Southern Branch of the University of California Los Angeles This book is DUE on the last Form L-9-157tt-8,'26 KEY NATURAL PHILOSOPHY. BY ISAAC SHARPLESS, Sc.D., PRESIDENT OP HAVEKFORD COLLEGE, AND GEO. MORRIS PHILIPS, PH.D., PRINCIPAL OF STATE NORMAL SCHOOL, WEST CHESTER, PA. GTATEHOBMALSCHOOL 1.06 AWOttfES. --,., PHILADELPHIA: J. B. LIPPINCOTT COMPANY. 1893. Copyright, 1884, by J. B. LIPPINCOTT & Co Copyright, 1893, by J. B. LIPPINCOTT COMPANY. 749 1 . .c 2 3 5 $3 PREFACE. THIS Key is written for the convenience of teachers, to save them the time necessary to work out all the Exercises of the Philosophy. It will assist them also in determining which of these Exercises to omit. While all of them will be useful for a class of good or well- developed students, it is undesirable to exact them all of the weaker ones: in general, the authors think it unwise to require any which the average member of the class cannot work out for himself, at least with a few hints from the teacher. Properly used, anything which requires original thinking of the students is the most beneficial part of a treatise; but examples too difficult, which have to be fully explained by the teacher, have no especial value. We think these considerations will be a guide to the teacher in the selection of the exercises which he assigns to his students. KEY TO NATURAL PHILOSOPHY. fage 17. Questions. 1. It is doubled. 2. It is quadrupled. 3. It is reduced to one-fourth. 4. It is not changed. Page 19. Exercise 1. Matter is never destroyed. "Water changes to invisible vapor; gunpowder to invisible gases; house-gas combines with oxygen in the air and forms new substances. 2. Blotting-pads owe their utility to porosity; rubber bands, to elasticity; watch-springs, to elasticity; pop-guns, to elasticity of air; putty, to adhesion ; hammers, to inertia ; piano-strings, to elasticity ; water-filters, to porosity. 3. Because the sugar goes between the pores of the water. 4. Inertia. 5. Because the little air in it tends to expand, and the outward pressure is not resisted. It collapses because the air leaks out through the pores. 6. The divisibility of matter into very minute portions. 7. By seeing which would scratch the other. 8. The molecules are separated. The mass is unchanged ; the density diminished ; the weight unchanged ; the volume increased. 9. Water. 10. Divide the mass by the volume. Divide the mass by the density. 11. Graphite, or the " lead" of lead-pencils. 12. Because the earth pulls all the particles of each the same, and draws each particle just as if it were not con- nected with the others. 1* 6 6 KEY TO NATURAL PHILOSOPHY. 13. It is three times as far from the centre of the earth, and, as attraction varies inversely as the square of the dis- tance, it will weigh one-ninth of its weight at the surface. 14. One thirty-six-hundredth of its weight at the surface, or -jgY^r Tff f a P oun d- 15. Since the volumes of spheres are as the cubes of their radii, the volume would be one-eighth the volume of the earth ; and, as they have the same density, the mass would be one-eighth also. Hence at the same distance the attrac- tion would be one-eighth. But, the body being only half as far from the centre, the attraction would be increased four- fold from the effect of nearness, if the mass remained the same. Both causes acting, the weight would be 100 X i X 4 = 50. Page 2O. Questions. Train starting accelerated. Train stop- ping retarded. Ball upward retarded. Ball falling accelerated. Hands of a watch uniform. Kiver and winds varied. Page 23. Question. Momenta equal. Velocities as 2 to 1. Page 26. Exercise 2. 16 X 200 = 3200 grammetres. 3200 -=- 1000 = 3.2 kilogrammetres (Ans.). Page 28. Exercise L 25 pounds per square inch, multiplied by 165, the number of square inches, gives 4125 pounds' pressure on the piston. This, multiplied by 4, the number of feet travelled by the piston in each revolution of the engine, and 150, the number of revolutions per minute, gives 2,475,000 foot-pounds per minute. 2,475,000 -=- 33,000 = 75 horse-power. 2. 20 X 10 = 200 foot-pounds (Ans.}. KEY TO NATURAL PHILOSOPHY. 3. 2 tons = 4000 pounds. 4000 X 600 = 2,400,000 foot- pounds per minute. 2,400,000 -=- 33,000 = 72^ horse- power (Ans.}. 4. 72^- -T- 2 = 36^. horse-power required to do the work in 2 minutes. 72^ r -=- 5 = 14^- horse-power to do it in 5 minutes. 72j 8 T -=- i = 145^- horse-power to do it in one-half minute. Question (Art. 71). As there are 980 ergs in 1 gram, 1000 grams in a kilogram, and 100 centimetres in a metre, 1 kilo- grammetre = 980 X 1000 X 100 = 98,000,000 ergs (Ans.). Page 34. 1. It will slide when the line downward from the centre of gravity falls within the base of support, and roll when it falls outside. 2. To bring the centre of gravity in front of the vertical line through the feet, so that our weight will assist the muscles in rising. 3. Because the stick enables us more quickly to place the centre of gravity directly over the fence when it gets a little out. 4. When it is at the bottom of its swing. 5. A cone on its apex is in unstable equilibrium ; on its base, in stable ; on its side, in neutral. 6. Because when he leans forward he brings his centre of gravity in front of his base of support, and he cannot place any part of himself backward to balance it. Hence he cannot rise. 7. In both cases it should be low, to avoid the danger of overturning. 8. Because in one case the centre of gravity hangs on the nail, in the other it is balanced upon it. In one case the equilibrium is stable, in the other unstable. 9. The man, being only one-fourth as far from the centre of the earth, and inside of it, would (Art. 41) weigh only one-fourth as much. KEY TO NATURAL PHILOSOPHY. Pages 35, 36. Exercise 1. s = %gt*. g = 32.2, \ g = 16.1, t = 5, ? 2 = 25. Hence space = 16.1 X 25 = 402.5 feet (Ans). 2. 16.1 X W= 16.1 X 42^ = 680.225 feet (Aw). 3. Here ^ = 9.8 metres. Hence s = 4.9 X 4 = 19.6 metres (Aw). 5. Here s = 300 metres. Hence t 3 = 300 -=- 4.9 = 61.2245, and t = 1/61.2245 7.82 seconds (Aw). 6. s = 16.1 X 9'= 1304.1 feet (Aw). Page 36. Exercise 1. The ball rose 2 seconds and fell 2 seconds. 16.1 X 2 s = 64.4 feet high (Ans). 32.2 X 2 = 64.4 feet per second, velocity given (Ans). 2. 257.6 -=- 32.2 = 8, the number of seconds required for gravity to stop it. 16.1 X 8 s = 1030.4 feet high (Aw). Page 39. Exercise 1. The square of % is %; therefore (converse of law 1), a pendulum to beat half seconds is one-fourth of 39, or 9| inches long. Similarly, to beat one-third seconds, it must be the square of , or of 39, which is 4^ inches long. To vibrate in two seconds it is 2* or 4 times 39 inches = 156 inches long. To vibrate in one minute it is 60' or 3600 times 39 inches = 140,400 inches = 11,700 feet = 2|f miles. 2. 208^ feet = 2502.4 inches. 2502.4 H- 39.1 = 64, which (law 2) is the square of the number of seconds. |/64 = 8 seconds (Ans). KEF TO NATURAL PHILOSOPHY. Page 41. Questions. Kind. Fulcrum. Balance. First. Point of support. Sec-saw. First. Point of support. Scissors. First. Pivot. Ladder. Third. Ground. Forearm. Third. Elbow. Tongs. Third. Joint. Pincers. First. Joint. Wheelbarrow. Second. Axle. Shears. Third. Upper end. Pump-handle. First. Pivot. Claw-hammer. First. Bearing of hammer. Eudder. First. Post. Page 43. Exercises. These may mostly be solved from the simple proportion given in Art. 105, by transposing and substi- tuting the values given for p, w, &c. 1. 2 inches : 10 inches : : 20 pounds : w, whence w 100 pounds (Ans.~). 2. Here p -f- w : p : : be (ab -\- be) : ac, or 440 : 40 : : 22 inches : ac, whence ac = 2 inches (Ans.). 3. 12 inches : 3 inches : : 40 pounds : 10 pounds. There- fore p = 10 pounds (J.ns.). 4. Leverage = 240 -h 16 = 15 (4ns.). 5. 20 : 4 : : 75 pounds : 15 pounds (Ans."). 6. 300 pounds : 100 pounds : : 24 inches : 8 inches. There- fore ac = 8 inches, and 6c = 24 8, or 16 inches (Ans.), 7. 1 inch : 12 inches : : 20 pounds : 240 pounds (Ans.~). 8. 2 inches : 8 inches : : 3 pounds : 12 pounds, p. By Art. 107, 2 : 8 : : 3 feet : 12 feet per second, the rate at which c moves. 10 KEF TO NATURAL PHILOSOPHY. Pages 46, 47. Exercises. The general formula is E X P=r X W. 1. 20 X 200 = 5 X W. W= 800. 2. 20X100 = ;- X 1000. r = 2. 3. 20 X P=i X 500. P=l2l 4. R X 40 = J X 1000. E = 121 5. Here 1000 X 2 = moment of weight = moment of the 2000 four men; hence ^ = 333.33 = force of the four men, o and 333.33 -=- 4 = 83.33, force of one man. 6. Each wheel is 1 foot in radius, and its pinion or axle is 2 inches in radius; therefore the power is increased six- fold by each wheel and axle, or 6 X 6 X 6 = 216 times, by the train of wheels. 216 times 5 pounds = 1080 pounds (Ant). 7. 10X50 = 10XT7. W= 50 kilograms. 50 kilograms : 10 kilograms : : 20 metres : 4 metres, dis- tance W is hoisted. Pages 49, SO. Exercise 1. There are two movable pulleys, each one doubling the power. Hence 20 X 4 80 pounds can be lifted. 2. The weight will move one-fourth as far as the power. 30-4-4 = 7^ feet. 3. The power required will be one-fourth of a kilogram, and it will move through four metres. 4. The power being increased 12 times, and each pulley increasing it two times, there must be 6 pulleys. KEY TO NATURAL PHILOSOPHY. \\ 5. 30 pounds applied at b will be equivalent to 30 X 4 = 120 at c. The first pulley is fixed, and gives no new power; the second is movable, and doubles the power. 120 X 2 = 240. 6. The wheel ac will have to turn as many times as to cause a point of its circumference to go through 6 feet to raise w through 3 feet. Its radius is 2 foot. Hence its circumference is 3.1416 feet. Hence it has to make nearly 2 turns. 7. At each turn the larger sheave gains 1 link on the smaller and raises the movable pulley \ link. By Art. 107, P: TF::|;31 :: 1 : 62. P = 10 pounds, therefore W= 620 pounds (Ans.). 8. P : W : : | : 20 : : 1 : 40. W 3000 pounds, therefore P= 75 pounds (Ans.). Page 51. Exercise 1. If DF = 20 and EF == 4, DE = 1/20' -f 4' = 20.4. W: P::DE : EF : : 20.4 : 4. Or 20.4 : 4 : : 500 : P. Whence P= 98+ Ibs. (Ans.~). 2. Here 4 : 20.4 : : 250 : W. Whence W= 1275 Ibs. (Ans.}. 3. If the boy can push but 100 pounds and must roll the 500-pound barrel, the length of the plane (skids) must be five times the height. 5 times 4 feet = 20 feet (Ans.). 4. As in the last case, the length of the plane must be 990,000 -j- 15,000 == 66 times the height, or the grade 1 in 66, which is 80 feet per mile (Ans.}. 5. In the lever CB the power is increased five times, in the movable pulley twice, and in the inclined plane as 20 : 8, or ^. Hence a power will lift a weight 5 X 2 X f = 25 times itself. A power of 100 being employed, it will lift a weight of 2500. 6. To lift a ton, or 2000 pounds, will be needed ^ of 2000 pounds, or 80 pounds. 7. Through 25 feet. 12 KEY TO NATURAL PHILOSOPHY. Pages 55, 56. Exercise 1. Since the second-hand goes around 60 times to the minute-hand once, if they were the same length the end of the second-hand would go sixty times as fast as the end of the minute-hand. But, as the hand is twice as long, its end has to go twice as far. Hence its end goes only % of 60, or 30 times as fast. 2. The formula for the space in terms of the time is s = fcf. In four seconds this would be 8 = J - x 32 2 X 16 = 257.6 feet. In five seconds, s=J X32 2 X 25 = 402.5. Hence in the fifth second it would fall 402.5 257.6 144.9 feet. 3. The formula for velocity in terms of the time is v = gt. At the end of 15 seconds this would be v = 3"2.2 X 15 = 483. 483 feet per second would be 483 X 3600 = 1,738,800 feet per hour. 1,738,800 -3- 5280 = 329+ miles per hour. 4 = i^. s = iX32x(f> 2 = 100. Note. In some examples g will be taken 32, instead of 32.2, for convenience. 5. The trains going in opposite directions, a point in one will pass a point in the other at the rate of 40 -f 20 = 60 miles per hour, which is one mile, or 1760 yards, per min- ute. Hence to pass over one yard will require Y^SV ^ a minute, and 110 yards -jV^ iV m inute. For the whole express-train to pass would require the front to go the length of the train, 66 yards farther. Hence the whole time required would be -J^ -(- i f | o iV m > nu t e - 6. The steamer could go through still water at the rate of 6 -|- 4 = 10 miles per hour. Hence when it goes down ,lt will move 10 + 4= 14 miles per hour. 7. From our formula for falling bodies, which can be KEY TO NATURAL PHILOSOPHY. 13 applied here by putting x) = x = %%a, instead of \a as it should be. Hence the other arm would be ^a, and the pi-oportion of their lengths is 16 : 17. 15. When placed on the short arm, if w represent the weight of the body, a the length of the balance, and x the short arm, we have A 4 x = w (a x). When placed on the long arm, 9 (a x) = wx. wa From the first, x = ^, w m Substitute this in the second, or, 36 = w 1 . w = 6. 16. The moment of each man's force will be 100 X 5 = 500, and of the four men 2000. The lever arm of the resistance is 1 foot ; hence there is a resistance of 2JtAQ. = 2000 pounds. 17. In three revolutions the wheel will have travelled KEF TO NATURAL PHILOSOPHY. 15 through 60 X 3 = 180 inches = 15 feet. Since the bucket rises one foot, the work done on the bucket is its weight, 30 X 1 = 30 foot-pounds. To do this work by a force acting through 15 feet will require 30 -=- 15 = 2 pounds. 18. The length of the plane will be \/ 5 2 -4- 12* 13. From proportion on page 56, Philosophy, Power : 65 : : 5 : 13. Hence power is 25. 19. The circumference described by the power will be the diameter 2 X 3.1416. Hence, from the law of the screw, Weight : Power : : 2 X 3.1416 : ^ : : 62.832 : 1. 20. Given time = ^f of a second. Required time =. 1 second. The converse of the second law, page 38, gives the proportion. Given length : Eequired length : : (if) 2 : I 1 : : |f $ : 1. Or the pendulum must be lengthened from J-|^ to i-||, which requires an addition of 1 ^j of its original length. 21. At 4000 miles above the earth's surface the distance from the centre would be doubled and the intensity of grav- ity would be ^ of surface intensity (Kule, Art. 41). Hence (third law, page 38) the time would be 2 seconds (Ans.). At 2000 miles under ground the intensity of gravity (Art. 41) is ^ the surface intensity. The square root of is .707 -f- ; hence the time is y^- of a second, or 1.41 seconds (Ans.). 22. The time of one vibration is ^ of a minute = 1 seconds. By law 2, page 38, I 1 : (1) 2 : : 39.1 inches : 87.975 inches (Ans.). 23. The time of one vibration is -^ of a minute = f seconds : therefore (converse of law 3, page 38), (H) 2 : I 2 :: *tt : l - fit of 32.2 = 30.1 (Ans.-). Page 67. Exercise 1. 2 2 : 12 2 : : 5 pounds : 180 pounds (Ans.). 2. I 2 : 12 1 : : 50 pounds : 7200 pounds (Ans.). 16 KEY TO NATURAL PHILOSOPHY. 3. The conditions are the same as in the last example, except that the 50 pounds is applied to the lever with a leverage of 6 (ratio of 3 feet to 6 inches). Therefore the weight lifted will be 6 times 7200 pounds, or 43,200 pounds (Ans.). 4. One peck = of 2150.4, or 537.6 cubic inches. One cubic foot of water weighs 62 pounds : therefore one cubic inch weighs 17 * 2 g, and 537.6 cubic inches weigh ^^-| of 62 pounds = 19 pounds (Am.). 5. Each cubic foot would contain 62|- pounds of water. 6. The bottom supports the weight of the water, and each of the four sides one-half the weight (Art. 140). Pages 78, 79. Exercise 1. 900 -=- 1000 = .900 (Ans). 2. The water weighs 300 200 = 100 grams. The acid weighs 320 200 = 120 grams. Therefore the specific gravity of the acid is 120 -f- 100 = 1.20 (Ans.). 3. The water buoys up the weight of 40 24 = 16 nails. The lye buoys up the weight of 40 18 = 22 nails. There- fore the lye is ^f or If times as heavy as the water. 4. That the water may exert its full effect in supporting them. 5. It requires 11 inches of ether to weigh as much as 8 inches of water : therefore the specific gravity of ether is8-5-ll| = .719 (Am.). 6. For the same reason as above, the specific gravity of the syrup is 8 -=- 6.4 = 1.25 (Ans.). 7. The 7 ounces of iron displace one ounce of water: therefore the specific gravity of the iron is 7 -=- 1 = 7 04ns.). 8. The egg displaces 260 220 = 40 cubic centimetres of water, which weighs 40 grams (Art. 26). But the egg weighs 44 grams. Hence its specific gravity is 44 -j- 40 = 1.1 (4ns.). KEY TO NATURAL PHILOSOPHY. 17 9. The dam is 12 feet in perpendicular height, and 15 feet in slant height : therefore its thickness at the base must be V 15" 12* = 9 metres. According to Art. 140, it there- fore supports half a column of water 12 by 9 by 1000 metres = 54,000 cubic metres, and each cubic metre weighs 1000 kilograms, which gives 54,000,000 kilograms (/Ins.). 10. From Art. 145 we learn that the earth curves 8 inches, or f of a foot, in 1 mile and increases as the square of the number of miles. Hence 200 feet divided by $ of a foot gives the square of the number of miles. 200 -f- = 300. 1/300 = 17.32 (Ans.). Similarly, from the mast the water surface could be seen 12.24 miles, for 100 -f- f = 150. 1/150 = 12.24, which added to 17.32 gives 29.56 miles (Ans.). 11. See Art. 147, III. 12. 2^- 1 = 1, weight of the stone in water compared with an equal bulk of water. Therefore gf of the real weight is what the boy can lift. And if 120 pounds = gf of the weight he can lift, that weight is 200 pounds (Ans.}. 13. 1100 grams 975 grams = 125 grams. 1100 -=- 125 = 8.8 (Ans.~). 14. The wood just buoys up the 2 ounces of lead. There- fore specific gravity = g--g = & (Ans.~). 15. The keys displace 6 X 3 X 1 = 4 } cubic inches of the water, which is the volume of the keys. The volume of the hand is 6 X 3 X = 9 cubic inches. 16. i| in cork = ^ in water. ^ in cork = yf^ in water. -5^5- in cork = ^ in water. If we divide the weight of ^ inch cork by the weight of -^ inch water, we shall have the specific gravity of the cork, or 2* 18 KET TO NATURAL PHILOSOPHY. 17. As the specific gravity of the body is 17 times as much as the specific gravity of water, the volume of a certain quantity of it is '^ of the volume of the same quantity of water. The volume of 89 ounces of water is yflfo. of 1728 cubic inches (Art. 138), and J ? of yflfo of 1728 cubic inches = 9^1^ cubic inches (Ans.~). 18. 16 ounces 6 ounces = 10 ounces, weight of water. 14J ounces 6 ounces = 8 J ounces, weight of coal oil. And 81 -s- 10 = .875 (Ans.). 19. In each case the weight of the displaced liquid is equal to the weight of the hydrometer. Therefore we have 11 cubic inches oil = 9 cubic inches water. 1 cubic inch oil = ^- cubic inches water. 9 cubic inches oil = fi cubic inches water. And H - 9 ==;&=:. Sift- (Ans.> 20. As in Ex. 19, 7800 cubic feet sea-water = 8000 cubic feet fresh water. 1 cubic foot sea-water = f $$ cubic feet fresh water. 8000 cubic feet sea-water = 94-qfflgM. CUD i e f ee t fresh water. And M..O|O.< t 10 = .78 second. ^/ 32. 2 . . the range = 35.9 feet X -78 = 28.002 feet (Ans.}. These two answers would agree exactly if the decimals had been carried out far enough all through the solution (see page 81). 4. Solved in the book. 5. | of 55f } = 34ff (Ans.}. 6. From the upper opening a cylindrical stream 1 inch in diameter and 60 limes 25.3 feet will flow out. Reducing the length to inches, and applying the rule for finding the volume of a cylinder, we have 25.3 X 12 X 60 X t X 3.1416 = 14,306.8+ cubic inches, or 61.9 gallons (Ans.}. And in the case of the lower opening we have 35.9 X 12 X 60 X 4 X 3.1416 = 20,301+ cubic inches, or 87.9 gal- lons (Ans.}. 7. I of 61.9 gallons = 38.7 gallons (Ans.}. * f of 87.9 gallons = 54.9 gallons (Ans.}. 20 KEY TO NATURAL PHILOSOPHY. 8. At the sides of a stream the water is retarded by the friction of the water against the banks. 9. In a flood the ordinary body of water in the stream is the bed of the extra water, which rushes over it with very little retardation from friction. 10. 62 X 1800 = 112,500 pounds of water per minute. 112,500 x 12 = 1,350,000 foot-pounds per minute, the potential energy of the water. 75 % of 1,350,000 = 1,012,500 foot-pounds per minute utilized by the overshot wheel (Art. 161). 1,012,500 -=- 33,000 = 30.7 horse-power (Ans). 11. 90 cubic feet per second = 90 X 60 = 5400 cubic feet per minute. 62| x 5400 = 337,500 pounds of water. 337,500 X 5 = 1,687,500 foot-pounds. 65 % of 1,687,500 = 1,096,875 (Art. 162). 1,096,875 -4- 33,000 == 33.2 horse-power (Ans.~). 12. 8 X 2500 = 20,000 pounds per minute. 20,000 X 18 = 360,000 foot-pounds per minute 80 % of 360,000 = 288,000 foot-pounds available (Art. 164). 288,000 -=- 33,000 = 8.7 horse-power (Ans). Page 92. No ; because the pressure of the 30 inches of mercury in the long arm will raise the mercury above a in the short arm. If the long column is 30 inches higher than the short one, there is a pressure of one atmosphere of mercury upon the air in the short tube, and the air must be compressed to half its former bulk. The short column must therefore have risen 3 inches, and, to be 30 inches higher, the high column must be 33 inches above c. There are, then, 3 inches more of mercury in the short ^arm, and 30 -{- 3, or 33, inches more in the long arm. Therefore 36 inches must be poured in. KEY TO NATURAL PHILOSOPHY. 21 When the mercury in one tube is 60 inches higher than that in the other, there is the pressure of three atmospheres upon the air in the short arm : therefore there are 64 inches of mercury above c, and 64 -(- 4, or 68, inches have been poured in. Pages 114, 115. 1. There is a pressure of four atmospheres : therefore the air is compressed into the space of i of its original bulk of 6 inches, or li inches (J.ns.). 2. 6 1J = 4 J. And 4 -f 4 + 90 = 99 (Ans.). 3. Solved in the book. 4. Since the pressure is 1, or f, as great as before, the air will occupy I as much space, or 4 inches (Ans.). And when one column is 45 inches higher than the other, the pressure is 2, or -|, as great as before, and the air will occupy ^ as much space, or 2^ inches (,4ns.). 5. One inch of mercury will support 13.6 -f- .8, or 17 inches of alcohol. And 30 inches of mercury, or the weight of the atmosphere, will support 17 inches X 30 = 42 feet 6 inches (Ans.^). 6. 13.6-=-1.8 = 7f And 7| inches X 30 = 18 feet lOf inches (Ans.). 7. The pressure must be twice as great as that of the atmosphere, or there must be one atmosphere of water, f.e., 34 feet (J.ns.). It is the upper surface of the water in the tumbler, or the middle of the tumbler, that is 34 feet deep (Art. 134). 8. There is a pressure of three atmospheres, one of air and two of water. Therefore we have 34 feet X 2 = 68 feet (Ans.~). And in the second case, as there are one atmosphere of air and four atmospheres of water, the depth must be 34 feet X 4 = 136 feet (Ans.~). 9. According to Art. 172, the mercury falls 1 inch for every 900 feet of ascent. Therefore the barometer would 22 KEY TO NATURAL PHILOSOPHY. stand 2 inches lower than at the level of the sea, or usually at about 28 inches. 10. As the pressure of the atmosphere is usually equal to that of 30 inches of mercury, when the gauge is 3 inches higher on one side than on the other, the pressure, or ten- sion, of the air in the pump is only ^j-, or y 1 ^, of the normal pressure: therefore there must be only ^ of the air left in the receiver. "When one side is J inch higher than the other, there must be -^ = g 1 ^ of the air left in the receiver. 11. If TsVfr of the air is left in the receiver, it will support 15 1 00 of 30 inches of mercury, or -fa inch (A/is.). 12. The force to be overcome is the pressure of the air upon a circle 3 inches in diameter. The area of the circle is (f ) 2 X 3.1416 = 7.0686 square inches. And 15 pounds X 7.0686 = 106+ pounds (Am.). It may be supposed that we ought to find the pressure on the surface of a sphere 3 inches in diameter. It is true that this is the whole pressure on the surface of the hemi- spheres, but the resistance to pulling one hemisphere away from the other is only the pressure on a section of the sphere. 13. The area of a circle 2 feet in diameter is 12 2 4 X 3.1416 = 452.39 square inches. And 15 pounds X 452.39 = 6785.85 pounds. 14. 3 2 X 3.1416 28.27+ square inches. And 15 pounds X 28.27 = 424+ pounds (Am). 15. Not less than 34 feet (Art. 192). 16. The atmosphere at Denver would sustain a column of mercury 24.7 inches high. Therefore it sustains a column of water 24.7 inches X 13.6 = 28 feet (nearly). 17. The specific gravity of sea-water is 1.026 (Appendix). Therefore 25 feet-- 1.026 = 24.3+ feet (Am). KEY TO NATURAL PHILOSOPHY. 3 Page 129. At 82 sound travels 1090 + 50 = 1140 feet per second (Art. 205) : therefore we have 1140 feet -r- 120 = 9| feet (Ans.). And in the second case we have 1140 feet -*- 300 = 3* feet (Ans.). Page 135. As in the questions on page 129, we have 1120 feet -r-480 = 2 feet 4 inches. And 2 feet 4 inches -=-4 = 7 inches (Am.}. 1120 feet -f- 280 = 4 feet. And 4 feet ~ 4 = 1 foot (Ana.). A column of hydrogen would have to be about four times as long as a column of air to be synchronous with a certain, tuning-fork. Pages 155, 156. 1. Touching the call-bell stops the vibrations, and there- fore stops the sound. 2. No sound could be heard unless the sounding body was connected with the ear by a solid body, or by several solid bodies touching one another. (Liquids are excluded, because it is believed that there are no liquids on the moon.) 3. ^ of 60 seconds = 3| seconds. 1090 feet X 3| == 40871 feet (Ana.). 4. ._*. of 60 seconds = 3 seconds. (1090 feet + 48 feet) X 3 = 3414 feet (Ana.). 5. They will speak the words as long after he does as it will take sound to go from him to them. But it will be just as long again before he hears them, for their sound must travel back to him. The problem, therefore, is to find how long it takes sound to travel 120 feet. In a temperature of 68 sound travels 1090 feet -f 36 feet, or 1126 feet, per second. Therefore we have -tffo or ^ seconds (Ana.). 24 KEY TO NATURAL PHILOSOPHY. 6. 1150 feet 1090 feet 60 feet. And 32 + 60 = 92 (Ans.). 7. 1142 feet 1090 feet = 52 feet. And 32 + 52 = 84 (Ans.). 8. -3-Q O f eo seconds = 2J seconds, the time that has elapsed. 1090 feet + 28 feet = 1118 feet. And 1118 feet X2J = 2515z feet, the distance the sound has travelled. But one-half of this, or 12571 feet, is the distance to the barn. 9. The distance that the sound travels is the bypothenuse of a right-angled triangle whose perpendicular is 300 feet and its base 400 feet. The distance is, therefore, 500 feet. [Draw the figure of the right-angled triangle, and work this out.] The time, therefore, is -fffg or ||| seconds (Ans.). 10. If one string vibrates 100 times per second, one twice as long will vibrate 50 times, and one of the same length and four times as heavy as the last one would vibrate 25 times per second (Ans.). (Arts. 222-3.) 11. If the string is lengthened, it vibrates slower ; if short- ened, it vibrates faster. When the tension is made greater, it vibrates faster; when made less, it vibrates slower. 12. From Art. 226 we learn to proceed as follows : 15 inches X 4 = 5 feet, the length of one vibration. And 1120 feet -i- 5 feet = 224, the number of vibrations and the answer. 13. Mi vibrates f of 264, or 330, times per second (Ans.). (Art. 241.) Sol vibrates f of 264, or 396, times per second (Ans.) Upper do vibrates twice 264, or 528, times per second (Ans). 14. 2 ::: 264 : re's vibrations. Therefore re's vibrations = 148 (Ans.). 2 : J^- : : 264 : si's vibrations. Therefore si's vibrations = 247| (Ans.). 2 : | : : 264 :/a's vibrations. Therefore fa's vibrations = 176 (Ans.). KEY TO NATURAL PHILOSOPHY. 25 15. Multiplying 30 inches by the numbers in the series given in Art. 240, we have the lengths of the strings as follows : do re mi fa sol la si do 30 - 26! 24 22| 20.- 18 16 15 16. 2 : f : : 272 : vibrations of 1st A below. Therefore 1st A below vibrates 226 times. 2d A below vibrates 226! -5-2, or 113J times. 3d A below vibrates 113 J -*- 2, or 56 j times. 4th A below vibrates 56f -J- 2, or 28J times (Am.*). 1 : | : : 272 : vibrations of 1st A above. Therefore 1st A above vibrates 453J times. 2d A above vibrates 453| X 2, or 906| times. 3d A above vibrates 906| X 2, or 1813J times. 4th A above vibrates 1813J X 2, or 36261 times (Ans.~). 17. 1st C above vibrates 272 X 2, or 544 times. 2d C above vibrates 544 X 2, or 1088 times. 3d C above vibrates 1088 X 2, or 2176 times. 4th C above vibrates 2176 X 2, or 4352 times. And 4352 36261 = 725 J (Ans.). 18. | : 1 : : 216 : do's, vibrations. Therefore do's vibra- tions are 192 (Ans.*). 216 -j- 2 = 108, vibrations of re below. | : -| : : 216 : la's vibrations. Therefore la's vibrations are 320 (Ans.~). 19. A little more than eleven octaves. Multiply 16 by 2, then that product by 2, and so on. Eleven multiplica- tions will give us a product a little below 38,000. 20. Middle C vibrates 272 times per second. G above vibrates 408 times. Therefore 408 272 == 136, first answer (Art. 246). And 510 272 = 238, second answer. 21. At the upper, or right-hand, end of the key -board the pitch is higher, the vibrations are more numerous, and the differences, and therefore the number of beats, are greater. In the lower end of the kej--board the beats are fewer. 26 KEY TO NATURAL PHILOSOPHY. Page 167. 1. The wall is 6 feet from the board, and, hence, 3 times as far from the light as the board is. The shadow will be 3 feet square, or contain 9 square feet. 2. An ellipse. 3. The lamp is 4 times as far from the wall as the candle is, and, since light is proportional to the squares of the dis- tances, it must be 16 times as bright. 4. Light travels 186,380 miles per second. In 3J years there are 3J X 365 X 24 X 60 X 60 = 110.366.000 seconds. 110,366,000 X 186,380 = 20,569,014,080,000 miles. 5. No; because they may have gone out of existence, and the light be still coming to us. 6. About 1^ seconds. Less than \ second. Pages 173, 174. 1. When we look into a mirror perpendicularly, the reflections from the two faces will coincide as AB, and we will not see double reflec- tion. If the ray strikes the mirror a little obliquely, as CD, the reflected light from the two surfaces will come out much closer together than if it strikes very obliquely, as EF. 2. Images will come into the eye from various reflec- tions from the mirrors. KEY TO NATURAL PHILOSOPHY. 27 3. The man CD can see his head reflected perpendicu- larly from A, and his feet from B. 4. The tremors of the atmosphere A make the light from the stars to quiver, and so give them apparent size, and the reflections in the air make the whole sky somewhat light. For the same reason light is reflected into shadows. We see only rays which enter the eye, but a beam of light going through a dark room lights up particles of dust, and these reflect the light to the eye. Page 217. 2. 122 F. is 90 above freezing, and 90 F. = 50 C. 3. 10 C. = 18 F., and 18 above freezing is 50 above zero. 4. _4Q F. is 72 below freezing, and 72 F. = 40 C. 5. Since 772 units are required to raise one pound through 1 F., and 1 F. = f C., to raise one pound through 1 C. would require f of 772 = 1389.6. Page 220. Question. 273 C. 491.4 F. below the freezing-point, or 491.4 32 = 459.4 below zero. Page 224. 1. Because the water is dried out. This more than makes up for the expansion due to heat. 28 KEY TO NATURAL PHILOSOPHY. 2. Because the wires are expanded by heat and rendered longer. 3. Because the tire is larger when hot. When it cools it contracts, and binds the wheel together. 4. It will. Because some heat is taken from the coffee to dissolve the sugar. Page 227. 3. Since the volume of a gas increases ^ its volume at zero for each degree, for 546 it will expand fff its volume at zero, or twice. Add this to the original volume, and we have the new volume, three times the old. Page 231. 1. Stoves should not be bright, for that makes them poorer radiators of heat. Teapots should be, for we do not wish them to radiate. The same is true of the cylinders of steam-engi ncs. 2. Dark colors will absorb more luminous heat than light colors will. Hence light colors are cooler in sunlight. But by a dark stove there is no difference. 3. Very little heat will go through the alum, and nearly all through the rock-salt. Hence the focus of the latter will be the hotter. 4. Four times. There is the same law as with light and gravitation and all such forces. 5. The luminous waves; for, being nearer the violet end, it is shown that thej^ are refracted most, and hence must bo more rapid. 6. The heat from an open fire will penetrate a glass screen much more readily than dark heat will : hence it is less effective. Page 234. 1 1. Because the vessel of better conductive power more quickly carries the heat through its mass, and thus there is not unequal expansion and contraction in its different parts. 2. Because these have poor conductive power. KEY TO NATURAL PHILOSOPHY. 29 3. Because it will not so readily carry away the heat of the body. 4. Because there is around his hand a layer of vapor from the perspiration of the skin, which, being a poor conductor, prevents for a little time the heat of the iron from being conveyed to the nerves. 5. The brass rapidly conducts the heat away, and the wood does not. Hence the heat accumulates in the latter case and scorches the paper. 6. They surround the house with a partition of air, which is not a good conductor of heat, and so retain it in the house. 7. Not so effective. Because dark heat will not penetrate glass. 8. They reflect the heat back to the earth and retain it there. 9. They are not always. If our hand is cooled for any reason, any object warmer than itself will feel warm to it. In the case mentioned, the tepid water will feel warm to the hand previously in the cold water, and cold to the other. Pages 241, 242. 1. 113 F. is 81 above freezing. 81 F. =-f of 81 C. = 45 C., or 45 above 0. 140 F. is 108 above freezing. 108 F. =f of 108 C. = 60 C. 2. 15 C.=| of 15 F. = 27 F. above freezing, or 32 -f 27 = 59 F. 35 C. =f of 35 F. = 63 F. above freezing, or 32 -f 63 = 95 F. 3. It would expand tfft of its volume at 0. |ff of 30 = 11. 30 + 11 = 41 (Art. 366). 4. 185 F. = 153 above = 85 C. Hence 98 cubic inches = f^f + -^fa times the volume at 0. If 98 is f ff , the volume at will be f|f of 98, and the volume at 10 30 KEY TO NATURAL PHILOSOPHY. will be |^| -j- JT% = |ff times the volume at 0. Therefore the volume at 10 = fff of ff of 98 = 77+ cubic inches (Am). 5. If 50 is fff the volume at 0, then the volume at is !| x 50, and at 15 the volume would be f|f X 50 X fff = 53.8 nearly. 6. Since it expands ^rs f r eacn degree, the expansion of 20 cubic inches, or of its volume, at must represent i.-j-^i^ or -2-p 546 of temperature. 7. The pound of water if lowered to C. will lose 50 units of heat, and this is given to the ice. Now, it requires 80 units of heat to melt 1 pound of ice (Par. 368). The 50 units will melt f of a pound, or 10 ounces. 8. The 3 pounds at 12 will have 36 units of heat, and the 3 pounds at 16 will have 48 units; in all, 84 units. If 6 pounds have 84 units, each pound must be of the tem- perature of 14. 9. 4 X 7 = 28, and 6 X 12 = 72. 72 + 28= 100 units of heat in the mixture of 10 pounds. 100 -r- 10 = 10, the temperature. 10. The water evaporates rapidly through its pores. This evaporation takes heat from the water, and so keeps it cool. 11. The evaporation, or conversion into a gas, is so rapid that heat is taken from the adjacent particles, and they freeze. 12. If the specific heat of iron is .1, it takes -^ unit to raise 1 pound through 1. The iron therefore loses 10 to give the water 1. They thus become 11 nearer the same temperature each time the water is raised 1. To agree, the water will have to be raised through 99 -f- 11 = 9 degrees. 13. The air on a high mountain is rare and dry. The sun-rays pass through it without heating, and the radiation from the earth is not retained by the cloak of moisture which exists in the valleys. KEY TO NATURAL PHILOSOPHY. 31 14. Silver-foil being a poor absorber, the mercury will not rise so fast (Art. 389). 15. One way would be to coat them all with wax and place one end in a vessel of hot water. The distance to which the wax would melt would determine the relative conductive power. 16. Platinum is a poor conductor of heat, and copper is a good one. 17. 386 X 2000 = 772,000 foot-pounds of work done. A unit of heat is 772 foot-pounds : hence 1000 units are re- quired. 18. 1544 X 68 = 104,992 foot-pounds^of work. This would raise 1 pound of water through 104,992 -=- 772 = 136, or 68 pounds through 2. 19. It requires ^ as much heat to raise a pound of iron as a pound of water. To raise 1 pound of water 1 re- quires 772 foot-pounds. To raise it 100 requires 77,200 foot-pounds. To raise 1 pound of iron through 100 requires 7720 foot-pounds. This would lift 7720 pounds, or about 3 z tons, through one foot. Page 279. 1. The fur will become positively electrified by friction against the boy's clothing, and the positive electricity will be conducted through the hand to the boy who holds it, thus giving him a positive charge. The other boy will at the same time be negatively electrified through his clothing. If the two boys bring their knuckles near to each other, a spark will pass. 2. She will be positively electrified, just as the prime con- ductor is positively electrified by presenting a row of points to the plate of the machine. 3. It will be positively charged. 4. The end nearest the prime conductor will be , that nearest the negative conductor will be -f-. 5. In the first case, they should dance vigorously be- 32 KEY TO NATURAL PHILOSOPHY. tween the rod and the table. In the second case, they should be attracted to the rod, then repelled to the pane of glass, and there remain. 6. By Art. 450, 1 X -1 X 2 2 = 4 dynes (Ans.). 7. As above, 4 X 2 X I 2 = 8 dynes, attraction (Am.). When they touch, the two units neutralize two of the -f- units and there are left two -(-units, which distribute themselves one on each ball. Then 1 X 1 X I 2 = 1 dyne, repulsion (Ans.). 8. 16 (dynes) = 32(+unit8) 4 r ( ~ U '"'" ) . Clearing of fractions and transposing, * 32 a; = 256. x = 8 units (Ans). 9. By Art. 61, 1 gram = 980 dynes. 980 X 5 =4900 dynes, repulsion. 490 (-f units) X x (+ units) = 4900. 490 x = 4900. x = 10 -{- units (Ans.~). Page 334. 1. 12 X 24 = 288 ohms, resistance of line wire. 1.8 X 12 = 21.6 volts, E. M. F. of 12 cells. By Ohm's law the current is 21.6 -4- 288 = .075 amperes, or 75 milliamperes (Ans.). 2. Let x = the number of cells required. 288 ohms, as above, = resistance of line wire. 104.25 X 4 = 417 ohms, resistance of relays. .5 x = resistance of battery. 288 -f 417 -\- .5 x = total resistance of the line. 1.8 x = E. M. F. of whole battery. Ohm's law, cleared of fractions and transposed, becomes E = CE. Or, 1.8 x = .075 (288 + 417 + .5 *). 1. 8 x = 52.875-1-. 0375 x. 1.7625 a; = 52.875. x = 30 ( Ans). KEY TO NATURAL PHILOSOPHY. 33 3. By Art. 501, a copper wire ^ of an inch in diameter has a resistance of 1 ohm in 962 feet. The half-inch wire has -fa the resistance per foot of the one-tenth-inch wire (Art. 496). Therefore 962X25 = 24,050 feet of half inch wire for each ohm of resistance. 5280 X 400 = 2,112,000 feet in 400 miles. 2,112,000 -4- 24,050 = 87.8 ohms (Ans.~). 3000 -f- 87.8 = 34.17 amperes (Ans.). (By Art. 495) 34.17 X 3000 = 102,510 watts = 102.51 kilowatts (Ans.). 4. Current in the shunt is always f$ = 5 amperes (.4ns.). One lamp takes ^^ = ^ ampere (Ans.). 200 lamps take % X 200 = 100 amperes (Ans.). 5. I X 600 = 300 amperes for lamps. 300 -f 12.05 + 5 (for field magnet) = 317.05 amperes, whole current required. 80 X 317.05 = 25,364 watts. (Art. 495) 25,364 -=- 746 = 34 horse-power (Ans). amflORMALSCHOOL, 319* UC SOUTHERN REGIONAL LIBRARY F 000 947 352 1 UNIVERSITY of CALIFORNIA AT L06 ANGELES LIBRARY