of 2 cms. radius. [Flexural rigidity=Young's modulus x square of radius of gyration x cross sectional area |