Q 
 
 TO THE 
 
 K 
 
 PROBLEMS 
 
 IN I* 
 
 SCHOOL 
 PHYSICS 
 

 If 
 
 1 J - 
 
 I MEMORIAL 
 Henry Senger 
 
 
KEY TO THE PROBLEMS 
 
 IN 
 
 AVERTS SCHOOL PHYSICS 
 
 NEW YORK AND CHICAGO 
 
 SHELDON AND COMPANY 
 
M 6 
 
 COPYRIGHT, 1896, 
 BY SHELDON AND COMPANY, 
 
 INMEMORIAM 
 
 TYPOGRAPHY BY J. S. CCSHING & Co., NORWOOD, MASS. 
 
TO THE TEACHER. 
 
 IF you find any error in any of these solutions, you will 
 confer a favor on the publishers and the author by carefully 
 writing out the correction and sending it to Dr. Elroy M. 
 A very, No. 657 Woodland Hills Avenue, Cleveland, Ohio. 
 
 926494 
 
KEY 
 
 TO THE 
 
 PROBLEMS IN AVEKY'S SCHOOL PHYSICS. 
 
 CHAPTER I. 
 
 Page 2O. 
 
 8. See Exercise 11, page 95. 
 
 9. The number of inches in a meter is given on page 18 ; 
 39.37 inches. The number of millimeters in an inch is the 
 reciprocal of the value of a millimeter as given on the same 
 page ; 1 -=- 0.03937 = 25.39. 
 
 10. Four times 250 cu. cm. equals 1 liter. 
 
 11. See 21. 
 
 1.0567 qts. = 1,000 cu. cm. 
 
 1 qt. = 1,000 cu. cm. -=- 1.0567 = 946.34 cu. cm. 
 
 Page 23. 
 
 1. A liter = 1 cu. dm. = 1,000 cu. cm. One cu. cm. of pure 
 water weighs 1 gram ; 1,000 cu. cm. of water weigh 1,000 
 grams, or 1 Kg. 
 
 2. 1,000 g. x 1.8 = 1,800 g. 
 
 3. 1,250 cu. cm. of water weigh l,250g. 
 1,250 g. x 0.8 = 1,000 g. or 1 Kg. 
 
 4. Since a liter of water weighs 1,000 g., 250 g. of water is 
 i of a liter of water. 
 
 5 
 
6 KEY TO THE PROBLEMS 
 
 5. 1 cu. dm. = 1,000 cu. cm. 1 cu. dm. of water, therefore, 
 'weighs. 1,000 g. .- 1 Kg. 
 
 6. 1 liter = 1,000 cu. cm. ; 1 dl. = 100 cu. cm. 1 dl. of water 
 weighs 100 times 1 g. = 100 g. = 1 Hg. 
 
 7. 18 cu. in. x 19 x 20 = 6,840 cu. in. 
 6,840 -r- 231 = 29.6+, the number of gals. 
 
 8. 25 cu. cm. x 35 x 75 = 65,625 cu. cm. 
 65,625 -T- 1,000 = 65.625, the number of liters. 
 
 NOTE. Exercises 7 and 8 are intended to illustrate the advantage of 
 the decimal scale as used in the metric measures. It is easier to divide 
 by 1,000 than to divide by 231. 
 
 Page 37. 
 
 11. Be sure that the pupil understands that he is testing 
 the stiffness and not the strength of the beams used, (a) The 
 deflection is proportional to the load. (&) The deflection is 
 proportional to the cube of the length, (c) The deflection 
 is inversely proportional to the breadth, (d) The deflection is 
 inversely proportional to the cube of the thickness, i.e., depth. 
 
 13. The elongation is proportional to the length of the wire, 
 and inversely proportional to the cross-section of the wire, i.e., 
 to the square of the diameter. 
 
 14. The angle of torsion is proportional to the force of tor- 
 sion. 
 
 15. The angle of torsion is proportional to the length of the 
 rod. 
 
 Page 42. 
 
 4. Air held in solution is largely expelled by boiling. Fewer 
 air bubbles will, therefore, accumulate on the inner wall of the 
 tumbler that contains the boiled water. See the next exercise. 
 
IN AVERY'S SCHOOL PHYSICS. 7 
 
 Page 54. 
 
 6. Porosity. 
 
 7. No. Scientifically speaking, pores are the minute inter- 
 stices between molecules ; intermolecular spaces. 
 
 8. Fluids include liquids. Both terms imply freedom of 
 molecular motion, but the tendency of the molecules to cohere 
 pertains necessarily to liquids, and not necessarily to fluids. 
 
 9. Water and steam are identical substances, varying only 
 in respect to their physical condition ( 41). The nature of 
 a substance is determined by the nature of its molecules ( 6) 
 and a physical change, as from water to steam, does not change 
 the molecule ( 9). As there is no chemical change in passing 
 from water to steam, no molecule is changed in any way ; they 
 remain equal in size. 
 
 10. A cubic inch of water makes about a cubic foot of steam. 
 As there is no increase in the size or the number of the mole- 
 cules, the increased volume must be due to an increase in the 
 distances between them. 
 
 12. See 45. 
 
 13. See Experiment 36. 
 
 14. See 33 (a). Extension, impenetrability, weight, inde- 
 structibility, inertia, etc., pertain to all matter and are, there- 
 fore, universal properties. Hardness, malleability, ductility, 
 etc., pertain to only certain forms of matter and are, therefore, 
 characteristic of those that possess them. 
 
 15. See 52 (6). Dialysis and evaporation. 
 
 16. Divide the number of pounds by 2.2046. See" 25. 
 
 18. n cc 
 
 I 
 
 19. See 33 (a). Tenacity is proportional to sectional area, 
 and that is proportional to the square of the diameter. 
 
 (14.19) 2 : (17.90) 2 : : 15 : x. 
 
8 KEY TO THE PROBLEMS 
 
 See Appendix 4. The squares of the diameters in mils may 
 be found in the column headed circular mils, and the pro- 
 portion written directly from the table, thus : 
 
 201.5 : 320.4 : : 15 : x. 
 
 20. From the table of wire dimensions, it appears that No. 27 
 wire has a sectional area twice that of No. 30 wire. Conse- 
 quently, No. 27 iron wire has a breaking strength of 10 pounds. 
 Comparing these brass and iron wires of the same size (No. 27), 
 it appears that the tenacity of spring-brass is 1.5 times that of 
 annealed iron. 
 
 Page 55. 
 
 1. Steam passes through the cardboard and condenses upon 
 the inner surface of the upper tumbler, thus illustrating the 
 porosity of the cardboard. 
 
IN AVERY'S SCHOOL PHYSICS. 9 
 
 CHAPTER II. 
 
 Page 63. 
 
 1. 500x500 = 250,000. 
 
 2. 321.6 x 200 = 64,320. 
 
 3. | X 9 ~ 2() | Their momenta are equal. 
 
 4. Uniformly accelerated motion, like that of a falling body 
 which is acted upon by the attraction of the earth. 
 
 5. There are 5,280 feet in a mile. 
 
 5,280 x 15 x 1 _ g 
 
 1,320 x 12 
 The momentum of the ball will be 5 times that of the stone. 
 
 ( 50,000 x 2 = 100,000 ) 
 , 6 ' {loioOOx 10 = 100,000} Their momenta are equal. 
 
 7. 25 x 60 = 1,500, the momentum of the first, and, conse- 
 quently, of the second. 1,500 -*- 40 = 37.5. The velocity of 
 the second is 37.5 ft. per second. 
 
 8. 100x20 = 2,000. 
 
 2,000 -r- 500 = 4, the number of kilograms. 
 
 9. From a point, (7, draw three lines 3^ inches, 5 inches, 
 and 6 inches respectively in length. Draw the first line 
 toward the right hand; the second line downward; and the 
 third line downward and toward the left so as to make an 
 angle of 45 with the second line. Eemember the old 
 geographical rule: "The top of the map is north; the bottom, 
 south; the right hand, east; and the left hand, west." The 
 three arrowheads point from C. 
 
10 KEY TO THE PROBLEMS 
 
 10. 4 ^ x 5? ff = 58|, the velocity in feet per second. 
 bO X t>0 
 
 ^ 92,390,000 x 2 
 (16 x 60) + 36* 
 
 12. See Formula (3) on page 59. 
 
 J= Jo* 8 = 10 ft. x6 2 = 360ft., 
 the distance traversed in 6 seconds. 
 See Formula (2) on page 58. 
 
 v = at = 20 ft. x 6 = 120 ft., the final velocity. 
 
 13. The load of 50 Kg. = 2.2046x50 -110.23 Ibs. (see 25). 
 This load is 7.35 times as great as the load that can be carried 
 by the No. 27 wire. We must, therefore, have ajwire that is 
 7.35 times as great in sectional area. From the table of wire 
 dimensions, we find that the diameter of No. 27 wire is 0.01419 
 of an inch. 1 : 7.35 : : (0.01419) 2 : ^ 
 
 x = 0.03847 of an inch, the diameter required. This is 
 greater than the diameter of No. 19 wire and less than that of 
 No. 18 wire. An easier solution for practical purposes is to 
 multiply 201.5, the number of circular mils for No. 27 wire, 
 by 7.35, the ratio between the given loads. This gives 1,481 
 circular mils, which is greater than that for No. 19 wire and 
 less than that for No. 18 wire. No. 18 wire must be used. 
 
 Page 64. 
 
 1. Draw two diameters at right angles, thus dividing the 
 circumference into arcs of 90 each. From the extremities of 
 these diameters as centers, draw arcs cutting the circumference 
 into arcs of 30 each. Use the protractor only for the trisec- 
 tion of the 30 arcs. 
 
 Page 68. 
 
 1. The locomotive must travel 360 ft. + 120 ft. = 480 ft. 
 It moves ^ mile or 2,640 ft. per minute, or 44 ft. per second. 
 480 -r- 44 = 11 nearly, the number of seconds. 
 
IN AVERY'S SCHOOL PHYSICS. 11 
 
 2. See 66 (a). 32.16 x 25 = 804, the number of poundals. 
 
 3. 980 x 5,000 = 4,900,000, the number of dynes. 
 
 4. 13,825 dynes. Multiply the number of grams in a 
 pound by the number of centimeters' in a foot. 
 
 5 64 x 16 x 1,300 __ 1 109 i The momentum of the cannon- 
 
 1 x 1,200 
 ball is 1,1091 times that of the bullet. 
 
 6. See solution to Laboratory Exercise 11, page 37. The 
 deflection is proportional to the cube of the length. 
 
 3 3 : 4 3 : : 0.5 : x. 
 
 7. Inertia. 
 
 Page 79. 
 
 1. Adopt any convenient scale, as 0.1 of an inch to the 
 pound. Then the 100-pound force will be represented by a 
 line 10 in. long, and the other force by a line 15 in. long. See 
 69 (1). The resultant will be represented by a line 25 in. 
 long. 
 
 2. See 69 (2). Using the same scale as in Exercise 1, 
 the resultant will be represented by a line 5 in. long, and the 
 motion will be in the direction of the greater force. 
 
 3. Suppose we adopt the scale of an inch to the mile. 
 Draw a horizontal line 4 inches long to represent the force 
 of the oars. From one end of this line, draw a vertical line 
 3 inches long to represent the force of the current. Join the 
 free- ends of these lines; the hypotenuse thus formed will 
 represent the resultant of these two forces. 
 
 3 2 + 4 2 = 25. V25 = 5. 
 
 See 70 (a). The boat will move in the direction indicated 
 by the hypotenuse and with a velocity of 5 miles per hour. 
 Of course, the problem means that the boat is headed directly 
 across the stream. 
 
 4. Draw a vertical line, 64 units (as mm., or 16ths of an 
 inch) long. From the foot of this line draw a horizontal line 
 
12 
 
 KEY TO THE PROBLEMS 
 
 24 units long. (Use the same kind of units that you adopted 
 for the vertical line.) Join the free ends of these lines. The 
 hypotenuse will be the graphic representation of the re- 
 sultant. 
 
 64 2 + 24 2 = 4,672. V4,672 = 68 + . 
 
 5. Draw, as before, a vertical line (3 x 20 = ) 60 units long, 
 and a horizontal line (12 x 20 = ) 240 units long. Draw the 
 hypotenuse. 
 
 60 2 + 240 2 = 61,200. V6pOO = 247 + . 
 
 6. Draw BN=10 units of length. From B, draw BE at 
 right angles to BN and make it 15 units long. Complete the 
 
 parallelogram and draw the partial resultant, Br. From B, 
 draw BS at an angle of 45 from BE, and make it 25 units 
 long. Complete the parallelogram, and draw the diagonal, 
 BM, which will represent the complete resultant. The scale 
 here adopted is 2 mm. per pound. The line, BR, being 67 mm. 
 long, represents a force of 33|- pounds. The direction of the 
 resultant is a little to the north of west, as is indicated by 
 the angle, EBR. 
 
IN AVERY'S SCHOOL PHYSICS. 13 
 
 7. See 60 (3). The momentum of the gun is equal to 
 the momentum of the projectile. As the gun is heavier than 
 the projectile, its velocity must be less than that of the pro- 
 jectile, in order that the products of the numbers representing 
 the weight and velocity in each case may be equal. 
 
 8. The width of the river is represented by a line 4 units 
 long ; the actual course of the boat by a line 5 units long. If 
 the 4 units represent 1 mile, the 5 units will represent 1^ miles, 
 the distance that the boat moves. It takes no longer. See 62. 
 
 9. See Fig. 94, in which LM represents the plank, and MN 
 the distance that one end of it is raised. WC represents the 
 weight of the cask. From T7, draw WD perpendicular to LM. 
 From (7, draw CD parallel to LM. Complete the parallelogram, 
 WBCD. The force of gravity represented by WC may be 
 resolved into two components, represented by WD and WB. 
 WB represents the force with which the cask tends to roll 
 down the plank. This tendency may be successfully resisted 
 by a force represented by WB', equal to WB and opposite in 
 direction. WB = ^ WC, as may be seen by direct measure- 
 ment or as may be proved geometrically, the triangles WBC 
 and LNM being similar. Hence, the muscular force needed is 
 25 Ibs. or more. 
 
 10. See 66 (a). 32.16 x 60 = 1,929.6, the number of 
 poundals. 
 
 11. 60 Kg. = 60,000 g. 
 
 980 x 60,000 = 58,800,000, the number of dynes. 
 
 12. See 65, and 66 (a). 
 
 32.16 x 10 = 321.6, the number of poundals. 
 
 13. 1,000 -f- 20 = 50, the number of grams. 
 
 14. 12 x 6 = 72, the number of dynes. 
 
 15. 490 -j- 70 = 7, the number of centimeters. 
 
 16. The wind strikes obliquely upon the faces of the blades 
 and is resolved by the blades into two components, as it is in 
 
14 KEY TO THE PROBLEMS 
 
 the case of the obliquely set sails of vessels. One of these 
 components tends to move the blade in a direction perpendic- 
 ular to that of the wind. The corresponding blade on the 
 opposite side of the wheel is similarly urged in the opposite 
 direction, so that the force of the wind upon each such pair of 
 blades constitutes a couple that helps to turn the wheel about 
 its axis. 
 
 17. See 78. 
 
 r 
 
 18. Compute the length of the hypotenuse of an imaginary 
 right-angled triangle with sides of 5 and 12 respectively. See 
 70 (a). 52 + 122 = 0. 169 = a-2. 13 = 3.. 
 
 The resultant has a force of 13 Kg. 
 
 19. In this case, the hypotenuse of the right-angled triangle 
 is 30, and one of the other sides is 18. 
 
 30 2 - 18 2 = x\ 
 x 24, the numerical value of the component sought. 
 
 20. See Experiment 45, and 69 (3). 
 
 8 + 11 = 19. 
 19: 8::57:24. 
 19 : 11 : : 57 : 33. 
 
 The point will be 24 inches from the line of the 11-pound force, 
 and 33 inches from the line of the 24-pound force. 
 
 21. See 58. The southward and the eastward motions 
 are equal. The parallelogram is a square with a diagonal of 
 30 units. 
 
 V450 = 21+, the number of miles. 
 22. See 78. 
 
 G.F. = ^10,000 x (2,000)* = 40,000,000, the nmnbe r of 
 
 r 1,000 
 
 dynes. 
 
IN AVERY'S SCHOOL PHYSICS. 15 
 
 23. (a) See 60 (3). The action on the sail will be balanced 
 by the reaction on the bellows and the boat will not be moved 
 by the operation. (6) The reaction on the bellows will push 
 it and the boat ahead. 
 
 Page 81. 
 
 1. See 69 (3). 
 
 2. From the ends of a vertical line 5 inches long, draw 
 toward the right hand horizontal lines 3^- and 5^ inches long 
 respectively. An arrowhead on each line should point to the 
 right. From a point on a vertical line, T 7 of its length from 
 the 5|-inch line and -j-J- of its length from the 3 J -inch line, 
 draw a horizontal line to the left and 9 inches long. This line 
 with an arrowhead pointing to the right represents the result- 
 ant. The same line with an arrowhead pointing to the left 
 represents the equilibrant. 
 
 3. The line drawn downward from with a length equal 
 to OD represents the gravity of the unknown weight. 01) 
 represents the equilibrant of that weight. 
 
 4. ZT represents the equilibrant of the 10-pound pull of 
 the weight. The scale adopted is -J- inch = 1 pound. See 70. 
 
 7. Draw an equilateral triangle as follows : lay off a line 
 AB, of convenient length. From A as the center and with a 
 radius equal to AB, describe an arc. From B as a center and 
 with the same radius, describe a second arc that cuts the first 
 arc. Mark the point of intersection of the two arcs, C, and 
 join AC and BC. 
 
 8. The angle measures 127 3'+. 
 
 Page 9O. 
 
 1. See 81. 8 ' 250 x 176 = n the number of H. P. 
 
 550 x 60 x 4 
 
 SUGGESTION : Reduce by cancellation. 
 
 2 . gee 85. 192 ' 96 x 10 ' 000 = 30,000. 
 
 64.32 
 
16 KEY TO' THE PROBLEMS 
 
 3. The direction makes no difference. 
 50 x 19.6 x 19.6 
 
 = 50 x 19.6 = 980, the number of kilo- 
 A x y.o 
 
 gram-meters. 
 
 4. 50 x 750 = the momenta. 
 
 Kf) v 7KA 
 
 * = 500, the required velocity in feet. 
 
 75 
 
 5. The velocity is 500 m. per minute, or 8^ m. per second. 
 40 x 81 x 81 = 141 |^ the num ber of kilogram-meters. 
 
 X t/.o 
 
 6. 1?50 X 2 ' 876 = 36, the number of H. P. 
 550 x 60 x 3 
 
 7. That quantity of water weighs about (62.5 Ibs. x 300 =) 
 18,750 pounds. 
 
 = 75, the number of H. P. necessary. 
 
 oo 
 00,000 
 
 8. A velocity of 20 miles per hour is one of 29J feet per 
 second. 
 
 K.E. = 1QO X 29 ^ X 29 ^ = the number of foot-pounds. 
 u4.o^ 
 
 9. (a) 6,000 x 50 = 300,000, the number of foot-pounds. 
 (6) 300,000 -- 16,500 = the number of H. P. 
 
 10. 2 = 10 > QQQ X 1QQ . ... m = 15 5 the num ber of minutes. 
 
 33,000m 
 
 11. 2= 10 > 000 ^. a? = 3.3, the number of feet. 
 
 o50 x 30 
 
 12. It is often convenient to remember that 550 foot-pounds 
 per second is equivalent to 33,000 foot-pounds per minute. 
 1,650,000 -*- 33,000 = 50, the number of H. P. 
 
IN AVERY'S SCHOOL PHYSICS. 17 
 
 14. K. E. = 2 * X f Q = 1,250, the number of foot-pounds 
 
 o4.oJ 
 
 that the moving sphere can perform. This working power is 
 the exact measure of the work performed upon it. 
 
 15. A resistance of 8 pounds per ton signifies that to move 
 a ton one foot on the rails involves as much work as to lift 8 
 pounds one foot high, or 8 foot-pounds. To move 10 tons 50 
 ft. on the rails would require 4,000 foot-pounds. The addi- 
 tional work done in giving kinetic energy to the car will be 
 measured by the kinetic energy of the car. The velocity of 
 the car is 4.4 ft. per second. 
 
 KE-= 20,000^ 44 x 4.4 = 6>019+> 
 
 the number of foot-pounds that the car can perform, or the 
 amount of work done in giving the car the velocity specified. 
 4,000 + 6,019 = 10,019, the total amount of work, in foot- 
 pounds, that was done on the car. 
 
 16. The triangle or parallelogram will not "close." None 
 of the given forces can be the equilibrant of the other two. 
 
 17. See 46 (a). 
 
 18. The velocity along the track is 29^ ft. per second. 
 The problem is to find the hypotenuse of a right-angled 
 triangle, the sides of which measure respectively 29J and 20 
 ft. This magnitude of 35.5 ft. represents the velocity per 
 second. 
 
 19. 15 x 20 -f- 4 = 75, the number of dynes. 
 
 20. The momenta of gun and projectile = 250 x 1,420 = 
 355,000. This divided by 34,000, the weight of the gun, gives 
 14.79 as the required velocity in feet. 
 
 21. 6 x 3.14159 -r- 5 = 3.769, the velocity in feet per second. 
 
 The C. F. = 2Q x ( 3 - 769 ) 2 . gee 78. 
 3 
 
18 KEY TO THE PROBLEMS 
 
 22. 50 x 10 = 500, the number of foot-pounds. 
 
 23. The force is resolved into two components, one parallel 
 to the track and producing motion ; the other perpendicular to 
 the track and producing no motion. The first component is 
 the only one that overcomes resistance or does work. See 79. 
 In the parallelogram constructed for the resolution of the 
 force, we have 50 as the diagonal of a square, the length of 
 one side of which is to be determined. 
 
 KA2 
 
 = 35.355, the force in pounds, 
 
 35.355 x 10 = 353.55, the work in poundals. 
 
 24. His push contributes nothing to the motion of the car, 
 and so it does no work on the car. See 79. 
 
 25. The man does not push the car or help to push it ; the 
 car pushes the man. The man does no work upon the car; 
 the car does work upon the man. The exercise is the converse 
 of Exercise 23. 
 
 26. (a) See 85. 50 x 30 - foot-pounds. 
 
 32.16 x 2 
 
 (6) See 86. 50 * 3 2 foot-poundals. 
 
 1,056,000 = 1 
 ' 550 x 60 x 60 x 8 15* 
 
 Page 92. 
 
 1. It is better to take a tenth of the number of revolutions 
 made by the spindle while the large wheel revolves 10 times. 
 Eecord this ratio on the whirling table. 
 
 2. The greater the speed, the greater the centrifugal force 
 of the ball, and the greater the length to which the cord is 
 stretched. 
 
 3. The mercury gets as far as possible from the center of 
 rotation, and thus rises to the upper ends of the tubes. 
 
IN AVERY'S SCHOOL PHYSICS. 
 
 19 
 
 4. In the geologic-long-ago, when the earth was in a plastic 
 condition, its rotation upon its axis would give it the spheroidal 
 form that it now has ; it would flatten out at the poles. 
 
 7. See 69 (2). The components are equal and the result- 
 ant is zero. 
 
 8. This exercise shows that centrifugal force varies directly 
 as the mass and inversely as the radius, as is indicated by the 
 formula given in 78. 
 
 10. The momenta produced by the elastic force of the cord 
 are equal ; the lighter block will move twice as fast as the 
 other block. 
 
 11. 
 
 Page 98. 
 
 2. As the assumed distance from the earth's center is only 
 one-fourth of the surface distance, the required weight will be 
 only one-fourth of the surface weight. 550 -*- 4 = 137^. Or, 
 4,000 : 1,000 : : 550 : x. 
 
20 KEY TO THE PROBLEMS 
 
 NOTE. The teacher will be fortunate who finds that all of his pupils 
 are able to handle a proportion intelligently and easily. Secure this 
 ability if possible. Frequently vary the form of the statement from 
 
 a : b = C : d, to 2 = 1 
 
 = = 
 
 50 2 2,500 
 
 4. If the first and second were at equal distances from the 
 third, the first would have, on account of its lesser mass, only 
 -- or -| as great an attraction as the second. But being only 
 half as far distant, its attraction is four times as great as it 
 would be if it were at an equal distance. -f x 4 = -|. The 
 smaller ball exerts 2J as much force upon the third ball as 
 the larger one does. 
 
 50 2 8 (9:6) 
 
 5. Beckon distances from the earth's center. 
 
 12,000 2 : 4,000 2 : : 900 : x. 
 
 Or, the distance from the earth's center being increased three- 
 fold, the weight will be divided by 3 2 or 9. 
 900 Ibs. -s- 9 = 100 Ibs. 
 
 6. 1 : 16 : : 4,000 2 : or 8 . .-. a = 16,000, the number of miles 
 from the earth's center. 16,000 - 4,000 = 12,000, the number 
 of miles from the earth's surface. 
 
 7. 7,000 2 : 4,000 2 : : 200 : x. The man would weigh 65.3 Ibs. 
 7,000 2 : 4,000 2 : : 100 : y. The boy would weigh 32.65 Ibs. 
 
 8. 4,000 2 : 6,000 2 : : 180 : x. Ans. 80 Ibs. 
 
 9. Work as in preceding examples, or as follows: 
 
 50 Ibs. x if = 32 Ibs., the weight 1,000 miles above the 
 surface. 
 
 50 Ibs. x f = 371 Ibs., the weight 1,000 miles below the 
 surface. 
 
 It would weigh 5J Ibs. more when below the surface. 
 
IN AVERY'S SCHOOL PHYSICS. 21 
 
 10. It would be ^ as great in either case. 
 
 11. See solution to Exercise 23, page 92. The effective 
 component is a force of 565.685 pounds. 
 
 565.685 x 12 = 6,788.22, the number of foot-pounds. 
 
 12. 1,200 : 2,700 : : 4,000 2 : x 2 . .-. x = 6,000. 
 6,000 - 4,000 = 2,000. 
 
 13. It would increase the weight fourfold. See 90. 
 
 14. In such cases of mutual action between free bodies, the 
 two bodies experience equal changes of momentum. Just 
 before collision, the small ball had a momentum of 3,750 in 
 a certain direction. Just after collision, it had a momentum 
 of 1,250 in the opposite direction. One might be called posi- 
 tive and the other negative. The change of momentum of the 
 small ball was 5,000. This change was produced by the 
 reaction of the collision. The equivalent action must have 
 given the large ball a momentum of 5,000. This divided by 
 the weight of the large ball gives the velocity of the large ball. 
 5,000 -r- 200 = 25, the required velocity in feet per second. 
 
 Page 1O3. 
 
 1. See 97, 100. 
 
 2. See 97. 
 
 3. See 100 (6). 
 
 4. The center of mass is lower. See 100. 
 
 5. The distance from the center of the earth being multi- 
 plied by 60, the attraction will be divided by 60 2 . See 90. 
 
 32.16 ft. -s- 3,600 = 0.0089 + ft. 
 
 6. Draw two lines, respectively 10 and 7.5 cm. in length, 
 making them include an angle of 45 degrees. Complete the 
 parallelogram and draw the diagonal from an acute angle. 
 Its length of 12.147 cm. represents a value of 121.47 feet per 
 second. The direction is 25 53' to the east of north. 
 
22 KEY TO THE PROBLEMS 
 
 7. 5,280 ft. -j- 60 = 88 ft., the velocity per second. This 
 velocity was developed in 600 seconds. 88 -r- 600 = 0.1466, 
 the acceleration in feet. Force = mass x acceleration; / ma. 
 66 (6) and 86. 
 
 0.1466 x 100 = 14.66, the number of poundals. 
 
 8. See 81. 
 
 Or, 
 
 1 H. P. = 550 foot-pounds in 1 second. 
 100 H. P. = 55,000 foot-pounds in 1 second. 
 
 = 3,300,000 foot-pounds in 1 minute. 
 = 198,000,000 foot-pounds in 1 hour. 
 = 1,980,000,000 foot-pounds in 10 hours. 
 Each gallon requires (8 x 200 =) 1,600 foot-pounds. 
 1,980,000,000 -f- 1,600 = 1,237,500, the number of gallons. 
 
 9. Because the base is large, and the center of mass is low. 
 
 10. Because the base is larger. See 97. 
 
 11. One is at the center of the axle, and the other should be. 
 
 12. In the first case, the center of mass was so low that the 
 line of direction fell within the base ; in the other case, it was 
 so high that the line of direction fell without the base. 
 
 Page 1O4. 
 
 1. The center of mass. 
 
 2. No; it is about midway between the two sides of the 
 cardboard. 
 
 3. The three lines should intersect at a common point. 
 
 4. The difference represents the distance that the center of 
 mass must be raised to throw the line of direction outside the 
 base. The products represent the amount of work that must 
 be done in each case to overturn the box. Foot-pounds. 
 
 5. The red patch will circle around the black patch as the 
 latter moves through the air. 
 
IN AVERY'S SCHOOL PHYSICS. 
 
 23 
 
 Page 113. 
 
 4. l' = $ g(2 t-l)= 16.08 ft. x 5 - 80.4 ft. 
 
 5. I = $gt 2 = 16.08 ft. x 100 = 1,608 ft. 
 
 6. I = $ gt 2 = 16.08 ft. x i = 4.02 ft. 
 
 7. I = %gt 2 = 16.08 ft. x 2.25 = 36.18 ft. 
 
 8. 2,512ft. 
 
 9. I =%gt 2 -, 787.92 = 16.08 Z 2 ; 49 = **; 7 = t. 
 
 10. v=gt = 32.16 ft. x 7 = 225.12 ft. 
 
 11. v=gt = 32.16 ft. x 15.5 = 498.48 ft. 
 
 12. 7J oz. + 7 oz. + 1 oz. = 16 oz., the total weight to be 
 moved. To move this load we have the weight of the rider, a 
 force of 1 oz. This force can give to an ounce of matter a 
 velocity of (</=)32.16 ft. per second; the same force can give 
 to 16 oz. of matter only T ^ of this velocity. 
 
 32.16 ft. -s- 16 = 2.01 ft. 
 
 13. I = gt- ; 257.28 = 16.08 1 2 ; 4 = t. 
 
 The ball will reach the ground at the end of 4 seconds. 
 During that time it will move from the tower (60 ft. x 4=) 
 240 ft. See 62 and 111. 
 
 14. Suppose the sinker to be at A. Gravity pulls or pushes 
 it toward B. The current pushes it toward C. The resultant 
 
 of these tends to move it to D. The equilibrant of AD is AE, 
 which represents the tension or pull of the line. 
 
24 KEY TO THE PROBLEMS 
 
 15. v = gt; 80.4 = 32.16 1; 2.5 = t. 
 
 The body will rise for 2 seconds. At the time specified, it 
 will have been falling second, and will have a velocity of 
 16.08 ft. 
 
 16. Fl represents the distance that gravity will move the 
 body during the first second. Fa represents the velocity due 
 to the horizontal impelling force, or the distance that force 
 would move it in the first second. Aa represents the amount 
 of deviation from a horizontal plane, as a consequence of the 
 pull of gravity during the first second. It is equal to Fl. Fc 
 represents the horizontal distance the projectile will move in 3 
 seconds. It is 3 times the initial velocity. Dd represents the 
 total pull of gravity for four seconds from starting. 
 
 17. I = i gt 2 = 16.08 ft. x 16 = 257.28 ft. 
 
 This is the distance the body would have moved in the given 
 time if it had had no initial velocity. But it moved (357.28 
 257.28=) 100 feet further in the 4 seconds. The initial 
 velocity must, therefore, have been (100 -5- 4 =)25 feet. 
 
 18. During that time, gravity alone moved it 2,512.5 ft. The 
 additional force moved it (35 x 12 =) 437.5 ft. Together they 
 moved it (2,512.5 + 437.5 =) 2,950 ft. Its final velocity due to 
 gravity is (32.16 x 12.5 =)402 ft., to which we must add the 
 initial velocity, making a velocity of 437 ft. 
 
 19. (a) 3,216 -r- 32.16 = 100, the number of seconds. 
 
 (b) The end of the 4th second of the ascent corresponds to 
 the end of the 96th second of the descent. 
 
 v = gt = 32.16 ft. x 96 = 3,087.36 ft. 
 
 Do not forget that this result disregards the resistance of 
 the air; that it is true only for a freely falling (or rising) 
 body. 
 
 (c) The end of the 7th second of the ascent corresponds to 
 the end of the 93d second of the descent, 
 
 v = gt = 32.16 ft. x 93 = 2,990.88 ft. 
 
IN AVERY'S SCHOOL PHYSICS. 25 
 
 20. The ball has to fall from a height of 257.28 ft. Whether 
 the force of gravity draws it vertically downward as a freely 
 falling body, or acts with the projecting force to produce a 
 curved path, it will reach the ground in the same time. It 
 would fall this distance in what time ? 
 
 I = $ gt 2 ; 257.28 = 16.08 t 2 ; 16 = t 2 ; 4 = t. 
 If it would take 4 seconds to fall from the highest point 
 reached, it would require 4 seconds for it to rise to that 
 height; it would be in the air 8 seconds. During each of 
 these seconds it has a horizontal motion of 1,000 ft. 
 
 21. l=gt* = 16.08 ft. x 25 = 402 ft., the distance that the 
 force of gravity would move the body in 5 seconds. The force 
 with which it was thrown moves it 10 ft. each second, or 50 
 ft. during the five seconds. 
 
 402 ft. + 50 ft. = 452 ft. 
 
 22. v = gt = 32.16 ft. x 5 = 160.8 ft. 
 
 160.8 ft. + 10 ft. = 170.8 ft. 
 
 23. See Experiment 54. (a) The value of each space is 7 ft. 
 In 5 seconds it will pass over (t 2 = ) 25 such spaces, or 175 ft. 
 
 (b) Its final velocity will be (2 1 = ) 10 times 7 ft. = 70 ft. 
 
 24. (a) See 107 (c). The distance passed over in the first 
 second will be half the velocity acquired during the first sec- 
 ond. Hence, the value of the spaces traversed in this case is 
 10 ft. In 10 seconds it will pass over (t 2 =) 100 such spaces, 
 or 1,000 ft. Or, we may say that the increment of velocity (g) 
 under these circumstances is 20 ft., instead of 32.16 ft. Then, 
 
 I = i gt 2 = 10 ft. x 100 = 1,000 ft. 
 
 (b) See 107 (a). The ratio between the height and length 
 of the plane is the same as that between 20 ft. and 32.16 ft. 
 20 : 32.16 : : 800 ft. : 1,286.40 ft. 
 
 25. (a) It would take just as long to rise 1,302.48 ft. as it 
 would to fall that distance. Then, 
 
 l = tgP; 1,302.48 = 16.08 1 2 j 81 = t 2 ; 9 = t. 
 
26 KEY TO THE PROBLEMS 
 
 (&) The initial velocity of a body that can rise for 9 seconds, 
 is the same as the final velocity of a body that has fallen for 9 
 seconds. Then, 
 
 v = gt = 32.16 ft. x 9 = 289.44 ft. 
 
 26. During 7 seconds, gravity would give it a velocity of 
 (gt = 32.16 ft. x 7 =) 225.12 ft. But as its velocity is 
 235.12 ft., it must have had an initial velocity of (235.12 
 225.12 = ) 10 ft. During 4 seconds, gravity would move it 
 (gt 2 = 16.08 ft. x 16 =) 257.28 ft. But during each of these 
 4 seconds, the force of the throw moves it 10 ft. more. This 
 amounts to 40 additional feet during the 4 seconds. 
 
 257.28 ft. + 40 ft. = 297.28 ft. 
 
 27. I = i gt- ; 787.92 = 16.08 t* ; 49 = P ; 7 = t. 
 
 This is the time that the second body was in falling 787.92 
 ft. But the first body fell 3 seconds more, or 10 seconds. 
 During 10 seconds it would fall 16.08 ft. x 100 = 1,608 ft. 
 
 28. During 4 seconds it will fall 257.28 ft. During 6 sec- 
 onds it will fall 578.88 ft. During the 5th and 6th seconds 
 it will fall 578.88 ft. - 257.28 ft. = 321.6 ft. Or we may 
 say that its average velocity during these two seconds will be 
 that attained at the end of the 5th second, which is 160.8 ft. 
 Moving at this average rate for 2 seconds, it will move 
 (160.8 ft. x 2 =) 321.6 ft. 
 
 29. Inversely ; as the time increases, the velocity decreases. 
 
 30. When they have been extended to an infinite distance. 
 
 Page 117. 
 
 4. For the sake of illustration, suppose that the origin of 
 co-ordinates is taken at a corner of one of the squares near the 
 upper left-hand corner of the sheet of cross-section paper. 
 Let each division on the axes represent 10 ft. ; then the suc- 
 cessive points will fall on successive vertical ruled lines. In 
 
IN AVERY'S SCHOOL PHYSICS. 
 
 27 
 
 the first half second, the ball will move 10 ft. toward the right 
 and 4.02 ft. downward. The corresponding point will be one 
 space to the right of the axis of ordinates and about 0.4 of a 
 space below the axis of abscissas. The co-ordinates for the 
 successive periods will be as follows : 
 
 SECONDS. 
 
 ABSCISSAS. 
 
 OKDINATES. 
 
 SECONDS. 
 
 ABSCISSAS. 
 
 ORDINATES. 
 
 0.5 
 
 1 
 
 -4.02 
 
 3.0 
 
 6 
 
 -144.72 
 
 1.0 
 
 2 
 
 -16.08 
 
 3.5 
 
 7 
 
 -196.98 
 
 1.5 
 
 3 
 
 -36.18 
 
 4.0 
 
 8 
 
 -257.28 
 
 2.0 
 
 4 
 
 -64.32 
 
 4.5 
 
 9 
 
 -325.62 
 
 2.5 
 
 5 
 
 -100.50 
 
 5.0 
 
 10 
 
 -402.00 
 
 The mapping of the curve will require 11 or 12 spaces to the 
 right of the axis of ordinates, and 49 or 50 spaces below the 
 axis of abscissas. The same scale being used for both ordi- 
 nates, the curve will not be distorted, but will accurately repre- 
 sent the actual path of the ball. Find the point on the curve 
 that has an abscissa of 6.5 ; i.e., a point that is 6.5 spaces to 
 the right of the axis of ordinates. This point corresponds to 
 a period of 3.25 seconds. Measure its ordinate ; i.e., the num- 
 ber of spaces that it is below the axis of abscissas. Remember 
 that the scale adopted is 1 space = 10 feet, and determine the 
 magnitude represented by the ordinate. It should be nearly 
 170 feet. Next, find the point on the curve that has an abscissa 
 of 11 ; i.e., a point that is 11 spaces to the right of the axis of 
 ordinates. This point corresponds to a period of 5.5 seconds. 
 Measure its ordinate ; i.e., the number of spaces that it is below 
 the axis of abscissas. Determine the magnitude represented 
 by this ordinate. It should be about 486 feet. 
 
 5. Divide the several values of I by the square of the corre- 
 sponding number of time intervals. For instance, divide the 
 value of I as shown in Fig. 64, by the square of 15. The 
 several quotients will be equal to the value of / when t = 1. 
 
28 KEY TO THE PROBLEMS 
 
 Page 124. 
 
 NOTE. Take 39.1 inches or 99.33 centimeters as the length of a 
 seconds pendulum. 
 
 1. The period is 3 seconds. See 115 (3). 39.1 in. x 3 2 = 
 351.9 in., the length of the pendulum. 
 
 2. The period is 2 seconds. 39.1 in. x 2 2 = 156.4 in. 
 
 3. We have two pendulums to compare ; the seconds pendu- 
 lum, the length and period of which are known, and the given 
 pendulum, the length of which is 30 inches, but the period of 
 which we are to find. 
 
 39.1 in. : 30 in. : : I 2 : 2 ; * = 0.87. Since the period is 0.87 
 seconds, there will be as many oscillations in 60 seconds as 0.87 
 is contained times in 60, which is 68.9, the number of oscilla- 
 tions per minute. 
 
 4. 39.1 : 16 = I 2 : t 2 ; t 0.64 . The number of vibrations 
 is 93.7 + . 
 
 5. 60 -T- \ = 240, the number of oscillations per minute. 
 39.1 : 1 : : I 2 : (|)-; I = 2.44+ in. Or, we may say that, as 
 
 the time is J that of the seconds pendulum, the length will be 
 T L that of the seconds pendulum, or 2.44+ in. 
 
 6. 60 sec. -f- 15 sec. = 4, the number of oscillations per 
 minute. 
 
 39.1 : 1 : : I 2 : 15 2 . Or, since the time is 15 times that of the 
 seconds pendulum, the length will be (15*=) 225 times the 
 length of the seconds pendulum. 
 
 7. 39.1 : 39.37 : : I 2 : t 2 ; t = 1+ sec. 
 
 Notice how closely the meter corresponds to the length of 
 the seconds pendulum. 
 
 8. The period is 6 seconds. 
 
 39.1 in. x 6 2 = 1,407.6 in. = 117.3 ft. 
 
 9. 39.1 in. : 10 in. : : I 2 : 2 ; t = 1+. If the period is | 
 second, the number of oscillations per second is 2. 
 
IN AVERY'S SCHOOL PHYSICS. 29 
 
 10. The period is 60 seconds. 39.1 in. x 60 2 is equivalent 
 to 11,730 ft., or more than 2^- miles. 
 
 11. The length of this pendulum is that of the seconds pen- 
 dulum. Hence the number of oscillations is 1 per second, and 
 the period is 1 second. 
 
 12. 99.33 cm. x 2 2 = 397.32 cm. = 3.9732 m. 
 
 13. 99.33 cm. x 120 2 = 1,430,352 cm. = 14.3+ Km. 
 
 14. (b) 99.33 : 24.83 :: I 2 it 2 -, 
 
 15. The period is second. 99.33 cm. x (|) 2 = 1.552 cm. 
 
 16. 99.33 : 397.32:: I 2 :t 2 ; t = 2. 
 
 17. 99.33 : 11.03 : : I 2 : t 2 ; t = nearly. 
 
 18. 99.33 cm. x 10 2 = 9,933 cm. = 99.33 m. 
 
 19. 99.33 : 2,483.25 : : I 2 : t 2 ; t = 5. 
 SUGGESTION. Divide the 1st couplet by 99.33. 
 
 20. (a) 99.33 cm. x 4 2 = 1,589.28 cm. = 15.8928 m. 
 
 21. See 115 (3). pip': : V4: V49; i.e., the periods will 
 be as 2 : 7. 
 
 22. 1: I':: 70 2 : 80 2 ; i.e., the lengths will be as 49 : 64. 
 
 23. The period of the longer pendulum will be twice that 
 of the shorter one. 
 
 24. 39.1 in. x (i) 2 = 1.564 in. 
 
 25. (a) 39.1 in. x 8 2 = 2,502.4 in., or 99.33 cm. x 8 2 
 
 = 6,357.12 cm. = 6,357+ m. 
 (b) 39.1 in. x (i) 2 = 0.61 in., or 993.3 mm. x (i) 2 = 15.5 mm. 
 
 26. 39.1 in. x (3.5) 2 = 478.975 in., nearly 40 ft. 
 
 27. Time of vibration = 0.8 seconds. 
 
 39.1 in. x (0.8)* = 25.02+ in. 
 
30 KEY TO THE PROBLEMS 
 
 28. 60 : 601 : : w * : 400 2 ; n = 398+. 
 
 29. See 115. 
 
 1 = ir~l-L- ; I = 99.396, the number of centimeters. 
 
 30. The period is f|f . 
 
 300 150 83 150 2 7r 2 x83 
 
 Page 125. 
 
 i. a = j ; 2/ = i ; = T V; 2V; sV 
 
 2. Substitute the known values in the formula given in 
 115. t = 1, and I is determined by measurement. 
 
 3. The period is 0.5 of a second, or half as long as that of 
 the other pendulum. Its length should be one-fourth as great. 
 
 4. The period is \ of a second. 
 
 7. The upper bullets correspond to shorter pendulums, and 
 tend to move faster than the lower bullets. 
 
 Page 134. 
 
 1. See 124. 
 
 (a) 5Q * 15 = 3, the number of feet. 
 
 KA v 1 K 
 
 (&) r u x = 10, the number of pounds. 
 75 
 
 2. (a) 1QQ x 1Q = 8, the number of pounds. 
 
 (6) 1 QQ x 10 = 5, the number of feet per second. 
 200 
 
 3. The lever has equal arms; i.e., it is a balance. The 
 power and load are equal. 
 
 4. (a) 15 feet, (b) 10 feet, (c) Such a lever cannot be 
 of the 3d class, for in such levers the weight-arm is always 
 greater than the power-arm. 
 
IN AVERY'S SCHOOL PHYSICS. 31 
 
 5. (a) First and second classes. (6) If of the 1st class, 30 
 Ibs. ; if of the 2d class, 40 Ibs. 
 
 6. Figure the lever. It must be of the 1st class, with arms 
 of 4 and 6 feet. See 131. 
 
 40 x 4 = 160 ; 1,000 x 2 = 2,000 ; 160 + 2,000 = 2,160, 
 
 the sum of the moments for the short, and. consequently, the 
 moment of the force to be applied at the end of the long arm 
 to produce equilibrium. But in the case of the long arm, 6, 
 the length of the arm, is one factor of this 2,160 ; there- 
 fore (2,160-5-6=) 360, the number of pounds, is the other 
 factor. Or, more briefly, 
 
 6 x = 2,160 ; x = 360. 
 
 7. The lever is of the first kind, with arms of 8 and 12 
 feet respectively. Figure the lever. Eepresent the force 
 at the end of the short arm by x. Then will the force at the 
 end of the other arm be represented by 1,200 x. Placing 
 the moments of the two forces equal to each other, we have, 
 
 8 x = 12(1,200 - a?)= 14,400 - 12 x. 
 .-. 20 x = 14,400; a = 720; 1,200 - x = 480. 
 
 Ans. 720 Ibs. at the end of the short arm, and 480 Ibs. at 
 the end of the long arm. 
 
 Proof {720 + 480 = 1,200. 
 
 '(720x 8= 480x12. 
 
 8. Figure the lever, with the first three forces acting. 
 
 40 
 300 
 
 100 
 
 340 = 100 + 240. 
 
 We thus see that an additional moment of 240 is needed 
 with the 100, to produce equilibrium. Of this 240, one factor 
 is 80 ; then the other factor is 3, the number of feet from the 
 fulcrum, on the same side as the force of 100 Ibs. Or, we may 
 
32 KEY TO THE PROBLEMS 
 
 represent the four forces acting upon the lever, calling the dis- 
 tance of the fourth from the fulcrum, x. Then we shall have 
 
 40 
 300 
 
 100 
 
 80 x 
 
 340 = 100 + 80 x; x = 3. 
 
 SUGGESTION. In a case like this, simple inspection will generally 
 show on which arm to figure the fourth force as acting. If it is errone- 
 ously represented, either as to position or direction, the result will be a 
 negative quantity. For instance, suppose that in this case we represent 
 the force of 80 Ibs. as acting on the same arm as the forces of 10 and of 
 100 Ibs., and in a downward direction. We should then have 
 
 40 
 300 
 
 80 x 
 
 100 
 
 340 + 80 x = 100 ; 80 x = - 240 ; x = - 3. 
 
 Here the 3 would indicate that the force was acting at a distance of 3 ft. 
 from the fulcrum ; the minus sign would indicate that the force was 
 tending to turn the lever in a direction opposite to that assumed. This 
 contrary tendency might result from an upward force of 80 Ibs. acting 
 3 ft. from the fulcrum and on the same arm as the 10 Ibs., or from a 
 downward force of 80 Ibs. acting 3 ft. from the fulcrum and on the same 
 arm as the 100 Ibs. Due regard for the algebraic sign of the result will 
 correct any error of this kind. In this problem, a downward force is 
 specified ; the problem is therefore definite. 
 
 9. Draw a line, a&, to represent, on any convenient scale, 
 as 1 inch to the foot, a lever 6J- feet long. Then this line 
 will be 61 inches long. Measure on the line f of an inch from 
 b, for the position of the fulcrum, c. The lever is evidently 
 of the 1st class, with arms of 5J feet and f feet respectively. 
 Represent a downward force of 60 pounds at a (Fig. 76). 
 Locate d, 2| inches (feet) from c. Represent a downward 
 force of 75 pounds acting at" d. Represent a downward force 
 of x pounds at b. 
 
 51 x 60 = 330 
 
 2J x 75 = 2061 
 
 536} = | x; z = 
 
IN AVERY'S SCHOOL PHYSICS. 33 
 
 10. Figure the lever as in Exercise 9. Place the moments 
 tending to move a upward in one column, and those tending 
 to move it downward in the other column. 
 
 40 x 10 = 400 
 
 28 x 
 
 364 = 56 x 6 
 288 = 96 x 3 
 
 28 x + 400 = 652 
 
 28 x = 252 
 
 x= 9 
 
 The force of 28 Ibs. must act at a distance of 9 ft. from the 
 fulcrum. It must, therefore, act on the long arm, as the short 
 arm is only 3 ft. long. We assumed that the force would tend 
 to draw a downward. This assumption was not contradicted 
 by a negative result. The force will therefore act downward, 
 at a distance of 1 ft. from a. 
 
 11. Draw a line 18 cm. long to represent the beam. (Scale : 
 1 cm. = 1 ft.) Letter its extremities a and b. At a point 3 
 cm. from a, represent a downward force of 2,000 Ibs. At a 
 point 8 cm. from &, represent a downward force of 1,400 Ibs. 
 Consider either end as the power, and the other end as the 
 fulcrum. 
 
 WITH POWER AT b. 
 
 2,000 x 3= 6,000 
 1,400 x 10 = 14,000 
 
 20,000 =18 a; 
 1,1111=" x 
 
 Pressure at b = IjH-H Iks. 
 
 Pressure at a = (3,400 - 1,111^ =) 2,288| Ibs. 
 
 WITH POWER AT a. 
 
 11,200 = 1,400 x 8 
 x 30,000 = 2,000 x 15 
 
 18 x = 41,200 
 
 x= 2;288f 
 
 Pressure at a = 2,288f Ibs. 
 
 Pressure at b = (3,400 - 2,288f =) 1,111J- Ibs. 
 3 
 
34 KEY TO THE PROBLEMS 
 
 12. Eepresent the power-arm by x and the weight-arm by 
 3 x. Call 40 the power, and 200 the weight. 
 
 P: W: : WF: PF, or 40 : 200 : : 3 - x : x. 
 
 .-. 40 x = 600 -200 x; 240^=600; 
 a = 21; 3-ic = f 
 
 The fulcrum must be 2^ ft. from the 40 Ibs., and 6 in. from 
 the 200 Ibs. 
 
 Or, we may proceed as follows : The weight is 5 times the 
 power; consequently, the power-arm must be 5 times the 
 weight-arm. Dividing 36 inches, the length of the lever, 
 into two parts, so that one shall be 5 times as great as the 
 other, we have 30 inches and 6 inches. 
 
 13. See 132 (a). Eepresent the true weight by x, and 
 remember that 1 Ib. 9 oz. = 25 oz., and that 2 Ibs. 4 oz. = 36 oz. 
 
 25 :x::x: 36 ; x 2 = 900 ; x = 30. 
 The true weight is 30 oz., or 1 Ib. 14 oz. 
 
 14. A ought to carry 200 Ibs. ; B, 100 Ibs. Call the 6 ft. 
 bar a lever of the second class, fulcrum at either end, as A's 
 end. Then the weight is 300 Ibs. ; the power, 100 Ibs. ; the 
 power-arm, 6 ft. ; the weight-arm = x. 
 
 P:W::WF:PF; 100 : 300 : : x : 6; 
 300^ = 600; x = 2. 
 
 The weight should be hung two feet from the fulcrum ; or two 
 feet from A, and four feet from B. 
 
 15. As the beef weighs 450 pounds, A should carry 250 
 pounds, and B, 200 pounds. Suppose B to act as fulcrum, 
 and A as power. 
 
 P:W::WF:PF; 250 : 450 : : x : 8; 
 9a = 40; # = 4f 
 
 The beef should hang 4f ft. from the fulcrum ; or 4J ft. from 
 B, and 3| ft. from A. 
 
IN AVERY'S SCHOOL PHYSICS. 35 
 
 16. The lever is of the 1st class, with arms of 3J ft. and 
 (16 _ 31 =) 121 ft. respectively. 
 
 100 x 121 =1,250 = 3Ja; x = 357f 
 
 It would require a power of 357-f Ibs. to balance the 100 
 Ibs. On the supposition that the machine is free from fric- 
 tion and other hindrances to motion, anything more than 
 357^- Ibs. would move the load. In practice, an allowance, 
 determined by experience, is made for these necessary hin- 
 drances. 
 
 17. (a) 
 
 (6) 50 x 40 x 33 = x x 6 x 5 ; x = 2,200, the number of 
 pounds. 
 
 (c) x x 40 x 33 = 4,400 x 6 x 5 ; x = 100, the number of Kg. 
 
 SUGGESTION. Theoretically, the parts of a lever or other simple 
 machine have no weight, suffer no loss by friction, from the stiffness of 
 ropes, the resistance of the air, or similar causes. In practical mechanics, 
 these have to be taken into consideration. 
 
 18. The weight of the bar, 4 Ibs., may be considered as con- 
 centrated at the center of mass, 5 in. from either end. See 
 94 (a). Let x represent the distance of the fulcrum from 
 the end at which the 6-lb. weight is hung. 
 6x = 4(5-z); a? = 2. 
 
 Figure the lever, and notice the equality of the moments; 
 3x4 = 2x6. 
 
 Page 136. 
 
 2. The reading of the dynamometer includes the weight of 
 the bar. 
 
 10. In every case, the weight of the rod acts as if it was 
 concentrated at the same point, viz., at the center of mass. 
 
36 KEY TO THE PROBLEMS 
 
 Page 14O. 
 
 1. P: W:: d: D; x : 180 :: 6: 36; x = 30. 
 Any power greater than 30 Ibs. will move the rudder. 
 
 2. x : 2,000 : : 8 : 80 ; x = 200. 
 200 -r- 4 = 50 Ibs. Ans. 50+ Ibs. 
 
 3. See Fig. 85. P: W: : d : D; x: 1,100 :: 1 : 10; x = 110; 
 
 110 -7- 4 = 27-J-, the number of pounds. 
 
 4. The circumference of the axle is ^ of the circumference 
 of the wheel ; 1 of 85 Ibs. is 14J Ibs. 
 
 5. 70 : 300 : : x : 120 ; x = 28. 
 
 Or, the power is -j^ of the weight ; hence, the diameter of the 
 axle must be -^ of 120 inches 28 inches. 
 
 6. The diameter of the circle described by the power is 36 
 inches. 
 
 (a) x : 62.5 : : 10 : 36; x= 17.36 + , the number of pounds. 
 (6) 20 liters of water weigh 20 kilograms. 
 
 x : 20 : : 10 : 36 ; x = 5.55, the number of Kg. 
 
 7. We have two sets of tw r o men each. The first set move 
 in a circle 14 feet in diameter ; the second set in a circle 10 
 feet in diameter. Find the effect produced by each set and 
 add. 
 
 1st set 60 : x : : 14 : 12 x 14 ; x = 720 
 
 2dset 80: a;':; 14; 12x10; x' = 685^ 
 
 Total effect = 1,405 Ibs. 
 
 8. The diameter of the circle traversed by the horse is 
 14 feet. The diameter of the capstan barrel is 14 inches ; i.e., 
 the first diameter is 12 times the second diameter. Hence, the 
 circumference of the circle traversed by the horse is 12 times 
 as great as the circumference of the barrel of the capstan. 
 From this it follows that the horse must move 12 times as far 
 as the house. (This may be made clear by computing the two 
 
IN AVERY'S SCHOOL PHYSICS. 
 
 37 
 
 circumferences ; one of them will show the distance the horse 
 travels in turning the barrel around once ; the other will show 
 how much rope was wound up, or how far the house moved for 
 that revolution of the capstan barrel.) Then the horse must 
 travel (5 miles x 12 =) 60 miles. At the given rate, this will 
 require (60 -f- 2| =) 24 hours. 
 
 Page 145. 
 
 1. 50 pounds. The fixed pulley gives no mechanical advan- 
 tage. See 139. 
 
 2. It may be 25 Ibs. (see Fig. 88), or 16 1 Ibs. (see Fig. 89). 
 
 3. 75 Ibs. x 7 525 Ibs. The friction, etc., help to sup- 
 port the load, so that the power may be considered as 90 Ibs. 
 90 Ibs. x 7 = 630 Ibs. 
 
 4. 2,000 -r- 10 = 200, the number of pounds. 
 
 5. 2,000 -r- 11 = 181 T 9 T +, the 7. 
 number of pounds. Friction 
 resists the motion, and may 
 
 be considered as added to the 
 load. 2,500 Ibs. -=- 11 = 227y\, 
 the number of pounds. 
 
 6. There may be 12 cords 
 or 13 cords. The friction will 
 support \ of the weight, leav- 
 ing 1,350 Ibs. to be otherwise 
 supported. 
 
 1,350 Ibs. -5- 13 = 10311 Ibs. 
 The power will move 13 times 
 as fast as the weight, if either 
 moves at all. 
 
 8. The center of mass of the 125-pound board is 10 feet 
 from each end of the board. When the boy lifts at one end of 
 the board, the other end becomes the fulcrum of a lever of the 
 
38 KEY TO THE PROBLEMS 
 
 second class, and the power required is only half the load to 
 be raised. He is, therefore, strong enough to lift one end of 
 the plank as required. 
 
 20 : 5 : : 196 : x; x = 49. 
 
 9. It involves just as much work and requires as much 
 energy to move it up the inclined plane as it would to lift it 
 outright, namely, 600 foot-pounds. This is involved in the 
 very definitions of work and energy, but it may be well to 
 illustrate it. The power required to support the weight on 
 the plane (ignoring friction) is determined according to the 
 law given in 141 (1) : 
 
 10 ft. : 6 ft. = 100 Ibs. : 60 Ibs. 
 
 But this power of 60 pounds must act along the whole length 
 of the plank, i.e., through a distance of 10 feet, in order to 
 lift the weight the required distance. 
 
 60 x 10 = 100 x 6. 
 
 The result is 600 foot-pounds, whether it is reached in one way 
 or in the other. But the use of the inclined plane introduces 
 friction and increases the work to be done by one-fifth, or 120 
 foot-pounds, making a total of 720 foot-pounds. 
 
 10. 4,000 Ibs. X T 3 o - 1,200 Ibs. ; or P : W: : h : b; 
 
 P: 4,000:: 3: 10; P= 1,200. 
 
 11. (a) If the power acts horizontally, 
 
 50 Ibs. x LO = 125 Ibs. Ans. Nearly 125 Ibs. 
 
 (6) If the power acts in the direction of the plane, first find 
 the length of the plane : 
 
 yi0 2 + 4 2 = 10.77 + . 
 Then, P: W: : h : I; W= nearly 134.625 Ibs. 
 
 12. 1 of 40 Ibs. is 5 Ibs. 
 
 13. The power is (g 2 -^ =) -fa of the weight. Consequently, 
 the base is 32 times as long as the height, or 256 ft. The 
 
IN AVERY'S SCHOOL PHYSICS. 39 
 
 length is the hypotenuse of a right-angled triangle, having 
 sides of 8 and 256 ft. respectively. 
 
 V8 2 + 256 2 = 256.1. 
 
 SUGGESTION. If you have pupils in the class who can do original 
 work, let them try to solve the problem by construction. 
 
 14. 75 Ibs. : 4,000 Ibs. = 1 yd. : 531 yds., or 
 75 Ibs. : 4,000 Ibs. = 3 ft. : 160 ft. 
 
 15. PiWiih: 1; 250 : 1,500 : : 5 : Z; I = 30 ft. or more. 
 
 16. The load is 400 Ibs. + 10 Ibs. = 410 Ibs. The movable 
 pulley divides the rope into 4 parts, and the load is, therefore, 
 4_p_ = 102 1. Ibs. The support of the upper block has to carry 
 both blocks, the power, and the load, or 20 Ibs. + 1021 Ibs. -f 
 400 Ibs. = 5221 ibs. 
 
 Page 148. 
 
 1. 8 ft. = 96 in. 96 in. -5- 1 in. = 288. The power moves 
 96 in. while the weight moves J in., or it moves 288 times as 
 fast. Hence, the theoretical pressure will be 288 times 15 Ibs. or 
 4,320 Ibs. Deducting for friction, we have 4,320 Ibs. - 240 Ibs. 
 = 4,080 Ibs., the actual pressure. 
 
 2. Diameter = 16 in. ; circumference = 16 in. x 3.1416 = 
 
 50.2656 in. 
 
 50 Ibs. x 50.2656 x 11 = 27,646.08 Ibs. 
 
 Deduct i of this for friction, 3,455.76 Ibs. 
 Available power 24,190.32 Ibs. 
 
 Any resistance less than 24,190.32 Ibs. may be overcome. The 
 friction may be provided for by deducting 1 of the 50 Ibs. before 
 multiplying by 50.2656 x 11. The problem may also be solved 
 
 as follows : 
 
 W X T V = 50 Ibs. x 50 x 50.2656. See 124 (1). 
 
 SUGGESTION. The distances here used are those traversed during one 
 revolution of the screw. 
 
40 KEY TO THE PROBLEMS 
 
 4^x 2x12 x 3.1416 = 2nA + 
 
 Of course this ignores friction. 
 
 4. 15 Ibs. x 8 - 120 Ibs. 
 
 P:W::r:E; 25 : W : : 41 : 30; W= 166|. 
 
 The boy could support a load of 166| Ibs. with the wheel 
 and axle, without the aid of the inclined plane. With the 
 inclined plane, he could support 20 times as great a load, or 
 3,3331 Ibs. He could lift anything less than 3,333J Ibs. 
 
 6. 5001bs.x7^x2xl2x6 = ^^ i bs . 
 
 Deducting \ of this for loss by friction . 11,454.54 Ibs. 
 we have, for the force exerted 34,363.64 Ibs. 
 
 Page 149. 
 
 3. P:TF::1:4. 
 
 4. There is no difference in the work performed. The 
 power is \ as large, but it moves 4 times as far. 
 
 5. The ratio is 16. In general terms, P : W : : 1 : 2% or 
 WPx 2 n , in which n represents the number of movable 
 pulleys. 
 
 6. Construct the figure very carefully. The larger it is, 
 
 the better. Represent the base, 
 AB, by a line at least 20 cm. 
 long, and make AC one-fifth as 
 long. Of course, AB is supposed 
 to be horizontal, so that the angle 
 B measures the inclination of the 
 plane with reference to the hori- 
 zon. Anywhere on BC, take the 
 point W to represent the position 
 of the globe. Represent the grav- 
 ity of the globe by a vertical line. If you adopt the scale of 
 1 mm. = 2 Kg v this line, WG, will be 12.5 cm. long. From W, 
 
IN AVERY'S SCHOOL PHYSICS. 
 
 41 
 
 draw WX, so that it shall make an angle of 45 with AB. 
 This may be done by taking oi = o W, and drawing a line 
 from W through i. From W, draw WZ perpendicular to BG. 
 From G, draw GP parallel to WZ, and GE parallel to WP. 
 WE represents the pressure of the globe upon the plane, and 
 WP the tendency to move in the direction WX. This ten- 
 dency must be resisted by the supporting force which equals 
 it in intensity and is opposite to it in direction. PW will 
 measure 30+ mm., and represent a force of 60+ Kg. 
 
 7. Draw ABC, WG, and WZ, as in the last problem. 
 From W, draw WX' perpendicular to WX. From G, draw 
 a line parallel to WZ, and prolong it until it intersects WX' 
 at P'. Complete the parallelogram by drawing GE 1 parallel to 
 WP'. Measure GP', and find its value according to the scale 
 adopted. 
 
 SUGGESTION. In the 6th problem, the supporting force tends to lift 
 the globe from the plane, and thus to make its pressure on the plane less 
 than its weight ; in the 7th, the sup- 
 porting force tends to press the globe 
 against the plane, and thus to make 
 its pressure on the plane greater 
 than its weight. 
 
 9. Draw the right-angled 
 triangle, ABC, with its sides 
 as 3 : 4 : 5, as shown in the ac- 
 companying figure. Then will 
 AC represent the base, and 
 BC the length of the plane. 
 From W, the point where the 
 ball rests on the plane, draw 
 the vertical line, WG, 1\ units 
 long to represent the weight 
 of the ball. From W, draw a 
 line perpendicular to BC, and 
 from G draw a line parallel to BC, so that they will intersect 
 at P. Complete the parallelogram WPGE. WE represents 
 
42 KEY TO THE PROBLEMS 
 
 the tendency of the ball to roll down the plane, and WP repre- 
 sents the pressure of the ball on the plane. As the triangle, 
 WPG, is similar to the triangle BAG, it follows that 
 
 PG:WG::3:5- y 
 PW: TF#::4:5; PTF:71::4:5; PW= 6. 
 
 But PG (or its equal, WE) represents the tendency of the ball 
 to roll down the plane, and its value should be indicated by 
 the reading of the dynamometer that resists that tendency. 
 Similarly, PW represents the pressure of the ball on the plane, 
 and its value should be indicated by the reading of the dyna- 
 mometer that measures that pressure. 
 
 Page 158. 
 
 1. See 151 (2). The imaginary column has a base of 
 30 x 20 and an altitude of 10 ft. Cubic contents = 6,000 cu. 
 ft. The weight of 6,000 cu. ft. of water is 62.42 Ibs. x 6,000 
 = 374,520 Ibs. 
 
 SUGGESTION. Allow the pupil to use 62 1 Ibs. as the weight of a cu. ft. 
 of water, if he prefers to do so. That value is more easily remembered, 
 and nearly enough accurate for non-professional purposes. 
 
 2. Bulk of imaginary column = (6 x 10 x 3 =) 180 cu. m. 
 = 180 Kl. = 180,000 liters. 21. Each liter of water weighs 
 1 Kg. Hence the pressure is 180,000 Kg. 
 
 3. 5 x 12 x 6 = 360, the number of cu. ft. 
 62.42 Ibs. x 360 = 22,471.2 Ibs. 
 
 4. 2x4x2 = 16, the number of cu. m. 
 
 16 cu. m. = 16,000 1., which weigh 16,000 Kg. 
 
 5. 2 cu. yds. = 54 cu. ft. 
 
 62.42 Ibs. x 54 = 3,370.68 Ibs. 
 
 6. 2 cu. m. = 2,000 1. 
 
 2,000 1. of water weigh 2,000 Kg. 
 
 7. 25 cu. ft. of water weigh (62.42 Ibs. x 25 =) 1,560.5 Ibs. 
 
IN AVERY'S SCHOOL PHYSICS. 43 
 
 8. 30 x 30 x 800 = 720,000, the number of cu. cm. This 
 quantity of water weighs 720,000 g. or 720 Kg. 
 
 9. 3 x 3 x 7 = 63, the number of cu. ft. 
 62.42 Ibs. x 63 = 3,932.46 Ibs. 
 
 10. The dimensions of the vessel are 12 x 12 x 12. The 
 acid stands 8 in. deep. 12 x 8 x 4 = 384, the number of cu. 
 in. in the imaginary column. This is | of a cu. ft. If the 
 imaginary column were of water, it would weigh -| of 62.42 Ibs. 
 = 13.871+ Ibs. Such a column of acid would weigh 1.8 times 
 13.871+ Ibs. = 24.9678+ Ibs. 
 
 11. 237 x 35 = 8,295, the number of cu. cm., and consequently 
 the number of grams. 
 
 12. 237 x 35 = 8,295, the number of cu. in. 
 62.42 Ibs. x ff f| = 299.7 Ibs. 
 
 13. The water must stand (- 2 7 -=)4i ft. deep. The sides 
 subjected to lateral pressure have an area of (10 x 4-J-=) 45 
 sq. ft. 45 x 21 = 101 J, the number of cu. ft. in the imaginary 
 column producing lateral pressure. There are 27 cu. ft. (2x3 
 X 41) in the column producing pressure on the bottom. 
 101.25 + 27 = 128.25. 
 
 62.42 Ibs. x 128.25 = 7,999.365 Ibs. 
 
 14. If the lever of the press is of the second class, as shown 
 in Fig. 108, the piston of the pump will go down with a force 
 of (75 Ibs. x 6 =) 450 Ibs. The area of the cylinder being 
 200 times as great as that of the piston, the weight will be 200 
 times 450 Ibs., or 45 tons. If the" lever is of the first class, the 
 weight will be (75 Ibs. x 5 x 200 =) 75,000 Ibs. = 37 tons. 
 
 15. 2 x 3 x 5 = 30, the number of cu. ft. 
 62.42 Ibs. x 30 = 1,872.6 Ibs. 
 
 16. See Appendix. 
 
 Area = *& = 3.1416 x 100 = 314.16. 
 314.16 x 48 = 15,079.68, the number of cu. in. 
 15,079.68 cu. in. = 8.72 cu. ft. 
 62.42 Ibs. x 8.72 = 544.3+ Ibs. 
 
44 KEY TO THE PROBLEMS 
 
 17. 1,000 cm., or 10 m. 
 
 18. Draw a sectional view of the vessel and tube. The 
 water in the tube is. 35 cm. higher than the upper end of the 
 cylinder; the pressure is 35 grains per square centimeter. 
 The water in the tube is 65 cm. higher than the lower end 
 of the cylinder; the pressure is 65 grams per square centi- 
 meter. 
 
 19. Draw a sectional view. 
 
 (a) The jar contains 2,000 cu. cm., and the tube 30 cu. cm. 
 The 2,030 cu. cm. of water weigh 2,030 grams. 
 
 (6) The base is 50 cm. below the surface of the water, and 
 the pressure is 50 grams per square centimeter, 
 (c and d) The pressure on the base is uniform. 
 
 20. (a) The surface described is 30 cm. below the surface 
 of the water, and the pressure is 30 grams per square centi- 
 meter. 
 
 (b) The pressure on the cover is uniform. 
 
 (c) 99 x 30 = 2,970, the number of cubic centimeters in the 
 imaginary column,- and the number of grams pressure. 
 
 21. The pressure on the base is (100 x 50 =) 5,000 grams. 
 The pressure on the cover is ( 99 x 30 =) 2,970 grams. 
 
 The weight of water is 2,030 grams. 
 
 Page 16O. 
 
 3. The difference in the mercury levels measures the 
 upward pressures at the several depths of the water levels in 
 the tube below the water levels outside the tube. The upward 
 pressure varies as the depth. 
 
 4. If the line is straight, it shows that m and w vary alike, 
 i.e., in the same ratio. If the line is curved, it shows that one 
 varies more rapidly than the other. 
 
IN AVERY'S SCHOOL PHYSICS. 45 
 
 Page 166. 
 
 1. It will lose the weight of 1 cu. dm. of water. 153. 
 This is the weight of a liter of water, 1,000 g. or 1 Kg. 
 
 2. It would lose the weight of 1 cu. dm., or 1 liter of the 
 liquid (mercury), which would weigh 13.6 times 1 Kg., or 
 13.6 Kg. = 13,600 g. 
 
 3. The iron is supposed to be immersed in the mercury. 
 The mercury pushes it up with a force of 13,600 g. Gravity 
 pulls it down with a force of 7,780 g. It must be held down 
 by an additional force of 5,820 g. It can carry a load of 5,820 g. 
 If the iron is allowed to float on the mercury, it will displace 
 only 7,780 g. of mercury, thus losing its own weight. 154. 
 
 4. The weight of a cu. ft. of water, or 62.42 Ibs. 
 
 5. It will displace 100 cu. cm. of water. It will there- 
 fore lose 100 g. in weight. The remaining weight will be 
 1,035 g. 
 
 6. It loses 10 g. in weight when placed in water. It there- 
 fore displaces 10 g. or 10 cu. cm. of water; its own bulk is 
 10 cu. cm. 
 
 7. Draw a sectional view. 
 
 (a) The cube contains (20 x 20 x 20 =) 8,000 cu. cm. 
 The tube contains (2 x .2 x 10 =) 40 cu. cm. 
 The 8,040 cu. cm. of water weigh 8,040 g. 
 
 (5) The imaginary column (see 151) contains (20 x 20 x 
 30=) 12,000 cu. cm., and weighs 12,000 g. The downward 
 pressure is 12,000 g. 
 
 (c) The top of the vessel that is subjected to an upward 
 pressure is (400 4 =) 396 sq. cm. The imaginary column 
 measures 3,960 cu. cm. The upward pressure is 3,960 g. The 
 downward pressure minus the iipward pressure represents the 
 net downward tendency or weight. 1 2,000 g. - 3,960 g. = 8,040 g. 
 
46 KEY TO THE PROBLEMS 
 
 8. In Fig. 117, the weight of the particle is represented by 
 B W. This force is resolved into two components, of which BE 
 is perpendicular to the surface of the water. This component 
 is met by the resistance of the water. The other component, 
 BD, pulls the water particle to a lower level. When the sur- 
 face becomes level, BE coincides with BW, and BD vanishes. 
 
 Page 173. 
 
 1. As it displaces only half of its own volume of water, it 
 weighs only half as much as its own volume of water (see 154). 
 Consequently, its density is one-half, or 0.5. 
 
 2. The volume of a sphere = 1 TrD 3 (see Appendix 1) = 
 8,181.2 cu. cm. This quantity of water would weigh 8,181.2 g. 
 This quantity of aluminium would weigh 2.6 times as much, or 
 21,271.12 g. =21.27+ Kg. 
 
 3. 52.35 g. -r- 5 g. = 10.47, the density of silver. 
 
 4. (a) 695 g. -j- 83 g. = 8.37+, the density of brass. 
 (6) 83 g. x 0.792 = 65.736 g. 
 
 695 g. - 65.736 g. = 629.264 g. 
 
 5. 708 gr. -=- 1,000 gr. = 0.708, the density of the benzoline. 
 
 6. 2.4554 oz. - 2.0778 oz. = 0.3776 oz. 
 2.4554 oz. -r- 0.3776 oz. = 6.5+. 
 
 7. 4.6764 oz. -r- 0.2447 oz. = 19.11, the density of the gold. 
 
 8. See 160 (2). 970 gr. - 895 gr. = 75 gr. 
 970 gr. - 910 gr. = 60 gr. 
 
 60 gr. -f- 75 gr. == 0.8. 
 
 9. 23 gr. -r- 25 gr. = 0.92, the density of the oil. 
 
 19 gr. -=- 25 gr. = 0.76, the density of the alcohol. 
 
 10. 1,536 g. -1,283 g. = 253 g. 
 1,536 g. -f- 253 g. = 6.07. 
 
 11. Subtract the weight of bottle from the other two weights. 
 4.2544 g. -* 4.1417 g. = 1.027 + . 
 
IN AVERY'S SCHOOL PHYSICS. 47 
 
 12. (1) Weight of wood and sinker in air, 14 g. 
 
 (2) " " water, 8.5 g. 
 
 (3) " " water displaced by both, 5.5 g. 
 
 (4) " " " sinker 
 
 (10 g. ---10.5=) .954 g. 
 
 (5) Weight of water displaced by wood, 4.546 g. 
 
 (6) Density of the wood (4 g. -=- 4.546 g. =) 0.897. 
 
 .13. Density of the metal is 1.73 ; of the unknown liquid, 0.67. 
 
 14. 2,160 gr. - 1,511.5 gr. = 648.5 gr. 
 2,160 gr. -*- 648.5 gr. = 3.33 + . 
 
 15. (1) Weight of ice and lead in air . . 24 Ibs. 
 
 (2) " " " water . 13.712 Ibs. 
 
 (3) " " water displaced by ice 
 
 and lead 10.288 Ibs. 
 
 (4) Weight of water displaced by lead, 1.4 Ibs. 
 
 (5) " " ice . 8.888 Ibs. 
 
 (6) Density of ice (8 Ibs. -f- 8.888 lbs. = ) 0.9+. 
 
 16. (a) 600 g. - 545 g. = 55 g. 
 
 600 g. -55 g. = 10.9+. 
 (6) 600 g. - 557 g. = 43 g. 
 
 43 g. -55 g. =0.78 + . 
 (c) 600 cu. cm. -f- 10.9 = 55+ cu. cm. 
 
 17. The simple question is, " How much does that stone 
 weigh in water ? " 
 
 156 2.5 = .-. w' = 1801bs. 
 300 - w' 
 
 18. An equal bulk of water weighs 1,000 g. 
 
 870 g. -=- 1,000 g. = 0.87, the density of the turpentine. 
 
 19. The volume of the fragments was 1,000 cu. cm. - 
 675 cu. cm. = 325 cu. cm. The fragments weighed 1,487.5 g. 
 675 g. = 812.5 g. An equal bulk of water would weigh 
 
 325 g. 
 
 812.5 g, 325 g. = 2.5, the density of the mineral. 
 
48 KEY TO THE PROBLEMS 
 
 20. The 800 cu. cm. of water weigh 800 g. The 200 cu. cm. 
 of sand weigh 1,350 g. - 800 g. = 550 g. An equal bulk of 
 water weighs 200 g. 550 g. -- 200 g. = 2.75, the density of the 
 sand. 
 
 21. 1?OOQ ~ 4Q = 8, the density of the brass. 157. 
 
 22. + ' =1.8, the density of the acid. See 160 (3). 
 ZjUUU -f~ ijuuu 
 
 23. The given body weighs 10 g. It displaces 2.5 g. of 
 water (an equal bulk of water weighs 2.5 g.). 
 
 10 g. -r- 2.5 g. = 4, the density. 
 
 24. The coal would weigh 2.4 times as much as a cubic foot 
 of water or (62.5 Ibs. x 2.4 =) 150 Ibs. It would displace 
 1 cu. ft. of the solution, which would weigh (62.5 Ibs. x 1.2 ) 
 75 Ibs. It will lose 75 Ibs. weight when in the saline solution. 
 150 Ibs. - 75 Ibs. = 75 Ibs. 
 
 Or, we may say that the coal will weigh as much as 2.4 cu. ft. 
 of water, and that the solution displaced by it will weigh as 
 much as 1.2 cu. ft. of water. The weight less the loss by 
 buoyancy will be the weight of (2.4 cu. ft. 1.2 cu. ft. =) 
 1.2 cu. ft. of water, or 75 Ibs. 
 
 A comparison of the given densities shows that the coal is 
 twice as heavy as the solution. Under such conditions, the 
 weight of the coal in the solution will equal the weight of the 
 solution displaced. 
 
 25. With a force equal to the weight of the iron. 
 
 500 g. x 7.8 = 3,900 g. 
 
 26 (1) Weight of both in air ...... 41.2 g. 
 
 (2) " " " water ..... 26.2 g. 
 
 (3) " lost by both in water . . . 15 g. 
 
 (4) " iron " " . . . . _5 g. 
 
 (5) " " cork " " .... 10 g. 
 
 (6) Density of cork (2.3 -=- 10 =) . . . . 0.23 
 
IN AVERY'S SCHOOL PHYSICS. 49 
 
 27. (a) See 156. 
 
 11.35 = 60 = 547.13+. 
 600 - w' 
 
 The lead weighs 547. 13 +.gr. in water. 
 (6) (1) Weight of both in air .... 900 gr. 
 
 (2) " " water . . . 472.5 gr. 
 
 (3) " lost by both in water . . 427.5 gr. 
 
 (4) " " lead " . . 52.87 gr. 
 
 (5) " " wood " . . 374.63 gr. 
 
 (6) Density of wood (300 gr. -5- 374.63 gr. = 0.8 + . 
 
 28. 618 gr. 
 
 31 gr. 
 
 618 gr. 
 93 gr. 
 
 649 gr. -5- 711 gr. = 0.9-K 
 
 29. (a) 330 g. - 315 g. = 15 g. 
 
 330 g. - 15 g. = 22. 
 
 (b) 330 g. - 303 g. = 27 g. 
 27 g. + 15 g. = 1.8. 
 
 (c) It displaces 15 g. or 15 cu. cm. of water. 
 Its volume is 15 cu. cm. 
 
 30. Its volume must be at least that of 1 Kg. of water. The 
 volume of 1 Kg. of water is 1 liter, 1 cu. dm., or 1,000 cu. cm. 
 
 31. See 160. 2.8 = 
 
 37 w' 
 .-. w' = 23.785+ g. 
 Or, we may say that the body being 2.8 times as heavy as 
 
 Q rT 
 
 water, an equal bulk of water would weigh '- g. 13.2142+ g. 
 
 This is what the body would lose in water. 
 
 37 g. - 13.2142 + g. = 23.785+ g. 
 
 Page 186. 
 
 1. 15 Ibs. x 20 x 144 = 43,200 Ibs. 
 
 2. TrZ) 2 = 3.1416 x 16 = 50.2656, the number of square 
 inches of surface. 15 Ibs. x 50.2656 = 753.984 Ibs. 
 
50 KEY TO THE PROBLEMS 
 
 3. The room contains 6,000 cu. ft. or 10,368,000 cu. in. of 
 air. This weighs 0.31 gr. x 10,368,000 = 3,214,080 gr., or 
 459.154 Ibs. Avoirdupois. 
 
 4. If the barometer-tube had a sectional area of 1 sq. cm., 
 the atmospheric pressure per sq. cm. would support a mercury 
 column containing 76 cu. cm. Such a column of water would 
 weigh 76 g. ; such a column of mercury would weigh 76 g. x 
 13.6 = 1,033.6 g. 
 
 Each side of the cube has a surface of 100 sq. cm. The six 
 sides have a surface of 600 sq. cm. The atmospheric pressure 
 being 1.0336 Kg. per sq. cm., the total pressure is 1.0336 Kg. 
 x 600 = 620.16 Kg. 
 
 5. It loses the weight of a cubic foot of air, or about an 
 ounce and a quarter. 164. 
 
 6. (a) The trunk has an horizontal section of (2| x 3J =) 
 8f sq. ft., or 1,260 sq. in. 15 Ibs. x 1,260 = 18,900 Ibs., the 
 downward pressure on the top of the trunk. If the trunk has 
 &flat top, its area will be the same as that of the horizontal 
 section (or bottom) of the trunk ; if it has an arched top, the 
 total pressure on the upper surface will be more than here 
 given, the excess being lateral pressure, which would not at all 
 interfere with opening the trunk, even if the air was exhausted 
 from it. The downward pressure would not be affected by the 
 shape of the top. 
 
 (b) The upward pressure on the under surface of the trunk 
 top is equal to the downward pressure on the upper surface. 
 
 7. (a) None. 
 
 (6) See solution to Exercise 4. 
 
 8. (a) The capacity of the room is 320 cu. m., or 320,000 1. 
 See 164. 1.3 g. x 320,000 - 416,000 g., or 416 Kg. 
 
 (6) and (c) The surface is 80 sq. m., or 800,000 sq. cm 
 The atmospheric pressure being about 1 Kg. to the sq.cm., 
 the pressure on each of these surfaces is about 800,000 Kg. 
 
IN AVERY'S SCHOOL PHYSICS. 51 
 
 (d) The surface is 32 sq. m., or 320,000 sq. cm. The pressure 
 on each of these surfaces is 320,000 Kg. 
 
 (e) The surface is 40 sq. m., or 400,000 sq. cm. The pressure 
 on each of these surfaces is 400,000 Kg. 
 
 (/) The total surface of the room is 3,040,000 sq. cm. The 
 total pressure is 3,040,000 Kg. 
 
 (g) The outward pressure from within, is counterbalanced 
 by the inward pressure from without. 
 
 9. A liter of hydrogen weighs 0.0896 g. ; 10 liters weigh 
 0.896 g. The balloon and hydrogen weigh 5.896 g. The 10 1. 
 of displaced air weigh 12.93 g. 
 
 12.93 g. - 5.896 g. = 7.034 g. 
 
 Page 19O. 
 
 1. (a) Under a pressure of two atmospheres. 
 (b) Under a pressure of half an atmosphere. 
 
 2. Under ordinary circumstances, that quantity of air would 
 weigh l-i- ounces. 164. To get twice that quantity of air 
 into that space, the pressure upon it must be doubled, i.e., the 
 air must be subjected to a pressure of two atmospheres. 
 
 3. (a) 500 cu. cm., or half a liter. 
 
 (b) 1 liter. It makes no difference what gas is used. 
 
 4. Half of it. 
 
 5. (a) 15 Ibs. + 5 Ibs. = 20 Ibs. 168. 
 
 (b) The pressure is 30 Ibs., or 2 atmospheres. The 
 pressure being doubled, the volume will be halved, i.e., 12 
 cu. in. 
 
 6. (a) In the short arm. The rising of the barometer 
 indicates an increase of atmospheric pressure. This increase 
 of pressure will push down the mercury in the open arm, and, 
 consequently, push it up in the closed arm. 
 
52 KEY TO THE PROBLEMS 
 
 Page 196. 
 
 1. 30 in. x 13.6 = 408 in. = 34 ft. 
 
 2. 28 ft. -*- 0.8 = 35 ft. 
 
 3. 76 cm. x 13.6 = 1,033.6 cm. = 10.336 m., the height to 
 which atmospheric pressure will lift water. 
 
 (a) and (b). It cannot, because atmospheric pressure is not 
 sufficient to force the water up to that height. 
 
 (c) It can, because the water is lifted by muscular or some 
 similar form of energy not subject to the limitations placed 
 upon atmospheric pressure. 
 
 4. At the bottom, because the atmospheric pressure being 
 greater, the resultant of the resultant forces (see 171) is 
 greater there. 
 
 5. 34 ft. -r- 1.8 = 18.8 ft. 
 
 6. The stone is pushed up by the pressure of the air on 
 the under side of the stone, when that pressure is not counter- 
 balanced by the weight of the stone and the downward pressure 
 of the air on the upper side of the stone. That downward 
 pressure of the air is largely carried by the string that carries 
 the sucker. 
 
 (6) There being only -|||- as much air in the receiver 
 as there was at the beginning, when its elastic force was the 
 same as that of the external air, its density and, hence, its 
 elastic force are only fff as great as those of the exter- 
 nal air. 
 
 8. 29.5 in. x 13.6 -f- 1.35 - 223.11 in. 
 
 9. 69 cm. x 13.6 = 938.4 cm. = 9.384 m. 
 
 10. (a) 15 Ibs. x 3.1416 x 2 2 = 188.496 Ibs. 
 (6) 1 Kg. x 3.1416 x 4 2 = 50.2656 Kg. 
 
IN AVERY'S SCHOOL PHYSICS. 53 
 
 11. By placing the pump within 28 ft. of the surface of the 
 water and extending the spout (Fig. 140) to the top of the 
 well. The cylinder of the pump may be a tube extending 
 to the top of the well, the piston rod running down the inside 
 of such long cylinder. 
 
 12. The elastic force of the air, Avheii you blow in at /, lifts 
 the water to li, and fills the tube. The tube then constitutes 
 a siphon, and continues to deliver water into g. The rising of 
 the water in g reduces the air space, and thus increases the 
 elastic force of the air in g, i and a. This increased pressure 
 exerted by the air on the surface of the water in a is trans- 
 mitted to the water contained in a, and forces it out in a jet 
 at n. 
 
54 KEY TO THE PROBLEMS 
 
 CHAPTER III. 
 
 Page 213. 
 
 1. See 183. 
 
 2. 1,145 -f- 458 = 2.5, the number of feet. 
 
 3. The water is a much better conductor of sound than the 
 air is. See 179. 
 
 4. The sound first heard is propagated through the iron; 
 the other, through the air in the pipe. The iron transmits the 
 sound waves more rapidly than the air. 
 
 5. The first refers to the vibrations of the sounding body, 
 or to vibrations caused by them in the air or other medium 
 that may carry them to the ear ; the second refers to the sensa- 
 tions produced by said waves through the ear or organ of 
 hearing. 
 
 6. See 180 (a). 
 
 7. In a vibratory motion, each moving particle has a to- 
 and-fro motion so that it periodically returns to the starting 
 point; translatory motion is progressive so that the moving 
 particle does not return to the starting point. The wind may 
 swing a thistlehead to and fro, giving it a vibratory motion; 
 it may also loosen some of the winged seeds, and carry them 
 to another field with a translatory motion. 
 
 8. Because the velocity of the particle changes like that 
 of a common pendulum. See Experiment, 108. 
 
 9. Nowise; they are the same. 
 10. Sinusoidal. 
 
IN AVERY'S SCHOOL PHYSICS. 55 
 
 Page 222. 
 
 1. 1,090 ft. x 18 = 19,620 ft. 
 
 2. (1,090 ft. + 2 ft. x 15) x 7 = 7,840 ft. See 184 (c). 
 
 3. Exercise 2 shows that the velocity of sound at the given 
 temperature is 1,120 ft. per second. 1,120 ft. -f- 224 = 5 ft. 
 See 184 (6). 
 
 1,150 - 1,090 
 
 4. - ^ =60) the number 01 centigrade degrees 
 
 above the freezing-point, i.e., 30 C. 
 
 5. It acts as a reflecting surface to the ear, and cuts off 
 confusing sound waves from sources other than the speaker. 
 
 6. The velocity is 1,120 ft. per second. The sound reached 
 the cliff in 3 seconds. 1,120 ft. x 3 = 3,360 ft. 
 
 7. The velocity must be 1,100 ft. per second. 
 
 1,100-1,090 
 
 - = 5, the number of centigrade degrees above 
 
 z 
 
 the freezing-point, i.e., 5 C. 
 
 8. 1,120 ft. -f- 4 ft. = 280, the number of vibrations. The 
 given velocity implies a temperature of 15 C. See solutions 2 
 and 3 above. 
 
 9. 332 m. = 33,200 cm. 33,200 cm. + 830 = 40 cm. 
 
 10. 1,128 ft. -s- 12 ft. = 94, the number of vibrations per 
 second. 
 
 11. It took the sound T 3 ^ of a second to reach the cliff. The 
 velocity of sound at the ordinary temperature of the air 
 (15 C) is 1,120 ft. T 3 g- of 1,120 ft. is 210 ft. 
 
 12. It is a case of echo. It took the sound 1^ seconds to 
 reach the reflecting surface. 1,090 ft. x 1 J = 1,362J ft. 
 
56 KEY TO THE PROBLEMS 
 
 13. Let x represent the depth of the mine. From the third 
 
 formula in 107, we derive t 2 , and t = \/ . Hence, 
 
 2x~ 9 x 9 
 
 : represents the time of falling, represents the 
 
 u 1,1^U 
 
 time required for the sound to pass from the bottom to the top 
 of the shaft. Adding the times required for these two motions, 
 
 we have 
 
 2x x _ ~ 
 
 ~ 
 
 32.16 1,120 
 
 The algebraic solution of this equation for the value of x, the 
 required depth of the mine, is somewhat tedious, and may well 
 be waived by the teacher, the mere statement of the problem as 
 above being accepted as an answer. 
 
 14. See 184 (d). Compare Exercise 4, page 214. 
 
 15. The prison was a whispering gallery. See 185. 
 
 16. The usual temperature of 15 C., and the corresponding 
 velocity of 1,120 ft. per second, are assumed. The strokes fol- 
 low at intervals of 1 of a second ; i.e., the pulse caused by any 
 stroke is i of 1,120 ft., or 224 ft. in advance of the pulse caused 
 by the succeeding stroke. When the bells are at the same 
 distance from the hearer, the pulses arrive at the listening ear 
 at the same time, and blend as a single pulse caused by a 
 single stroke. When one bell is 224 ft. further away than the 
 other, the first pulse from the further bell arrives at the same 
 time as the second pulse from the nearer bell, and blends with 
 it as a single pulse caused by a single stroke; succeeding 
 pulses from the two bells similarly blend. But when the 
 distances of the bells differ by 112 ft., the first pulse from the 
 further bell arrives y 1 ^- of a second after the first pulse from 
 the nearer bell ; yL of a second later, the second pulse from the 
 nearer bell arrives ; T V ^ a second later, the second pulse from 
 the further bell arrives, etc., in succession, the pulses arriving 
 at intervals of yL of a second instead of blending in pairs as 
 before. Pulses that arrive at intervals of y 1 ^ of a second give 
 the impression of 10 strokes per second. 
 
IN AVERY'S SCHOOL PHYSICS. 57 
 
 Page 242. 
 
 1. See 194. 
 
 144 x = 172f , the number of vibrations for the minor third. 
 144 x -f = 180, the number of vibrations for the major third. 
 144 x -| = 216, the number of vibrations for the fifth. 
 144 x 2 = 288, the number of vibrations for the octave. 
 
 2. The octave above is produced by (264 x 2 =) 528 vibra- 
 tions. The fifth above the octave is produced by (528 X f =) 
 792 vibrations. 
 
 3. The disk revolves 211 times per second. The disk has 
 24 holes in the inner row. The number of puffs emitted per 
 second is 24 x 21^ 512. The unison indicates that the given 
 tone is due to 512 vibrations per second. This is twice the 
 number of vibrations assigned to middle C, so that the tone is 
 just one octave above middle C ; i.e., it is C 4 . See 196. 
 
 4. Evidently the vibrations of the two forks are as 9 : 15. 
 Referring to the relative vibration-numbers as given in 196, 
 we see that 27 corresponds to the given lower tone D. The 
 proportion 9 : 15 : : 27 : 45 shows that the higher tone is B. 
 
 5. Multiply 261 successively by the vibration-ratios given 
 in 196. 
 
 6. The whistle has a certain vibration-number correspond- 
 ing to its pitch. When the locomotive and the listener are 
 standing still, the number of pulses that come to the ear is 
 the same as the vibration-number. When the locomotive is 
 approaching the observer, the number of pulses per second 
 that arrive at the ear exceeds the vibration-number, and the 
 pitch is therefore raised. When the locomotive is receding 
 from the observer, the number of pulses per second that arrive 
 at the ear is less than the vibration-number, and the pitch is 
 therefore lowered. The change of pitch is very noticeable if 
 the whistle is blowing when the locomotive passes by the 
 observer. 
 
58 KEY TO THE PROBLEMS 
 
 7. See the relative vibration-numbers in 196. 
 
 36 : 40 : : x : 440 ; x = 396, the number of vibrations. 
 
 8. Because the resistance of the board reduces the velocity 
 of the saw, and the rapidity of the successive blows of the 
 saw-teeth. 
 
 9. The rapidity with which the pulses would strike his ear 
 would be doubled, and the pitch would be raisad an octave. 
 
 10. The rapidity with which the pulses would strike his ear 
 would be lessened. The tone would be flatted. We might 
 imagine the velocities to differ so little that the number of 
 pulses that overtook the ear per second would be so small that 
 they would be recognized as separate sounds instead of blend- 
 ing into a continuous tone. 
 
 11. The observer would overtake in inverse order the pulses 
 that started ahead of him ; the music would bo heard again as 
 if played backward. 
 
 12. The reed has a certain vibration-number. When the 
 observer is in the prolongation of the axis of rotation, his dis- 
 tance from the reed is constant, and the number of pulses that 
 come to his ear per second is the same as the vibration-number 
 of the reed. As the number of pulses does nol. vary, the pitch 
 is constant. When the observer stands in th-3 plane of rota- 
 tion, the reed approaches him, and recedes from him once for 
 every revolution of the axis. When the reed is approaching 
 the ear, the succession of pulses is more rapid, and the pitch 
 rises. When the reed is moving away from 1he ear, the suc- 
 cession of pulses is less rapid, and the pitch falls. 
 
 Page 254. 
 
 1. See 202. 
 
 2. Three beats per second. 
 
 3. The strings are not in unison, or there would be no 
 beats. The vibration-number of the second string is 398 or 
 402 per second. 
 
IN AVERY'S SCHOOL PHYSICS. 59 
 
 4. 308-297 = 11. 
 
 5. See 205 (a). As the prong swings outward, it sends 
 a condensation down the jar. If the air column in the jar 
 is ^ of a wave-length, the condensation will travel to the lower 
 end of the air column and back in f, or |, the period of the 
 fork. At the end of this half-period, the prong is beginning 
 its swing in the opposite direction, so that the reflected pulse 
 will coincide with the direct outward pulse. If the length of 
 the air column is f of a wave-length, the reflected pulse will 
 coincide with the direct pulse of the prong on its second return 
 swing; if it is j of a wave-length, the reflected pulse will 
 coincide with the direct pulse of the prong on its third return 
 swing; etc. 
 
 6. See 205 (d). 
 
 7. (a) 15 in. x 4 = 60 in., or 5 ft. 
 
 (6) Assume that the temperature is such that the veloc- 
 ity of sound in air is 1,120 ft. 1,120 -j- 5 = 224, the number 
 of pulses per second. As the vibration-number is 224, the 
 wave-period is 2-0-4 of a second. 
 
 8. The velocity of the sound is 34,160 cm. See 184 (c). 
 The wave-length is 64.8 cm. x 4 = 259.2 cm. 34,160 -5-259.2 = 
 131.8, the computed vibration-number. Under the conditions 
 assumed in the text, we must allow for experimental error, 
 and take the nearest whole number for the actual vibration- 
 number. 
 
 9. Knowing the velocity of sound per second at a given 
 temperature, and the number of vibrations per second that 
 the sounding body undergoes, it is evident that the wave- 
 length may be determined by dividing the velocity by the 
 vibration-number; i.e., 
 
 n 
 
60 KEY TO THE PROBLEMS 
 
 From 205 (6), we also know that the wave-length equals 
 four times the length of the resonant air column increased by 
 ^ of its diameter; i.e., 
 
 =<<+!} 
 
 Equating these values of w, we have 
 
 -V-; whence v = (l + $}n. 
 
 4:J n V 4 / 
 
 10. w = 4f 50 + - ) = 204, the wave-length in centimeters. 
 
 11. There are two beats per second. 512 2 = 510, the 
 vibration-number of the loaded fork. 
 
 Page 266. 
 
 1. Ignore the number of vibrations, and see 210. 
 
 2. (a) The vibration-number is 50. See 210 (1). 
 ' (6) The vibration-number is 200. 
 
 3. See relative vibration-numbers, 196, and 210 (1). 
 
 27 : 24 : : 18 : x; x = 16, the length in inches. 
 
 1 00 
 
 4. = = 25, the vibration-number. 
 
 2xV4 
 
 5. See 196 and 210 (2). 
 
 3:4::V4:Vz; = 7|lbs. 
 
 6. 3 ft. : 5 ft. : : 256 : x; x = 426f, the vibration-number. 
 
 7. 2:1:: Vl6 : V#; x = 4, the number of pounds. 
 
 8. (a) The wave-length = 1,120 ft. -5-32= 35 ft. The length 
 of the air-column is half the wave-length. See 212 (a). 
 35 ft. -5- 2 = 171 ft., the length of the tube. 
 
 (6) The wave-length = 1,120 ft. -=- 4,480 = J- ft., or 3 in. The 
 length of the air-column is half the wave-length, or 1^ in. 
 
IN AVERY'S SCHOOL PHYSICS. 
 
 61 
 
 9. (a) The length of the air-column in an open pipe is 
 half the wave-length. 4 ft. -5- 2 = 2 ft., the length of the open 
 pipe. 
 
 (6) The length of the air-column in a stopped pipe is a 
 fourth of the wave-length. 212 (a). 4 ft. -s- 4 = 1 ft., the 
 length of the stopped pipe. 
 
 10. The catgut string will vibrate three times as rapidly. 
 
 V9 = 3. 210 (4). 
 
 11. There would be marked differences in pitch. The ve- 
 locity of sound is greater for hydrogen than it is for air, and 
 greater for air than it is for carbon dioxide. See 184. For 
 a given frequency of vibration, the greater the velocity, the 
 greater the wave-length. The length of the air-column has a 
 definite relation to the wave-length. See 212 (a). Conse- 
 quently, a change of gas that increases the wave-length must 
 be accompanied by an increase in the length of the organ- 
 pipe, or by a rise of pitch. As the velocity of sound in 
 hydrogen is almost four times as great as it is in air, the 
 tone emitted by a pipe filled with 
 
 hydrogen is nearly two octaves 
 above that produced by the same 
 pipe filled with air. For a similar 
 reason, the tone yielded by the pipe 
 when filled with carbon dioxide is 
 of lower pitch. 
 
 12. Drive the handle of a pocket 
 tuning-fork into a hole bored in a 
 block of wood, and incline the fork 
 as shown in the accompanying fig- 
 ure. Suspend a varnished cork-ball, 
 the size of a pea, by a silk fiber or 
 
 very fine thread, sound the fork and at once bring the ball 
 against the face of a prong. Hunt for a position in which 
 the ball will remain at rest touching the face of the prong. 
 
62 KEY TO THE PROBLEMS 
 
 One or more such places may be found just above the crotch 
 of the fork. These places of rest are nodes. Then slowly 
 raise the ball, keeping it in contact with the fork; it will 
 begin to tremble. Raise it higher until, near the end of the 
 fork, it is driven violently away. 
 
IN AVERY'S SCHOOL PHYSICS. 63 
 
 CHAPTER IV. 
 Page 274. 
 
 1. 36 -r- 1.8 = 20, the difference in centigrade degrees. 
 
 2. 35 x 1.8 = 63, the difference in Fahrenheit degrees. 
 
 3. (a) 68-32=36; 36 -=-1.8=20, the centigrade reading. 
 
 (6) 68 F. 
 
 4. 10 C. 
 
 5. The mercury would stand higher, and the readings 
 would indicate temperatures higher than the real tempera- 
 tures. 
 
 6. The error o>: the readings would increase as the tempera- 
 ture became higher. 
 
 7. 15 x 0.8 = 12, the Eeaumur reading. 
 
 8. It would indicate that the paint increased the thermal 
 emission of the tin. 
 
 9. The expansion of the flask increases its capacity before 
 the contained liquid expands. 
 
 10. If the bulb remained as large, a diminution of the bore 
 would increase the sensitiveness of the instrument. 
 
 11. Perhaps a little vapor of mercury ; otherwise it is a high 
 vacuum. 
 
 12. The flow from the warmer to the cooler body will vary 
 with the difference of the temperatures. 
 
 13. There are none. See 222. 
 
 14. See 222 (a). 
 
64 KEY TO THE PROBLEMS 
 
 Page 282. 
 
 1. Because the cloth is a poor conductor of heat. 
 
 2. Because the oil-cloth is a better conductor of heat than 
 the carpet is. 
 
 3. Non-conductor in each case. 
 
 4. The confined air is a good non-conductor. 
 
 5. Because of the convection currents thus set up. See 
 227 and 228. 
 
 6. See answer to Exercise 4. The animal heat is better 
 retained thereby. 
 
 7. Copper is a better conductor than tin. 
 
 Page 286. 
 
 1. See 230 (a). The contraction of the cooling iron forces 
 the parts of the wheel closely together ; the cooled iron holds 
 them there. 
 
 2. Such walls generally bulge outward. See 230 (a). 
 
 3. To provide for the inevitable expansion and contraction 
 caused by variations of temperature. 
 
 4. Because the contraction of the alcohol in cold weather 
 exceeds the diminution in the capacity of the measure. 
 
 5. See 228. 
 
 6. The surface temperature is 0. The water at the bottom 
 of the pond at the same time may be 4. See 232 (a). 
 
 7. 273. See 232. 
 
 12. 273 : 273 + 300 : : 900 : x. 
 
 13. 273 : 273 + 100 : : 1,000 : 1,366.3 ; 1,366.3-1,000=366.3. 
 
 14. 273 + 50 : 273 + 15 : : 1 5,000 : x. 
 
IN AVERY'S SCHOOL PHYSICS. 65 
 
 15. 185 F. =85C. 
 
 273 + 85 : 273 + 10 : : 98 : x. 
 
 16. 273 + 10:273 + 18.7) ,^. x 
 
 590 : 530 j 
 
 x = 143.5 + , the number of cu. cm. 
 
 17. 273: 273 + 60:: 231:*. 
 
 18. The water contracted, and the level in the tube fell as 
 the temperature fell to 4. As the water cooled from 4 to 0, 
 it expanded, and the level rose. See 232 (a). 
 
 Page 295. 
 
 1. By confining the steam, thus increasing the pressure on 
 the liquid and raising its temperature. See 240 (a). A 
 vessel designed for this purpose is called a digester. 
 
 2. See answer to Exercise 1. 
 
 3. By pumping out the steam as fast as it is formed, and 
 thus reducing the pressure on the liquid. See 240 (3). A 
 vessel designed for this purpose is called a vacuum-pan. 
 
 4. No. 
 
 5. It expands or it would not float. It resembles ice in 
 this respect. As it expands in the mold, it fills every part of 
 the mold, in consequence of which the face of the casting is 
 sharply defined. Lead and iron when similarly cast lack the 
 clear-cut outlines of ordinary type. 
 
 6. They are due to differences in atmospheric pressure. 
 See 240 (3). 
 
 7. See 235 (3). The melting-point is lowered by the pres- 
 sure, and. the ice melts at the surfaces of contact. The water 
 thus produced has the temperature of the ice from which it 
 came. When the pressure is removed, the melting-point 
 becomes higher than the temperature of the water, which, 
 consequently, freezes. 
 
 5 
 
titi KEY TO THE PROBLEMS 
 
 8. The air is said to be too dry ; i.e., the quantity of watery 
 vapor should be increased as the temperature is increased. 
 This may be done by keeping an open vessel of water on the 
 stove or in the hot-air chamber of the furnace. 
 
 9. At 115. See 235 (1). 
 
 10. See 241. 
 
 11. See 240 (c). 
 
 12. See 240 (6). 
 
 13. Some of the water passes through the vessel and evapo- 
 rates. The heat that produces this change of condition dis- 
 appears in the process. See 236. This withdrawal of heat 
 lowers the temperature of the vessel and its contents. This 
 exercise and the one following may well be reviewed after the 
 study of latent heat. 
 
 14. When the water freezes it gives out heat. See 234. 
 When the ice melts the heat that does the work disappears in 
 the process. See 233. 
 
 ' 15. They are cases of sublimation. See 240 (d). 
 16. An increase of the humidity lowers the dew-point. 
 
 Page 3O3. 
 
 1. Find the number of calories that may be furnished by 
 the several quantities of water in cooling to any given tem- 
 perature, as C. 
 
 1 Kg. at 40 gives 40 large calories. 
 
 2 " " 30 " 60 " " 
 
 3 20 " 60 " " 
 
 4 <c 10 40 " " 
 10 " " 200 " " 
 
 These 200 heat-units would warm the 10 Kg. of water 20 
 above the assumed temperature, or to 20. It makes no differ- 
 
IN AVERY'S SCHOOL PHYSICS. 67 
 
 ence what temperature is chosen for the reduction. If, e.g., 
 we try a temperature of 10 we shall have (30 +- 40 + 30 + =?) 
 100 heat-units. This quantity of heat will warm 10 Kg. of 
 water 10 above the chosen temperature, or to 20. If a still 
 higher temperature is chosen for the reduction, care being 
 given to the algebraic signs, the result will still be the 
 same. 
 
 2. Instead of the water-gram-degree unit, or calory, we 
 may use the water-pound-degree unit. Let x represent the 
 specific heat of the mercury. The mercury was cooled from 
 20 to O.G34 , or 19,366. 1 x 19.3G6 x x = the number of 
 heat-units given up by the lead ; 1 x 0.634 x 1 = the number 
 of heat-units received .by the water. 19.366 x = 0.634 ; x = 
 0.0327. 
 
 3. Let x represent the number of pounds of water required. 
 The water will be cooled to and will part with 85 x heat- 
 units. The melting of the ice will require 80 x 15 heat- 
 units. 85 -x = 1,200 ; x = 14.117 + . 
 
 4. The specific heat of ice being 0.5, it will require 50 heat- 
 units to warm the ice to the melting-point. It will require 
 800 more to melt it. 95 x = 850 ; x = 8.94. 
 
 5. The specific heat of steam is 0.48. Each pound of 
 steam, in cooling to 100, would yield 12 heat-units. These, 
 with the heat from condensation and cooling to 25, would 
 yield (12 + 537 + 75=) 624 heat-units. The required quan- 
 tity of steam must furnish 624 x heat-units. To warm the 
 ice to will require 20 heat-units. This, with the heat 
 required for fusion and warming the melted ice to 25 C., 
 amounts to 545 heat-units. 
 
 G24aj = o45; a = 0.87 + . 
 
 6. The gram of steam will yield (537 + 100 =)637 calories. 
 Each gram of ice will require 80 calories. 637 -=- 80 = 7.96, 
 the number of grams of ice. 
 
68 KEY TO THE PROBLEMS 
 
 7. Assume any quantity of ice, as 1 gram. Let x represent 
 the initial temperature of the water. The melting of the ice 
 requires 80 calories, which must be supplied by 1 gram of 
 water in cooling to 0. For 1 gram of water to part with 
 80 calories, it must cool 80. Consequently, the initial tem- 
 perature must be 80 higher than its final temperature. 
 
 8. Let x represent the temperature. Then, 80 + x = 10 
 (20 -a?); a; = 10.9 + . 
 
 9. The water can furnish 60 heat-units for melting ice. 
 This will melt f of a pound of ice. The result will be J of a 
 pound of ice and 5| Ibs. of ice-cold water. 
 
 10. The steam can furnish 6,370 heat-units in cooling to 0. 
 This heat must warm 1,010 g. of water, and can raise it to 
 (6,370 -r- 1,010 =) 6.3+. Or, we may let x represent the tem- 
 perature. Then 10 (637 - x)= 1,000 x; x = 6.3 + . 
 
 11. Let x represent the resulting temperature. 537 4-100 
 x = 10 x. 537 +- (100 - x) = 10 x; x = 58 nearly. 
 
 12. Iron. Water. 
 
 Specific heat, 0.1138 
 
 Weights, 200. 
 
 Changes of temperature, 300 x 
 
 1. 
 1,000. 
 
 x 
 
 6,828 - 22.76 x = 1,000 x; x = 6.67 + . 
 
 13. Let x represent the specific heat. The temperature of 
 the 80 grams falls 80. 
 
 6,400 x = 200 x (20 - 10) ; x = 0.31 + . 
 
 14. The ice requires 8,000 heat-units. Each pound of steam 
 will supply 637 heat-units. 8,000 -j- 637 = 12.55. 
 
 15. It will require (80 x 5=) 400 water-ouiice-degree heat- 
 units to melt the snow. The water can furnish 460 such units. 
 Hence the snow will all be melted, and the water warmed. 
 Let x represent the resultant temperature of the water. 
 
 5 (80 -f x) = 23 (20 - x) ; x = 2.14+. 
 
IN AVEKY'S SCHOOL PHYSICS. 69 
 
 16. The steam must supply 5,000 water-pound-degree heat- 
 units. Each pound of steam can supply (537 -j- 90 =) 627 such 
 units. 5,000 -j- 627 = 7.97. 
 
 17. Let x represent the number of grams of mercury. 
 From Exercise 2, it appears that the specific heat of mer- 
 cury is 0.0327. 
 
 150 x 299 x 0.0314 = 0.0327 x; x = 43,066.97. 
 
 18. Let x represent the resultant temperature. 
 
 4 [537 + (100 - x)-] = 200 (x - 10) -, x = 22.29. 
 
 19. That the specific heat of sulphur is 0.2. 
 
 20. The specific heat of a body represents the number of 
 calories required to raise the temperature of 1 gram of that 
 substance 1 degree, and will be the same whatever the mass of 
 the body. The thermal capacity of the body introduces mass 
 as a factor. See 248. 
 
 21. Much less, on account of the high specific and latent heat 
 of water. As will appear further on (see the answer to Exer- 
 cise 12, on page 389), the watery vapor in the earth's atmos- 
 phere acts as a trap for solar heat, or as a huge terrestrial 
 blanket. 
 
 22. Work is done upon the gas; it is liquefied by the 
 pressure put upon it. The latent heat, not being employed in 
 maintaining the aeriform condition, appears as sensible heat. 
 When the liquefied gas expands and resumes its aeriform con- 
 dition, it does work and the reverse thermal effects follow. 
 
 23. See 234. 
 
 24. Work is done upon the air, as in Exercise 22, and the 
 compressed air is heated thereby. The pump is heated by the 
 heated air. Exercises 22 and 24 will be better understood after 
 the section on the relation between heat and work has been 
 studied. 
 
70 KEY TO THE PROBLEMS 
 
 Page 311. 
 
 1. Because of this great latent heat of water, the processes 
 of melting ice and freezing water are necessarily slow. Other- 
 wise the waters of our northern lakes might freeze to the bottom 
 in a single night, while "the hut of the Esquimaux would 
 vanish like a house in. a pantomime," or all the snows of winter 
 be melted in a single day, with inundation and destruction. 
 
 3. The work done by the engine is (8,540 x 50=) 427,000 
 kilogrammeters, or 427,000,000 grammeters. As the mechan- 
 ical equivalent of a calory is 427 grammeters, the work per- 
 formed is equivalent to 1,000,000 calories, or to 1,000 large 
 calories. 
 
 4. See 252. 34,462 -=- 1,000 = 34.462. The number of 
 calories developed would raise the temperature of 1,000 g. of 
 water 34.462. 
 
 5. 80 x 1,390 = 111,200, the number of feet (in vacuo) that 
 the ice must fall. 
 
 6. See 247 (6). 0.1138 x 100 = 11.38, the number of 
 water-pound-degree heat-units required. Each such unit is 
 equivalent to about 1,400 foot-pounds, a total of (1,400 x 
 11.38 =) 15,932 foot-pounds, or nearly enough to raise 8 tons 
 a foot high. 
 
 7. 8,080 x 5 x 1,400 -3- 2,000 = 28,280, the number of feet. 
 
 8. 88.42% of 5 Ibs. = 4.421 Ibs., the quantity of carbon. 
 
 5.61% of 5 Ibs. = 0.2805 Ibs., the quantity of hydrogen. 
 8,080 x 4.421 x 1,400 = 50,010,352. 
 
 34,462 x 0.2805 x 1,400 = 13,533,227.4. 
 Total number of foot-pounds, 63,543,579.4. 
 63,543,579.4 -=- 2,000 = 31,771.8, the number of feet. 
 
 9. The weight makes no difference, as an increase in the 
 weight would increase the working power and the work to be 
 done at the same rate. It may be called 1 gram (or anything 
 else). 
 
IN AVERY'S SCHOOL PHYSICS. 71 
 
 That weight of water heated 100 would require 100 calories. 
 That weight of lead heated 100 would require (0.0314 x 100 =) 
 3.14 calories, equivalent to 1,340.78 grammeters. See 85. 
 
 K.E.=; 1,340.78 = :-; v = 162.108, the velocity in 
 
 LI o j.y.t) 
 
 meters. 
 
 10. 6(442 - 374) x 0.056 = 22.848, the number of water- 
 pound-Fahrenheit degree heat-units required for heating the 
 tin to its melting-point. 25.6 x 6 = 153.6, the number of 
 such units required for melting the tin. The total of such 
 units required is 176.448. The mechanical equivalent of each 
 such unit is 778 foot-pounds ( 251). 778 foot-pounds x 176.448 
 = 137,276.544 foot-pounds. 
 
72 
 
 KEY TO THE PROBLEMS 
 
 CHAPTER V. 
 Page 323. 
 
 1. (a) The shadow is the frustum of a cone, the curved 
 surface of which is bounded by rays that are tangent to the 
 ball. There is no penumbra, and the crosS-section increases as 
 the distance from the ball increases. See Fig. 233. 
 
 (b) The rays that come from each luminous point and that 
 are tangent to the ball form a limiting surface as above- 
 described. The umbra is the frustum of a cone, and is sur- 
 rounded by a penumbra. See Fig. 234, and the left-hand end 
 of the accompanying figure. The cross-section increases with 
 the distance from the ball. 
 
 (c) The shadow, including umbra and penumbra, is the frus- 
 tum of a cone as before. The umbra is a cylinder. Have 
 each pupil draw a figure resembling Fig. 234, replacing the 
 lamp by a ball at L, with diameter equal to that of the ball, AB. 
 
 (d) The shadow, including umbra and penumbra, is the frus- 
 tum of a cone as before. The umbra is a cone with its base at 
 the intercepting ball. Have each pupil draw a figure as for 
 (c), but making the ball at L larger than the ball at AB, as in 
 the right-hand end of the figure on this page. 
 
IN AVERY'S SCHOOL PHYSICS. 73 
 
 2. Water waves ; because the motion of the particles 
 involved in the wave are transverse as it is in the case of 
 light, while in sound they are longitudinal. 
 
 3. See 254. 
 
 4. See 265. 
 
 6. The diameter of the shadow will be 5 times that of the 
 coin. The area of the shadow will be 25 times that of the 
 coin. 
 
 7. The wall is 100 times as far from the eye as the screen 
 is. The screen has an area of 9 sq. in. 9 sq. in. x 100 2 = 
 90,000 sq. in. 
 
 8. See Experiment 211. The distance of the lamp is three 
 times that of the candle. The lamp is (3 2 =)9 times as power- 
 ful ; i.e., it has 9 c. p. 
 
 9. See Experiment 212. One is four times as powerful as 
 the other. 
 
 10. The spot is translucent, and transmits more light to the 
 eye than the rest of the paper. 
 
 11. As the translucent spot transmits more light than the 
 rest of the paper, it reflects less ; the same light cannot be 
 transmitted and reflected. 
 
 12. There is no penumbra. 
 
 13. See answer to Exercise 1 (d). 
 
 14. The umbra has a finite length when the luminous body 
 is larger than the intercepting body; under other circum- 
 stances, its length is infinite. 
 
 15. (a) It will be in the umbra near the apex. 
 
 (6) It will be in the penumbra at one side of the apex 
 of the umbra. 
 
 16. Small density and high elasticity. The formula given 
 in 184 applies to the medium of any vibratory motion. 
 
74 
 
 KEY TO THE PROBLEMS 
 
 1. 45. 
 
 Page 34O. 
 
 2. In the accompanying figure, the rays that form the 
 image for the right eye are marked with single arrowheads ; 
 
 , E those that form the image for 
 the left eye, with double arrow- 
 heads. 
 
 3. (a) See 280 (5) ; 
 
 (b) See 280 (6) ; 
 
 (c) See 280 (4) ; 
 
 (d) See 280 (2) ; 
 
 (e) See 280 (3). 
 
 6. See the lines with double 
 arrowheads in Fig. 256. 
 
 7. See 275. 
 
 8. (a) See 283 ; (b) see 281. 
 
 9. He cannot in either case. Whatever his position, the 
 image will appear as far back of the mirror as the man is in 
 front of it, and the part of the mirror used to give a complete 
 image will be half the length of the man. If the mirror is not 
 half the length of the man, it cannot give a full-length image 
 of him, no matter what his distance from it. Let MN repre- 
 
 sent the mirror, AB the man, and ab his image. The triangles, 
 Aei and Aab, are similar and Ae = Aa; therefore, ei \A~B. 
 But ei is the part of the mirror used in forming the image. 
 Suppose now the man to move either way, as to CD. The 
 image appears at cd, and Ce 1 Cc. Hence, ei = cd or J CD. 
 
IN AVERY'S SCHOOL PHYSICS. 
 
 75 
 
 \v 
 
 Wherever he stands, he cannot see his complete image unless 
 the length of the mirror is half his own length ; if it is too 
 short in one case, it will be too short in every such case. 
 
 10. 60. 
 
 11. Let MN represent the mirror, EE' the two eyes, and ee' 
 the two images of the eyes. Of course, EE' are in the same 
 horizontal line. When E' is 
 
 closed, E sees the image of E' 
 
 at e'. Placing the wafer at W, 
 
 hides e'. When E is closed, 
 
 E' would see the image of E 
 
 at e, were it not for the wafer 
 
 which hides it. See 276. 
 
 By drawing the parallelogram, 
 
 EE'e'e, and remembering that 
 
 MN is midway between EE' 
 
 and ee', it may be proved, geometrically, that Ee' and E'e will 
 
 intersect at W as represented in the figure. 
 
 12. Construct the figure, and measure the lengths of object 
 and image. Modify Fig. 256, by placing OB as far at the left 
 of the mirror as C is at the right. 
 
 13. f = ' Either conjugate focal distance equals the 
 
 product of the other conjugate focal distance into twice 
 the principal focal distance, divided by the difference between 
 twice the other focal distance and twice the principal focal 
 distance. Notice that r is twice the principal focal distance. 
 
 14. Since the sum of the reciprocals of / and /' is constant, 
 
 2 
 i.e., -, it necessarily follows that if one increases the other 
 
 must decrease. When either conjugate focal distance, as /, 
 becomes infinite, its reciprocal becomes zero. Substituting 
 this value in the formula given in 280 (c), we have 
 
 1919 v 
 
 + - = -: - = -: f = -- 
 
 V r' f r' J 2 
 
76 
 
 KEY TO THE PROBLEMS 
 
 15. Such a negative value indicates that the focus is. back of 
 the mirror ; i.e., that it is virtual instead of real. 
 
 16. Locate the three points on paper, and letter them. Draw 
 the lines, AC and BC. Bisect the angle, ACB, by the line, 
 CD. Through C, draw the line, mr, perpendicular to CD. 
 This line, mr, indicates the position of the mirror. The angle, 
 ACD, is the angle of incidence. The angle, BCD, is the angle 
 of reflection. 
 
 17. If the middle of a plane mirror is placed opposite the 
 corner of a building so that the line of the face of the mirror 
 makes an angle of 45 with the prolongation of the sides of 
 the house, as shown in the accompanying figure, a person at A 
 might see an image of a person at B. A rectangular box may 
 
 be made with openings, as shown at a and b, in the second 
 figure accompanying. Mirrors may be placed so as to make 
 angles of 45 with the sides of the box, as shown at mmmm. 
 An object at a may be seen by an eye at b, the light being re- 
 flected by each mirror. The experiment is made more interest- 
 ing by placing an opaque object, as a brick, at B. 
 
 18. It must be vertical. 
 
 19. Two mirrors at an angle of 60 with each other. 
 
 20. If the rays that are reflected by the outer parts of the 
 mirror are removed, the foci of the others will not vary much 
 from F. 
 
 21. Try it with a bit of looking-glass and a pin. The image 
 will be horizontal. 
 
IN AVERY'S SCHOOL PHYSICS. 
 
 77 
 
 22. See 270. The small particles of water suspended in 
 the air reflect light, but only those that are in certain positions 
 relative to the observer can reflect light to his eye. The circle 
 sometimes seen around the moon is similarly produced. Com- 
 pare 303 and 315 (a). 
 
 FIG. A. 
 
 Page 363. 
 
 1. In the accompanying figure, ab represents the plane of 
 the horizon for an observer at E. A ray of light coming to the 
 
 eye from the sun 
 
 tf .;n i^ r^^A *~ s^^^>^ 
 
 at $ will be grad- 
 ually bent, describ- 
 ing a curve. The 
 direction in which 
 the sun is seen is 
 that of a tangent 
 to this curve at the 
 eye. This makes the sun appear higher than its true position, 
 as at /S". Similar lines drawn toward the right from E will 
 represent the morning phenomenon. 
 
 2. See 290 (). The pupil is supposed to know that a 
 burning-glass is a single convex lens. If he does not know it, 
 the teacher should send him to the dictionary or other source 
 i of information. 
 
 4. (a) See 285, and Ex- 
 periments 233 and 234. 
 
 (b) See Tig. 271, especially 
 the limiting lines, Aa and Ee. 
 
 5. See 285 <7>) and 284 
 (c). The ratio of the two 
 radii is 4 : 3. The angle BAD 
 is the angle sought. 
 
 6. See 288 (6). 
 
 Fia. B. 
 
78 
 
 KEY TO THE PROBLEMS 
 
 7. Draw CC' through the centers of curvature. Draw AB, 
 the incident ray, and BC. The problem is to construct BD 
 and DA', the paths of the ray through the lens and beyond. 
 With B as a center and with radii of 2 and 3, draw the arcs, n 
 and m. From the intersection of n and AB, draw a line 
 parallel with BC, and cutting the arc, m. From the point 
 
 where this line 
 cuts m, draw a 
 line through 72 
 to D. The line, 
 BD, is the path of 
 the ray through 
 the lens. Draw 
 DC 1 . From D as 
 a center and with 
 radii of 2 and 3, 
 draw the arcs, y and x. Produce BD, and from its intersection 
 with x, draw a line parallel to DC' and cutting the arc, y. 
 Through the point 
 where this line cuts 
 y, draw DA', which 
 line marks the path 
 of the ray beyond 
 the lens. 
 
 -. 
 
 ^ 'f*T''=-fc-' > ->P/\p ---"*" ''\ 
 
 8. See Fig. D. 
 Draw AB parallel to 
 CC', and complete 
 the construction as 
 in Exercise 7. 
 
 9. See Fig. E. 
 Draw AB parallel to 
 the principal axis as 
 before, and complete 
 
 the construction. The focus at F is virtual. 
 
 See 292, b (1). 
 
AVERY'S SCHOOL PHYSICS. 
 
 79 
 
 10. The optical center. 
 
 11. See 291 (6). 
 
 (d) The method of construction is fully illustrated by 
 Fig. F. The effect of moving the object toward the principal 
 
 FIG. F. 
 
 focus and the lens, the image receding as the object ap- 
 proaches, is illustrated by Fig. Gr. 
 
 FIG. G. 
 
 (e) The emergent rays are parallel, as is illustrated by 
 Fig. H, and form no image. 
 
 H. 
 
80 KEY TO THE PROBLEMS 
 
 (/) Magnified, erect and virtual, as illustrated by Fig. I. 
 
 FIG. I. 
 
 12 (a) See 291 (6) (3). 
 
 (6) See 291 (ft) (2). Construct the image on the scale 
 of 1 : 1. 
 
 13. The focus is virtual. See Fig. J. 
 
 FIG. J. 
 
 14. Secondary foci ; they will be equal ; construct the image. 
 
 15. Inasmuch as the incident rays are diverging and the 
 focus is real instead of virtual, it is apparent that the lens is 
 convex. See 292 (7>) (2). The focal distance is 8 inches. 
 
 16. (a) Five feet from the flame. See 291 (6) (4). Notice 
 the principle of reversibility that prevails in optics. 
 
 (5) One will be five times as long as the other. Their 
 lengths are proportional to their distances from the lens. 
 
 17. Make the experiment, and see Experiment 230. 
 
IN AVERY'S SCHOOL PHYSICS. 81 
 
 18. Part of the incident light is reflected at each surface of 
 eat-h particle. What escapes reflection by one particle is 
 reflected by some other particle. All of the light being thus 
 reflected, none is transmitted, and the glass is, therefore, 
 opaque. 
 
 Page 389. 
 
 1. Because of the uniform angular distance from the axis 
 of the bow, as explained in 303 (c). 
 
 2. See 299 (a). 0.007 x 0.3937. 
 
 3. 186,000 x 5,280 x 12 - 0.00002. 
 
 4. See 294 and 298 (a). 
 
 5. Dispersion. See 368. 
 
 6. See 303 (a). 
 
 7. See 307 and 308 (3). 
 
 8. See 308. 
 
 9. Because the air is diathermanous ; i.e., it transmits the 
 radiant energy instead of absorbing it. 
 
 10. The glass is diathermanous to the short-wave radiations 
 that constitute the sunshine, and athermanous to the long-wave 
 radiations emitted by the objects within the greenhouse after 
 they have been heated by absorbing the solar energy. 
 
 11. Because the watery vapor in the atmosphere transmits 
 to the earth the short-wave radiations of the sun and absorbs 
 the long-wave radiations from the earth. See 309(6). 
 
 12. Because of the diathermancy of dry air and the 
 athermancy of moist air, as stated in 309 (b). When the 
 night is clear, the earth's heat is rapidly radiated into space ; 
 clouds act as a blanket and absorb the energy radiated from 
 the earth, and return some of it to the earth. 310. 
 
 13. The glass is diathermanous to short-wave radiations. As 
 the energy is freely transmitted, it cannot also be absorbed. 
 Only the absorbed energy heats the medium. 
 
82 KEY TO THE PROBLEMS 
 
 14. That water transmits short-wave radiations and absorbs 
 long-wave radiations. 
 
 15. " Radiant heat " is used as a synonym for radiant energy, 
 and " obscure heat " as a synonym for radiations of greater wave- 
 length than any that constitute a part of the visible spectrum, 
 neither of which is heat at all. See 225. 
 
 16. If the radiations from the lamp are passed through a 
 solution of iodine in carbon disulphide, the short-wave energy 
 will be absorbed and the long-wave energy transmitted. See 
 309 (b) and 313. If the radiations are passed through a 
 solution of alum in water, the long-wave energy will be absorbed 
 and the short-wave energy transmitted. 
 
 17. Because its polished surface is a poor radiator ( 311). 
 
 18. The convection currents of heated air have varying 
 densities and refractive powers. 
 
 19. See answer to Exercises 11 and 12. 
 
 Page 392. 
 
 10. A Plticker-tube is a Geissler tube (Fig. 409) with a capil- 
 lary part, by means of which the luminous intensity of feeble 
 electric discharges may be raised sufficiently to allow of spectro- 
 scopic investigation. 
 
IN AVERY'S SCHOOL PHYSICS. 
 
 CHAPTER VI. 
 Page 43O. 
 
 1. See Experiment 298. 
 
 2. See 329 (a). 
 
 3. See 329. 
 
 4. The opposite character of the electrifications is indicated 
 by the attraction between the nibbed body and the body with 
 which it is nibbed ; the equality of the two electrifications is 
 indicated by the fact that when the two bodies are brought 
 together, the two electrifications mutually neutralize each other. 
 
 5. To avoid the condensation of moisture from the air. See 
 326 (a). 
 
 6. Since the charges are of opposite signs, the force will be 
 attractive and not repellent. See 332. 
 
 Qxry_24x8_ 1? 
 
 d 2 4 2 
 
 Our units are all C.G.S. units. By recognizing the algebraic 
 signs, we have +24x(-8)_ 
 
 42 
 
 which also indicates an attractive force. 
 
 7. The 8 units will neutralize an equal number of the 
 + units, leaving -f- 16 units to be equally divided between the 
 two balls (which are assumed to be of equal capacity). 
 
 d 2 4 2 
 
 As the algebraic sign of the answer is + (the charges being 
 alike), the force is one of repulsion. 
 
KEY TO THE PROBLEMS 
 
 .'.32 = 
 
 28^<_56 
 d 2 
 
 Whence, d 2 = 28 x . 56 = 49. 
 
 32 
 
 = 7. 
 
 9. (a) See Fig. 321. Place the dozen globes in actual con- 
 tact. They will be polarized as a single body, and may all be 
 charged as described in Experiment 306. (6) Charge one 
 negatively by induction ; with it, charge another positively by 
 induction. For example, give a positive charge to (7 and bring 
 it near M, thus polarizing the latter as shown in the figure. 
 
 Touch M with the finger, thus leaving it negatively charged. 
 Remove (7, and bring N into position, as shown in the figure. 
 N will be polarized by the negative charge of M. Touch N 
 with the finger and its negative electrification will be repelled 
 to the earth, leaving the conductor positively charged. 
 
 xo. 
 
 11. By bringing the charged conductor into simultaneous 
 contact with two conductors of equal capacity. 
 
 12. See 333 (a). 
 
 13. See 343 (a). 
 
IN AVERY'S SCHOOL PHYSICS. 85 
 
 14. The shock will be greater when the jar is held in the 
 hand. When the outer coat is insulated, its charge "binds" 
 part of the charge of the inner coat. 
 
 15. The divergence will be greater when the jar is held in 
 the hand, for the reason just given. 
 
 16. When an electrified glass rod is brought near an electric 
 pendulum, the pith-ball is polarized, as shown in the figure. 
 As the distance between the opposite electrifications is less 
 than the distance between the similar elec- 
 trifications, the attraction is greater than 
 
 the repulsion. If the pith-ball is sus- 
 pended, not by a silk thread but by somo 
 good conductor, the attraction will be more 
 marked, for the + of the ball will escape to the earth through 
 the support, and thus the repelling component will be removed. 
 
 Page 442. 
 
 1. The table gives it as 1.023 ohms. 
 
 2. The resistance of 1,000 feet of No. 4 copper wire is 
 0.254 of an ohm ; that of 800 feet of such wire is 0.1932 of an 
 ohm. Multiplying this by the ratio between the resistivities 
 of copper and of German-silver as given in the Appendix (page 
 592), we have 0.1932 x 128.29-^-10.45 = 2.374, the number of 
 ohms. 
 
 3. The resistance of that length of No. 8 copper wire is 
 0.643 of an ohm x 0.750 = 0.48225 of an ohm. Multiply this by 
 
 f* Q Q ^ 
 
 '-;. the ratio between the resistivities of copper and of iron. 
 10.45 
 
 4. The resistance of that length of No. 14 copper wire is 
 
 O Q A 
 
 2.585 ohms x 0.35 = 0.90475 of an ohm. Multiply this by ~ 
 the ratio between the resistivities of copper and of silver. 
 
 5. 33.135 ohms x 6.050 = 200.46675 ohms. 
 
86 KEY TO THE PROBLEMS 
 
 6. The table shows that 386.8 ft. of No. 14 copper wire 
 has a resistance of 1 ohm. The given resistance of 1.75 ohms 
 indicates that the fault would be 386.8 ft. x 1.75 = 676.9 ft. 
 from the end of the line if the wire was of copper. Correct 
 
 this result by multiplying 676.9 ft. by : , the ratio between 
 
 63.35 
 the resistivities of iron and of copper. 
 
 7. The resistance of 3,590 ft. of No. 14 copper wire is 2.585 
 
 ohms x 3.59 = 9.28 ohms. Multiply this by 63 ' 35 , the ratio 
 
 10.45 
 between the resistivities of copper and of iron. 
 
 8. The length of a No. 14 copper wire that has a resistance 
 of 1.75 ohms is (386.8 x 1.75=) 676.9 ft., as given in the answer 
 to Exercise 6. 
 
 9. The table shows that the resistance of No. 13 copper 
 wire is very nearly 2 ohms per thousand feet, and that the 
 diameter of such a wire is 0.071961 of an inch, or 1.828 milli- 
 meters. For exact determination, apply the principle set forth 
 in 352 (2) : 2 ohmg . 2M8 ohmg . . (L828)2 . ^ 
 
 The value of x will represent the diameter in millimeters. 
 2 ohms : 2.048 ohms : : (71.961) 2 : y*. 
 
 r j?he value of y will represent the diameter in mils. 
 
 A shorter solution may be had by using the formula for 
 resistivity, given on page 592 : 
 
 2 = 10.45x1,000. d = 72+mils . 
 
 The difference in the two results is partly due to the fact that 
 the resistivity as given in the table was computed for the tem- 
 perature of 20, while the resistance was computed for the 
 temperature of 24. 
 
 10. See the formula for resistivity given on page 592. 
 
 ou 
 This resistivity indicates German-silver as the metal. 
 
IN AVERY'S SCHOOL PHYSICS. 87 
 
 Page 4-5O. 
 
 5. See 81 (a). 
 
 746 watts = 1 H.P. = 33,000 foot-pounds per minute. 
 33,000 -=- 746 = 44.236. 
 
 6. 9.6 x 2,900 -r- 746 = 37.32. 
 
 7. 37.32 - 0.90 = 41.46. 
 
 8. (a) 0=^; 21 = ; 72 = 0.07143. 
 
 R R 
 
 (b) 21 x 1.5 = 31.5. 
 
 (c) There will be 1,600 ft. of line wire, the total resist- 
 ance of which is indicated by the answer to (a). This repre- 
 sents a resistance per 1,000 ft. that is less than that of the 
 largest wire mentioned in the table. 
 
 9. (a) 110 x 0.5 -r- 16 = 3.4375. 
 (6) 45x10-2,000 = 0.225. 
 
 10. There is no difference : 
 
 900 x 10 = 1,800 x 5. 
 
 11. 110 x 0.5 x 50 -5- 0.90 -r- 746 = 4.09. 
 
 12. See 361 (6) ; W= R x C 2 = 23 x 3.5 2 = 281.75. 
 
 13. TT= = = 327.02. 
 
 14. ( a ) 350,000 = 35 the number of amp eres. 
 
 10,000 
 
 (b) 35 x 14 = 490, the number of volts. 
 
 (c) 490 x 35 = 17,150, the number of watts. 
 
 (d) The 525 H.P. expended on the dynamo is equivalent 
 to (746 x 525 =) 391,650 watts. 
 
 350,000 -5- 391,650 = 0.893+ . Ans. 89.3+ per cent. 
 
88 KEY TO THE PROBLEMS 
 
 Page 476. 
 
 1. The non-active part that is about midway between the 
 poles. See 367. 
 
 2. See 369 (d). 
 
 3. By the rapidity of oscillation after the needle has been 
 displaced from its chosen position. The greater the intensity 
 of the field, the greater the rapidity of oscillation. " When- 
 ever a body oscillates under the action of a force, the square of 
 the time of a single oscillation is directly proportional to the 
 moment of the inertia of the body about the axis of oscillation, 
 and inversely proportional to the directive force." As the 
 number of oscillations is the reciprocal of the time or period 
 of an oscillation, it follows that the directive force is directly 
 proportional to the square of the number of oscillations. Com- 
 pare the formula in 115. 
 
 4. See Laboratory Exercise 2, on page 476. No magnet can 
 be found that tends, as a whole, to drift toward the north or 
 south. The attraction of the north pole of the earth for the 
 marked pole of the needle is counterbalanced by its repulsion 
 for the unmarked pole of the needle. The accompanying 
 figure shows the way in which the needle is placed in a north 
 
 and south line. The arrow a 
 represents the attraction of 
 the north magnetic pole of the 
 earth, and c represents the re- 
 pulsion of the south magnetic 
 pole of the earth acting upon 
 the marked pole of the needle ; 
 b represents the repulsion of the north pole, and d represents 
 the attraction of the south pole of the earth acting upon the 
 unmarked pole of the needle. The combined effect of these 
 four forces is to place the needle in the magnetic meridian. 
 When the needle is thus placed, since the two ends of the 
 needle are practically equidistant from either pole of the 
 
IX AVERY'S SCHOOL PHYSICS. 89 
 
 earth, a + c = b + d. The two couples thus counterbalance 
 each other. 
 
 5. See 376. 220x3-0.7958 = 829+. 
 
 6. (a) 40 x 5 -T- 0.7958 = 251.3, the number of gilberts. 
 
 (b) 251.3 -r- 0.00593 = 42,378, the number of webers. 
 
 (c) See 369 (a). 43,378 -=- 3 = 14,126, the number of 
 
 gausses. 
 
 (d) See 378. T1 L = 0.000593, the reluctivity. 
 
 (e) 251.3 -r- 30 = 8.37, the number of gausses. 
 
 Page 482. 
 
 = 2-+0.1^ = 
 
 o 
 
 '0.5 
 
 + 500^ = 0.004 nearly. 
 
 = 10 -s- (2.5 + 0.1)= 3.846+. 
 
 = 10 -(2.5 + 500)== 0.0199+. 
 nr + R 
 
 3. (7= 2 -f- (1 + 16) =0.1176. 
 
 4. C= 16-*- (64 + 16) =0.2. 
 
 5. 0=8-5- (16 + 16) = 0.25. 
 
 6. C=l-f- 5.001 = 0.19996+. 
 
 7. C= 1-5- (0.5 + 0.001) =1.996+. 
 
 8. C = 10 -r- (50 + 0.001) = 0.19999+. 
 
 9. C = 1 -- (5 + 1,000) = 0.00099502. 
 
 10. <7 = 1 -&- (0.5 + 1,000) = 0.0009995. 
 
 11. C = 10 -=- (50 + 1,000) = 0.00952, the number of amperes. 
 
 NOTE. Compare the results in Exercises 6, 7, and 8, where we have 
 a small external resistance. Then compare the results in Exercises 9, 
 10, and 11, where we have a high external resistance. 
 
90 KEY TO THE PROBLEMS 
 
 12. (a) Consult the table in the Appendix. 0.643 x 0.006 
 = 0.00386 ; 0.8 x 6 + 0.00386 = 4.80386. 
 
 (b) C = 12 -- 4.80386 = 2.498. 
 
 13. ( a ) 03 + 0.00386 = 0.13719. 
 
 6 
 
 (6) 0=2-5- 0.13719 = 14.6 nearly. 
 
 14. The pupil must judge from the evident analogy, "tan- 
 dem" referring to grouping in series, and "abreast" referring 
 to grouping in parallel. 
 
 15. Test them with a high resistance. 
 
 16. Apply the " rule of thumb " given in Experiment 347, 
 and the " cork-screw rule " given on page 468 to determine the 
 direction of the lines of force. Consider the rectangle as a 
 disk magnet (see Experiment 348), and compare its action with 
 that of a magnetic needle. Consider the solenoid as a bar 
 magnet, determine its polarity, and see 371. 
 
 Page 496. 
 
 1. See 388. 
 
 2. See 395. 
 
 3. See Experiments 315 and 347. 
 
 4. See Experiments 353 and 357. 
 
 5. The cast iron has a residual magnetism that is helpful 
 in starting the machine ( 396 a), and it is cheaper and more 
 easily worked than wrought iron or pure soft iron is. In most 
 other respects, a pure soft iron core would be vastly preferable. 
 
 6. It may depend upon the intensity of the magnetic field, 
 the number of coils on the armature, or the rapidity of revolu- 
 tion of the armature. 
 
 7. The field magnets of one are permanent 5 those of the 
 other are temporary. 
 
 a It diminishes it, by increasing the resistance of the wire. 
 9. See Fig. 388, 
 
IN AVERY'S SCHOOL PHYSICS. 91 
 
 Page 518. 
 
 1. (a) See 401. (6) and (c) See 400 (a). 
 
 2. See 407. 
 
 3. See 400. It weakens the direct current so that it may 
 be unable to force its way through the dielectric, as in the case 
 of the sparks from an induction coil. 
 
 4. (a) See 378 and 379; oersteds = g l] k erts . 
 
 webers 
 
 (6) Draw a right-angled triangle; mark the hypotenuse 
 " impedance," and the other two sides " reactance " and " re- 
 sistance." See 401 (a). 
 
 5. See Fig. 399 and 403(6). 
 
 6. See 404. 
 
 Page 536. 
 
 1. See Fig. 425. Represent the current that flows through 
 nx by the letter c; and the current that flows through mp by 
 c'. See 361 ; E=Cx R. 
 
 The fall of potential from A to C = en; represent it by v. 
 The fall of potential from A to D = c'm; represent it by v'. 
 But when the resistances are adjusted so that there is no 
 flow through G, C and D are at the same potential ; i.e., v = v'. 
 
 Hence, en = c'm, and = 
 m c 
 
 x c' 
 Similarly, it may be proved that - = - Equating these 
 
 i P c 
 
 values of -, we have x , or m : n : : p : x. Q.E.D. 
 
 G m p 
 
 2. That they may absorb only a small part of the energy. 
 If either instrument absorbed much, it would change the value 
 of the function that it is to measure. A voltmeter receives 
 the full voltage ; it should use a small current ; i.e., the resist- 
 ance should be high. An ammeter receives the full current ; 
 its resistance should be low so that the voltage required to 
 force the current through it may not be excessive. 
 
92 KEY TO THE PROBLEMS 
 
 3. (a) C = =- = n/> ' = 0.001423, the number of amperes. 
 R 26,000 
 
 (6) E=CR = 0.003 x 26,000 = 78, the number of volts. 
 
 amperes. 
 
 (6) 40,000 : 25,000 : : 110 : 68.75, the number of volts. 
 (c) 40,000 : 15,000 : : 110 : 41.25, the number of volts. 
 
 R 3,500 + 2 
 peres. 
 
 (b) E=CR = 0.004 x 3,500 = 14, the number of volts. 
 
 (c) 110-14 = 7.85. 
 
 (d) 117 -r- 27,500 = 0.004254, the number of amperes. 
 
 (e) The voltmeter indicates (.#= (7/2=0.004254x117=) 
 14.89 volts. 117 -r- 14.89 = 7.85, the same multiplier as before. 
 
 6. (10 - 1) X 3,500 = 31,500, the number of ohms. 
 
 7. 110 x 0.002 = 0.22, the number of watts. 
 
 8. 6.435 : 0.065 : : 6 : x; x = 0.0606 of an ohm. 
 
 9. Pass a current of known strength through the coil to be 
 measured. Take the readings of a voltmeter at the terminals 
 of the coil. Substitute the known values in the formula, 
 
 B = , and solve for the value of R. 
 G 
 
 10. See Fig. 387. Connect the terminals of the voltmeter 
 to the main-circuit wire each side of the field magnets. 
 
 Page 575. 
 
 1. C= = 838 ' 44 = 10.04. The problem ignores 
 
 R 4.56 x 16 + 10.55 
 
 the resistance of the line, i.e., assumes that the circuit is short 
 and of inconsiderable resistance. 
 
 2. See 400. 
 
IN AVERY'S SCHOOL PHYSICS. 93 
 
 3. See 414(. 
 
 9.925 : x : : tan 60 : tan 74. 
 9.925 : x : : 1.73 : 3.49 ; x = 20+. 
 
 4. R = = = 30, the number of ohms. 
 
 C 1 
 
 5. (a) a = = = 0.46 nearly. 
 
 (&) 110 x 0.46 = 50.6. 
 (c) 50.6 -=- 16 = 3.16. 
 
 6. The resistance of the series of lamps will be 250 ohms. 
 That of the wire may, then, be 5 ohms. This is at the rate of 
 25 ohms per 1,000 ft. No. 24 is the nearest to the size desired, 
 but as the line resistance " must not be more " than 5 ohms, 
 we must use the next larger size, or No. 23, copper wire. 
 
 7. The resistance of the lamp circuit will be 2.5 ohms, and 
 that of the 200 ft. of wire 0.05 of an ohm. This is at the 
 rate of 0.25 of an ohm per 1,000 ft. The difference between 
 this desired resistance and that of 200 ft. No. 4 of copper wire 
 is inconsiderable and may be ignored. In practice, a differ- 
 ence like that of Exercise 6 would probably be ignored, and 
 No. 24 wire used. 
 
 8. E = CR = 10 x 3.8 = 38. 
 
 9. The total resistance of the lamp, including the arc, is 
 3.8 ohms + 0.62 of an ohm = 4.42 ohms. 
 
 E = CR = 10 x 4.42 = 44.2. 
 
 10. 206 -r- (1.6 + 25.4) = 7.63, the number of amperes. 
 
 11. (2.8 + 1.1 -f 9.36) x 14.8 = 196.248. 
 
 12. It is not running .fast enough. With the given resist- 
 ances, a 25-ampere current will require an E.M.F. of 212.5 
 volts. The dynamo must be " speeded up " so as to give the 
 additional 12.5 volts. 
 
94 KEY TO THE PROBLEMS 
 
 13. 81.58 -r- 29.67 = 2.75, the total number of ohms. 
 2.75 ohms 1.14 ohms = 1.61 ohms. 
 
 14. The total resistance of the circuit is (157.5 -* 17.5 =) 
 9 ohms. 9 ohms 4.58 ohms = 4.42 ohms. 
 
 15. (39.3 x 3 + 11.2) x 1.2 = 154.92. 
 
 16. As the lamps were in series, the strength and E.M.F. of 
 the current must have been the same in both lamps. 
 
 97 x 2 + 12 = 206, the number of ohms. 
 E = CR = 1 X 206 = 206, the number of volts. 
 
 17. The armature wire must carry 4.8 amperes. 4.8 -f- 2,000 
 = 0.0024, the required area in square inches. The nearest 
 size, as given in the fifth column of the table on page 593, is 
 No. 15. 
 
 18. The E.M.F. required to overcome the re- 
 sistance of the lamps is 45 volts x 60 = 2,700 volts. 
 
 The E.M.F. required to overcome the other 
 resistance is 9.6 (16 + 5) = 201.6 volts. 
 
 Total voltage required, 2,901.6 volts. 
 
 7,200,000 x 120 x 780 = 112 Q 2 
 60 x 100,000,000 
 
 20. It would lessen it. 
 
 21. The a,rmature that has twice as many bobbins will develop 
 twice as great an E.M.F. 
 
 22. As the external resistance is lessened, the current will 
 be increased ; if such increase is continued too far, the current 
 will become greater than the No. 8 wire can safely carry, the 
 wire will become red-hot, and its insulation destroyed ; i.e., the 
 armature will " burn out." Hence the danger of short-circuit- 
 ing a dynamo. 
 
 23. An increase of the resistance in the field circuit of a 
 shunt dynamo lessens the current that passes around the field 
 
IN AVERY'S SCHOOL PHYSICS. 95 
 
 magnets, weakens the magnetic field, and lessens the number 
 of lines of force that pass through the armature. The E.M.F. 
 of such dynamos is often thus controlled. 
 
 24. To send a 2-ampere current through a 2-ohm coil requires 
 a pressure of 4 volts ; E = CR. The dynamo must be run at 
 a speed that will give this voltage. The added resistance must 
 be such that (9.6 2=) 7. 6 amperes will flow through it at 
 this pressure. ^ . 
 
 or 7.6: 2:: 2: a;; x = 0.526. 
 
 25. (a) 110-5-0.96 = 114.58 + . 
 
 (6) The fall of potential in the line = 114.58 volts - 110 
 volts = 4.58 volts. 
 
 Current required = 127 x 0.5 = 63.5, the number of amperes. 
 
 = 0.0721 +, the number of ohms. 
 
 63. 5 
 
 (c) 0.0721 -f- 1,240 = 0.0000581, the number of ohms. 
 
 (d) No. 000, i.e., "three naught" wire. 
 
 C 
 
 26. R = . The resistance of the wire must be (110 -5- 15 =) 
 
 7.333 ohms. The table shows that the resistance of a foot of 
 No. 14 wire is 0.002585 of an ohm. 7.333 -s- 0.002585 = 2,837, 
 the number of feet. 
 
 27. (a) The two miles of wire have a resistance of 
 
 (0.051 x 5.28 x 2 =) 0.53856 of an ohm. 
 
 (b) 175 -j- 350 = 0.5, the number of amperes required for 
 each lamp. 
 
 0.5 x 100 = 50, the number of amperes required for the 100 
 lamps. 
 
 E = CR = 50 x 0.53856 = 26.928, the fall of potential or 
 loss of voltage due to the resistance of the wire. 
 
 110 + 26.928 = 136.928, the voltage required at the dynamo. 
 
96 KEY TO PROBLEMS IN AVERY'S PHYSICS. 
 
 28. As either sounder adds little relatively to the high resist- 
 ance of the line, the current strength is nearly as great with 
 the high resistance sounder in the circuit as it is with the 
 other. On the other hand, the difference in the number of the 
 turns of the wire does materially affect the ampere-turns 
 ( 376) in favor of the high resistance instrument. Do not let 
 the pupil get the idea that the high resistance of the instru- 
 ment is the cause of the greater efficiency. The resistance is 
 an incidental and not a desirable factor in the case, but in 
 order to get the great number of turns in a small space, a long 
 fine wire is necessary, and this involves a high resistance. 
 
 29. See 424. 
 
 30. The difficulty in the latter case is due to the alternating 
 nature of the telephonic current, and to the greater resistance 
 and impedance of the circuit. 
 

 
YB 16958 
 
 926494 
 
 THE UNIVERSITY OF CALIFORNIA LIBRARY