GIFT OF WER DlVISldff TEACHERS' MANUAL TO ACCOMPANY A TEXT-BOOK IN PHYSICS WILLIAM X. MUMPER, Ph.D. ROFE: MODEL SCHOOL- :SEY \niXXATI :- If 1C A < JO AMERICAN BOOK COMPANY UNIVERSITY OF CALIFORNIA LIBRARY OF THE Received.. Accessions No. Book No. /. r*/6 LOWER DIVISION TEACHERS' MANUAL TO ACCOMPANY A TEXT=BOOK IN PHYSICS BY WILLIAM N. MUMPER, Ph.D. PROFESSOR OF PHYSICS IN THE STATE NORMAL AND MODEL SCHOOLS OF TRENTON, NEW JERSEY NEW YORK - : CINCINNATI : CHICAGO AMERICAN BOOK COMPANY M 7 NOTE Slight disagreements between the numerical answers to some of the problems as here given and those obtained by others will probably in all cases be due to the fact that many of the physical quantities used are, of necessity, approximate values, and to the additional fact that some of the solutions were obtained by means of logarithms. COPYRIGHT, 1909, BY WILLIAM N. MUMPER. w. P. i ; f\ TEACHERS' MANUAL FOR MUMPER'S PHYSICS INTRODUCTORY Page 19 1. The capacity may be found by counting the number of gallons required to fill it. This is a direct measurement. It may also be found by making certain linear measurements and from these computing the volume. This is an indirect measurement. 2. 1 m.= 1000 mm., 3 m. = 3000 mm. 1 mm. = .001 m., 35 mm. = .035 m. 1 m.= 100 cm., 48 m. = 4800 cm. 1 in. = 2.54 cm., 10 in. = 25.4 cm. 1 cm. = .3937 in. 25 cm. = 25X.3937 in. = 9.8+ in. 3. 1 k. = 2.21b., 12 k.= 12X2.21b. = 26.41b. 1 oz. = 28.3+ gm. 4. 1 ft. = 30.48 cm., J^l = 32.1 + ft. per sec. 5. It is not correct. The second statement means that when the age of the tree becomes any multiple of its present age the height will become the same multiple of its present height. MATTER Page 25 1. Its number of units of volume and its number of units of mass. 2. The mass may be found by weighing. See page 24 (1). The vol- ume of a block of wood may be found by measuring the volume of the liq- uid it displaces when immersed. The surface of the wood should be made impervious by a coat of paraffin or varnish. The volume of a liquid may be found by measuring it in a graduated vessel. Its mass may be found by weighing. 3. The volume of the liquid is measured directly and the other quan- tities are measured indirectly. 4. See Table, page 25. 6. 1 c. c. of water weighs 1 gm., 1 liter=1000 c. c., hence 1 1. of water weighs 1000 gm., 45 c. c. of water weigh 45 gm. 6. 1 c. c. is merely a volume and a volume has no weight. Objects which have a volume of 1 c. c. may have different weights depending upon their densities. M^.MUM.PH.S.-l 32^825 ' MANUAL FOE MUMPER'S PHYSICS 7. If 32 grii. niass fill 66 : c: c., the density is 32-^-60= .53+ gm. per c. c. 8. See page 16, par. 9. The kilogram mass is the same everywhere though the earth's attraction for it, that is, its weight may change with a change of location. 9. The assumption is that at a given place equal masses have equal weights, that is, they are attracted equally by the earth. 10. The volume is 5X6X7X1 c. c. = 210 c. c. The density is 1638 -=-210= 7.8 gm. per c. c. 11. 12 c. c. holds 12X 13.6 gm.= 163.2 gm. of mercury. 1 liter of mercury contains 1000X13.6 gm.= 13600 gm. 1 k.= 1000 gm., 1000-1-13.6=73.5 c. c. 12. Volume of tank= 2X3X4X1 cu. ft. = 24 cu. ft. 1 cu. ft. of water weighs 62.4 Ib. 24 cu. ft. of water weigh 24X62.4 lb. = 1497.6. 1 cu. ft. of mercury weighs 848 Ib. 24 cu. ft. weigh 24X848 Ib. = 20352 Ib. MATTER Page 38 1. By showing that air occupies space, has weight, elasticity and other characteristics common to all matter. 2. The gas molecules occupy a part of the space between the molecules of the liquid. 3. Let the pupil take different substances and demonstrate by his experiments and his descriptions of them that he knows the proper terms to use in describing his observations. 4. The surface tension of the liquid wax makes it approach as nearly as possible to the spherical shape. See page 35, line 6. 6. The spaces between the fibers of the towel act like capillary tubes, or like the glass plates. See Fig. 22. 6. Essentially the same as 5. THE MECHANICS OF LIQUIDS Page 43 1. The intensity is 64-f-16=4 Ib. per sq. in. External pressure, be- cause produced by a body acting on the fluid. 2. The intensity of external pressure is the same throughout the fluid; it is 18 gm. per sq. cm. Area is 8X12X1 sq. cm. = 96 sq. cm. The total pressure is 96X18 gm.= 1728 gm. 3. The intensities are equal. The total pressures and the areas are directly proportional; hence the areas are to each other as 3:16. 4. 7200 fb. -^30= 240 Ib. on 1 sq. in.; being external pressure it is also 240 Ib. per sq. in. on the small piston. THE MECHANICS OF LIQUIDS 3 5. 40 ft>. on 2.5 in. The intensity is 40-1-2.5= 16. Ib. per sq. in. (Pas- cal's law). 6. A total pressure of 2000 Ib. is to be produced on the piston. The intensity is 40 Ib. on 1 sq. in. 2000 H- 40= 50 sq. in. to produce 2000 Ib. To lift the elevator the pressure must exceed the computed amount. Just how much greater depends upon friction, speed desired, etc. THE MECHANICS OF LIQUIDS Pages 53, 54 1. Laws 1 and 2, page 48. Page 48, last two lines. 20 ft. deep the pressure intensity is 20X62.4 ft>.= 1248 Ib. per sq. ft. When water is 1 cm. deep the pressure is 1 gm. per sq. cm., hence for 18 cm. deep it is 18X1 gm. = 18 gm. per sq. cm. 2. See page 49, line 10 from bottom. Since the intensity is 13.6 gm. per sq. cm. when the mercury is 1 cm. deep, when it is 7 cm. deep the intensity is 7X13.6 gm. = 95.2 gm. per sq. cm. When alcohol is 1 cm. deep the intensity pressure is .79 gm. per sq. cm. When 4 cm. deep it presses 4X.79 gm. = 3.16 gm. per sq. cm. 3. We must use an average depth when the water does not have the same depth at all points of the surface in question. The average depth is here \ (0, least depth + 40 cm., greatest depth) = 20 cm. Hence the average pressure intensity is 20 gm. per sq. cm. When containing mer- cury, average depth is 20 cm., pressure intensity =20 XI 3. 6 gm. = 272 gm. per sq. cm. 4. The area of the bottom is 224 sq. cm. (a) Pressure intensity is 12 X 1 gm.= 12 gm. per sq. cm. Total pressure is 224X12 gm.= 2688 gm. (6) Pressure intensity is 12 X 13.6 gm.= 163.2 gm. per sq. cm. Total pressure is 224X163.2 gm. = 36556.8 gm. 5. Area of 1 side, 12X16X1 sq. cm. = 192 sq. cm. Average depth is (0+12) = 6 cm. (a) Average pressure intensity is 6 gm. per sq. cm. Total pressure is 192X6 gm.= 1152 gm. (6) Average depth is 6 cm. and aver- age pressure intensity is 6X 13.6 gm. = 81.6 gm. per sq. cm. Total pressure is 192X81.6= 15667.2 gm. 6. Area of the surface which is exposed to the pressure of the water on only one side is 10X20X1 sq. ft. = 200 sq. ft. 4 MANUAL FOR MUMPER'S PHYSICS The average depth on this area is 5 ft. The intensity of pressure is 5X62.4=312 Ib. per sq. ft. The total excess of pressure is 200X312 = 62400 Ib. 7. (a) When pistons are on same level, only external pressure is concerned and the pressure intensities are the same at both pistons. (6) The weight pressure of the liquid is added to the external pressure at the large piston, hence the total pressure intensity is greater than that at the smaller piston, and the denser the liquid used the greater the difference between the intensities at the two pistons. 8. Since the depth at all points of the bottom is 16 cm. the pressure intensity is 16 gm. per sq. cm. The area of the bottom is 80 sq. cm., hence the total pressure on the bottom is 80X16 gm.= 1280 gm. The area of side abed is 160 sq. cm., the average depth is \ (OX 16) = 8 cm., and the average pressure intensity is 8 gm. per sq. cm. Total pressure on this side is 160X8 gm.= 1280 gm. Since the free surface is on the level of the top of the vessel there is no weight pressure there. 9. (a) When the tube also is full of water the depth at the level adfe is 20 cm. and the depth at all points of the bottom is 36 cm., hence the average depth on all points of the side abed is (20 +36) = 28 cm. (6) The average pressure intensity on side abed is 28 gm. per sq. cm. The area of abed is 160 sq. cm., hence the total pressure on abed is 160 X 28gm. = 4480gm. 10. A change in diameter of the tube has no effect when depth and density of liquid remain constant. A change of liquid would change the results because the density is different in this case less than that of water. 11. (a) The pressure intensity is 12 gm. per sq. cm. on the bottom of each. (6) Since the area of the bottom is in each case 100 sq. cm. the total pressure in each case is 100X12 gm.= 1200 gm. (c) Since sides of A are vertical, total pressure on the bottom of A is equal to the weight. (d) In vessels C and F. (e) In vessels B, D, and probably E. 12. When the water is at rest, the vertical depth of the water below the free surface of the water in the reservoir. The same for a pipe on second floor. The pressure will be less than the values thus computed in case the water is running, the decrease in pressure being caused by friction. 13. The head at any time is the vertical depth of water required to produce the pressure of the water at any point at the given time. 14. If the pressure at any point is 5 times as great as that at another THE MECHANICS OF GASES AND LIQUIDS 5 point in the same body of water the depth is 5 times as great, hence the depth of the reservoir is 5X4.5 ft.= 22.5 ft. THE MECHANICS OF GASES AND LIQUIDS Pages 58, 59 1. Air is so highly compressible that a given motion of the first piston may not produce any motion of the other. There is lost motion. 2. If the cap is on, air cannot enter there and the atmospheric pressure at the outlet keeps the oil in. An intermittent flow is due to the occa- sional entrance of air bubbles at the inlet. 3. The air pressure at A plus the weight pressure of the liquid above the level of equals in intensity the atmospheric pressure at 0. As the ink is used the level at will be lowered till an air bubble enters the bulb through the horizontal tube and an equal volume of ink flows to the outlet side. 4. The air pressure at b (Fig. 56) and at A (Fig. 57) are each less than the atmospheric pressure. In Fig. 58 the air pressure is greater than the atmospheric pressure. 5. Since the water is 10 cm. deeper in column b than it is in column a and the atmospheric pressure at a balances the air pressure in b plus the pressure of this depth of water, the excess is 10 gm. per sq. cm. When mercury is used the excess is 10 X 13.6 gm.= 136 gm. per sq. cm. 6. In charging the filler the atmospheric pressure drives in the ink as the rubber top expands. The action of the lungs decreases the pressure within the straw and the atmospheric pressure sends the liquid up the straw. 7. In breathing, the chest cavity is enlarged and the atmospheric pressure drives the air into the lungs. The air is not pulled or drawn in as is often assumed. 8. The water stops rising when the pressure at B plus the pressure of the water above the level of A equals the atmospheric pressure. The atmospheric pressure is 6 gm. per sq. cm. greater than the air pressure at B. 9. As the tumbler goes deeper the water pressure becomes greater, thus compressing the air. 10. When the finger is removed the atmospheric pressures at A and are practically equal and the weight pressure of the water makes it flow out. When the top is closed there is no atmospheric pressure at A, and the atmospheric pressure at is greater than the weight pressure of the water, hence the flow stops. THE MECHANICS OF GASES AND LIQUIDS Page 64 1. When mercury is 1 cm. deep the pressure intensity is 13.6 gm. per sq. cm. WTien 20 cm. deep it is 20X13.6 gm.= 272 gm. per sq. cm. When 76 cm. deep it is 76X 13.6 gm.= 1033.6 gm. per sq. cm. 6 MANUAL FOR MUMPER'S PHYSIOS 2. This means that the column of mercury (ab, Fig. 61) is 76 cm. deep, hence the pressure intensity is 1033.6 gm. per sq. cm. (last problem). To compute a total pressure the area must be stated. 3. In the barometer the atmospheric pressure counterbalances the weight pressure of the mercury. It has been shown that intensity of weight pressure does not depend upon either the quantity of the liquid or the shape of the vessel. Capillary action in a glass tube containing mercury gives depression, hence the actual reading is less than the true reading. It is the reverse in a water and glass barometer. 4. Since mercury is 13.6 times as dense as water a water column must be 13.6 times as deep as the mercury column to produce the same pres- sure 13.6X30 in. = 408 in. = 34 ft. 13.6X25 in. = 340 in. = 28^ ft. 5. Mercury 1 ft. deep exerts a pressure of 13.6X62.4 lb. = 848.6 Ib. per sq. ft. When 1 inch deep the pressure would be ^ of 848.6 lb.= 70.7+lb. per sq. ft. When 30 in. deep this pressure is 30X70.7 lb. = 2121 ft>. per sq. ft. or 2121 -=-144= 14.7 Ib. per sq. in. 6. Pressure intensity of mercury is 13.6 gm. per sq. cm. for each cm. of depth. To produce 980 gm. per sq. cm. the depth must be 980-M3.6= 72.0 cm. 7. The same as the height of the standard barometer, 76 cm. To produce the same pressure water must be 13.6X76. cm. = 1033. 6 cm. 8. Air would, if uniformly dense, exert a pressure of .001293 gm. for each cm. of depth, hence to produce 1033. gm. per sq. cm. it would have to be 1033 -f-. 00 1293 = 7989 18 cm. = 7989 meters=5 miles nearly. 9. See use of the barometer, page 62. THE MECHANICS OF GASES AND LIQUIDS Pages 66, 67 1. The weight pressure of the water at 480 cm. depth is 480 gm. per sq. cm. The atmospheric pressure when the barometer is 70 cm. is 70 X 13.6 gm. = 952 gm. per sq. cm. The pressure due to both is 480 gm.+ 952 gm.= 1432 gm. per sq. cm. 2. At 10 m. or 1000 cm. depth the weight pressure of the water is 1000 gm. per sq. cm. and the standard atmospheric pressure transmitted to it is 1033. gm. per sq. cm., hence the 12 c. c. of air is under a pressure of 1000 + 1033=2033 gm. per sq. cm. At the surface the bubble is under the atmospheric pressure only. Let z=the volume at the surface. Then, according to Boyle's law, 1033:2033= 12: re. z=23.6c.c. 3. Since the pressure of the remaining air is equal to that of 9 cm. of mercury and the atmospheric pressure is 72 cm. of mercury, the pressure of the remaining air is 9X13.6 gm.= 122.4 gm. per sq. cm.; 7 9 ^ of the original mass of air remains and ff= of the air was removed. 4. A difference of 1 cm. in level of mercury surfaces gives a pressure THE MECHANICS OF GASES AND LIQUIDS 7 of 13.6 gm. per sq. cm. To produce 200 gm. per sq. cm. the difference in level must be 200-M3.6=14.7 cm. 6. The animal enlarges the chest cavity and the external pressure fills the lungs. In exhalation the animal reduces the volume of the chest cavity and lungs, driving out the air. 6. As the air leaves the bell jar (B, Fig. 72) the pressure of the remain- ing air becomes less than that of the air in A, which as it expands drives the water over into vessel 0. 7. If the mercury has fallen \ of the distance which represents the whole pressure, according to Boyle's law, the remaining air which exerts f of the pressure must be of the original mass. THE MECHANICS OF GASES AND LIQUIDS Pages 71, 72 1. An immersed body is pressed upon by the fluid in all directions, perpendicular to the different surfaces of the body. The pressure in- tensity is the same at all points on the same level, but it is greater at that one of two points where the depth is greater, consequently the pressure intensity is greatest at the lowest points of the immersed body. 2. Since the greatest pressure is on the lowest points and the pressure there is upward, it follows that the upward pressure of a fluid on an im- mersed body always exceeds the downward pressure on the same body, hence a part or all of the weight of the body is supported by the fluid. This excess of upward pressure is called buoyancy. 3. 1 cu. ft. of water weighs 62.4 -|- Ib. Its weight is supported by the cu. ft. of water or the surface of the tank beneath it. The upward pressure on this cu. ft. must exceed the downward just enough to sustain the weight of 1 cu. ft. of water, 62.4 Ib. 4. A cubic foot of any other material immersed in the water would experience the same pressure as does the cu. ft. of water. The upward exceeds the downward pressure by just 62.4 Ib., the weight of the water displaced. This is less than the weight of 1 cu. ft. of iron or of marble, hence they would sink; but it is greater than the weight of 1 cu. ft. of wood and 1 cu. ft. of oil, hence they would rise in the water. 5. (a) 3X4X5X1 c. c. = 60 c. c., volume of the block. (6) When completely immersed it displaces 60 c. c. or (since 1 c. c. of water weighs 1 gm.) it displaces 60 gm. of water. (c) The buoyancy is equal to the weight of water displaced, 60 gm. 6. (a) 450 gm. 60 gm. = 390 gm. unsupported weight. The block sinks. (6) When released in mercury it displaces 60 c. c. of mercury which weigh 60X13.6 gm. = 8l6 gm. The buoyancy is 816 gm. and the ob- ject weighs 450 gm., hence the buoyancy exceeds the weight (816 450) and the block rises. 8 MANUAL FOR MUMPER'S PHYSICS 7. When the body has the same density as that of the fluid, it dis- places its own weight. When it is denser it displaces less than and when less dense it displaces more than its own weight of the fluid in which it is immersed. 8. Since a floating body always displaces its own weight of the fluid and this body displaces 320 c. c. of water, it must displace 320 gm. of water, and the body weighs 320 gm. 9. Air has weight and the presence of air in the water-tight com- partments would add to the total weight of the boat that much. Hence the boat could carry a slightly greater weight if these air-tight com- partments were empty. 10. (a) A balloon rises when its entire weight is less than that of the air it displaces. (6) It falls when the balloon's weight exceeds that of the displaced air. (c) It will remain at a given elevation when the two weights are equal. 11. Throwing out sand decreases the weight of the balloon without materially decreasing the weight of displaced air. Letting out some gas decreases the volume of the balloon and consequently decreases the buoyancy without materially decreasing the weight of the balloon. Be- cause the sand is several thousand times as dense as the gas, a loss of sand changes the weight of the balloon without affecting buoyancy appreciably, but a loss of gas decreases the weight very little and the buoyancy very much. 12. Net displacement here means that the weight of the vessel when equipped for sea is 20000 tons, not including the cargo. The displace- ment is now equal to the weight of the ship plus the weight of the cargo. 13. A body rises when it displaces more than its own weight of the air. Particles of dust and smoke are solid particles and they rarely, if ever, displace as much as their own weight of air. They generally fall very slowly through the air in which they are. If the air is rising it carries these particles with it. 14. A body is attracted by the earth as much in both cases, but when it is surrounded by air (or by any other fluid) a part of the weight is sus- tained by this air and the apparent weight is less. 15. When placed in a vacuum they will not balance, because the cork on account of its larger volume experienced a greater loss of weight when they were in the air and their apparent weights balanced. In the vacuum the cork weighs more than does the iron. 16. The true weight of the cotton is greater, and it has a greater mass than that of the iron. 17. Weight pressure depends upon depth and density of the fluid only. Hence if the boat is of nearly the same size and shape as the dry dock a very small quantity of water will produce the required depth. 18. The pressure compresses the air in both a' and a. When the air THE MECHANICS OF GASES AND LIQUIDS 9 is compressed at a the object displaces less than its own weight of water, hence it sinks. When the pressure of the finger is released the air at a expands and the weight of water displaced exceeds the weight of the diver, hence it rises. THE MECHANICS OF GASES AND LIQUIDS Pages 76, 77 1. It means that the weight of any piece of copper is 8.8 times as much as the weight of an equal volume of water. 1 c. c. of copper weighs 8.8 gm. In water it weighs 1 gm. less, the weight of the 1 c. c. of water it displaces. 2. 10 c. c. of water weigh 10 gm. 14.8-:- 10= 1.48, the specific gravity of chloroform. 3. 1 cu. ft. of water weighs 62.4 lb.= 58.8 -^62.4 =.94, the specific gravity of linseed oil. 1 c. c. of water weighs 1 gm. and because the linseed oil is .94 times as dense as the water, 1 c. c. of it will weigh .94 gm. 4. 904 gm. 791 gm.= 113 gm., the weight of an equal volume of water. 904-^-113= 8. sp. gr. of the brass. (See Archimedes' principle.) 6. (a) The coal weighs 1.5 times as much as an equal volume of water, hence 1 cu. ft. of coal will weigh in air 1.5X62.4 lb. = 93.6 Ib. (6) In water it will weigh 93.6 lb.-62.4 lb. = 31.2 Ib. 6. Let x = loss of weight in water. Then 1.5 =100 -hz. z=100-M.5= 66.6+ Ib. 10066.6= 33.3 Ib., the weight of the coal in water. 7. When floating a piece of ice displaces its own weight of water. In fresh water .9 of the volume of the ice will be immersed, for a volume of water .9 as large as the ice will weigh just as much as the ice. In salt water a smaller portion will be immersed. .9 -h 1.025= .88. Hence (a) .1 of the volume is above fresh water and (6) .12+ of the volume is above the surface of salt water. 8. 1 c. c. of cast iron (sp. gr. 7.4) weighs 7.4 gm. and 10 c. c. weigh 74. gm. In gasoline it will displace 10 c. c., or 10X-66 gm. = 6.6 gm. It will weigh in gasoline 746.6=67.4 gm. 9. Since 1 c. c. of mercury weighs 13.6 gm., the volume of 300 gm. will be 300-M3.6=22+c. c. 10. The body displaces 8 c. c. or 8 gm. of water. Its weight in air is 54.4 gm.+8 gm. = 62.4 gm. Its sp. gr. is 62.4^8=7.8. 11. 1 liter=1000 c. c. 1000 c. c. of cork weigh 1000X.24 gm.= 240 gm.; and it will displace 240 gm. of water when floating. 240 c. c. of water will be displaced. 240-MOOO= .24, or 24%. 12. 170 gm. 150 gm. = 20 gm., weight of water displaced. 170 gm. 152.6= 17.4, weight of turpentine displaced. Volume of nickel is 20 c. c. (a) Sp. gr. of nickel is 170^20=8.5. 10 MANUAL FOR MUMPER'S PHYSICS (6) The nickel displaces 20 c. c. of turpentine which weigh 17.4 gm. but an equal volume of water weighs 20 gm. Sp. gr. of turpentine is 17.4-^-20 =.87. 13. 100 c. c. of cast iron weigh 100X7.4 gm. = 740 gm. Since the iron floats it displaces its own weight, 740 gm. of mercury. 14. The block of iron now displaces both water and mercury, the joint weights of which are equal to the weight of the iron, 740 gm. Hence the block of iron rises a little when the water is poured in. The volume of water and mercury displaced are together equal to that of the iron, 100 c. c. Let x=vol. of mercury; then 100 x=vol. of water displaced. 1 c. c. of mercury weighs 13.6 gm: 13.6 x=no. of gm. of mercury displaced. 100 x = no. of gm. of water displaced. 13.6 x + 100 x = total weight of iron, 740 gm. 12.6 x= 740 100. re =50. 7 c. c., the volume of mercury displaced. 100 z= 49. 3 c. c., the volume of water displaced. Floating in mercury alone the block displaces 740-7-13.6=54.4 c. c. 54.4 c. c. 50.7 c. c. = 3.7 c. c., less mercury displaced after water is in. 15. 40 gm. 20 gm. = 20 gm., weight of alcohol displaced. Since 1 c. c. of alcohol weighs .79 gm., 20 gm. weight have a vol. of 20-r-. 79= 25.3 c. c., hence the same vol. of water as that of sugar weighs 25.3 gm. Sp. gr. of sugar= 40 -=-25.3= 1.58. 64 16. 64_i-85_33 == -55 sp. gr. of the wood. 17. 156 gm. 73 gm. =83 gm., weight of water required to fill bottle. 148 gm. 73 gm. = 75 gm., weight of oil required to fill bottle. Sp. gr. of oil=75-r-S3=.9+. 18. (a) The boat displaces its own weight or the weight of 580 cu. ft. of water, which is 580X62.4 lb.= 36192 Ib. It displaces the same weight of sea water the volume of which is 580-:- 1.025 cu. ft. = 565.8 cu. ft. It will be nearer the deck when in fresh water, for it requires a larger volume of the fresh water to equal the weight of the boat. 19. The vessel will hold an equal volume of mercury, which weighs 13.6X68 gm. = 924.8 gm. The volume of mercury equals the volume of water, which is 68 c. c. 20. Since 1 c. c. of mercury weighs 13.6 gm., the volume of mercury is here 480-f-13.6= 35.3 c. c. The same volume of alcohol weighs 35.3 X .79 gm. = 27.8 gm. 21. When the pressures of two liquid columns counterbalance, the relative densities or specific gravities of the two liquids are inversely proportional to the depths of the counterbalancing columns. THE MECHANICS OF GASES AND LIQUIDS 11 Sp. gr. of water: sp. gr. of benzine = depth of benzine: depth of water, 1:. 9=27: x. x=. 9X27 = 24.3 cm., the depth of the water. 22. 20% water means 1 part water to 4 parts alcohol. The weight of 5 parts, say 5 c. c., is 1 gm. + (4X-79 gm.) = 4.l6 gm. An equal vol. of water weighs 5 gm. The sp. gr. is 4.16-^-5= .83. Let z = percentage of alcohol, then 100 x = percentage of water. Each c. c. will weigh (zX-79 gm.) + (100 z) = .88 gm. .79x-x= -.12. -.21x=-. 12. 21z=12. z=.571. THE MECHANICS OF GASES AND LIQUIDS Pages 86, 87 1. The intensities are equal. The air originally in the receiver divides itself between the receiver and the cylinder in the ratio of their relative capacities, 3:1. | comes out and f remains. On the second double stroke of the first remainder comes out and f remains. That is, of f=y 3 g- comes out, and f of f or T \ of the original mass remains. 2. When the man who moves the wheelbarrow precedes it he is said to pull, when he follows he is said to push. Air and other fluids have so little cohesion that they cannot be pulled or drawn in a column, the object responsible for their motion must follow, hence push them. 3. The air pressure must equal the water pressure at the given depth. At 1 m. (100 cm.) the water pressure is 100 gm. per sq. cm., at 15 m. deep the pressure is 1500 gm. per sq. cm. The atmospheric pressure is 75 X 13.6 gm.= 1020 gm. per sq. cm. Since the air in the bell counterbalances both, its pressure is 1500+1020=2520 gm. per sq. cm. For 1 ft. depth the pressure is 62.4 Ib. per sq. ft. For 50 ft. depth the pressure is 50X62.4 lb. = 3120 Ib. per sq. ft. water pressure. Mercury 1 ft. deep exerts a pressure of 13.6X62.4 = 848 Ib. per sq. ft. 30 in. = ff=f ft. -|X848 lb. = 2120 Ib. per sq. ft. atmospheric pressure. 3120+2120=5240 Ib. per sq. ft. due to both. 4. Since the water rises 1 cm. for each gram per sq. cm., it will rise 1000 cm. or 10 m. when the pressure is 1000 gm. per sq. cm. 5. When the barometer is 80 cm. the atmospheric pressure is 80 X 13.6 gm.= 1088 gm. per sq. cm. The water will rise 1088 cm.= 10.88 m. Kerosene, which is .8 times as dense as water, will rise 1088 -f- .8= 1360 cm. = 13.6m. 6. The pressure is 70X13.6 gm. = 952 gm. per sq. cm., hence water 12 MANUAL FOR MUMPER'S PHYSICS would rise 952 cm. And a liquid 1.5 times as dense would rise 9524- 1.5= 634.6 cm. 7. Since intensity of weight pressure depends only upon the depth and density of the liquid, it will work just as well as though AT had a diameter the same as OT. 8. A pressure of 13 Ib. per sq. in. is equal to a pressure of 1872 Ib. per sq. ft. A column of water 1 ft. high has a weight pressure of 62.4 Ib. per sq. ft. Since 1872-^62.4=30, an atmospheric pressure of 13 Ib. per sq. in. will support a column of water 30 ft. high. 9. When o is closed with the finger and air is removed from b by means of a pump or the mouth, the atmospheric pressure at a drives the liquid over the high point T and fills the siphon. MOTION Page 91 1. See pages 88 and 89. 2. See page 89, subtopic, "Motion and Rest are Relative." 3. Generally speaking there is little or no relative motion. While turning from one street into another there is relative motion, which here consists in a change of direction from each other but not a change in distance. MOTION Pages 94, 95 1. The train is capable of the greater speed, but the horse, for a time at least, can gain speed more rapidly. The horse is capable of the greater acceleration. 2. 10 ft. per sec. gained in 1 sec. is the same rate of gain or accelera- tion as is 600 ft. per sec. gained in 1 min. The first car gains speed 60 times as fast as the second car gains it; its acceleration is 60 times as great. 3. 1 mi. per min. = 5280 ft. per min. 1 mi. in 3 min.= 1760 ft. per min. The speed of the train is 3 times that of the horse. Ace. of train 5280 -f- 120 =44 ft. per min. in each second. Ace. of horse 1760-:- 10= 176 ft. per. min. in each second. The horse has 4 times the acceleration of the train, but the train gains more speed because it keeps gaining for a longer time. 4. At the end of the first second the speed is 2 ft. per sec. At the end of 10 seconds the speed is 20 ft. per sec. At the end of 1 min. (60 sec.) the speed is 120 ft. per sec. 6. It will require 30-1-2=15 sec. 1 mi. or 5280 ft. per min. = 5280 -=-60 =88 ft. per sec. To gain speed of 88 ft. per sec. it will require 88-1-2=44 sec. 6. The speed at the end of 10 sec. is 20 ft. per sec., at the beginning MOTION 13 it is 0; hence the average speed is (0+20) or 10 ft. per sec. It will travel 10 X 10 ft.= 100 ft. in 10 sec. 7. It means that a falling body gains a velocity of 980 ft. per sec. for each second of its fall. 8. A rising body loses speed at the same rate that a falling body gains it. 32.16 ft. per sec. in each sec. or 980.2 cm. per sec. for each sec. at New York. 9. In 7 sec. it gains a velocity of 7X980.2=6861.4 cm. per sec. In the seventh second it gains as much velocity as it does in any other second. 10. d=$ at 2 = 98(X2 2 X(5)3 = 490.1X25= 12252.5 cm., distance fallen in 5 sees. d =(980.2 -f- 2) X 36 =17643.6 cm., dist. in 6 sec. 17643.612252.5=5391.1 cm., dist. fallen in 6th second. 11. In first 7 seconds d= 490.1X49= 24014.9 cm. In first 8 seconds d= 490.1X64= 31366.4 cm. 31366.4-24014.9=7351.5 cm., distance fallen in eighth second. 12. In 1 sec. the gain is 32.-}- ft. per sec., to gain 112 ft. per sec. it will require 112-^32=3^ sec. nearly. In 1 sec. the gain is 980+ cm. per sec.; to gain 4410 cm. per sec. it will require 4410-r-980=4.5 sec. 13. (a) It will lose 32. + ft. per sec. in each second, hence it will stop in 320^-32=10 seconds. (6) It rises as far as a body falls in 10 sec. d= 32X 100-:- 2= 1600 ft. (c) It will acquire the same velocity as it had when starting upward, 320 ft. per sec. (d) It will require the same time, 10 sec. 14. 30 mi. per hour = 30-f- 3600 = T ^ of a mile per sec. = 5280 -M 20= 44 ft. per sec. To stop the car or remove all its velocity in 10 sec.; in 1 sec. the velocity must be decreased -fa of 44 ft. per sec., which is 4.4 ft. per sec. To stop it in 4 sec. the velocity must be decreased at the rate of \ of 44= 11 ft. per sec. in each sec. 15. The distance = (4.4X10 2 ) = 220 ft. when stopped in 10 sec. Dis- tance = \ (11 X (4) 2 ) = 88 ft. when time is 4 sec. 16. At end of 1 sec. the velocity V=i+at= 10+32=42 ft. per sec. At end of 8 sec. V= 10 +8(32) = 266 ft. per sec. At end of 2 sec. V= 10+2(32) = 74 ft. per sec. At the beginning V is 10 ft. per sec. Hence average V = % (10+74) = 42 ft. per sec. D = it+$aP. Distance=(10X8) + (16X82) = 1104 ft. 17. When the weight is greater the mass to be moved is also greater; for example, if the mass of the second body is 2 times that of the first its weight or the earth pull is also twice as great. With double the mass we have double the force, hence the same acceleration. 14 MANUAL FOR MUMPER'S PHYSICS 18. 980.2 cm. per sec. at New York. Less at Washington and at Panama. More at Boston and at Quebec. 19. The weight of a body depends upon its mass, and the intensity of the earth's attraction at the place where the body is. The mass means how much there is of it and the acceleration measures the attraction in- tensity. Hence the weight varies as mass X ace. 20. (a) F = z+crf=20 + (12XlO) = 140 cm. per sec. (6) D = it -H at ^ = 20 + ^ - 26 cm. (c) D=(20XlO)+i (12X(10)2)=800 cm. (d) Dist. for 10 sec. dist. for 9 sec. = dist. for tenth sec. D for 10 sec.= (20X10)-KX(10)2)=800cm. D for 9 sec. = (20 X 9) + ( V 2 - X (9)2) = 666 cm. 800666=134 cm., dist. passed over in tenth second. MOTION Page 101 1. (1) It is equal to and opposite to the force on the breech. (2) The powder acts for the same time on each. (3) The directions are oppo- site. FIG. 2 (4) The velocity of the bullet is 160 times the velocity of the recoil of the gun, each being free to move. (5) 160 (mass of gun) X 1=160 (speed of bullet) XI. 2. The rapidly flowing water when suddenly checked exerts a pressure due to its momentum, which is added to the weight pressure of the water at rest. 3. A gram mass is a certain fixed quantity of material and a gram weight is the attraction of the earth for this mass. This attraction changes with certain changes in location, hence the weight called a gram weight is not always the same. The earth is not a true sphere and it is rotating; these are the conditions which lead to a change in weight with changes in location. See page 155. 4. A spring balance measures force directly, in this case weight. If the spring balance is being used in the same latitude in which it is gradu- ated the mass is also 80 gm. mass, since the earth's attraction varies for a fixed mass. No. 5. Both are equal, buoyancy of air neglected. Yes, because if a change in location changed the weight of one it would produce an equal change in the weight of the other. 6. We infer that something else is acting as well as the body known. In common speech, the second body is not free to move. 7. Let /= the pull of the engine. MOTION 15 Let a=the acceleration of the first train. Then ^ = the acceleration of the second train. Let ra=the mass of the first train. Let m 1 =the mass of the second train. Then /= ma. m = 2~* That is, the mass of the first train is half the mass of the second. 8. Let/ = the pull of the earth on a 1-lb. mass. Let 2/ = the pull of the earth on a 2-lb. mass. Let a = the acceleration of a 1-lb. mass. Let a c =the acceleration of a 2-lb. mass. Let m=the acceleration of a 1-lb. mass. Let 2m = the acceleration of a 2-lb. mass. Then 2f=ma. or 2/= 2 ma. and 2/=2 ma c . 2 ma=2 ma c . a=a c . That is, the accelerations are equal. 9. 1 gm. weight i. e., the weight of 1 gm. mass. It has different values. See question 3. 1 dyne (see line 18, page 98) has same value everywhere because the other units (cm., gm. mass, sec.) used in its definition, have the same value everywhere. 10. 1 gm. wt. = 980.2 dynes. 7 gm. = 6861.4 dynes. 7 gm. wt. in Panama are equivalent to a less number of dynes because the earth at- tracts a gram mass less there than it does at New York. MOTION Pages 109, 110 1. See Fig. 3. 2. See Fig. 4. 16z=6 (44-x). 22x=246. 3=12. 4ft 44 a;- 32. 12 Ib. 3. When A is on pan x and B is on y (Fig. 5) . Am=Bn (1). When A is on y and 5 is on x An>Bm (2). 4ft. 24 Ib. 16 MANUAL FOR MUMPER'S PHYSICS Multiplying (2) by (1) A2mn>&mn (3). Dividing (3) by mn AI>& (4). A>B (5). Dividing (2) by (1) An^Bm FIG. 4 n. m or m>n Multiplying by mn, n 2 >w 2 (7) or w>/w. 4. Horse A must have a force arm \ as long as that of horse B. Horse A must have J of the en- tire length and horse B f of it. In the first case A is 16 in., B, 32 in. from point of attach- ment. When A's force is f of B's, then B has 5 units of force to A's 3, and A has f of the dist., while B has $ of dist. A's dist. = 30 in., B's =18 in. 5. 10 lb. = 160 oz. Let x=dist. of sliding wt. from support. 160Xi=8:r. 8z=80. 151b. = 240oz. 8z=120. FIG. 5 6. At the start he lifts $ the total weight=i of 250=125 Ib. The rotational moment is 125X6=750, the ground end being the axis. To rotate it around same axis he must produce the same moment no matter where he takes hold. If the arm is 5 and the moment 750 the force = 750 -^-5= 150 Ib. 7. See Fig. 6. Let a;=dist. of force 24 to point of application. Then 12 x= dist. of force 18 to point of application. 12-X -> 42 24 FIG. 6 42x=2l6. v The resultant has a magnitude of 42 gm. and its point of application is 5f cm. from that of 24 gm. MOTION 17 36 8. See Fig. 7. 84 gm. 36 gm. = 48 gm., second force. 36X20=48*. 2t 48x=720. I 84 x=l5. 9. See Fig. 8. * Each boy has 15 ft>. when the bucket is 3 ft. JT IG 7 from each boy's hand. When x has 20 Ib. y has 10 Ib. Then the dist. of x is \ as much as that of y from the bucket, i. e., x has 2 ft. and y 4 ft. Whenx has 24 Ib. y has 6 J? Ib. Then the load at x is 4 ffft. times the load at y, hence y's dist. is 4 times x's, that is, x has | and y f of the whole. | of 6ft.= l=dist. ofx. And 4X H = 4|ft. = dist. of y from bucket. 10. There is an increased tension on the other rope of 160-48=112 Ib. 48 -f- 160 X 18=5.4 ft., dist. from man to the rope having 112-lb. pull. 185.4= 12. 6 ft. dist. from man to the rope having 48-lb. pull. MOTION Page 113 1. Relative to the earth its velocity is 12 5=7 mi. per hr. up stream. ca is path of boat relative to bed of the stream and cb is the path through the water. Let x=no. hrs. required to cross. Then xX 12= dist. cb the boat travels through the water and xX5=dist. ab the water moves in same time; but in the right triangle, abc, (c6)2= (ca) 2 + (a&)2; substituting, (12x)2=(2)2+(5x)2. 144x2=25x2+4. 119x2=4. 4 * 2= TI9- 2 FIG. 8 11 min. nearly. MAN. MUM. PHYS. 2 18 MANUAL FOR MUMPER'S PHYSICS 2. See Fig. 10. & = (30)2 + (40)2. #2 =900 + 1600 =2500. #=50 Ib. 30 76. 3. See Fig. 11. #2= (20)2 + (4)2= 416. R = \/41G> #=20.4 ft. per sec. 4. See Fig. 12. #2= (25)2 + (35)2= 1850. #2=V1850=43.0+ ib. 5. Since the direction is N. W. the two components are equal. Then the square of the northwesterly speed is equal to twice the square of the northerly speed. (16)2 40Ib. FIG. 10 4ft = 128. JV=. When lifting the box, work =24X25 =600 ft. Ib. additional. 6. Z>=7ft. Force= 100+50= 150 Ib. Work= 7X150= 1050 ft. lb. 20 MANUAL FOR MUMPER'S PHYSICS The force the boy must exert in pulling the wagon containing the trunk is much less than the combined weight of trunk and wagon, hence he does less work per foot. The force the boy exerts, measured by a spring balance, and the dis- tance from one street corner to the next must be known. 7. Three cu. ft. of water weigh 3X62.4 lb.= 187.2 Ib. 1 yd. = 3 ft. Work= 3X187.2 = 561. 6 ft. Ib. The amount of work required to lift the water a given distance is not affected by the direction provided no other factors are introduced thereby and the vertical elevation is the same. 8. (When is work or energy said to be transferred?) When energy is given from one body to another. It is transformed when it is changed from one kind into another, as for example, mechanical energy into heat. 9. Distance lifted is 1 ft. in 12, hence 5 ft. in 60 ft. Work= 5X 1000 = 5000 ft. Ib. (not counting waste). The arrangement of the trunks af- fects the amount the man must lift but not the force with which he pushes, hence the amount of work is not affected. 10. (a) Work= 3X400 =1200 ft. Ib. (6) Work=3X650=1950 ft. Ib. 11. The energy of his moving hand is called kinetic; that of the wound spring is called potential. It is called potential because it is not energy of motion but may be converted in energy of that type. 12. We must know the force with which the powder acts and the dis- tance through which it acts upon the bullet. (Approximately the length of the barrel.) The weight of the bullet and the distance it rises before it comes to a stop. At the instant it stops, in the air, its energy is called potential. At the instant of leaving the gun its energy is kinetic and it may be computed by knowing the mass and the velocity. E=\ MV 2 . 13. E=^p(pagel26). = g^ =2 ft. Ib. 250X(60)2 E== 2(980-}-) = 459 ' 18 g m - cm - 14. The most kinetic energy when it is at the lowest point where its speed is greatest and the most potential at the highest point of the arc where it comes to rest. When at rest at its lowest point it has no energy as a pendulum, though it is said to have potential energy as a body elevated above the surface of the earth. 16. Kinetic energy= MV 2 . Since motion and speed are relative to some body; hence energy as thus computed, depends upon the standard of motion. Potential energy computed from position must be relative for position is relative. See pages 88-89. 16. We usually select a point on the earth's surface vertically below the object. 17. Work = energy rendered potential =800 XI 500 =1200000 gm. cm. ENERGY AND WORK 21 The potential energy will have become kinetic, hence the body when strik- ing has 1200000 gm. cm. of energy. d=' v=gt (page 93) H = V2X 1.500X980 = \/2940000 = 1714.6 cm. per sec. 18. A pound force changes slightly with a change in latitude, but a dyne has the same value everywhere. In consequence a foot pound varies with latitude, but an erg has the same value everywhere. The latter is based on the gm. (mass), cm. (length), and second (time). ENERGY AND WORK Pages 131, 132 1. Work to be done = force Xdist. Force is 1000X8 lb. = 8000 lb. Work= 8000X20= 160000 ft. lb. The amount of work is the same in each case. A does it in 40 min. or 1600004-40=4000 ft. lb. per min. B does it in 60 min. or 1 60000 -h 60 =2666f ft. lb. per min. B's power is only that of A. 2. A one-horse power engine can do 33000 ft. lb. per min., hence a 4 h. p. engine can do 4X33000= 132000 ft. lb. per min. 3 hours= 3X60 =180 min. A 4 h. p. engine can do 180X132000=23760000 ft. lb. in 3 hours. 3. The work to be done is 160000 ft. lb. A 2 h. p. engine can do 66000 ft. lb. in 1. min. To do 160000 ft. lb. it will require 160000 -H 66000= 2.4+ min. 4. 10000 cu. ft. of water weigh 624000 lb. Work to be done in 1 hour =60X624000 =37440000 ft. lb. Work to be done in 1 min.= 5 V of 37440000=624000 ft. lb. 1 h. p. can do 33000 in 1 min., hence to do 624000 ft. lb. in 1 min. will require 624000-1-33000=18.9 h. p. 6. 1 gm. wt. = 980.+ dynes. To lift 1 gm. 4 cm. requires 4X980= 3920 ergs. See page 129. 6. 1 kilowatt=1000 watts = 1000 joules per sec. 10 kilowatts= 10000 joules per sec. 15 min. = 900 sec. 10 kilowatts for 15 min. = 9000000 joules. 1 kilowatt^ H h. p. 10 kilowatts^ 13$ h. p. 22 MANUAL FOR MUMPER'S PHYSICS 7. The wasted work appears chiefly as heat at the axles and platform on which the truck runs. Oiling the axle decreases the waste work but does not affect the useful work. Ball bearings decrease the waste work or heat at the axles, hence decreases the total amount of work the man must do. 8. When roads are graded and smooth a given power can do more useful work, hence a horse, for example, can move more material with the same expenditure of energy. The saving in power usually more than makes up for the cost of improvement. 9. To do work upon itself as a whole, a body would have to create energy. One part may do work upon another part of the same body. The loss of energy in one part equals the gain in the other. A man may use one hand to lift the other. ENERGY AND WORK Page 136 1. It is no evidence of another body acting on the earth. (First law.) It shows attraction between the parts of the earth (cohesion) or each particle would move in a straight line tangential to the circumference. 2. Yes, otherwise the earth would move in a straight line. 3. To do so would be an object putting itself in motion or creating its own energy. (First law.) 4. When the shoe stops, the adhesion of the snow to the shoe is not strong enough to overcome the inertia of the snow, which therefore moves on away from the shoe. MACHINES Pages 139, 140 1. It is said to furnish a gain in force when by its use the force upon a load is made greater than it is at the agent. It furnishes a gain in speed when the speed of the load is greater than that of the agent. A change in direction means that the load moves in a different direction from that of the agent. A gain in force is accompanied by a loss in distance moved by the load, that is, loss in speed of load. See sec. 85, page 139. 2. Simple machines are intended only to transmit energy. The energy they transform is wasted, generally as heat. 3. It can give out only the energy given to it, and since it gives as fast as it receives, it is in no sense a storehouse of energy. 4. Since they transmit energy and nothing more they cannot add to the quantity given them per unit of time (by any agent as a man or engine). They may multiply force or speed. See ans. to ques. 1. 5. See 3, page 138. Oiling decreases the amount of the waste energy and also diminishes the wear on the bearings. MACHINES 23 MACHINES Pages 144, 145, 146 1. See Fig. 16. AB = 28 in. AC=U in. DC =7 in. BC=U in. ACXW=BCXx. A C D B 14X40= 14Xz. i A ' I The agent at B must act down- ward with a force of 40 Ib. 2d case. AC X 40= DC X?/. 14X40=7X2/. 7j/=560. 7/=80. An agent at D must act downward with a force of 80 Ib. 2. First class. Place the load at D acting downward and agent at B acting upward, and it shows second class. Place load at B acting upward and agent at D acting downward, and it shows third class. 3. See page 139, first 5 lines. 4. The agent must do an amount of work (varying with circum- stances) more than^ijiat computed as necessary to move the load. 6. Since AO, agent arm, is shorter than BO, load arm, the force at A must be greater than that produced at B, hence there is a loss in force. But the load moves a greater distance in same time, hence it has a greater speed than that of agent. 6. The load is 2 ft. from axis, the agent is 2+3=5 ft. from axis. 2X225= 5Xz. 5z=450. z=90 Ib. at man's hands. 7. It will increase the amount the man must lift. 160X1 = 5Xy. 5^=160. ?/=32 Ib., due to wt. of wheelbarrow. 90 Ib. (previous prob- lem) +32= 122 lb. = total force at man's hands, 61 Ib, on each hand. 8. (a) At the axle, 225-90=135 Ib. (6) Due to wt. of barrow, 160 -32= 128 Ib. Due to both load and barrow, 135 lb.+128 lb. = 263 Ib. 9. Sugar tongs, agent at T, fulcrum at B, load at S. Scissors, agent at F, fulcrum at A, load at C. Claw hammer, agent at end of handle, fulcrum where hammer head touches the wood, load at nail. 10. Agent at keys, load the type. Nut cracker, agent at C, fulcrum at A, load at B. 11. Not counting waste, amounts are equal. See page 139. When grasped nearer the head the agent must act with more force to produce the required force at the nail or load. The work is the same, a change in direction of action would change force but not the useful work required, though on account of increased friction it would increase the wasted work. 24 MANUAL FOR MUMPER'S PHYSICS 30/b, 12. See Fig. 17. FB=S ft. FA is to be found. FAX 5Q=FBX30. FIG ft. To find FA (not counting the wt. of plank); FA is f and BF f of the A ^ whole, 7^4 = 6 ft. To find FA (counting the wt. of plank) . Center of gravity (C) is in mid- dle of plank. See Fig. 19. FA = 4. 8 ft. See Fig. 18. AB=16 = 3QXBF, hence 50/b. 3Olb. FIG. 18 A F C B 50/b. A \30Ib. Let x = dist. from C to F. Sx = dist. from A to F. = dist. from B to F. FIG. 19 50 X^ CF. 50 (8 x) = 30 (8 +x) +40z. 120z=160. 13. The arm is a lever having its axis at y and agent, the muscle, acting at x. The force on the load w is much less than the force or tension of the muscle at x, but there is a corresponding gain in the speed of the load, a very important advantage. Not counting the waste the work done by the muscle is equal to that done upon the load W. FIG. 20 MACHINES Page 148 1. The speed of the load is here equal to that of the agent. It furnishes a change in the direction of motion of the load in relation to that of the agent. 2. See Fig. 20. The force at the load is greater. The distance the agent moves is greater. F a XD a =F l XD l (waste neglected). 3. See Figs. 21 and 22. 4. In the first case (Fig. 21) force A : force L= 1 : 2. In the FIG. 21 MACHINES 25 FIG. 22 FIG. 23 second case (Fig. 22) force A: force L=l: 3. In first case the force at load is 2X15 ft. =301b.andwork= 8X30= 240 ft. lb. In second case force at load=3Xl5= 45 lb. and work= 8X45= 360 ft. ft>. These results give the work upon the load alone. The agent must do an ad- ditional amount to make up for waste. 5. See Fig. 23. Force at L is 5 times the force at A, hence force L= 5X210 gm. = 1050 gm. If fixed end of cord is attached to fixed block of pulleys there will then be only 4 parts of cord to movable block (invert the figure shown), hence force at L= 4X210 = 840 gm. (not counting weight of pulleys). MACHINES Page 150 1. The radii are 10 cm. and 3 cm. Force at AX 10 = force at LX3. LX3=300X10. 3L = 3000. L=1000 gm. 2. The distance the agent moves, hence the speed of the agent, is 3 times as great as that of the load. Work on L= 15X1000=15,000 gm. cm.; neglecting waste, the quantities of work are equal according to the law of conservation of energy. 3. 96X^=4.5X480 4.5X480 96 = 22.5 in., length of crank. 4. The large wheel has 5 times the diam. and 5 times the circum- ference of the small wheel. The chain moves enough in 1 rotation of the large wheel to rotate the small wheel 5 times. 5. There is a gain in speed with an accompanying loss in force. MACHINES Pages 153, 154 1. The weight of the wagon is lifted through the vertical height of the hill, but the horse exerts his force throughout the entire length of the hill. 26 MANUAL FOR MUMPER'S PHYSICS 2. The force with which a body acts in a given direction and the distance through which it moves another body in that direction. When a body is being lifted the weight is the required force and the vertical dis- tance is the required distance. 3. We must know the pull (in Ibs., dynes, etc.) exerted by the horse and the distance through which the horse moves the wagon measured on the slope of the hill. Or if we know the weight of the wagon and the vertical distance we may compute the work from them. Neglecting waste the results should be equal. 4. Distance lifted, 20 ft.Xwt. of wagon, 500 lb.= 10000 ft. lb. = work. 5. The horse must do 10000 ft. Ib. in moving the wagon 100 ft., hence his force is 10000 -M00= 100 Ib. 6. Here " easier" means less force required. Work is the same. 7. In this case " easier " means less work, for the oiling decreases the waste, hence the total work the horse must do is decreased. 8. (a) 250 Ib. (6) T 5 7 of 250= 83 Ib. (c) Work, when moved directly =250X5= 1250 ft. Ib. Work, when moved on the plank =83^X15 =1250 ft. Ib. 9. Area of small piston, 20 sq. cm. Area of large piston, 180 sq. cm. Total pressure on small piston =60 gm. . Total pressure on large piston =540 gm. Work at small, piston =45X60 =2700 gm. cm. Work at large piston is also 2700 gm. cm. (waste neglected). Distance large piston moves =2700-^540 =5 cm. MOTION IN A CURVED LINE Page 158 1. When the speed reaches a certain amount the cohesion of the particles of the wheel is no longer able to pull the parts of the wheel into curved paths, hence they move off in (approximately) straight lines. 2. When changing direction there must be an action and reaction between the wheels and the roadbed; if on account of mud, etc., this is insufficient at any part of the curve the machine, or a part of it, may suddenly resume the straight-line motion. 3. It makes the weight of a body less or apparently less than it other- wise would be. It is greatest at the equator and zero at the poles. , mi in (massX veloc. 2 ) . 4. The upward pull * Q - is equal to or greater than the downward pull (weight of the mass). 5. The same as 4. 6. It has a large amount of kinetic energy. This stock of energy may be drawn upon or added to in the working of the machine without materially affecting the speed, because the mass is so large. HEAT 27 THE PENDULUM Pages 161, 162 1. The earth pull or the weight of the bob. It vibrates in less time when near the poles because the value of g is greater. 2. The mainspring is the storehouse of the energy given to the clock when it is being wound; it produces the motion of the wheel work. The pendulum regulates the motion, that is, it prevents the wheels from going too fast. If it runs too fast make the pendulum longer, so that more time is used in making each vibration. 3. * 2 =f sec. ti:t2=VTi- Vh. squaring 1 : f = 39. 1 : 2. Z 2 =17.3+in. 4. tj?:W=li:h. (3)2; (4)2=25: h 9Z 2= 400. Z2=44.4 cm. 5. \~ = Y- or ^ = ^ , substituting * ^92 0i gz . 99.3^=98020. <72= 987.1. 6. W:t#=h:l2. 1: 16=36: Z 2 . 1 2 = 16X36= 576 in. 7. An increase in earth's attraction or in the value of g would decrease the time required by a given pendulum to make one vibration. This suggests that a given pendulum, of fixed length, may be used to de- termine the relative values of g at different places. HEAT Page 175 1. 4 C. are equivalent to 4Xf F. = 7i F. 2. The body is above zero and it is warmer than freezing water. 3. It is above zero F., but it is 28 F. degrees below the temperature of freezing water. 4. See sec. 105, page 170. 5. 68 F. 32 F. = 36 F.. This means- that the room is 36 F. degrees above the freezing point. 36 F.= f of 36= 20 C. 6. f of 45 = 81 F. This is the equivalent number of degrees. But the reading will be 81+32=113 F. on the scale. 7. 4C. = fX4+32=39-f F. 10 C. = |X10+32=50 F. -40 C.= X -40+32= -40 F. 28 MANUAL FOR MUMPER'S PHYSICS 98 F.= f(98-32) = 36.+ C. 10 F. = f(10 32)= 12. + C. -10 F.= f(-10-32)=-23. + C. 8. 273 C.= |X -273+32= 459.+ F. 9. -182 C.= |X -182+32= -295+ F. 10. The bulb must be relatively large and the bore of the stem relatively small. A large bulb heats and cools slowly. The instrument is not re- sponsive. When the bore of the tube is very small it is difficult to see the mercury. HEAT Pages 179, 180 1. 10X1 gm. cal. = 10 gm. cal. 10X25 gm. cal. = 250 gm. cal. 17 11 = 6 rise in temperature. 6X65 gm. cal. = 390 gm. cal. 95 15= 80 fall in temperature. 30X 80 gm. cal. = 2400 gm. cal. 2. The quantity of heat lost by the warmer equals that gained by the colder. 3. (a) The temperature is the mean of the two others; that is, it lies halfway between them. (6) The temperature of the colder and smaller mass rises twice as many degrees as that of larger mass. Hence the rise is and the fall ^ of "the difference between the original temperatures. (c) The smaller changes f and the larger f of the number of degrees difference in their original temperatures. 4. Because, being different substances they have different specific heats and when one gains as much heat as the other loses they do not undergo equal temperature changes. 5. The mass, the specific heat and the temperature change. 6. The rise in temperature equals the fall in temperature, hence they come to a final temperature halfway between the original temperatures (0+56) = 28 C. 7. 87 23 = 64 C., temperature difference. The rise in tempera- ture equals the fall in temperature, and the final is 23 +32= 55. 8. Temperature changes are inversely as the masses. 58 14 =44, whole difference in temperature. Rise in temperature is T 8 T and fall is T 6 of the whole difference, 44. Rise= T 8 of 44=25f. Final temperature = 14 +25 j = 39| C. 9. It takes .11 of a gm. cal. to warm 1 gm. of iron 1 C., or more gen- erally, it takes .11 times as much heat to warm any mass of iron 1 as it does to warm an equal mass of water 1. 10. It does not, because specific heat is a ratio. 11. Heat given=25Xl gm. cal. = 25 gm. cal. 1 gm. of iron requires .11 gm. cal. for each degree. HEAT 29 10 gm. of iron require 1.1 gm. cal. for each degree. 25-M. 1 = 22.7 + C. 12. 80 15 = 65, the whole temperature difference. Let 3= fall of temperature of iron. 65 3 = rise in temperature of water. Quantity of heat lost by iron = quantity gained by water. .11X3203=320(653). .113=65-3. x = 58.5, fall in temperature of iron, hence final temperature =80 58. 5 = 21.5. 13. 75 15 = 60, total temperature difference. Let 3= rise in temperature of water and of glass vessel. 60 3= fall in temperature of copper. Heat received by water and vessel = heat lost by copper. 2003 + .16(403) = .09X450(60 3). 2003+6.43 = 243040.53. 246.93=2430. 3=9.8 C. final temperature=15 +9.8 = 24.8 C. 14. Let 3= specific heat of metal. 100 30 = fall in temperature of metal=70. 30 16 = rise in temperature of water =14. 720X703=120X14. 504003=1680. 3= .033 + = specific heat of metal. 16. It would warm 1 gm. of water through .091. The same amount of heat would warm .091 gm. of water through 1. 16. To warm 100 gm. of brass 1 C. require 100X.091 = 9.1 gm. cal. The same quantity of heat would warm 9.1 gm. of water through 1. The water equivalent of 100 gm. of brass is 9.1 gm. of water. 17. 200 gm. of lead require 200X-031 = 6.2 gm. cal. to warm it 1. 6.2 gm. of water would require the same amount of heat to warm it 1. Water equivalent of 200 gm. of lead is 6.2 gm. of water. HEAT Pages 185, 186 1. (a) The longer increases in length three times as much as the other increases. (6) Because they are in contact with each other. It could be deter- mined by means of a thermometer. (c) Since the longer increases three times as much as the other, the increase of each is the same part of the original length of each. (d) Since the expansion of each is the same part of the original length 30 MANUAL FOR MUMPER'S PHYSICS of each and they are warmed through the same number of degrees, the increase in length per unit length, called the coefficient of expansion, is the same. (e) This expansion would be larger when the degree is a centigrade degree, than it would be if it were a Fahrenheit degree. 2. Original length, original temperature, final temperature and amount of expansion of body. Yes, because if the longer rail were used both the original length and the amount of expansion would be three times as large and their quotient would be the same as that obtained with the shorter rail. 3. Any piece of glass, of that kind, increases or decreases in length .0000089 of its original length for each degree C. change in tempera- ture. 4. (a) 200 X- 0000231 =.00462 cm. for 1 C. (6) .00462X25= .1155 cm. for 25 C. 200 + . 1155= 200.1155 cm. length at 25 C. 5. Let x = coefficient of expansion 654.55 654.00=. 55 cm. increase for 100 654XlOOz=.55 65400* =.55 65400 = .0000084. 6. Expansion decreases and contraction increases the density of a body. Between C. and 4 C. water expands as it cools and contracts as it gets warmer, hence warming water between these temperatures in- creases and cooling decreases its density. 7. 1 m.= 100 cm. 100 X 120 X. 0000089 =.1068 cm. 100+.1068 cm.= 100.1068 cm., length at 120 C. 8. From 8 to +37 C. is a temperature change of 45 C. 180.2X45X- 0000168=. 136+ ft. 180.2+.136 ft. = 180.336 ft., length at 37. C. 00001 87 9. Brass expands : Qoool2T = 1-54 times as much as iron per unit length. Iron rod must be 1.54X26.8=41.27 cm. long to equal the expansion of brass rod 26.8 cm. long. 10. Temperature change, 86 5 = 81 F. or f of 81 = 45 C. 300 X 45 X. 0000109=. 147 ft., expansion of span. 11. .0000297 .0000 168 =.0000 129, hence each cm. of zinc expands .0000129 cm. more for 1 C. 600 X 60 X. 0000 129 =.46+ cm,, the excess expansion of 600 cm. of zinc for 60. HEAT 31 ---- HEAT Page 192 1. The coefficient of expansion of gases is 273- f r eacn 1 C. The expansion in this case is 1 c. c. and the volume becomes 274 c. c. In the second case the expansion is ^W of 273=40 c. c., and the volume becomes 273 +40= 313 c. c. at 40 C. 2. If originally at C. it must be warmed 273 to expand f^f or to double its volume. In cooling to 63 it contracts $fo of its volume, hence its volume is then Iff ^ = fM of its magnitude at C. 3. The pressure of the gas increases. See 189, sec. 119, and last 4 lines on page 190. 4. C. = 273 on absolute scale. 273+91 = 364 on absolute scale. Pressure becomes fff of original=l$. In terms of barometer height the pressure is 1^X76=101.3 cm. 6. The pressure being constant the volume varies directly as the absolute temperature, hence at 20 C., or 293 absolute, the pressure is If | of the original pressure; and the density, which varies inversely as the volume, is f$f of 1.293= 1.204 gm. per liter. 6. 60C. = 333 absolute. 180 C. = 453 absolute. 453 : 333 = 300 : x. 453z= 99900. x= 220.5 c. c. vol. at 60 C. 7. See page 64, sec. 39. 1X76 F 2 X70 273 ~ 293 ' V 2 = 1.16 liters. Since 1.293 gm. now have a vol. of 1.16 liters the density is T 1.114 gm. per liter. FiPi F 2 P 2 500X75 80XF 2 283 293 * F 2 =485.+ c. c. 9. He inhales a greater weight of air when the pressure is high; a greater weight when the temperature is low; a greater weight when at the sea level; a greater weight in winter, other conditions being the same in all the cases. HEAT Pages 197, 198 1. The water loses as much heat as the ice receives. The temperature of the water falls, but that of the ice does not change because the heat 32 MANUAL FOR MUMPER'S PHYSICS received by the ice does the work of melting and hence ceases to be heat. 2. 1 gm. of lead at its melting point requires 5.9 gm. cal. to melt it. The thermometer makes a difference because a heat unit as defined by means of a C. thermometer does not have the same value as a unit de- nned in terms of the F. thermometer. 3. 80 gm. cal. per gm. of ice, hence 10 gm. of ice require 800 gm. cal. 10 gm. of lead require 10X5.9=59. gm. cal. 4. It is losing heat without becoming colder. The molecular motion of the water furnishes the heat. The heat which leaves the water is taken in by the surrounding colder air. 5. 60 20 = 40. In cooling 40 the water gives out 40 X 250= 10000 gm. cal. 1 gm. of ice requires 80 gm. cal. to melt it. 10000 gm. cal. will melt 10000-1-80=125 gm. of ice. 6. 120 gm. cal. 80 gm. cal. = 40 gm. cal., the amount of heat left to warm the water; this will warm 1 gm. of water 40 C. As the ice melts the volume decreases and as the water is warmed from it decreases until a temperature of 4 C. is reached. It will ex- pand from this temperature up. 7. The ice is better because it requires a large amount of heat to melt it in addition to the heat required to warm the ice water afterward up to the highest temperature permitted in the refrigerator. 8. It takes 4 times as much heat to melt a gm., or any other mass of ice, as it does to warm an equal mass of water 20, hence 1 Ib. when melted and warmed 20 would require as much heat as 5 Ib. of ice water require in being warmed from to 20. 9. Heat lost by water and calorimeter = heat used to melt ice and warm water. Let z=fall in temperature of water and calorimeter. Then 200z +. 09 X40z= 60X80 +60X10. z=26.5. 10 +265 = 36.5, original temperature of the water and calorimeter. HEAT Pages 208, 209 1. From liquid to vapor state. It is receiving heat, but it is not getting warmer. The heat received is used in doing the work of vaporization. 2. 1 gm. of water, 536 gm. cal. 8 gm. of water, 4288 gm. cal. 1 gm. of ether, 90 gm. cal. 12 gm. of ether, 1080 gm. cal. 3. 1 gm. of steam by condensation alone furnishes 536 gm. cal. 16 gm. HEAT 33 of steam furnishes 16X536=8576 gm. cal. To warm 400 gin. of water 1 requires 400 gm. cal., hence 8576 gm. cal. will warm the water -^-=21.4. The temperature of the water must have been 100 21.4 = 78.6 C. before the steam was admitted. 4. The water is not warmed, all the heat given out by the iron is used in vaporizing water. The iron cools 800 100 = 700 . It gives out 700 X 1000 X.I 1 = 77000 gm. cal. This heat will vaporize 77000-^-536=143.+ gm. of water. 5. (1) 100 C. (2) 40 C. (3) The heat is furnished both by the con- densation of the steam and the cooling of the water formed from the steam, to 40. 6. Liquid air boils at (or about) 182 C., hence all the heat received by the liquid air at this temperature is used to convert it into gaseous air. 7. The good conducting metal vessel transmits heat from the ice, originally at C., to the liquid air which is at 182, far below the tem- perature of the ice. This heat from the ice boils the liquid air. Finally the ice in contact with the vessel reaches approximately 182 C., when the action nearly ceases. 8. (1) The specific heat of water is very large. (2) The heat of fusion and the heat of vaporization are also very large, hence the absence of water means the absence of these checks to sudden heating and cooling of the air. 9. Water vapor into liquid water. This condensation of water vapor means the generation of 536 gm. cal. per gm. of vapor condensed and this heat is given to the air. 10. The condensation of the steam furnishes the heat. HEAT Pages 221, 222 1. The blanket conducts heat poorly, hence it keeps in the heat of the human body and equally well keeps out the heat of the warm air from the ice. 2. The statement suggests that clothes produce more or less heat instead of merely keeping in the heat which the body produces. 3. (1) The circulation will keep the temperature very nearly uni- form. (2) The lower parts of the water, because they receive heat so rapidly, will be hotter. 4. Glass is a very poor conductor and also a very rigid substance. A sudden change in the temperature of one part of a thick piece of glass causes a change in the molecular motion in that part while the condition of the rest of the glass is not changed. The strain thus produced between MAN. MUM. PHYS. 3 34 MANUAL FOR MUMPER'S PHYSICS the parts having different molecular motion is often sufficient to over- come cohesion and the glass breaks. 5. When a room is heated by means of hot air the air is the warmest. When heated by steam or hot water the air is also warmest, but not so much warmer as in the case of hot air. When heated by a grate fire the air is cooler than furniture and other solid bodies. HEAT Pages 229, 230 1. 1 gm. cal. = 41900000 ergs. 1 gm. cm. at New York=980.2 ergs. 1 gm. cal. = 98Q 2 = 42 ?47 S m - cm. 1 joule =10000000 ergs. . 41900000 1 gm. cal.= 10000QQO =4.19 joules. 2. 1 gm. cal. = 42747 gm. cm. of mechanical energy to produce which a body having a weight of 1 gm. must fall 42747 cm. Specific heat of copper is .09. To warm 1 gm. of copper 5 requires 5 X. 09 =.45 gm. cal. .45 gm. cal. = .45X42747 gm. cm.= 19236 gm. cm. The gm. of copper must fall 19236 cm. 3. It lowers the temperature since some of the heat is used to do the work of expansion. 4. 80X250=20000 Ib. pressure against piston. 20000X1^ = 30000 ft. Ib. Work done. It lowers the temperature of the steam, the heat energy going into the mechanical work. 5. It means that 10% of the heat developed in the furnace is con- verted into mechanical work. 6. 1 gm. of ice requires 80 gm. cal. to melt it. 1 gm. eal. = 42747 gm. cm. 80X42747=3419760 gm. cm. To produce this the ice must fall 3419760 cm. Any mass of ice must fall the same distance, g being constant. 7. 10% of 7800= 780 gm. cal. of heat used for each gm. of coal burned. 10 kilowatts = 10000 watts. 1 watt= 1 joule per sec. = 3600 joules per hr. 10 kilowatts =3600X10000 joules per hr. = 36000000 joules per hr. 1 gm. cal. = 4. 19 joules. = gm Q prouce g = gm. of coal must be used. 20 horse power = 15 kilowatts. 15 kilowatts for 2 hr. = 30 kilowatts for 1 hr. Hence three times as much coal will be used as that computed in first part of problem. SOUND 35 SOUND Pages 256, 257, 258 1. The vibrations are so intense and the pitch of some of them so low that the window panes are able to take up the vibrations. They are sympathetic vibrations. 2. In the open air the hearer receives only the direct waves; there is little or no reflection as it occurs in rooms, streets, etc. 3. The reflection from side to side prevents the waves from spreading, hence their intensity decreases very slowly with the increase in distance from the source. 4. The sound waves travel to the woods and back, a total distance of \ mile or 2640 ft. Assuming the temperature 32, the speed is 1090 ft. 2640 per. sec., hence it will require jQ9o = 2.4+ sec. 6. The speed of a compressional wave (sound wave) varies directly as the \/elasticity and inversely as the \/density. A rise in the tem- perature of the atmosphere generally decreases the density without chang- ing the elasticity. 6. A uniform medium undisturbed by currents or by the relative motion of its parts serves the purpose best. 7. Speed of sound waves at 80 is (48 ft. per sec. faster than it is at 32), 1138 ft. per sec. In 2 sec. the waves travel from the source to the hearer. 2X1138 = 2276 ft. 8. See text, pages 231, 234. 9. If 5 sec. are required for the waves to travel to the hill and back, 2.5 sec. are required for the distance to the hill. Speed is 1090+38= 1128 ft. per sec. In 2.5 sec. the distance will be 2.5X1128=2820 ft. 10. Wave length= =- = 5 ' 5 ft ' The P eriod is of a sec " 2>0 11. Wave lengthXfrequency= velocity. 4X256=1024 ft. per sec. Because it increases the speed, by decreasing the density, it will increase the wave length. 12. An octave above has twice the frequency, hence half the wave length of the fundamental. A change of the medium transmitting the waves of a given note changes the speed and wave length without chang- ing the pitch, hence the relation between the wave lengths gives the true relation between the pitch of two notes only when the waves are in the same medium. 13. Tones having same pitch have same frequency, but when the media are different they may have different velocities, hence different wave lengths. f =725.+ per sec. Period -^ sec. 15. Because the waves are constantly enlarging spherical shells there 36 MANUAL FOR MUMPER'S PHYSICS will be less energy per unit, surface, even though we do not count the sound energy which is wasted by being converted into heat as the wave advances. 16. See page 246, lines 2-5. It may also be demonstrated by means of Fig. 266, page 265. The wasted energy makes the intensity decrease much faster than it would according to the law of inverse squares alone. Sound waves lose their energy more rapidly while traveling a given distance in sawdust than they do when traveling the same distance in the wood. 17. Mixed materials, especially those containing a great deal of air, are put into the spaces between the walls and floors, etc. These substances transform or use up the energy of sound waves very rapidly. 18. They reflect the waves so that the hearer receives them along with the direct waves, thus making the total effect or intensity greater than it would otherwise be. 19. Intensity (energy per unit surface of wave). Pitch (frequency of vibration of particles in wave). Quality (number and character of the overtones associated with the fundamental). 20. See page 250, sec. 177. 21. They interfere (254-250) 4 times per sec. 22. Law 2, page 254. When the length is 29 cm. the frequency will be 87 -f- 29 =3 times the frequency which it has when 87 cm. long, other conditions being the same. 23. An octave higher means that the upper note will have twice the frequency of the lower, to produce which the tension must be made 4 times as great. 4X41b.= 161b. Law 1, page 254. 24. \ the length under the same tension would give a note an octave higher. To reduce the pitch back to the original, an octave lower, the ten- sion must be made \ of 40 kil.= 10 kil. 25. Faster at noon because the temperature is generally higher and the density of the air less then. Faster at the sea level because generally warmer. It is true that the density is greater at sea level, but as the elas- ticity is correspondingly greater, they neutralize. 26. The compressed air has both a greater density and a greater elas- ticity, consequently the loss in one respect is counteracted by a gain in the other. LIGHT Pages 269, 270 1. The spherical character of the waves, on account of which each point on a wave front advances along a straight line, thus giving straight rays. LIGHT 37 . See Fig. 24. Let bg represent 4 ft., the distance of the screen from the pin hole 6, and bf, 100 ft., the distance of the tree. Triangle abc is similar to triangle bde. abf is similar to beg and bcf similar to bdg. Then ac: bf=de: bg; substituting, ac: 100= : 4. 4ac=50. ac=12.5 ft. FIG. 24 3. When moved farther from the pin hole the image becomes larger and, because that means the same amount of light distributed over more surface, the image becomes less bright. 4. Light acts upon the organ of sight. In total darkness there is no light to do the acting. In blindness there is not a proper organ to re- ceive light energy. 6. It is sharpest at the end of the shadow nearest the pole. See page 264. The bird is so small, compared to the sun, and its distance from the earth so great that its true shadow or umbra does not extend to the earth. See page 264, Fig. 265. 6. A flame is seen by means of the light it generates. 7. It is seen by its own light when the current is on. In the other case it is seen by the light it reflects; and being nearly black it reflects little or no light and is not easily seen. 8. Light is generated by the flame and by the glowing coal. We cannot see the ashes unless they are illuminated from some outside source. 9. The intensities at the two sources are to each other directly as the squares of the two distances from the surface on which they produce equal illumination. 38 MANUAL FOR MUMPER'S PHYSICS Let x= candle power of the lamp, then l:x= (8)2; (60)2 = 64 : 3600 z=56i- c. p. 10. Let x= distance from 16 c. p. lamp. 16: 75 =z 2 : (450)2 16 X (450)2=3240000 x= 207.8+ cm., distance from 16 c. p. lamp. 11. See Fig. 25. When held in the position ab the book receives all the light between the lines Sa and Sb. When it is in the position cd it receives FIG. 25 only the light between the lines Sc and Sb. The source has the same in- tensity and the average distance of the book from the course is the same as before. LIGHT Pages 276, 277 1. Regular reflection occurs at a smooth or practically smooth surface, essentially all the light going off from each point in only one direction. Irregular reflection occurs at a rough surface, the light from each point being sent off in practically all directions. A point as here used really means a very small surface and not a mathematical point. 2. It could not be seen since it would neither diffuse nor absorb any of the light it receives. 3. See page 271, par. 196. 4. When looking at his own image, the distance of the image of the observer is twice the distance of the mirror from the observer. It makes the image appear smaller and in consequence a person can see his entire image in a mirror his height. 5. See page 273, sec. 198. 6. In the case of a real image the light from the object reaches the place where the image is. In the case of a vertical image the light from the object does not reach the place where the virtual image seems to be. 7. The object must be at a distance from the mirror which exceeds the principal focal length. LIGHT 39 LIGHT Page 282 Index of glass 1.5240 ,. . . , .. Index of water= O34CT L1424 ' mdex of water to g lass ' 2. The index of refraction of air may be taken as 1 : X 1860QO = 112650 mi. per sec. = 139430 mi. per sec. 3. See page 280. The index of refraction varies with the speed of light in the medium. 4. It is refracted more when going from air into a glass prism, other things being the same. 1 524 1 (nearl ) = l-^ 24 > index of refraction, air to glass. 1 524 1 SS4() = 1-1 424 > m dex of refraction, water to glass. Hence the first is 1.1424 times the second. LIGHT Page 290 1. Drawings based on sec. 204, page 282. 2. The convex type. The difference is chiefly in the brilliancy of the image and the clearness of definition or of outline. 3. * U -i 1-1+-! / di d ' 12 20 do Clearing of fractions, 20d = 12d +240, d =30 cm., dist. of object from the lens. Since object is more distant than image, the object is larger, in ratio of 30: 20, or 1^ times. 4. ~ f =i^+l- Clearing, 486= 9/+54/, 63/=486, /=7.7 inches. 6. -.= 1+1. Which means that the image is also twice the focal length distant from the lens. The image and object are consequently of the same size. 40 MANUAL FOR MUMPER'S PHYSICS LIGHT Page 302 1. Strictly it is a characteristic of the sensation due to the kind of light which the object sends to the eye. 2. The difference which produces the difference in color effects is due to the difference in wave lengths or in the vibration frequency. 3. A prism, by separating the light into its constituents, shows that white is a mixture of light that contains essentially all the wave lengths which act upon the eye. White or gray color sensations may be pro- duced by certain combinations of two or more colors known as com- plementary, such as yellow and blue. 4. A pure black means a complete absence of light; hence there is no light sensation and no color. Not in the same sense, for white is not a color but a combination of colors or color effects. 5. In sound waves a similar difference would produce a difference in pitch. 6. It depends upon the kind of light with which the object is illumi- nated and how the object treats the light it receives; that is, what part it transmits, what it reflects and what it absorbs. MAGNETISM Page 314 1. For (a) and (6) see Fig. 322, page 309; for C see Fig. 26. 2. See Fig. 323. The iron armature serves best as the medium to com- plete the magnetic cir- cuit, so that the poles of the magnet cannot be neutralized or reversed by another magnetic in- N fluence from without. FIG. 26 3. Magnetic induction always produces an opposite pole next to the inducing pole, and opposite poles attract. 4. The dipping needle shows more nearly the true direction, provided its axis is at right angles to the magnetic meridian. If not at right angles, the dip indicated is greater than the true dip. 5. Fig. 324, page 309. They blend when the poles are unlike. 8X24 16 6. f a\-> = "o~=5^. The poles repel with a force of 5^ dynes. (6) 2 ELECTROSTATICS Pages 329, 330 1. Positive electrification is produced on the glass when it is rubbed with silk. It is produced whenever any two different substances are CURRENT ELECTRICITY 41 rubbed together, but this is the most convenient way of identifying it. The name positive is merely intended to contrast it with the other kind. The names might have been reversed. 2. Repulsion is more reliable, for a body electrified either way will attract an unelectrified body, but each will repel only the kind like itself. 3. Let us call the electrified bodies A and B. The suspended ball is first attracted by the nearer body, say A, and touches it. The ball shares the charge of A, is repelled by A and attracted by B until it touches B, when its charge is reversed, and now it is repelled by B and attracted by A, etc. 4. The hand and body being relatively good conductors remove the electrification as fast as it is produced. Support the rod on glass, rubber or silk. 6. See sec. 236, page 316. 6. The positively electrified glass rod will attract the negative charge in the electroscope and the leaves will come together, provided the glass rod is not brought so near as to produce an induced charge. 7. A further divergence of the leaves shows that the charge in the electroscope was repelled by that on the sulphur. From this it is evident that the sulphur was electrified negatively, the same as the electroscope. 8. The charge on the outside is distant from the inside charge by just the thickness of the glass walls of the jar. Since the attraction varies inversely as the square of the distance between the charges, neither charge can hold the other completely unless they are at the same place. A part of each charge on the jar is consequently not bound or held by the opposite charge. This excess can be removed by alternately touching the sides, this always leaving an excess on the side not touched last. CURRENT ELECTRICITY Pages 338, 339 1. A containing vessel, a fluid (semifluid or paste) and two con- ductors, the latter being placed in contact with the fluid but not with each other. The fluid must act chemically upon only one of these plates, or upon one more than upon the other. 2. See page 332, line 17. Note that the term pole has a very different meaning here from that which it has in connection with magnets. 3. (a) From the zinc through the liquid to the carbon. (6) From the carbon through the connecting wire to the zinc. Both are the direc- tions of the positive discharge. 4. The carbon is said to have the higher and the zinc the lower po- tential. 5. See page 332, line 21. 6. See page 335, sec. 254. 7. See page 334, sec. 253. 42 MANUAL FOR MUMPER'S PHYSICS 8. See page 333, lines 7-21. If a circuit is left open when not in use it is said to be an open circuit system, but if kept closed when not in use it is a closed circuit system. 9. See page 335, sec. 255, and page 361, sec. 275. 10. Uniform materials and no polarization. High E. M. F. and low resistance. CURRENT ELECTRICITY Pages 346, 347 1. A charge refers to an electrical condition which for the time is fixed, while a current conveys the idea of a continuous discharging or flow- ing from one place to another. 2. By sending the charge through a coil of well insulated wire which surrounds a piece of steel. 3. See page 308, line 8, etc. The magnetic field can be shown by means of a magnetic compass placed in different positions with reference to the conductor. 4. (a) Toward the west. (6) Upward, (c) Toward the east, (d) Down- ward. 6. The north end of the needle is deflected toward the west. 6. The current flows toward the south. 7. Clockwise as you look up the pole. 8. If steel is used the magnet made is permanent, though it may vary in strength with a variation of the current. 9. From the observer. Consult Fig. 362, page 342, where the reverse is shown. 10. The direction of the lines of force in the magnetic field, produced by the second and additional turns, agrees with that of the first turn, hence the total intensity is greater when the number of turns is increased. CURRENT ELECTRICITY Page 358 1. See page 354, line 8, etc. 2. The metal appears at the cathode. 3. The one, the cathode, gains as much as the other loses. 4. Within the cell the current is said to flow toward the plate, gaining in weight, the cathode. 6. Because there is a break in the circuit at the push button, which can be closed only by pushing the button. 6. Because the vibrating clapper alternately opens and closes the circuit at the bell, thus giving a succession of taps. 7. If air were present the carbon would burn. 8. Platinum is the only metal which has the same coefficient of ex- pansion as that of glass. 9. Because the storage cell is itself " charged " by means of a dynamo, CURRENT ELECTRICITY 43 and because of the necessary loss of energy in the transformation, the cell cannot furnish quite so much energy as the dynamo puts into it. Then in addition to that the cost of the cell itself must be included. 10. They attract each other because they are parallel and flow in the same direction around the coil. CURRENT ELECTRICITY Page 363 1. 6.5 -4-1000=. 0065 ohm per ft. 88 X. 0065 =.572 ohm, resistance of 88 ft. 2. Since iron is a substance with less conductance (more resistance), it must have a larger cross section. It must be 6.63 -f- 1.08= 6.1 -f- times as large. 3. They must be connected end to end. They must be connected side by side, that is, each wire must have its two ends connected to the same bodies as those to which the others are connected. 4. Their joint resistance is 3+6-j-9= 18 ohms. Their joint conductance is 4+i-J-$=H ohms. 5. The resistance varies directly as the length and inversely as the cross section. Resistance= 1250 -=-106.3-^4= 2.93+ ohms. 6. It decreases the length and hence the resistance of the conducting liquid between the plates, hence the current becomes stronger. 7. Lifting the plates when they are at a constant distance decreases the area of the cross section of the conducting liquid, hence increases the resistance in the circuit and the current is weakened. CURRENT ELECTRICITY Page 368 1. A water current is due to a difference between the pressures at the two points. An electric current is due to a difference between the potentials of the two points. There may be no current in either case if the medium between the two points is not a sufficiently good conductor, that is, if it offers too much resistance to the flow. 2. The so-called current strength is determined by the quantity of electricity transferred per unit of time and not the energy of this elec- tricity, just as the strength of a water current depends upon the quantity of water flowing and not upon its energy. 3. The unit current strength is defined in terms of the chemical effect produced. See page 359, sec. 274. 4. The current strength increases with the increase of the conduct- ance (decrease of resistance) of the conducting body and vice versa. 44 MANUAL FOR MUMPER'S PHYSICS 5. Page 362, sec. 276. The substance will have -3- of a unit con- ductance. 6. See page 365, sec. 278. See page 367, sec. 281. 7. It would furnish more than 1 ampere of current. 8. It would produce more than 1 ampere through 1 ohm resistance. The power would likewise be more than 1 watt. 9. See page 367, sec. 281. 10. See page 129, sec. 76.' 11. (a) Po wer = volts X amp. = 60X10 = 600 watts. (6) 600 watts would furnish 600 candle power. (c) 600 watts =^=.6 kilo watt. .6 kilowatt, at 6 cents per kilowatt hour, give .6X6 cents = 3.6 cents per hour. 12. H=.24C2Rt. gm. cal.= .24X (12)*X 1.5X300= 15552. CURRENT ELECTRICITY Page 375 1. See text. volts 6 3. Volts= amp. Xohm= 10X20= 120 volts. volts 110 __ , 4. Ohms = = = 220 ohms. 5. Volts=amp.Xohms=12X.8=9.6 volts. 6. 1 amp. sets free .001118 gm. per sec. 10 min=600 sec. 1 amp. in 600 sec. sets free 600X. 001118 gm.= .6708 gm. of silver. 6 amp. set free 6 X. 6708 =4.0248 gm. of silver in 10 min. volts= amp. Xohms= 6X24 =144 volts. 7. Resistance =50 ohms. Amp.Xohms=volts=.2X50=10 volts. To produce 10 volts it will require f$=6$; seven cells will be required. Because external resistance is large the cells are arranged in series. 8. 6X2=12 volts, E. M. F. 6X1.5 ohms =9 ohms, battery resistance. 9. When abreast the total E. M. F. is the same as that of a single cell, in this case 2 volts. The resistance is that of a single cell. of 1.5 ohms= .25 ohms. 10. See page 370. (1) For external resistance of 90 ohms. CURRENT ELECTRICITY 45 Series, Am.= - . ^= = - 121 am P- ex. ohms 4-inter . ohms 90 + (6 X 1 .5) 99 The series current is nearly 6 times as strong as the parallel. (2) For external resistance of .2 ohms. 1O Paralle1 ' .2+KT5) ==4 - 4 amp ' The parallel arrangement gives nearly 4 times the current. 11. Power in watts = amp. X volts. See page 367. Volts = amp . X ohms, hence watts = amp . X amp . X ohms . Power required for each lamp= (.4)2X60=. 16X60= 9.6 watts. ^=520 lamps. y.o THIS BOOK IS DUE ON THE LAST DATE STAMPED BELOW AN INITIAL FINE OF 25 CENTS WILL BE ASSESSED FOR FAILURE TO RETURN THIS BOOK ON THE DATE DUE. THE PENALTY WILL INCREASE TO SO CENTS ON THE FOURTH DAY AND TO $1.OO ON THE SEVENTH DAY OVERDUE. LD 21-95*n-7,'37 Gaylord Bros. Makers Tt> 1/7001 Syracuse N.Y. T Y\Aj^vv>Jkx-fl->v-~ U)WER DIVISION UNIVERSITY OF CALIFORNIA LIBRARY