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 (: 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 ^ 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