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respect to the x-axis. 11 x-component 4.41 m/s y-component 5.07 m/s 13 (a) 1.56 km (b) 120 m east 15 North-component 87.0 km, east-component 87.0 km 17 30.8 m, 35.8 west of north 19 (a) 30.8 m, 54.2º south of west (b) 30.8 m, 54.2º north of east 21 18.4 km south, then 26.2 km west(b) 31.5 km at 45.0º south of west, then 5.56 km at 45.0º west of north 23 7.34 km, 63.5º south of east 25 = 1.30 m×102 = 30.9 m. 27 (a) 3.50 s (b) 28.6 m/s (c) 34.3 m/s (d) 44.7 m/s, 50.2° below horizontal 29 This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key (a) 18.4° (b) The arrow will go over the branch. 31 = 0 sin2θ0 For = 45°, = 0 1563 = 91.8 m for 0 = 30 m/s ; = 163 m for 0 = 40 m/s ; = 255 m for 0 = 50 m/s. 33 (a) 560 m/s (b) 8.00×103 m (c) 80.0 m. This error is not significant because it is only 1% of the answer in part (b). 35 1.50 m, assuming launch angle of 45° 37 = 6.1° yes, the ball lands at 5.3 m from the net 39 (a) −0.486 m (b) The larger the muzzle velocity, the smaller the deviation in the vertical direction, because the time of flight would be smaller. Air resistance would have the effect of decreasing the time of flight, therefore increasing the vertical deviation. 41 4.23 m. No, the owl is not lucky; he misses the nest. 43 No, the maximum range (neglecting air resistance) is about 92 m. 45 15.0 m/s 47 (a) 24.2 m/s (b) The ball travels a total of 57.4 m with the brief gust of wind. 49
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− 0 = 0 = 0 − 1 22 = (0 sin ) − 1 22, so that = 2(0 sin ) − 0 = 0 = (0 cos ) = and substituting for gives: = 0 cos 20 sin = 20 2 sin cos since 2 sin cos = sin 2θ the range is: = 0 2 sin 2θ. 52 (a) 35.8 km, 45º south of east (b) 5.53 m/s, 45º south of east 1564 Answer Key (c) 56.1 km, 45º south of east 54 (a) 0.70 m/s faster (b) Second runner wins (c) 4.17 m 56 17.0 m/s, 22.1º 58 (a) 230 m/s, 8.0º south of west (b) The wind should make the plane travel slower and more to the south, which is what was calculated. 60 (a) 63.5 m/s (b) 29.6 m/s 62 6.68 m/s, 53.3º south of west 64 (a) average = 14.9km/s Mly (b) 20.2 billion years 66 1.72 m/s, 42.3º north of east Test Prep for AP® Courses 1 (d) 3 We would need to know the horizontal and vertical positions of each ball at several times. From that data, we could deduce the velocities over several time intervals and also the accelerations (both horizontal and vertical) for each ball over several time intervals. 5 The graph of the ball's vertical velocity over time should begin at 4.90 m/s during the time interval 0 - 0.1 sec (there should be a data point at t = 0.05 sec, v = 4.90 m/s). It should then have a slope of -9.8 m/s2, crossing through v = 0 at t = 0.55 sec and ending at v = -0.98 m/s at t = 0.65 sec. The graph of the ball's horizontal velocity would be a constant positive value, a flat horizontal line at some positive velocity from t = 0 until t = 0.7 sec. Chapter 4 Problems & Exercises 1 265 N 3 13.3 m/s2 7 (a) 12 m/s2. (b) The acceleration is not one-fourth of what it was with all rockets
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burning because the frictional force is still as large as it was with all rockets burning. 9 (a) The system is the child in the wagon plus the wagon. (b This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1565 Figure 4.10. (c) = 0.130 m/s2 in the direction of the second child’s push. (d) = 0.00 m/s2 11 (a) 3.68×103 N. This force is 5.00 times greater than his weight. (b) 3750 N; 11.3º above horizontal 13 1.5×103 N, 150 kg, 150 kg 15 Force on shell: 2.64×107 N Force exerted on ship = −2.64×107 N, by Newton’s third law 17 a. b. 0.11 m/s2 1.2×104 N 19 (a) 7.84×10-4 N (b) 1.89×10–3 N. This is 2.41 times the tension in the vertical strand. 21 Newton’s second law applied in vertical direction gives = − 2 sin = 0 = 2 sin =. 2 sin () () () 23 1566 Answer Key Figure 4.26. Using the free-body diagram: net = − − =, so that = − − = 1.250×107 N − 4.50×106 − (5.00×105 kg)(9.80 m/s2) 5.00×105 kg = 6.20 m/s2. 25 1. Use Newton’s laws of motion. Figure 4.26. 2. Given : = 4.00 = (4.00)(9.80 m/s2 ) = 39.2 m/s2 ; = 70.0 kg, Find:. 3. ∑ =+ − =, so that = + = + = ( + ). = (70.0 kg)[(39.2 m/s2 ) + (9.80 m/s2)] down on the ground, but is up from the ground and makes him jump. = 3.43×103N. The force exerted by the high-jumper is actually 4. This result is reasonable, since it is quite possible for a person to exert a force of the magnitude of 103 N. This content is available for free at http
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://cnx.org/content/col11844/1.13 1567 Answer Key 27 (a) 4.41×105 N (b) 1.50×105 N 29 (a) 910 N (b) 1.11×103 N 31 = 0.139 m/s, = 12.4º north of east 33 1. Use Newton’s laws since we are looking for forces. 2. Draw a free-body diagram: Figure 4.29. 3. The tension is given as = 25.0 N. Find app. Using Newton’s laws gives: = 0, so that applied force is due to the y-components of the two tensions: = 2 sin = 2(25.0 N)sin The x-components of the tension cancel. ∑ = 0. 15º = 12.9 N 4. This seems reasonable, since the applied tensions should be greater than the force applied to the tooth. 40 10.2 m/s2, 4.67º from vertical 42 1568 Answer Key Figure 4.35. 1 = 736 N 2 = 194 N 44 (a) 7.43 m/s (b) 2.97 m 46 (a) 4.20 m/s (b) 29.4 m/s2 (c) 4.31×103 N 48 (a) 47.1 m/s (b) 2.47×103 m/s2 (c) 6.18×103 N. The average force is 252 times the shell’s weight. 52 (a) 1×10−13 (b) 1×10−11 54 102 Test Prep for AP® Courses 1 This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1569 Figure 4.4. Car X is shown on the left, and Car Y is shown on the right. i. Car X takes longer to accelerate and does not spend any time traveling at top speed. Car Y accelerates over a shorter time and spends time going at top speed. So Car Y must cover the straightaways in a shorter time. Curves take the same time, so Car Y must overall take a shorter time. ii. The only difference in the calculations for the time of one segment of linear acceleration is the difference in distances. That shows that Car X takes longer to accelerate. The equation = corresponds to Car Y traveling for a time at
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4 top speed. Substituting = 1 into the displacement equation in part (b) ii gives = 3 2 1. This shows that a car takes less time to reach its maximum speed when it accelerates over a shorter distance. Therefore, Car Y reaches its maximum speed more quickly, and spends more time at its maximum speed than Car X does, as argued in part (b) i. 3 A body cannot exert a force on itself. The hawk may accelerate as a result of several forces. The hawk may accelerate toward Earth as a result of the force due to gravity. The hawk may accelerate as a result of the additional force exerted on it by wind. The hawk may accelerate as a result of orienting its body to create less air resistance, thus increasing the net force forward. 5 (a) A soccer player, gravity, air, and friction commonly exert forces on a soccer ball being kicked. (b) Gravity and the surrounding water commonly exert forces on a dolphin jumping. (The dolphin moves its muscles to exert a force on the water. The water exerts an equal force on the dolphin, resulting in the dolphin’s motion.) (c) Gravity and air exert forces on a parachutist drifting to Earth. 7 (c) 9 Figure 4.14. The diagram consists of a black dot in the center and two small red arrows pointing up (Fb) and down (Fg) and two long red arrows pointing right (Fc = 9.0 N) and left (Fw=13.0 N). In the diagram, Fg represents the force due to gravity on the balloon, and Fb represents the buoyant force. These two forces are equal in magnitude and opposite in direction. Fc represents the force of the current. Fw represents the force of the wind. The net force on the balloon will be − = 4.0 N and the balloon will accelerate in the direction the wind is blowing. 11 1570 Answer Key Since = /, the parachutist has a mass of 539 N/9.8 km/s2 = 55 kg. For the first 2 s, the parachutist accelerates at 9.8 m/s2. • 2s = = 9.8 m s2 = 17.6m s Her speed after 2 s is 19.6 m/s. From 2 s to 10 s, the net force on the parachutist is 539 N – 615 N, or 76 N upward. = = −
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76 N 55 kg = −1.4 m s2 Since = 0 +, = 17.6 m/s2 + ( − 1.4 m/s2)(8s) = 6.5 m/s2. At 10 s, the parachutist is falling to Earth at 8.4 m/s. 13 The system includes the gardener and the wheelbarrow with its contents. The following forces are important to include: the weight of the wheelbarrow, the weight of the gardener, the normal force for the wheelbarrow and the gardener, the force of the gardener pushing against the ground and the equal force of the ground pushing back against the gardener, and any friction in the wheelbarrow’s wheels. 15 The system undergoing acceleration is the two figure skaters together. Net force = 120 N – 5.0 N = 115 N. Total mass = 40 kg + 50 kg = 90 kg. Using Newton’s second law, we have that = = 115 N 90 kg = 1.28 m s2 The pair accelerates forward at 1.28 m/s2. 17 The force of tension must equal the force of gravity plus the force necessary to accelerate the mass. = can be used to calculate the first, and = can be used to calculate the second. For gravity: = = (120.0 kg)(9.8 m/s2) = 1205.4 N For acceleration: = = (120.0 kg)(1.3 m/s2) = 159.9 N The total force of tension in the cable is 1176 N + 156 N = 1332 N. 19 (b) This content is available for free at http://cnx.org/content/col11844/1.13 1571 Answer Key 21 Figure 4.24. The diagram has a black dot and three solid red arrows pointing away from the dot. Arrow Ft is long and pointing to the left and slightly down. Arrow Fw is also long and is a bit below a diagonal line halfway between pointing up and pointing to the right. A short arrow Fg is pointing down. Fg is the force on the kite due to gravity. Fw is the force exerted on the kite by the wind. Ft is the force of tension in the string holding the kite. It must balance the vector sum of the other two forces for the kite to float stationary in the air. 23 (b) 25 (
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d) 27 A free-body diagram would show a northward force of 64 N and a westward force of 38 N. The net force is equal to the sum of the two applied forces. It can be found using the Pythagorean theorem: net = 2 + = (38 N)2 + (64 N)2 = 74.4 N Since =, = 74.4 N 825 kg = 0.09 m/s2 The boulder will accelerate at 0.09 m/s2. 29 (b) 31 (b) 33 (d) Chapter 5 Problems & Exercises 1 5.00 N 4 (a) 588 N (b) 1.96 m/s2 Answer Key (5.30) (5.58) (5.59) 1572 6 (a) 3.29 m/s2 (b) 3.52 m/s2 (c) 980 N; 945 N 10 1.83 m/s2 14 (a) 4.20 m/s2 (b) 2.74 m/s2 (c) –0.195 m/s2 16 (a) 1.03×106 N (b) 3.48×105 N 18 (a) 51.0 N (b) 0.720 m/s2 20 115 m/s; 414 km/hr 22 25 m/s; 9.9 m/s 24 2.9 26 28 0.76 kg/m ⋅ s 29 [] = s [][] = kg ⋅ m/s2 m ⋅ m/s = kg m ⋅ s 1.90×10−3 cm 31 (a)1 mm (b) This does seem reasonable, since the lead does seem to shrink a little when you push on it. 33 (a)9 cm (b)This seems reasonable for nylon climbing rope, since it is not supposed to stretch that much. 35 8.59 mm 37 1.49×10−7 m 39 (a) 3.99×10−7 m (b) 9.67×10−8 m 41 4×106 N/m2. This is about 36 atm, greater than a typical jar can withstand. 43 1.4 cm This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1573 Test Prep for AP® Courses 1 (b) 3 (c)
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Chapter 6 Problems & Exercises 1 723 km 3 5×107 rotations 5 117 rad/s 7 76.2 rad/s 728 rpm 8 (a) 33.3 rad/s (b) 500 N (c) 40.8 m 10 12.9 rev/min 12 4×1021 m 14 a) 3.47×104 m / s2, 3.55×103 b) 51.1 m / s 16 a) 31.4 rad/s b) 118 m/s c) 384 m/s d)The centripetal acceleration felt by Olympic skaters is 12 times larger than the acceleration due to gravity. That's quite a lot of acceleration in itself. The centripetal acceleration felt by Button's nose was 39.2 times larger than the acceleration due to gravity. It is no wonder that he ruptured small blood vessels in his spins. 18 a) 0.524 km/s b) 29.7 km/s 20 (a) 1.35×103 rpm (b) 8.47×103 m/s2 (c) 8.47×10–12 N (d) 865 21 (a) 16.6 m/s Answer Key 1574 (b) 19.6 m / s2 (c) Figure 6.10. (d) 1.76×103 N or 3.00, that is, the normal force (upward) is three times her weight. (e) This answer seems reasonable, since she feels like she's being forced into the chair MUCH stronger than just by gravity. 22 a) 40.5 m / s2 b) 905 N c) The force in part (b) is very large. The acceleration in part (a) is too much, about 4 g. d) The speed of the swing is too large. At the given velocity at the bottom of the swing, there is enough kinetic energy to send the child all the way over the top, ignoring friction. 23 a) 483 N b) 17.4 N c) 2.24 times her weight, 0.0807 times her weight 25 4.14º 27 a) 24.6 m b) 36.6 m / s2 c) c = 3.73 This does not seem too large, but it is clear that bobsledders feel a lot of force on them going through sharply banked turns. 29 a) 2.56 rad/s b
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) 5.71º 30 a) 16.2 m/s b) 0.234 32 a) 1.84 b) A coefficient of friction this much greater than 1 is unreasonable. c) The assumed speed is too great for the tight curve. 33 a) 5.979×1024 kg b) This is identical to the best value to three significant figures. This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 35 a) 1.62 m / s2 b) 3.75 m / s2 37 a) 3.42×10–5 m / s2 b) 3.34×10–5 m / s2 1575 The values are nearly identical. One would expect the gravitational force to be the same as the centripetal force at the core of the system. 39 a) 7.01×10–7 N b) 1.35×10–6 N, 0.521 41 a) 1.66×10–10 m / s2 b) 2.17×105 m/s 42 a) 2.94×1017 kg b) 4.92×10–8 of the Earth's mass. c) The mass of the mountain and its fraction of the Earth's mass are too great. d) The gravitational force assumed to be exerted by the mountain is too great. 44 1.98×1030 kg 46 48 a) 7.4×103 m/s = 316 b) 1.05×103 m/s c) 2.86×10−7 s d) 1.84×107 N e) 2.76×104 J 49 a) 5.08×103 km b) This radius is unreasonable because it is less than the radius of earth. c) The premise of a one-hour orbit is inconsistent with the known radius of the earth. Test Prep for AP® Courses 1 (a) 3 (b) Answer Key (7.8) (7.9) 3.00 J = 7.17×10−4 kcal 3.14×103 J 1576 5 (b) Chapter 7 Problems & Exercises 1 3 (a) 5.92×105 J (b) −5.88×105 J (c) The net force is zero. 5 7 (a) −700 J (b) 0 (c) 700 J (d) 38.6 N (e) 0 9 1
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/ 250 11 1.1×1010 J 13 2.8×103 N 15 102 N 16 (a) 1.961016 J (b) The ratio of gravitational potential energy in the lake to the energy stored in the bomb is 0.52. That is, the energy stored in the lake is approximately half that in a 9-megaton fusion bomb. 18 (a) 1.8 J (b) 8.6 J 20 22 = 2 + 0 2 = 2(9.80 m/s2)( − 0.180 m) + (2.00 m/s)2 = 0.687 m/s 7.81105 N/m (7.45) (7.60) 24 9.46 m/s 26 4104 molecules 27 Equating ΔPEg and ΔKE, we obtain = 2 + 0 29 (a) 25×106 years 2 = 2(9.80 m/s2)(20.0 m) + (15.0 m/s)2 = 24.8 m/s This content is available for free at http://cnx.org/content/col11844/1.13 1577 (7.81) Answer Key (b) This is much, much longer than human time scales. 210−10 30 32 (a) 40 (b) 8 million 34 $149 36 (a) 208 W (b) 141 s 38 (a) 3.20 s (b) 4.04 s 40 (a) 9.46107 J (b) 2.54 y 42 Identify knowns: = 950 kg, slope angle = 2.00º, = 3.00 m/s, = 600 N Identify unknowns: power of the car, force that car applies to road Solve for unknown: = = = = where is parallel to the incline and must oppose the resistive forces and the force of gravity: = + = 600 N + sin Insert this into the expression for power and solve: = = + sin 600 N + = 2.77×104 W 950 kg 9.80 m/s2 sin 2º (30.0 m/s) About 28 kW (or about 37 hp) is reasonable for a car to climb a gentle incline. 44 (a) 9.5 min (b) 69 flights of stairs 46 641 W, 0.860 hp 48 31 g 50 14.3% 52 (a) 3.21104 N (
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b) 2.35103 N (c) Ratio of net force to weight of person is 41.0 in part (a); 3.00 in part (b) 54 Answer Key 1578 (a) 108 kJ (b) 599 W 56 (a) 144 J (b) 288 W 58 (a) 2.501012 J (b) 2.52% (c) 1.4104 kg (14 metric tons) 60 (a) 294 N (b) 118 J (c) 49.0 W 62 (a) 0.500 m/s2 (b) 62.5 N (c) Assuming the acceleration of the swimmer decreases linearly with time over the 5.00 s interval, the frictional force must therefore be increasing linearly with time, since = −. If the acceleration decreases linearly with time, the velocity will contain a term dependent on time squared ( 2 ). Therefore, the water resistance will not depend linearly on the velocity. 64 (a) 16.1×103 N (b) 3.22×105 J (c) 5.66 m/s (d) 4.00 kJ 66 (a) 4.65×103 kcal (b) 38.8 kcal/min (c) This power output is higher than the highest value on Table 7.5, which is about 35 kcal/min (corresponding to 2415 watts) for sprinting. (d) It would be impossible to maintain this power output for 2 hours (imagine sprinting for 2 hours!). 69 (a) 4.32 m/s (b) 3.47×103 N (c) 8.93 kW Test Prep for AP® Courses 1 (b) 3 (d) 5 (a) 7 This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1579 The kinetic energy should change in the form of –cos, with an initial value of 0 or slightly above, and ending at the same level. 9 Any force acting perpendicular will have no effect on kinetic energy. Obvious examples are gravity and the normal force, but others include wind directly from the side and rain or other precipitation falling straight down. 11 Note that the wind is pushing from behind and one side, so your KE will increase. The net force has components of 1400 N in the direction of travel and 212 N perpendicular to the direction of travel. So the net force is 14
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20 N at 8.5 degrees from the direction of travel. 13 Gravity has a component perpendicular to the cannon (and to displacement, so it is irrelevant) and has a component parallel to the cannon. The latter is equal to 9.8 N. Thus the net force in the direction of the displacement is 50 N − 9.8 N, and the kinetic energy is 60 J. 15 The potato cannon (and many other projectile launchers) above is an option, with a force launching the projectile, friction, potentially gravity depending on the direction it is pointed, etc. A drag (or other) car accelerating is another possibility. 17 The kinetic energy of the rear wagon increases. The front wagon does not, until the rear wagon collides with it. The total system may be treated by its center of mass, halfway between the wagons, and its energy increases by the same amount as the sum of the two individual wagons. 19 (d) 21 0.049 J; 0.041 m, 0.25 m 23 20 m high, 20 m/s. 25 (a) 27 (d) 29 (c) 31 (b) 33 (c) 35 (c) 37 (c), (d) 39 (a) 41 (b) Chapter 8 Problems & Exercises 1 (a) 1.50×104 kg ⋅ m/s (b) 625 to 1 (c) 6.66×102 kg ⋅ m/s 3 (a) 8.00×104 m/s (b) 1.20×106 kg · m/s 1580 Answer Key (c) Because the momentum of the airplane is 3 orders of magnitude smaller than of the ship, the ship will not recoil very much. The recoil would be −0.0100 m/s, which is probably not noticeable. 5 54 s 7 9.00×103 N 9 a) 2.40×103 N toward the leg b) The force on each hand would have the same magnitude as that found in part (a) (but in opposite directions by Newton’s third law) because the change in momentum and the time interval are the same. 11 a) 800 kg ⋅ m/s away from the wall b) 1.20 m/s away from the wall 13 (a) 1.50×106 N away from the dashboard (b) 1.00×105 N away from the dashboard 15 4.69×105 N in the boat’s
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original direction of motion 17 2.10×103 N away from the wall 19 = 2 p = v ⇒ 2 = 22 ⇒ 2 ⇒ 2 = 1 2 = 2 2 22 = (8.35) 21 60.0 g 23 0.122 m/s 25 In a collision with an identical car, momentum is conserved. Afterwards f = 0 for both cars. The change in momentum will be the same as in the crash with the tree. However, the force on the body is not determined since the time is not known. A padded stop will reduce injurious force on body. 27 22.4 m/s in the same direction as the original motion 29 0.250 m/s 31 (a) 86.4 N perpendicularly away from the bumper (b) 0.389 J (c) 64.0% 33 (a) 8.06 m/s (b) -56.0 J (c)(i) 7.88 m/s; (ii) -223 J 35 (a) 0.163 m/s in the direction of motion of the more massive satellite This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1581 (b) 81.6 J (c) 8.70×10−2 m/s in the direction of motion of the less massive satellite, 81.5 J. Because there are no external forces, the velocity of the center of mass of the two-satellite system is unchanged by the collision. The two velocities calculated above are the velocity of the center of mass in each of the two different individual reference frames. The loss in KE is the same in both reference frames because the KE lost to internal forces (heat, friction, etc.) is the same regardless of the coordinate system chosen. 37 0.704 m/s –2.25 m/s 38 (a) 4.58 m/s away from the bullet (b) 31.5 J (c) –0.491 m/s (d) 3.38 J 40 (a) 1.02×10−6 m/s (b) 5.63×1020 J (almost all KE lost) (c) Recoil speed is 6.79×10−17 m/s, energy lost is 6.25×109 J. The plume will not affect the momentum result because the plume is still part of the Moon system. The
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plume may affect the kinetic energy result because a significant part of the initial kinetic energy may be transferred to the kinetic energy of the plume particles. 42 24.8 m/s 44 (a) 4.00 kg (b) 210 J (c) The clown does work to throw the barbell, so the kinetic energy comes from the muscles of the clown. The muscles convert the chemical potential energy of ATP into kinetic energy. 45 (a) 3.00 m/s, 60º below -axis (b) Find speed of first puck after collision: 0 = ′1 sin 30º−′2 sin 60º ⇒ ′1 = 2 sin 60º sin 30º = 5.196 m/s KE = 1 KE = 1 21 2′1 2 = 18 J 2 + 1 2′2 KE KE′ 2 = 18 J = 1.00 Verify that ratio of initial to final KE equals one: 47 (a) −2.26 m/s (b) 7.63×103 J (c) The ground will exert a normal force to oppose recoil of the cannon in the vertical direction. The momentum in the vertical direction is transferred to the earth. The energy is transferred into the ground, making a dent where the cannon is. After long barrages, cannon have erratic aim because the ground is full of divots. 49 (a) 5.36×105 m/s at −29.5º (b) 7.52×10−13 J 51 1582 Answer Key We are given that 1 = 2 ≡. The given equations then become: and Square each equation to get 1 = 1 cos 1 + 2 cos 2 0 = ′1 sin 1 + ′2 sin 2. 1 0 2 = ′1 = ′1 2 cos2 1 + ′2 2 sin2 1 + ′2 2 cos2 2 + 2′1 ′2 cos 1 cos 2 2 sin2 2 + 2′1 ′2 sin 1 sin 2. 1 Add these two equations and simplify: 2 + ′2 2 + ′2 2 + ′2 2 = ′1 = ′1 2 + 2′1 ′2 cos 1 cos 2 + sin 1 sin 2 1 2 + 2′1 ′2 2 2 + 2′1 ′2 cos 1 cos cos. 1 + 2 + 1 2 cos 1 − 2 − 1 2 cos 1 + 2 (8.107) (8.
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108) (8.109) (8.110) Multiply the entire equation by 1 2 to recover the kinetic energy: 2 + 1 2 = 1 2′1 2′2 1 21 2 + ′1 ′2 cos 1 − 2 (8.111) 53 39.2 m/s2 55 4.16×103 m/s 57 The force needed to give a small mass Δ an acceleration Δ is = ΔΔ. To accelerate this mass in the small time interval Δ at a speed e requires e = ΔΔ, so = e Δ in magnitude to the thrust force acting on the rocket, so thrust = e Δ Newton’s second law to the rocket gives thrust − = ⇒ = e and unburnt fuel. 60 2.63×103 kg 61 −, where is the mass of the rocket, where all quantities are positive. Applying Δ Δ Δ Δ. By Newton’s third law, this force is equal (a) 0.421 m/s away from the ejected fluid. (b) 0.237 J. Test Prep for AP® Courses 1 (b) 3 (b) 5 (a) 7 (c) (based on calculation of = Δ Δ ) 9 (c) 11 (d) 13 (b) This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1583 15 (d) 17 (b) 19 (c) 21 (b) 23 (c) 25 (b) 27 (a) 29 (c) 31 (b) 33 (a) 35 (b) 37 (a) 39 (a) 41 (d) 43 (c). Because of conservation of momentum, the final velocity of the combined mass must be 4.286 m/s. The initial kinetic energy is (0.5)(2.0)(15)2 = 225 J. The final kinetic energy is (0.5)(7.0)(4.286)2 = 64 J, so the difference is −161 J. 45 (a) 47 (d) 49 (c) 51 (b) Chapter 9 Problems & Exercises 1 a) 46.8 N·m b) It does not matter at what height you push. The torque depends on only the magnitude of the force applied and the perpendicular distance of the force's application from the hinges. (Children don't have a tougher time opening a door because they push
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lower than adults, they have a tougher time because they don't push far enough from the hinges.) 3 23.3 N 5 Given: 1 = 26.0 kg, 2 = 32.0 kg, s = 12.0 kg, 1 = 1.60 m, s = 0.160 m, find (a) 2, (b) p (9.26) a) Since children are balancing: Answer Key (9.27) (9.28) (9.29) (9.30) 1584 So, solving for 2 gives: net cw = – net ccw ⇒ 1 1 + s s = 22 26.0 kg)(1.60 m) + (12.0 kg)(0.160 m) 32.0 kg b) Since the children are not moving: = 1.36 m So that net = 26.0 kg + 32.0 kg + 12.0 kg)(9.80 m / s2) = 686 N 6 wall = 1.43×103 N 8 a) 2.55×103 N, 16.3º to the left of vertical (i.e., toward the wall) b) 0.292 10 B = 2.12×104 N 12 a) 0.167, or about one-sixth of the weight is supported by the opposite shore. b) = 2.0×104 N, straight up. 14 a) 21.6 N b) 21.6 N 16 350 N directly upwards 19 25 50 N 21 a) MA = 18.5 b) i = 29.1 N c) 510 N downward 23 1.3×103 N 25 a) = 299 N b) 897 N upward 26 This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1585 B = 470 N; 1 = 4.00 cm; a = 2.50 kg; 2 = 16.0 cm;b = 4.00 kg; 3 = 38.0 cm 16.0 cm 9.80 m / s2 4.0 cm 38.0 cm 9.80 m / s2 4.00 cm – 1 – 1 = 2.50 kg + 4.00 kg = 407 N 28 1.1×103 N = 190º ccw from positive axis 30 V = 97 N, = 59º 32 (a) 25 N downward (b) 75 N upward 33 (a) A
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= 2.21×103 N upward (b) B = 2.94×103 N downward 35 (a) teeth on bullet = 1.2×102 N upward (b) J = 84 N downward 37 (a) 147 N downward (b) 1680 N, 3.4 times her weight (c) 118 J (d) 49.0 W 39 a) 2 = 2.33 m b) The seesaw is 3.0 m long, and hence, there is only 1.50 m of board on the other side of the pivot. The second child is off the board. c) The position of the first child must be shortened, i.e. brought closer to the pivot. Test Prep for AP® Courses 1 (a) 3 Both objects are in equilibrium. However, they will respond differently if a force is applied to their sides. If the cone placed on its base is displaced to the side, its center of gravity will remain over its base and it will return to its original position. When the traffic cone placed on its tip is displaced to the side, its center of gravity will drift from its base, causing a torque that will accelerate it to the ground. 5 (d) 7 a. FL = 7350 N, FR = 2450 N b. As the car moves to the right side of the bridge, FL will decrease and FR will increase. (At exactly halfway across the bridge, FL and FR will both be 4900 N.) 9 1586 Answer Key The student should mention that the guiding principle behind simple machines is the second condition of equilibrium. Though the torque leaving a machine must be equivalent to torque entering a machine, the same requirement does not exist for forces. As a result, by decreasing the lever arm to the existing force, the size of the existing force will be increased. The mechanical advantage will be equivalent to the ratio of the forces exiting and entering the machine. 11 a. The force placed on your bicep muscle will be greater than the force placed on the dumbbell. The bicep muscle is closer to your elbow than the downward force placed on your hand from the dumbbell. Because the elbow is the pivot point of the system, this results in a decreased lever arm for the bicep. As a result, the force on the bicep must be greater than that placed on the dumbbell. (How much greater? The ratio between the bicep and dumbbell forces is equal to the inverted ratio of their distances from
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the elbow. If the dumbbell is ten times further from the elbow than the bicep, the force on the bicep will be 200 pounds!) b. The force placed on your bicep muscle will decrease. As the forearm lifts the dumbbell, it will get closer to the elbow. As a result, the torque placed on the arm from the weight will decrease and the countering torque created by the bicep muscle will do so as well. Chapter 10 Problems & Exercises 1 = 0.737 rev/s 3 (a) −0.26 rad/s2 (b) 27 rev 5 (a) 80 rad/s2 (b) 1.0 rev 7 (a) 45.7 s (b) 116 rev 9 a) 600 rad/s2 b) 450 rad/s c) 21.0 m/s 10 (a) 0.338 s (b) 0.0403 rev (c) 0.313 s 12 0.50 kg ⋅ m2 14 (a) 50.4 N ⋅ m (b) 17.1 rad/s2 (c) 17.0 rad/s2 16 3.96×1018 s or 1.26×1011 y This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 18 1587 2 = + 42 = 1 2 32 − 1 42 = 1 122 Thus, = − 1 19 (a) 2.0 ms (b) The time interval is too short. (c) The moment of inertia is much too small, by one to two orders of magnitude. A torque of 500 N ⋅ m is reasonable. 20 (a) 17,500 rpm (b) This angular velocity is very high for a disk of this size and mass. The radial acceleration at the edge of the disk is > 50,000 gs. (c) Flywheel mass and radius should both be much greater, allowing for a lower spin rate (angular velocity). KErot = 434 J (10.104) 21 (a) 185 J (b) 0.0785 rev (c) = 9.81 N 23 (a) 2.57×1029 J (b) KErot = 2.65×1033 J 25 27 (a) 128 rad/s (b) 19.9 m 29 (a) 10.4 rad/s2 (b) net
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= 6.11 J 34 (a) 1.49 kJ (b) 2.52×104 N 36 (a) 2.66×1040 kg ⋅ m2/s (b) 7.07×1033 kg ⋅ m2/s The angular momentum of the Earth in its orbit around the Sun is 3.77×106 times larger than the angular momentum of the Earth around its axis. 38 22.5 kg ⋅ m2/s 40 25.3 rpm 43 Answer Key 1588 (a) 0.156 rad/s (b) 1.17×10−2 J (c) 0.188 kg ⋅ m/s 45 (a) 3.13 rad/s (b) Initial KE = 438 J, final KE = 438 J 47 (a) 1.70 rad/s (b) Initial KE = 22.5 J, final KE = 2.04 J (c) 1.50 kg ⋅ m/s 48 (a) 5.64×1033 kg ⋅ m2 /s (b) 1.39×1022 N ⋅ m (c) 2.17×1015 N Test Prep for AP® Courses 1 (b) 3 (d) 5 (d) You are given a thin rod of length 1.0 m and mass 2.0 kg, a small lead weight of 0.50 kg, and a not-so-small lead weight of 1.0 kg. The rod has three holes, one in each end and one through the middle, which may either hold a pivot point or one of the small lead weights. 7 (a) 9 (c) 11 (a) 13 (a) 15 (b) 17 (c) 19 (b) 21 (b) 23 (c) 25 (d) 27 A door on hinges is a rotational system. When you push or pull on the door handle, the angular momentum of the system changes. If a weight is hung on the door handle, then pushing on the door with the same force will cause a different increase in angular momentum. If you push or pull near the hinges with the same force, the resulting angular momentum of the system will also be different. 29 Since the globe is stationary to start with, This content is available for free at http://cnx.org/content/col11844/1.13 1589 Answer Key = Δ Δ �
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� Δ = Δ By substituting, 120 N•m • 1.2 s = 144 N•m•s. The angular momentum of the globe after 1.2 s is 144 N•m•s. Chapter 11 Problems & Exercises 1 1.610 cm3 3 (a) 2.58 g (b) The volume of your body increases by the volume of air you inhale. The average density of your body decreases when you take a deep breath, because the density of air is substantially smaller than the average density of the body before you took the deep breath. 4 2.70 g/cm3 6 (a) 0.163 m (b) Equivalent to 19.4 gallons, which is reasonable 8 7.9×102 kg/m3 9 15.6 g/cm3 10 (a) 1018 kg/m3 (b) 2×104 m 11 3.59×106 Pa ; or 521 lb/in2 13 2.36×103 N 14 0.760 m 16 units = (m) = kg/m3 kg ⋅ m/s2 m/s2 1/m2 = kg ⋅ m2 / m3 ⋅ s2 (11.30) 18 (a) 20.5 mm Hg = N/m2 (b) The range of pressures in the eye is 12–24 mm Hg, so the result in part (a) is within that range 20 1.09×103 N/m2 22 24.0 N 24 1590 Answer Key 2.55×107 Pa ; or 251 atm 26 5.76×103 N extra force 28 (a) = ii = oo ⇒ o = i i o. Now, using equation: Finally = oo = i o i i i o = ii = i. (11.32) (11.33) In other words, the work output equals the work input. (b) If the system is not moving, friction would not play a role. With friction, we know there are losses, so that out = in − f ; therefore, the work output is less than the work input. In other words, with friction, you need to push harder on the input piston than was calculated for the nonfriction case. 29 Balloon: = 5.00 cm H2 O, g abs = 1.035×103 cm H2 O. Jar: = −50.0 mm Hg,
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g abs = 710 mm Hg. 31 4.08 m 33 Δ = 38.7 mm Hg, Leg blood pressure = 159 119. 35 22.4 cm2 36 91.7% 38 815 kg/m3 40 (a) 41.4 g (b) 41.4 cm3 (c) 1.09 g/cm3 42 (a) 39.5 g (b) 50 cm3 (c) 0.79 g/cm3 It is ethyl alcohol. This content is available for free at http://cnx.org/content/col11844/1.13 1591 Answer Key 44 8.21 N 46 (a) 960 kg/m3 (b) 6.34% She indeed floats more in seawater. 48 (a) 0.24 (b) 0.68 (c) Yes, the cork will float because obj < ethyl alcohol(0.678 g/cm3 < 0.79 g/cm3) 50 The difference is 0.006%. 52 net = fl − 1fl 2 − 1 = 2 − 1 fl where fl = density of fluid. Therefore, net = (2 − 1)fl = flfl = fl = fl where is fl the weight of the fluid displaced. 54 592 N/m2 56 2.23×10−2 mm Hg 58 (a) 1.65×10−3 m (b) 3.71×10–4 m 60 6.32×10−2 N/m Based on the values in table, the fluid is probably glycerin. 62 w = 14.6 N/m2 a = 4.46 N/m2 sw = 7.40 N/m2. Alcohol forms the most stable bubble, since the absolute pressure inside is closest to atmospheric pressure. 64 5.1º This is near the value of = 0º for most organic liquids. 66 −2.78 The ratio is negative because water is raised whereas mercury is lowered. Answer Key 1592 68 479 N 70 1.96 N 71 −63.0 cm H2 O 73 (a) 3.81×103 N/m2 (b) 28.7 mm Hg, which is sufficient to trigger micturition reflex 75 (a) 13.6 m water (b) 76.5 cm water 77 (a) 3.98×106 Pa (b) 2.1×10−3 cm 79 (a
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) 2.97 cm (b) 3.39×10−6 J (c) Work is done by the surface tension force through an effective distance / 2 to raise the column of water. 81 (a) 2.01×104 N (b) 1.17×10−3 m (c) 2.56×1010 N/m2 83 (a) 1.38×104 N (b) 2.81×107 N/m2 (c) 283 N 85 (a) 867 N (b) This is too much force to exert with a hand pump. (c) The assumed radius of the pump is too large; it would be nearly two inches in diameter—too large for a pump or even a master cylinder. The pressure is reasonable for bicycle tires. Test Prep for AP® Courses 1 (e) 3 (a) 100 kg/m3 (b) 60% (c) yes; yes (76% will be submerged) (d) answers vary 5 (d) Chapter 12 Problems & Exercises 1 This content is available for free at http://cnx.org/content/col11844/1.13 1593 Answer Key 2.78 cm3 /s 3 27 cm/s 5 (a) 0.75 m/s (b) 0.13 m/s 7 (a) 40.0 cm2 (b) 5.09×107 9 (a) 22 h (b) 0.016 s 11 (a) 12.6 m/s (b) 0.0800 m3 /s (c) No, independent of density. 13 (a) 0.402 L/s (b) 0.584 cm 15 (a) 127 cm3 /s (b) 0.890 cm 17 = Force Area () units = N/m2 = N ⋅ m/m3 = J/m3 = energy/volume 19 184 mm Hg 21 2.54×105 N 23 (a) 1.58×106 N/m2 (b) 163 m 25 (a) 9.56×108 W (b) 1.4 27 1.26 W 29 (a) 3.02×10−3 N (b) 1.03×10−3 31 1.60 cm3 /min 1594 33 8.7×10−11 m3 /s 35 0.316 37 (a) 1.52 Answer Key (b)
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Turbulence will decrease the flow rate of the blood, which would require an even larger increase in the pressure difference, leading to higher blood pressure. 225 mPa ⋅ s 0.138 Pa ⋅ s, (12.98) (12.99) 39 41 or Olive oil. 43 (a) 1.62×104 N/m2 (b) 0.111 cm3 /s (c)10.6 cm 45 1.59 47 2.95×106 N/m2 (gauge pressure) 51 R = 1.99×102 < 2000 53 (a) nozzle: 1.27×105, not laminar (b) hose: 3.51×104, not laminar. 55 2.54 << 2000, laminar. 57 1.02 m/s 1.28×10–2 L/s 59 (a) ≥ 13.0 m (b) 2.68×10−6 N/m2 61 (a) 23.7 atm or 344 lb/in2 (b) The pressure is much too high. (c) The assumed flow rate is very high for a garden hose. (d) 5.27×106 > > 3000, turbulent, contrary to the assumption of laminar flow when using this equation. 62 1.41×10−3 m 64 1.3×102 s 66 This content is available for free at http://cnx.org/content/col11844/1.13 1595 Answer Key 0.391 s Test Prep for AP® Courses 1 (c) 3 (a) 5 (a) 7 (a) 9 (d) Chapter 13 Problems & Exercises 1 102ºF 3 20.0ºC and 25.6ºC 5 9890ºF 7 (a) 22.2ºC Δ(ºF) = 2 (ºF) − 1(ºF) (bºC) + 32.0º − 5 = 9 5 2 (ºC) − 1(ºC) 1 (ºC) + 32.0º Δ(ºC) 9 169.98 m 11 5.4×10−6 m 13 Because the area gets smaller, the price of the land DECREASES by ~17000. 15 = 0 + Δ = 0(1 + Δ) (13.25) = (60.00 L) 1 + 950×10−6 /
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ºC (35.0ºC − 15.0ºC) = 61.1 L 17 (a) 9.35 mL (b) 7.56 mL 19 0.832 mm 21 We know how the length changes with temperature: Δ = 0Δ. Also we know that the volume of a cube is related to its length by = 3, so the final volume is then = 0 + Δ = 0 + Δ 3. Substituting for Δ gives = 0 + 0Δ 3 = 0 3(1 + Δ)3. Now, because Δ is small, we can use the binomial expansion: ≈ 0 3(1 + 3αΔT) = 0 3 + 3α0 3Δ. (13.26) (13.27) 1596 So writing the length terms in terms of volumes gives = 0 + Δ ≈ 0 + 3α 0Δ and so Δ = 0Δ ≈ 3α 0Δ or ≈ 3α. Answer Key (13.28) 22 1.62 atm 24 (a) 0.136 atm (b) 0.135 atm. The difference between this value and the value from part (a) is negligible. 26 (a) = (mol)(J/mol ⋅ K)(K) = J (b) = (mol)(cal/mol ⋅ K)(K) = cal (c) = (mol)(L ⋅ atm/mol ⋅ K)(K) = L ⋅ atm = (m3)(N/m2) = N ⋅ m = J 28 7.86×10−2 mol 30 (a) 6.02×105 km3 (b) 6.02×108 km 32 −73.9ºC 34 (a) 9.14×106 N/m2 (b) 8.23×106 N/m2 (c) 2.16 K (d) No. The final temperature needed is much too low to be easily achieved for a large object. 36 41 km 38 (a) 3.7×10−17 Pa (b) 6.0×1017 m3 (c) 8.4×102 km 39 1.25×103 m/s 41 (a) 1.20×10−19 J (b) 1.24×10−17 J 43 458 K 45 1.95×107 K 47 6
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.09×105 m/s 49 This content is available for free at http://cnx.org/content/col11844/1.13 1597 Answer Key 7.89×104 Pa 51 (a) 1.99×105 Pa (b) 0.97 atm 53 3.12×104 Pa 55 78.3% 57 (a) 2.12×104 Pa (b) 1.06 % 59 (a) 8.80×10−2 g (b) 6.30×103 Pa ; the two values are nearly identical. 61 82.3% 63 4.77ºC 65 38.3 m 67 B / Cu B / Cu circumstances. 69 (a) 4.41×1010 mol/m3 = 1.02. The buoyant force supports nearly the exact same amount of force on the copper block in both (b) It’s unreasonably large. (c) At high pressures such as these, the ideal gas law can no longer be applied. As a result, unreasonable answers come up when it is used. 71 (a) 7.03×108 m/s (b) The velocity is too high—it’s greater than the speed of light. (c) The assumption that hydrogen inside a supernova behaves as an idea gas is responsible, because of the great temperature and density in the core of a star. Furthermore, when a velocity greater than the speed of light is obtained, classical physics must be replaced by relativity, a subject not yet covered. Test Prep for AP® Courses 1 (a), (c) 3 (d) 5 (b) 7 (a) 7.29 × 10-21 J; (b) 352K or 79ºC Chapter 14 Problems & Exercises 5.02×108 J 3.07×103 J 0.171ºC Answer Key (14.18) (14.19) (14.20) 1598 1 3 5 7 10.8 9 617 W 11 35.9 kcal 13 (a) 591 kcal (b) 4.94×103 s 15 13.5 W 17 (a) 148 kcal (b) 0.418 s, 3.34 s, 4.19 s, 22.6 s, 0.456 s 19 33.0 g 20 (a) 9.67 L (b) Crude oil is less dense than water, so it floats on top of the water, thereby exposing it
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to the oxygen in the air, which it uses to burn. Also, if the water is under the oil, it is less efficient in absorbing the heat generated by the oil. 22 a) 319 kcal b) 2.00ºC 24 20.6ºC 26 4.38 kg 28 (a) 1.57×104 kcal (b) 18.3 kW ⋅ h (c) 1.29×104 kcal 30 (a) 1.01×103 W (b) One 32 84.0 W 34 2.59 kg 36 (a) 39.7 W (b) 820 kcal 38 35 to 1, window to wall This content is available for free at http://cnx.org/content/col11844/1.13 1599 Answer Key 40 1.05×103 K 42 (a) 83 W (b) 24 times that of a double pane window. 44 20.0 W, 17.2% of 2400 kcal per day 45 10 m/s 47 85.7ºC 49 1.48 kg 51 2×104 MW 53 (a) 97.2 J (b) 29.2 W (c) 9.49 W (d) The total rate of heat loss would be 29.2 W + 9.49 W = 38.7 W. While sleeping, our body consumes 83 W of power, while sitting it consumes 120 to 210 W. Therefore, the total rate of heat loss from breathing will not be a major form of heat loss for this person. 55 −21.7 kW Note that the negative answer implies heat loss to the surroundings. 57 −266 kW 59 −36.0 W 61 (a) 1.31% (b) 20.5% 63 (a) −15.0 kW (b) 4.2 cm 65 (a) 48.5ºC (b) A pure white object reflects more of the radiant energy that hits it, so a white tent would prevent more of the sunlight from heating up the inside of the tent, and the white tunic would prevent that heat which entered the tent from heating the rider. Therefore, with a white tent, the temperature would be lower than 48.5ºC, and the rate of radiant heat transferred to the rider would be less than 20.0 W. 67 (a) 3×1017 J (b) 1×1013 kg (c) When a large meteor hits the ocean, it causes great tidal waves, dissipating large amount
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of its energy in the form of kinetic energy of the water. 69 (a) 3.44×105 m3 /s 1600 Answer Key (b) This is equivalent to 12 million cubic feet of air per second. That is tremendous. This is too large to be dissipated by heating the air by only 5ºC. Many of these cooling towers use the circulation of cooler air over warmer water to increase the rate of evaporation. This would allow much smaller amounts of air necessary to remove such a large amount of heat because evaporation removes larger quantities of heat than was considered in part (a). 71 20.9 min 73 (a) 3.96×10-2 g (b) 96.2 J (c) 16.0 W 75 (a) 1.102 (b) 2.79×104 J (c) 12.6 J. This will not cause a significant cooling of the air because it is much less than the energy found in part (b), which is the energy required to warm the air from 20.0ºC to 50.0ºC. 76 (a) 36ºC (b) Any temperature increase greater than about 3ºC would be unreasonably large. In this case the final temperature of the person would rise to 73ºC (163ºF). (c) The assumption of 95% heat retention is unreasonable. 78 (a) 1.46 kW (b) Very high power loss through a window. An electric heater of this power can keep an entire room warm. (c) The surface temperatures of the window do not differ by as great an amount as assumed. The inner surface will be warmer, and the outer surface will be cooler. Test Prep for AP® Courses 1 (c) 3 (a) 5 (b) 7 (a) 9 (d) Chapter 15 Problems & Exercises 1 1.6×109 J 3 -9.30108 J 5 (a) −1.0×104 J, or −2.39 kcal (b) 5.00% 7 (a) 122 W This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key (b) 2.10×106 J 1601 (c) Work done by the motor is 1.61×107 J ;thus the motor produces 7.67 times the work done by the man 9 (a) 492 kJ (b) This amount of
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heat is consistent with the fact that you warm quickly when exercising. Since the body is inefficient, the excess heat produced must be dissipated through sweating, breathing, etc. 10 6.77×103 J 12 (a) = Δ = 1.76×105 J (b) = = 1.76×105 J. Yes, the answer is the same. 14 = 4.5×103 J 16 is not equal to the difference between the heat input and the heat output. 20 (a) 18.5 kJ (b) 54.1% 22 (a) 1.32 × 109 J (b) 4.68 × 109 J 24 (a) 3.80 × 109 J (b) 0.667 barrels 26 (a) 8.30 × 1012 J, which is 3.32% of 2.50 × 1014 J. (b) –8.30 × 1012 J, where the negative sign indicates a reduction in heat transfer to the environment. 28 403ºC 30 (a) 244ºC (b) 477ºC (c)Yes, since automobiles engines cannot get too hot without overheating, their efficiency is limited. 32 (a) (b) 1 = 1 − c,1 h,1 = 1 − 543 K 723 K = 0.249 or 24.9% 2 = 1 − 423 K 543 K = 0.221 or 22.1% 1602 (c) 1 = 1 − c,1 h,1 ⇒ c,1 = h,1 1, −, similarly, c,2 = h,2 1 − 1 2 using h,2 = c,1 in above equation gives c,2 = h,,1 2 1 − overall 1 − 1 overall = 1 − (1 − 0.249)(1 − 0.221) = 41.5% 2 Answer Key 1 − overall (d) overall = 1 − 423 K 723 K = 0.415 or 41.5% 34 The heat transfer to the cold reservoir is c = h − = 25 kJ − 12 kJ = 13 kJ, so the efficiency is = 1 − c h = 1 − 13 kJ 25 kJ = 0.48. The Carnot efficiency is C = 1 − c h = 1 − 300 K 600 K = 0.50. The actual efficiency is 96% of the Carnot efficiency, which is much higher than the best-ever achieved of about 70%, so her
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scheme is likely to be fraudulent. 36 (a) –56.3ºC (b) The temperature is too cold for the output of a steam engine (the local environment). It is below the freezing point of water. (c) The assumed efficiency is too high. 37 4.82 39 0.311 41 (a) 4.61 (b) 1.66×108 J or 3.97×104 kcal (c) To transfer 1.66×108 J, heat pump costs $1.00, natural gas costs $1.34. 43 27.6ºC 45 (a) 1.44×107 J (b) 40 cents (c) This cost seems quite realistic; it says that running an air conditioner all day would cost $9.59 (if it ran continuously). 47 (a) 9.78×104 J/K (b) In order to gain more energy, we must generate it from things within the house, like a heat pump, human bodies, and other appliances. As you know, we use a lot of energy to keep our houses warm in the winter because of the loss of heat to the outside. 49 8.01×105 J 51 (a) 1.04×1031 J/K (b) 3.28×1031 J 53 199 J/K 55 This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key (a) 2.47×1014 J (b) 1.60×1014 J (c) 2.85×1010 J/K (d) 8.29×1012 J 1603 57 It should happen twice in every 1.27×1030 s or once in every 6.35×1029 s 1 y 6.35×1029 s 365.25 d 1 h 3600 s 1 d 24 h 2.0×1022 y = 59 (a) 3.0×1029 (b) 24% 61 (a) -2.3810 – 23 J/K (b) 5.6 times more likely (c) If you were betting on two heads and 8 tails, the odds of breaking even are 252 to 45, so on average you would break even. So, no, you wouldn't bet on odds of 252 to 45. Test Prep for AP® Courses 1 (d) 3 (a) 5 (b) 7 (c) 9 (d)
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11 (a) 13 (c) 15 (b) Chapter 16 Problems & Exercises 1 (a) 1.23×103 N/m (b) 6.88 kg (c) 4.00 mm 3 (a) 889 N/m (b) 133 N 5 (a) 6.53×103 N/m Answer Key 1604 (b) Yes 7 16.7 ms 8 0.400 s / beats 9 400 Hz 10 12,500 Hz 11 1.50 kHz 12 (a) 93.8 m/s (b) 11.3×103 rev/min 13 2.37 N/m 15 0.389 kg 18 94.7 kg 21 1.94 s 22 6.21 cm 24 2.01 s 26 2.23 Hz 28 (a) 2.99541 s (b) Since the period is related to the square root of the acceleration of gravity, when the acceleration changes by 1% the period changes by (0.01)2 = 0.01% so it is necessary to have at least 4 digits after the decimal to see the changes. 30 (a) Period increases by a factor of 1.41 ( 2 ) (b) Period decreases to 97.5% of old period 32 Slow by a factor of 2.45 34 length must increase by 0.0116%. 35 (a) 1.99 Hz (b) 50.2 cm (c) 1.41 Hz, 0.710 m 36 (a) 3.95×106 N/m (b) 7.90×106 J 37 a). 0.266 m/s b). 3.00 J 39 This content is available for free at http://cnx.org/content/col11844/1.13 1605 (16.75) (16.76) (16.77) (16.78) (16.79) Answer Key ± 3 2 42 384 J 44 (a). 0.123 m (b). −0.600 J (c). 0.300 J. The rest of the energy may go into heat caused by friction and other damping forces. = 9.26 d = 40.0 Hz w = 16.0 m/s = 700 m = 34.0 cm 46 (a) 5.00×105 J (b) 1.20×103 s 47 49 51 53 55 57 = 4 Hz 59 462 Hz, 4 Hz 61 (a) 3.33 m
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/s (b) 1.25 Hz 63 0.225 W 65 7.07 67 16.0 d 68 2.50 kW 70 3.38×10–5 W/m2 Test Prep for AP® Courses 1 (d) 3 (b) 5 The frequency is given by = 1 = 50 30 = 1.66 Time period is: 1606 Answer Key = 1 = 1 1.66 = 0.6 s 7 (c) 9 The energy of the particle at the center of the oscillation is given by = 1 ×0.2 kg×(5 m·s−1)2 = 2.5 J 22 = 1 2 11 (b) 13 19.7 J 15 (c) 17 = 2 2 − 2 = 50 N ⋅ m−1 = 0.06 = 0.5kg = 50N ⋅ m−1 2×0.06×9.8m ⋅ s−2 (0.2)2 − 0.06×0.5kg×9.8m ⋅ s−2) (50N ⋅ m−1)2 2 = 1.698 m 19 The waves coming from a tuning fork are mechanical waves that are longitudinal in nature, whereas electromagnetic waves are transverse in nature. 21 The sound energy coming out of an instrument depends on its size. The sound waves produced are relative to the size of the musical instrument. A smaller instrument such as a tambourine will produce a high-pitched sound (higher frequency, shorter wavelength), whereas a larger instrument such as a drum will produce a deeper sound (lower frequency, longer wavelength). 23 2 m 25 The student explains the principle of superposition and then shows two waves adding up to form a bigger wave when a crest adds with a crest and a trough with another trough. Also the student shows a wave getting cancelled out when a crest meets a trough and vice versa. 27 The student must note that the shape of the wave remains the same and there is first an overlap and then receding of the waves. 29 (c) Chapter 17 Problems & Exercises 1 0.288 m 3 332 m/s 5 7 0.223 9 (a) 7.70 m w = (331 m/s) 273 K = (331 m/s) 293 K 273 K = 343 m/s (17.12) This content is available for free at http://cnx.org/content/col11844/
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1.13 Answer Key 1607 (b) This means that sonar is good for spotting and locating large objects, but it isn’t able to resolve smaller objects, or detect the detailed shapes of objects. Objects like ships or large pieces of airplanes can be found by sonar, while smaller pieces must be found by other means. 11 (a) 18.0 ms, 17.1 ms (b) 5.00% (c) This uncertainty could definitely cause difficulties for the bat, if it didn’t continue to use sound as it closed in on its prey. A 5% uncertainty could be the difference between catching the prey around the neck or around the chest, which means that it could miss grabbing its prey. (17.23) (17.24) (17.25) (17.36) 3.16×10–4 W/m2 3.04×10–4 W/m2 1.45×10–3 J 3.79×103 Hz 12 14 16 106 dB 18 (a) 93 dB (b) 83 dB 20 (a) 50.1 (b) 5.01×10–3 or 1 200 22 70.0 dB 24 100 26 28 28.2 dB 30 (a) 878 Hz (b) 735 Hz 32 34 (a) 12.9 m/s (b) 193 Hz 36 First eagle hears 4.23×103 Hz Second eagle hears 3.56×103 Hz 38 0.7 Hz 40 0.3 Hz, 0.2 Hz, 0.5 Hz 42 (a) 256 Hz (b) 512 Hz 44 180 Hz, 270 Hz, 360 Hz Answer Key 1608 46 1.56 m 48 (a) 0.334 m (b) 259 Hz 50 3.39 to 4.90 kHz 52 (a) 367 Hz (b) 1.07 kHz 54 (a) = 47.6 Hz, = 1, 3, 5,..., 419 (b) = 95.3 Hz, = 1, 2, 3,..., 210 55 1×106 km (17.49) 57 498.5 or 501.5 Hz 59 82 dB 61 approximately 48, 9, 0, –7, and 20 dB, respectively 63 (a) 23 dB (b) 70 dB 65 Five factors of 10 67 (a) 2×10−10 W/m2 (b) 2×10−13 W/m2 69 2.5 71 1.
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26 72 170 dB 74 103 dB 76 (a) 1.00 (b) 0.823 (c) Gel is used to facilitate the transmission of the ultrasound between the transducer and the patient’s body. 78 (a) 77.0 μm (b) Effective penetration depth = 3.85 cm, which is enough to examine the eye. (c) 16.6 μm 80 (a) 5.78×10–4 m (b) 2.67×106 Hz This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 82 1609 (a) w = 1540 m/s = = 1540 m/s 100×103 Hz = 0.0154 m < 3.50 m. Because the wavelength is much shorter than ⇒ the distance in question, the wavelength is not the limiting factor. (b) 4.55 ms 84 974 Hz (Note: extra digits were retained in order to show the difference.) Test Prep for AP® Courses 1 (b) 3 (e) 5 (c) 7 Answers vary. Students could include a sketch showing an increased amplitude when two waves occupy the same location. Students could also cite conceptual evidence such as sound waves passing through each other. 9 (d) 11 (c) 13 (a) 15 (c) 17 (b) 19 (a), (b) 21 (c) Chapter 18 Problems & Exercises 1 (a) 1.25×1010 (b) 3.13×1012 3 -600 C 5 1.03×1012 7 9.09×10−13 9 1.48×108 C 15 (a) = 1.00 cm = − ∞ (b) 2.12×105 N/C (c) one charge of + 1610 17 (a) 0.252 N to the left (b) = 6.07 cm 19 Answer Key (a)The electric field at the center of the square will be straight up, since and are positive and and are negative and all have the same magnitude. (b) 2.04×107 N/C (upward) 21 0.102 N in the − 23 direction = 4.36×103 N/C 35.0º, below the horizontal. → (a) (b) No 25 (a) 0.263 N (b) If the charges are distributed over some area, there will
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be a concentration of charge along the side closest to the oppositely charged object. This effect will increase the net force. 27 The separation decreased by a factor of 5. 31 = |1 2| 2 = = 2 ⇒ 2 9.00×109 N ⋅ m2/ C2 = 1.60×10–19 m 2 1.67×10–27 kg 2.00×10–9 m = 3.45×1016 m/s2 2 32 (a) 3.2 (b) If the distance increases by 3.2, then the force will decrease by a factor of 10 ; if the distance decreases by 3.2, then the force will increase by a factor of 10. Either way, the force changes by a factor of 10. 34 (a) 1.04×10−9 C (b) This charge is approximately 1 nC, which is consistent with the magnitude of charge typical for static electricity 37 1.02×10−11 39 a. 0.859 m beyond negative charge on line connecting two charges b. 0.109 m from lesser charge on line connecting two charges 42 8.75×10−4 N 44 (a) 6.94×10−8 C (b) 6.25 N/C 46 (a) 300 N/C (east) This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1611 (b) 4.80×10−17 N (east) 52 (a) 5.58×10−11 N/C (b)the coulomb force is extraordinarily stronger than gravity 54 (a) −6.76×105 C (b) 2.63×1013 m/s2 (upward) (c) 2.45×10−18 kg 56 The charge 2 is 9 times greater than 1. Test Prep for AP® Courses 1 (b) 3 (c) 5 (a) 7 (b) 9 (a) -0.1 C, (b) 1.1 C, (c) Both charges will be equal to 1 C, law of conservation of charge, (d) 0.9 C 11 W is negative, X is positive, Y is negative, Z is neutral. 13 (c) 15 (c) 17 (b) 19 a) Ball 1 will have positive charge and Ball 2 will have negative charge. b) The negatively charged rod attracts
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positive charge of Ball 1. The electrons of Ball 1 are transferred to Ball 2, making it negatively charged. c) If Ball 2 is grounded while the rod is still there, it will lose its negative charge to the ground. d) Yes, Ball 1 will be positively charged and Ball 2 will be negatively charge. 21 (c) 23 decrease by 77.78%. 25 (a) 27 (d) 29 (a) 3.60×1010 N, (b) It will become 1/4 of the original value; hence it will be equal to 8.99×109 N 31 (c) 33 (a) 35 (b) 37 (a) 350 N/C, (b) west, (c) 5.6×10−17 N, (d) west. 39 (b) 1612 41 Answer Key (a) i) Field vectors near objects point toward negatively charged objects and away from positively charged objects. (a) ii) The vectors closest to R and T are about the same length and start at about the same distance. We have that / 2 = / 2, so the charge on R is about the same as the charge on T. The closest vectors around S are about the same length as those around R and T. The vectors near S start at about 6 units away, while vectors near R and T start at about 4 units. We have that / 2 = / 2, so / = 2/ 2 = 36/ 16 = 2.25, and so the charge on S is about twice that on R and T. (b) Figure 18.35. A vector diagram. (c) = − ( + )2 + 2 ()2 + ( − )2 (d) The statement is not true. The vector diagram shows field vectors in this region with nonzero length, and the vectors not shown have even greater lengths. The equation in part (c) shows that, when 0 < <, the denominator of the negative term is always greater than the denominator of the third term, but the numerator is the same. So the negative term always has a smaller magnitude than the third term and since the second term is positive the sum of the terms is always positive. Chapter 19 Problems & Exercises 1 42.8 4 1.00105 K 6 (a) 4104 W (b) A defibrillator does not cause serious burns because the skin conducts electricity well at high voltages, like those used in defibrill
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ators. The gel used aids in the transfer of energy to the body, and the skin doesn’t absorb the energy, but rather lets it pass through to the heart. 8 (a) 7.40103 C (b) 1.541020 electrons per second 9 3.89106 C 11 (a) 1.44×1012 V (b) This voltage is very high. A 10.0 cm diameter sphere could never maintain this voltage; it would discharge. (c) An 8.00 C charge is more charge than can reasonably be accumulated on a sphere of that size. 15 This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key (a) 3.00 kV (b) 750 V 1613 17 (a) No. The electric field strength between the plates is 2.5×106 V/m, which is lower than the breakdown strength for air ( 3.0×106 V/m ). (b) 1.7 mm 19 44.0 mV 21 15 kV 23 (a) 800 KeV (b) 25.0 km 24 144 V 26 (a) 1.80 km (b) A charge of 1 C is a very large amount of charge; a sphere of radius 1.80 km is not practical. 28 –2.22×10 – 13 C 30 (a) 3.31×106 V (b) 152 MeV 32 (a) 2.78×10-7 C (b) 2.00×10-10 C 35 (a) 2.96×109 m/s (b) This velocity is far too great. It is faster than the speed of light. (c) The assumption that the speed of the electron is far less than that of light and that the problem does not require a relativistic treatment produces an answer greater than the speed of light. 46 21.6 mC 48 80.0 mC 50 20.0 kV 52 667 pF 54 (a) 4.4 µF (b) 4.0×10 – 5 C 56 (a) 14.2 kV (b) The voltage is unreasonably large, more than 100 times the breakdown voltage of nylon. 1614 Answer Key (c) The assumed charge is unreasonably large and cannot be stored in a capacitor of these dimensions. 57 0.293 μF 59 3.08 µF in series combination
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, 13.0 µF in parallel combination 60 2.79 µF 62 (a) –3.00 µF (b) You cannot have a negative value of capacitance. (c) The assumption that the capacitors were hooked up in parallel, rather than in series, was incorrect. A parallel connection always produces a greater capacitance, while here a smaller capacitance was assumed. This could happen only if the capacitors are connected in series. 63 (a) 405 J (b) 90.0 mC 64 (a) 3.16 kV (b) 25.3 mC 66 (a) 1.42×10−5 C, 6.38×10−5 J (b) 8.46×10−5 C, 3.81×10−4 J 67 (a) 4.43×10 – 12 F (b) 452 V (c) 4.52×10 – 7 J 70 (a) 133 F (b) Such a capacitor would be too large to carry with a truck. The size of the capacitor would be enormous. (c) It is unreasonable to assume that a capacitor can store the amount of energy needed. Test Prep for AP® Courses 1 (a) 3 (b) 5 (c) 7 (a) 9 (b) 11 (b) 13 (a) 15 (c) 17 This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1615 (b) 19 (a) 21 (d) 23 (d) 25 (a) 27 (b) 29 (c) 31 (d) 33 (a) 35 (c) 37 (c) 39 (b) 41 (a) 43 (d) Chapter 20 Problems & Exercises 1 0.278 mA 3 0.250 A 5 1.50ms 7 (a) 1.67k Ω (b) If a 50 times larger resistance existed, keeping the current about the same, the power would be increased by a factor of about 50 (based on the equation = 2 ), causing much more energy to be transferred to the skin, which could cause serious burns. The gel used reduces the resistance, and therefore reduces the power transferred to the skin. 9 (a) 0.120 C (b) 7.501017 electrons 11 96.3 s 13 (a) 7.81 × 1014 He++ nuclei/s (b) 4
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.00 × 103 s (c) 7.71 × 108 s 15 −1.1310−4 m/s 1616 17 9.421013 electrons 18 0.833 A 20 7.3310−2 Ω 22 (a) 0.300 V (b) 1.50 V Answer Key (c) The voltage supplied to whatever appliance is being used is reduced because the total voltage drop from the wall to the final output of the appliance is fixed. Thus, if the voltage drop across the extension cord is large, the voltage drop across the appliance is significantly decreased, so the power output by the appliance can be significantly decreased, reducing the ability of the appliance to work properly. 24 0.104 Ω 26 2.810−2 m 28 1.1010−3 A 30 −5ºC to 45ºC 32 1.03 34 0.06% 36 −17ºC 38 (a) 4.7 Ω (total) (b) 3.0% decrease 40 2.001012 W 44 (a) 1.50 W (b) 7.50 W = V2 V/A 46 V2 Ω 48 1 kW ⋅ h= = AV = ×103 J 1 s (1 h) 3600 s 1 h = 3.60×106 J 50 $438/y 52 $6.25 54 1.58 h 56 $3.94 billion/year 58 25.5 W 60 (a) 2.00109 J This content is available for free at http://cnx.org/content/col11844/1.13 1617 Answer Key (b) 769 kg 62 45.0 s 64 (a) 343 A (b) 2.17103 A (c) 1.10103 A 66 (a) 1.23×103 kg (b) 2.64×103 kg 69 (a) 2.08×105 A (b) 4.33×104 MW (c) The transmission lines dissipate more power than they are supposed to transmit. (d) A voltage of 480 V is unreasonably low for a transmission voltage. Long-distance transmission lines are kept at much higher voltages (often hundreds of kilovolts) to reduce power losses. 73 480 V 75 2.50 ms 77 (a) 4.00 kA (b) 16.0 MW (c) 16.0% 79 2.40 kW 81 (a) 4.0 (b)
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0.50 (c) 4.0 83 (a) 1.39 ms (b) 4.17 ms (c) 8.33 ms 85 (a) 230 kW (b) 960 A 87 (a) 0.400 mA, no effect (b) 26.7 mA, muscular contraction for duration of the shock (can't let go) 89 1.20105 Ω 91 (a) 1.00 Ω 1618 (b) 14.4 kW 93 Temperature increases 860º C. It is very likely to be damaging. 95 80 beats/minute Test Prep for AP® Courses Answer Key 1 (a) 3 10 A 5 (a) 7 3.2 Ω, 2.19 A 9 (b), (d) 11 9.72 × 10−8 Ω·m 13 18 Ω 15 10:3 or 3.33 Chapter 21 Problems & Exercises 1 (a) 2.75 k Ω (b) 27.5 Ω 3 (a) 786 Ω (b) 20.3 Ω 5 29.6 W 7 (a) 0.74 A (b) 0.742 A 9 (a) 60.8 W (b) 3.18 kW 11 (a >>2 (b, so that >>2. 13 This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key (a) -400 k Ω (b) Resistance cannot be negative. 1619 (c) Series resistance is said to be less than one of the resistors, but it must be greater than any of the resistors. 14 2.00 V 16 2.9994 V 18 0.375 Ω 21 (a) 0.658 A (b) 0.997 W (c) 0.997 W; yes 23 (a) 200 A (b) 10.0 V (c) 2.00 kW (d) 0.1000 Ω 80.0 A, 4.0 V, 320 W 25 (a) 0.400 Ω (b) No, there is only one independent equation, so only can be found. 29 (a) –0.120 V (b) -1.4110−2 Ω (c) Negative terminal voltage; negative load resistance. (d) The assumption that such a cell could provide 8.50 A is inconsistent with its internal resistance. −22
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+ emf1 − 21 + 33 + 32 - emf2 = 0 3 = 1 + 2 emf2 - 22 - 22 + 15 + 11 - emf1 + 11 = 0 (21.69) (21.70) (21.71) 31 35 37 39 (a) I1 = 4.75 A (b) I2 = -3.5 A (c) I3 = 8.25 A 41 (a) No, you would get inconsistent equations to solve. (b) 1 ≠ 2 + 3. The assumed currents violate the junction rule. 42 30 44 1.98 k Ω 46 1.2510-4 Ω (21.75) Answer Key 1620 48 (a) 3.00 M Ω (b) 2.99 k Ω 50 (a) 1.58 mA (b) 1.5848 V (need four digits to see the difference) (c) 0.99990 (need five digits to see the difference from unity) 52 15.0 μA 54 (a) Figure 21.39. (b) 10.02 Ω (c) 0.9980, or a 2.0×10–1 percent decrease (d) 1.002, or a 2.0×10–1 percent increase (e) Not significant. 56 (a) −66.7 Ω (b) You can’t have negative resistance. (c) It is unreasonable that G is greater than tot (see Figure 21.36). You cannot achieve a full-scale deflection using a current less than the sensitivity of the galvanometer. Range = 5.00 Ω to 5.00 k Ω (21.82) 57 24.0 V 59 1.56 k Ω 61 (a) 2.00 V (b) 9.68 Ω 62 63 range 4.00 to 30.0 M Ω 65 (a) 2.50 μF (b) 2.00 s 67 86.5% 69 (a) 1.25 k Ω (b) 30.0 ms 71 (a) 20.0 s This content is available for free at http://cnx.org/content/col11844/1.13 1621 Answer Key (b) 120 s (c) 16.0 ms 73 1.73×10−2 s 74 3.3310−3 Ω 76 (a) 4.99 s
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(b) 3.87ºC (c) 31.1 k Ω (d) No Test Prep for AP® Courses 1 (a), (b) 3 (b) 5 (a) 4-Ω resistor; (b) combination of 20-Ω, 20-Ω, and 10-Ω resistors; (c) 20 W in each 20-Ω resistor, 40 W in 10-Ω resistor, 64 W in 4-Ω resistor, total 144W total in resistors, output power is 144 W, yes they are equal (law of conservation of energy); (d) 4 Ω and 3 Ω for part (a) and no change for part (b); (e) no effect, it will remain the same. 7 0.25 Ω, 0.50 Ω, no change 9 a. (c) b. (c) c. (d) d. (d) 11 a. I1 + I3 = I2 b. E1 - I1R1 - I2R2 - I1r1 = 0; - E2 + I1R1 - I3R3 - I3r2 = 0 c. d. I1 = 8/15 A, I2 = 7/15 A and I3 = -1/15 A I1 = 2/5 A, I2 = 3/5 A and I3 = 1/5 A e. PE1 = 18/5 W and PR1 = 24/25 W, PR2 = 54/25 W, PR3 = 12/25 W. Yes, PE1 = PR1+ PR2 + PR3 f. R3, losses in the circuit 13 (a) 20 mA, Figure 21.44, 5.5 s; (b) 24 mA, Figure 21.35, 2 s Chapter 22 Problems & Exercises 1 (a) Left (West) (b) Into the page (c) Up (North) (d) No force (e) Right (East) (f) Down (South) 3 (a) East (right) 1622 (b) Into page (c) South (down) 5 (a) Into page (b) West (left) (c) Out of page Answer Key 7 7.50×10−7 N perpendicular to both the magnetic field lines and the velocity 9 (a) 3.01×10−5
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T (b) This is slightly less then the magnetic field strength of 5×10−5 T at the surface of the Earth, so it is consistent. 11 (a) 6.67×10−10 C (taking the Earth’s field to be 5.00×10−5 T ) (b) Less than typical static, therefore difficult 12 4.27 m 14 (a) 0.261 T (b) This strength is definitely obtainable with today’s technology. Magnetic field strengths of 0.500 T are obtainable with permanent magnets. 16 4.36×10−4 m 18 (a) 3.00 kV/m (b) 30.0 V 20 0.173 m 22 7.50×10−4 V 24 (a) 1.18 × 10 3 m/s (b) Once established, the Hall emf pushes charges one direction and the magnetic force acts in the opposite direction resulting in no net force on the charges. Therefore, no current flows in the direction of the Hall emf. This is the same as in a current-carrying conductor—current does not flow in the direction of the Hall emf. 26 11.3 mV 28 1.16 μV 30 2.00 T 31 (a) west (left) (b) into page (c) north (up) (d) no force (e) east (right) (f) south (down) This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 33 (a) into page (b) west (left) (c) out of page 35 (a) 2.50 N (b) This is about half a pound of force per 100 m of wire, which is much less than the weight of the wire itself. Therefore, it does not cause any special concerns. 1623 37 1.80 T 39 (a) 30º (b) 4.80 N 41 (a) τ decreases by 5.00% if B decreases by 5.00% (b) 5.26% increase 43 10.0 A 45 A ⋅ m2 ⋅ T = A ⋅ m2 47 3.48×10−26 N ⋅ m 49 (a) 0.666 N ⋅ m west b) This is not a very significant torque, so practical use would be limited. Also, the current would need to be
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alternated to make the loop rotate (otherwise it would oscillate). 50 (a) 8.53 N, repulsive (b) This force is repulsive and therefore there is never a risk that the two wires will touch and short circuit. 52 400 A in the opposite direction 54 (a) 1.67×10−3 N/m (b) 3.33×10−3 N/m (c) Repulsive (d) No, these are very small forces 56 (a) Top wire: 2.65×10−4 N/m s, 10.9º to left of up (b) Lower left wire: 3.61×10−4 N/m, 13.9º down from right (c) Lower right wire: 3.46×10−4 N/m, 30.0º down from left 58 (a) right-into page, left-out of page (b) right-out of page, left-into page (c) right-out of page, left-into page Answer Key 1624 60 (a) clockwise (b) clockwise as seen from the left (c) clockwise as seen from the right 61 1.01×1013 T 63 (a) 4.80×10−4 T (b) Zero (c) If the wires are not paired, the field is about 10 times stronger than Earth’s magnetic field and so could severely disrupt the use of a compass. 65 39.8 A 67 (a) 3.14×10−5 T (b) 0.314 T 69 7.55×10−5 T, 23.4º 71 10.0 A 73 (a) 9.09×10−7 N upward (b) 3.03×10−5 m/s2 75 60.2 cm 77 (a) 1.02×103 N/m2 (b) Not a significant fraction of an atmosphere 79 17.0×10−4%/ºC 81 18.3 MHz 83 (a) Straight up (b) 6.00×10−4 N/m (c) 94.1 μm (d)2.47 Ω/m, 49.4 V/m 85 (a) 571 C (b) Impossible to have such a large separated charge on such a small object. (c) The 1.00-N force is much too great to be realistic in the Earth’
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s field. 87 (a) 2.40×106 m/s (b) The speed is too high to be practical ≤ 1% speed of light This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1625 (c) The assumption that you could reasonably generate such a voltage with a single wire in the Earth’s field is unreasonable 89 (a) 25.0 kA (b) This current is unreasonably high. It implies a total power delivery in the line of 50.0x10^9 W, which is much too high for standard transmission lines. (c)100 meters is a long distance to obtain the required field strength. Also coaxial cables are used for transmission lines so that there is virtually no field for DC power lines, because of cancellation from opposing currents. The surveyor’s concerns are not a problem for his magnetic field measurements. Test Prep for AP® Courses 1 (a) 3 (b) 5 (b) 7 (a) 9 (b) 11 (e) 13 (c) 15 (c) Chapter 23 Problems & Exercises 1 Zero 3 (a) CCW (b) CW (c) No current induced 5 (a) 1 CCW, 2 CCW, 3 CW (b) 1, 2, and 3 no current induced (c) 1 CW, 2 CW, 3 CCW 9 (a) 3.04 mV (b) As a lower limit on the ring, estimate R = 1.00 mΩ. The heat transferred will be 2.31 mJ. This is not a significant amount of heat. 11 0.157 V 13 proportional to 1 17 (a) 0.630 V (b) No, this is a very small emf. 19 2.22 m/s 25 1626 (a) 10.0 N (b) 2.81×108 J (c) 0.36 m/s Answer Key (d) For a week-long mission (168 hours), the change in velocity will be 60 m/s, or approximately 1%. In general, a decrease in velocity would cause the orbit to start spiraling inward because the velocity would no longer be sufficient to keep the circular orbit. The long-term consequences are that the shuttle would require a little more fuel to maintain the desired speed, otherwise the orbit would spiral slightly inward. 28 474 V 30 0.247 V 32
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(a) 50 (b) yes 34 (a) 0.477 T (b) This field strength is small enough that it can be obtained using either a permanent magnet or an electromagnet. 36 (a) 5.89 V (b) At t=0 (c) 0.393 s (d) 0.785 s 38 (a) 1.92×106 rad/s (b) This angular velocity is unreasonably high, higher than can be obtained for any mechanical system. (c) The assumption that a voltage as great as 12.0 kV could be obtained is unreasonable. 39 (a) 12.00 Ω (b) 1.67 A 41 72.0 V 43 0.100 Ω 44 (a) 30.0 (b) 9.75×10−2 A 46 (a) 20.0 mA (b) 2.40 W (c) Yes, this amount of power is quite reasonable for a small appliance. 48 (a) 0.063 A (b) Greater input current needed. 50 (a) 2.2 (b) 0.45 This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key (c) 0.20, or 20.0% 52 (a) 335 MV (b) way too high, well beyond the breakdown voltage of air over reasonable distances (c) input voltage is too high 1627 54 (a) 15.0 V (b) 75.0 A (c) yes 55 1.80 mH 57 3.60 V 61 (a) 31.3 kV (b) 125 kJ (c) 1.56 MW (d) No, it is not surprising since this power is very high. 63 (a) 1.39 mH (b) 3.33 V (c) Zero 65 60.0 mH 67 (a) 200 H (b) 5.00ºC 69 500 H 71 50.0 Ω 73 1.00×10–18 s to 0.100 s 75 95.0% 77 (a) 24.6 ms (b) 26.7 ms (c) 9% difference, which is greater than the inherent uncertainty in the given parameters. 79 531 Hz 81 1.33 nF 83 (a) 2.55 A (b) 1.53 mA 85 63.7 µH 87 1628 (a
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) 21.2 mH (b) 8.00 Ω 89 (a) 3.18 mF (b) 16.7 Ω 92 Answer Key (a) 40.02 Ω at 60.0 Hz, 193 Ω at 10.0 kHz (b) At 60 Hz, with a capacitor, Z=531 Ω, over 13 times as high as without the capacitor. The capacitor makes a large difference at low frequencies. At 10 kHz, with a capacitor Z=190 Ω, about the same as without the capacitor. The capacitor has a smaller effect at high frequencies. 94 (a) 529 Ω at 60.0 Hz, 185 Ω at 10.0 kHz (b) These values are close to those obtained in Example 23.12 because at low frequency the capacitor dominates and at high frequency the inductor dominates. So in both cases the resistor makes little contribution to the total impedance. 96 9.30 nF to 101 nF 98 3.17 pF 100 (a) 1.31 μH (b) 1.66 pF 102 (a) 12.8 kΩ (b) 1.31 kΩ (c) 31.9 mA at 500 Hz, 312 mA at 7.50 kHz (d) 82.2 kHz (e) 0.408 A 104 (a) 0.159 (b) 80.9º (c) 26.4 W (d) 166 W 106 16.0 W Test Prep for AP® Courses 1 (c) This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1629 Figure 23.6. 3 (c) 5 (a), (d) 7 (c) Chapter 24 Problems & Exercises 3 150 kV/m 6 (a) 33.3 cm (900 MHz) 11.7 cm (2560 MHz) (b) The microwave oven with the smaller wavelength would produce smaller hot spots in foods, corresponding to the one with the frequency 2560 MHz. 8 26.96 MHz 10 5.0×1014 Hz 12 14 0.600 m 16 = = 3.00108 m/s 1.201015 Hz = 2.5010 – 7 m () (a) = = 3.00108 m/s 110-10 m = 31018 Hz (b) X-rays 19 (a) 6.00×106 m (b)
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4.33×10−5 T 21 (a) 1.50 × 10 6 Hz, AM band (b) The resonance of currents on an antenna that is 1/4 their wavelength is analogous to the fundamental resonant mode of an air column closed at one end, since the tube also has a length equal to 1/4 the wavelength of the fundamental oscillation. 23 (a) 1.55×1015 Hz (b) The shortest wavelength of visible light is 380 nm, so that 1630 Answer Key () visible UV = 380 nm 193 nm = 1.97. In other words, the UV radiation is 97% more accurate than the shortest wavelength of visible light, or almost twice as accurate! 25 3.90×108 m 27 (a) 1.50×1011 m (b) 0.500 s (c) 66.7 ns 29 (a) −3.5102 W/m2 (b) 88% (c) 1.7 T 30 = = 2 0 0 2 () 3.00108 m/s 8.8510–12 C2 /N ⋅ m2 (125 V/m)2 2 = 20.7 W/m2 32 (a) = = 2 = 0.25010−3 W 0.50010−3 m 2 = 318 W/m2 ave = 2 0 2μ0 ⇒ 0 = 2μ0 1 / 2 = 410−7 T ⋅ m/A 2 318.3 W/m2 3.00108 m/s 1 / 2 = 1.6310−6 T 0 = 0 = 3.00108 m/s 1.63310−6 T = 4.90102 V/m (b) (c) 34 (a) 89.2 cm (b) 27.4 V/m 36 (a) 333 T (b) 1.331019 W/m2 (c) 13.3 kJ 38 This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key (a) = = 4π 2 ∝ 1 2 1631 (b 40 13.5 pF 42 (a) 4.07 kW/m2 (b) 1.75 kV/m (c) 5.84 T (d) 2 min 19 s 44 (a) 5.00103 W/m2 (b) 3.88×10
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−6 N (c) 5.18×10−12 N 46 (a) = 0 (b) 7.50×10−10 s (c) 1.00×10−9 s 48 (a) 1.01×106 W/m2 (b) Much too great for an oven. (c) The assumed magnetic field is unreasonably large. 50 (a) 2.53×10−20 H (b) L is much too small. (c) The wavelength is unreasonably small. Test Prep for AP® Courses 1 (b) 3 (a) 5 (d) 7 (d) 9 (d) 11 (a) Chapter 25 Problems & Exercises 1 1632 Answer Key Top 1.715 m from floor, bottom 0.825 m from floor. Height of mirror is 0.890 m, or precisely one-half the height of the person. 5 2.25×108 m/s in water 2.04×108 m/s in glycerine 7 1.490, polystyrene 9 1.28 s 11 1.03 ns 13 = 1.46, fused quartz 17 (a) 0.898 (b) Can’t have < 1.00 since this would imply a speed greater than. (c) Refracted angle is too big relative to the angle of incidence. 19 (a) 5.00 (b) Speed of light too slow, since index is much greater than that of diamond. (c) Angle of refraction is unreasonable relative to the angle of incidence. 22 66.3º 24 > 1.414 26 1.50, benzene 29 46.5º, red; 46.0º, violet 31 (a) 0.043º (b) 1.33 m 33 71.3º 35 53.5º, red; 55.2º, violet 37 5.00 to 12.5 D 39 −0.222 m 41 (a) 3.43 m (b) 0.800 by 1.20 m 42 (a) −1.35 m (on the object side of the lens). (b) +10.0 (c) 5.00 cm This content is available for free at http://cnx.org/content/col11844/1.13 1633 Answer Key 43 44.4 cm 45 (a) 6.60 cm (b) –0.333 47 (a) +7
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.50 cm (b) 13.3 D (c) Much greater 49 (a) +6.67 (b) +20.0 (c) The magnification increases without limit (to infinity) as the object distance increases to the limit of the focal distance. 51 −0.933 mm 53 +0.667 m 55 (a) –1.5×10–2 m (b) –66.7 D 57 +0.360 m (concave) 59 (a) +0.111 (b) -0.334 cm (behind “mirror”) (c) 0.752cm 61 63 6.82 kW/m2 = i o = − i o = − −25.61) Test Prep for AP® Courses 1 (c) 3 (c) 5 (a) 7 Since light bends toward the normal upon entering a medium with a higher index of refraction, the upper path is a more accurate representation of a light ray moving from A to B. 9 First, measure the angle of incidence and the angle of refraction for light entering the plastic from air. Since the two angles can be measured and the index of refraction of air is known, the student can solve for the index of refraction of the plastic. Next, measure the angle of incidence and the angle of refraction for light entering the gas from the plastic. Since the two angles can be measured and the index of refraction of the plastic is known, the student can solve for the index of refraction of the gas. 11 1634 Answer Key The speed of light in a medium is simply c/n, so the speed of light in water is 2.25 × 108 m/s. From Snell’s law, the angle of incidence is 44°. 13 (d) 15 (a) 17 (a) 19 (b) Chapter 26 Problems & Exercises 1 52.0 D 3 (a) −0.233 mm (b) The size of the rods and the cones is smaller than the image height, so we can distinguish letters on a page. 5 (a) +62.5 D (b) –0.250 mm (c) –0.0800 mm 6 2.00 m 8 (a) ±0.45 D (b) The person was nearsighted because the patient was myopic and the power was reduced. 10 0.143 m 12 1.00 m 14 20.0 cm 16
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–5.00 D 18 25.0 cm 20 –0.198 D 22 30.8 cm 24 –0.444 D 26 (a) 4.00 (b) 1600 28 (a) 0.501 cm (b) Eyepiece should be 204 cm behind the objective lens. 30 (a) +18.3 cm (on the eyepiece side of the objective lens) This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key (b) -60.0 (c) -11.3 cm (on the objective side of the eyepiece) 1635 (d) +6.67 (e) -400 33 −40.0 35 −1.67 37 +10.0 cm 39 (a) 0.251 μm (b) Yes, this thickness implies that the shape of the cornea can be very finely controlled, producing normal distant vision in more than 90% of patients. Test Prep for AP® Courses 1 (a) 3 (c) 5 (a) 7 (b) 9 (d) 11 (c) Chapter 27 Problems & Exercises 1 1 / 1.333 = 0.750 3 1.49, Polystyrene 5 0.877 glass to water 6 0.516º 8 1.22×10−6 m 10 600 nm 12 2.06º 14 1200 nm (not visible) 16 (a) 760 nm (b) 1520 nm 18 For small angles sin − tan ≈ (in radians). For two adjacent fringes we have, 1636 and Subtracting these equations gives sin m = sin m + 1 = ( + 1) ( + 1) − sin m + 1 − sin tan = = Answer Key (27.11) (27.12) (27.13) 20 450 nm 21 5.97º 23 8.99×103 25 707 nm 27 (a) 11.8º, 12.5º, 14.1º, 19.2º (b) 24.2º, 25.7º, 29.1º, 41.0º (c) Decreasing the number of lines per centimeter by a factor of x means that the angle for the x‐order maximum is the same as the original angle for the first- order maximum. 29 589.1 nm and 589.6 nm 31 28.7º 33 43.2
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º 35 90.0º 37 (a) The longest wavelength is 333.3 nm, which is not visible. (b) 333 nm (UV) (c) 6.58×103 cm 39 1.13×10−2 m 41 (a) 42.3 nm (b) Not a visible wavelength The number of slits in this diffraction grating is too large. Etching in integrated circuits can be done to a resolution of 50 nm, so slit separations of 400 nm are at the limit of what we can do today. This line spacing is too small to produce diffraction of light. 43 (a) 33.4º (b) No 45 (a) 1.35×10−6 m This content is available for free at http://cnx.org/content/col11844/1.13 1637 Answer Key (b) 69.9º 47 750 nm 49 (a) 9.04º (b) 12 51 (a) 0.0150º (b) 0.262 mm (c) This distance is not easily measured by human eye, but under a microscope or magnifying glass it is quite easily measurable. 53 (a) 30.1º (b) 48.7º (c) No (d) 2θ1 = (2)(14.5º) = 29º 2 − 1 = 30.05º−14.5º=15.56º. Thus, 29º ≈ (2)(15.56º) = 31.1º. 55 23.6º and 53.1º 57 (a) 1.63×10−4 rad (b) 326 ly 59 1.46×10−5 rad 61 (a) 3.04×10−7 rad (b) Diameter of 235 m 63 5.15 cm 65 (a) Yes. Should easily be able to discern. (b) The fact that it is just barely possible to discern that these are separate bodies indicates the severity of atmospheric aberrations. 70 532 nm (green) 72 83.9 nm 74 620 nm (orange) 76 380 nm 78 33.9 nm 80 4.42×10−5 m 82 The oil film will appear black, since the reflected light is not in the visible part of the spectrum. 84 45.0º 86 Answer Key 1638 45.7 mW/m2 88 90.0% 90 0 92 48.8º 94 41.2
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º 96 (a) 1.92, not diamond (Zircon) (b) 55.2º 98 2 = 0.707 1 100 (a) 2.07×10-2 °C/s (b) Yes, the polarizing filters get hot because they absorb some of the lost energy from the sunlight. Test Prep for AP® Courses 1 (b) 3 (b) and (c) 5 (b) 7 (b) 9 (b) 11 (d) 13 (b) 15 (d) 17 (b) Chapter 28 Problems & Exercises 1 (a) 1.0328 (b) 1.15 3 5.96×10−8 s 5 0.800 7 0.140 9 (a) 0.745 (b) 0.99995 (to five digits to show effect) This content is available for free at http://cnx.org/content/col11844/1.13 1639 Answer Key 11 (a) 0.996 (b) cannot be less than 1. (c) Assumption that time is longer in moving ship is unreasonable. 12 48.6 m 14 (a) 1.387 km = 1.39 km (b) 0.433 km (c) = Thus, the distances in parts (a) and (b) are related when = 3.20. 16 (a) 4.303 y (to four digits to show any effect) (b) 0.1434 y (c) Δt = γΔt0 ⇒ = Δt Δt0 = 4.303 y 0.1434 y = 30.0 Thus, the two times are related when 30.00. 18 (a) 0.250 (b) must be ≥1 (c) The Earth-bound observer must measure a shorter length, so it is unreasonable to assume a longer length. 20 (a) 0.909 (b) 0.400 22 0.198 24 a) 658 nm b) red c) / 9.92×10−5 (negligible) 26 0.991 28 −0.696 30 0.01324 32 ′ =, so = 1 + ( 2) = 1 + ( / 2) = 1 + ( / ) = () = 34 a) 0.99947 1640 b) 1.2064×1011 y c) 1.2058×1011 y (all to sufficient digits to
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show effects) 35 4.09×10–19 kg ⋅ m/s 37 (a) 3.000000015×1013 kg ⋅ m/s. Answer Key (b) Ratio of relativistic to classical momenta equals 1.000000005 (extra digits to show small effects) 39 2.9957×108 m/s 41 (a) 1.121×10–8 m/s (b) The small speed tells us that the mass of a proton is substantially smaller than that of even a tiny amount of macroscopic matter! 43 8.20×10−14 J 0.512 MeV 45 2.3×10−30 kg 47 (a) 1.11×1027 kg (b) 5.56×10−5 49 7.1×10−3 kg 7.1×10−3 The ratio is greater for hydrogen. 51 208 0.999988 53 6.92×105 J 1.54 55 (a) 0.914 (b) The rest mass energy of an electron is 0.511 MeV, so the kinetic energy is approximately 150% of the rest mass energy. The electron should be traveling close to the speed of light. 57 90.0 MeV 59 This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1641 (a) 2 − 1 2 = 22 + 24 = 22 4 so that 2 4, and therefore 2 2 = ()2 2 2 = 2 − 1 (b) yes 61 1.07×103 63 6.56×10−8 kg 4.37×10−10 65 0.314 0.99995 67 (a) 1.00 kg (b) This much mass would be measurable, but probably not observable just by looking because it is 0.01% of the total mass. 69 (a) 6.3×1011 kg/s (b) 4.5×1010 y (c) 4.44×109 kg (d) 0.32% Test Prep for AP® Courses 1 (a) 3 The relativistic Doppler effect takes into account the special relativity concept of time dilation and also does not require a medium of propagation to be used as a point of reference (light does not require a medium for propagation). 5 Relativistic kinetic energy is given as KErel = ( − 1)2 where = 1 1 − 2 2
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Classical kinetic energy is given as KEclass = 1 22 At low velocities = 0, a binomial expansion and subsequent approximation of gives: = 1 + 12 22 or − 1 = 12 22 Substituting − 1 in the expression for KErel gives KErel = 12 22 2 = 1 22 = KEclass 1642 Answer Key Hence, relativistic kinetic energy becomes classical kinetic energy when ≪. Chapter 29 Problems & Exercises 1 (a) 0.070 eV (b) 14 3 (a) 2.21×1034 J (b) 2.26×1034 (c) No 4 263 nm 6 3.69 eV 8 0.483 eV 10 2.25 eV 12 (a) 264 nm (b) Ultraviolet 14 1.95×106 m/s 16 (a) 4.02×1015 /s (b) 0.256 mW 18 (a) –1.90 eV (b) Negative kinetic energy (c) That the electrons would be knocked free. 20 6.34×10−9 eV, 1.01×10−27 J 22 2.42×1020 Hz 24 = 6.62607×10−34 J ⋅ s 2.99792×108 m/s 109 nm 1 m 1.00000 eV 1.60218×10−19 J (29.22) = 1239.84 eV ⋅ nm ≈ 1240 eV ⋅ nm 26 (a) 0.0829 eV (b) 121 (c) 1.24 MeV (d) 1.24×105 28 This content is available for free at http://cnx.org/content/col11844/1.13 1643 Answer Key (a) 25.0×103 eV (b) 6.04×1018 Hz 30 (a) 2.69 (b) 0.371 32 (a) 1.25×1013 photons/s (b) 997 km 34 8.33×1013 photons/s 36 181 km 38 (a) 1.66×10−32 kg ⋅ m/s (b) The wavelength of microwave photons is large, so the momentum they carry is very small. 40 (a) 13.3 μm (b) 9.38×10-2 eV 42 (a) 2.65×10−28 kg ⋅ m/s (
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b) 291 m/s (c) electron 3.86×10−26 J, photon 7.96×10−20 J, ratio 2.06×106 44 (a) 1.32×10−13 m (b) 9.39 MeV (c) 4.70×10−2 MeV 46 = 2 and =, so = 2 = 2. (29.35) As the mass of particle approaches zero, its velocity will approach, so that the ratio of energy to momentum in this limit is lim→0 = 2 = (29.36) which is consistent with the equation for photon energy. 48 (a) 3.00×106 W (b) Headlights are way too bright. (c) Force is too large. 49 7.28×10–4 m Answer Key 15.1 keV (29.42) 1644 51 6.62×107 m/s 53 1.32×10–13 m 55 (a) 6.62×107 m/s (b) 22.9 MeV 57 59 (a) 5.29 fm (b) 4.70×10−12 J (c) 29.4 MV 61 (a) 7.28×1012 m/s (b) This is thousands of times the speed of light (an impossibility). (c) The assumption that the electron is non-relativistic is unreasonable at this wavelength. 62 (a) 57.9 m/s (b) 9.55×10−9 eV (c) From Table 29.1, we see that typical molecular binding energies range from about 1eV to 10 eV, therefore the result in part (b) is approximately 9 orders of magnitude smaller than typical molecular binding energies. 64 29 nm, 290 times greater 66 1.10×10−13 eV 68 3.3×10−22 s 70 2.66×10−46 kg 72 0.395 nm 74 (a) 1.3×10−19 J (b) 2.1×1023 (c) 1.4×102 s 76 (a) 3.35×105 J (b) 1.12×10–3 kg ⋅ m/s (c) 1.12×10–3 m/s (d) 6.23×10–7 J This content is available for free at http://cnx.org/content/col11844/1.13
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Answer Key 78 (a) 1.06×103 (b) 5.33×10−16 kg ⋅ m/s (c) 1.24×10−18 m 80 (a) 1.62×103 m/s 1645 (b) 4.42×10−19 J for photon, 1.19×10−24 J for electron, photon energy is 3.71×105 times greater (c) The light is easier to make because 450-nm light is blue light and therefore easy to make. Creating electrons with 7.43 μeV of energy would not be difficult, but would require a vacuum. 81 (a) 2.30×10−6 m (b) 3.20×10−12 m 83 3.69×10−4 ºC 85 (a) 2.00 kJ (b) 1.33×10−5 kg ⋅ m/s (c) 1.33×10−5 N (d) yes Test Prep for AP® Courses 1 (b) 3 (c) 5 (b) 7 (c) 9 (c) 11 (a) 13 (a) 15 (c) 17 (d) 19 (d) Chapter 30 Problems & Exercises 1 1.84×103 3 50 km 1646 4 6×1020 kg/m3 6 (a) 10.0 μm Answer Key (b) It isn’t hard to make one of approximately this size. It would be harder to make it exactly 10.0 μm. i ⋅ f)2 2 2 − f i i = 2, f = 1, so that = m 1.097×107 (2×1)2 22 − 12 = 1.22×10−7 m = 122 nm, which is UV radiation6.626×10−34 J·s)2 4 2(9.109×10−31 kg)(8.988×109 N·m2 / C2)(1)(1.602×10−19 C)2 = 0.529×10−10 m 11 0.850 eV 13 2.12×10–10 m 15 365 nm It is in the ultraviolet. 17 No overlap 365 nm 122 nm 19 7 21 (a) 2 (b) 54.4 eV = 2 23 2 2 velocity, giving: = 2, so that = so that = 2 2 2. From the equation =
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2, we can substitute for the 2 2 4 2 = 2 B where B = 2 2 4π2. 25 (a) 0.248×10−10 m (b) 50.0 keV (c) The photon energy is simply the applied voltage times the electron charge, so the value of the voltage in volts is the same as the value of the energy in electron volts. 27 (a) 100×103 eV, 1.60×10−14 J (b) 0.124×10−10 m 29 (a) 8.00 keV This content is available for free at http://cnx.org/content/col11844/1.13 1647 Answer Key (b) 9.48 keV 30 (a) 1.96 eV (b) (1240 eV·nm) / (1.96 eV) = 633 nm (c) 60.0 nm 32 693 nm 34 (a) 590 nm (b) (1240 eV·nm) / (1.17 eV) = 1.06 μm 35 = 4, 3 are possible since < and ∣ ∣ ≤. 37 = 4 ⇒ = 3, 2, 1, 0 ⇒ = ±3, ± 2, ± 1, 0 are possible. 39 (a) 1.49×10−34 J ⋅ s (b) 1.06×10−34 J ⋅ s 41 (a) 3.66×10−34 J ⋅ s (b) = 9.13×10−35 J ⋅ s (c) = 12 3 / 4 = 4 43 = 54.7º, 125.3º 44 (a) 32. (b) 2 in 6 in 10 in and 14 in, for a total of 32. 46 (a) 2 (b) 3 9 48 (b) ≥ is violated, (c) cannot have 3 electrons in subshell since 3 > (2 + 1) = 2 (d) cannot have 7 electrons in subshell since 7 > (2 + 1) = 2(2 + 1) = 6 50 (a) The number of different values of overall factor of 2 since each can have equal to either +1 / 2 or −1 / 2 ⇒ 2(2 + 1). is ± ± ( − 1),...,0 for each > 0 and one for = 0 ⇒ (2 + 1
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). Also an (b) for each value of, you get 2(2 + 1) = 0, 1, 2,...,(–1) ⇒ 2 (2)(0) + 1 + (2)(1) + 1 +.... + (2)( − 1) + 1 = 2 1 + 3 +... + (2 − 3) + (2 − 1) ⏟ terms to see that the expression in the box is = 2 imagine taking ( − 1) from the last term and adding it to first term 1648 Answer Key = 2 1 + (–1) + 3 +... + (2 − 3) + (2 − 1)( − 1) = 2. Now take ( − 3) from penultimate term and add to the second term 2[ + +... + + ] + 3 +.... + (2 − 3) + = 22. ⏟ terms 52 The electric force on the electron is up (toward the positively charged plate). The magnetic force is down (by the RHR). 54 401 nm 56 (a) 6.54×10−16 kg (b) 5.54×10−7 m 58 1.76×1011 C/kg, which agrees with the known value of 1.759×1011 C/kg to within the precision of the measurement 60 (a) 2.78 fm (b) 0.37 of the nuclear radius. 62 (a) 1.34×1023 (b) 2.52 MW 64 (a) 6.42 eV (b) 7.27×10−20 J/molecule (c) 0.454 eV, 14.1 times less than a single UV photon. Therefore, each photon will evaporate approximately 14 molecules of tissue. This gives the surgeon a rather precise method of removing corneal tissue from the surface of the eye. 66 91.18 nm to 91.22 nm 68 (a) 1.24×1011 V (b) The voltage is extremely large compared with any practical value. (c) The assumption of such a short wavelength by this method is unreasonable. Test Prep for AP® Courses 1 (a), (d) 3 (a) 5 (a) 7 (b) 9 (a) 11 (d) 13 (d) 15 (a), (c) This content is available for free at http://cnx.org/content/col11844/1
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.13 Answer Key 1649 Chapter 31 Problems & Exercises 1 1.67×104 5 = = 3 ⇒ = 1 3 1/3 = 2.3×1017 kg 1000 kg/m3 = 61×103 m = 61 km 7 1.9 fm 9 (a) 4.6 fm (b) 0.61 to 1 11 85.4 to 1 13 12.4 GeV 15 19.3 to 1 17 19 21 23 25 27 29 3 H2 → 2 1 ¯ 3 He1 + − + 50 25 → 24 25 50 Cr26 + + + 7 Be3 + − → 3 4 7 Li4 + 210 Po126 → 82 84 206 Pb124 + 2 4He2 137 Cs82 → 56 55 ¯ 137 Ba81 + − + 232 Th142 → 88 90 228 Ra140 + 2 4He2 (31.17) (31.47) (31.48) (31.49) (31.50) (31.51) (31.52) (a) charge:(+1) + (−1) = 0 electron family number: (+1) + (−1b) 0.511 MeV (c) The two rays must travel in exactly opposite directions in order to conserve momentum, since initially there is zero momentum if the center of mass is initially at rest. 31 33 35 = ( + 1) − 1; = ; efn : 0 = (+1) + (−1) - 1 = − 1; = ; efn :(+1) = (+1) (31.53) (31.54) (a) 88 226 Ra138 → 86 222 Rn136 + 2 4He2 (b) 4.87 MeV 37 1650 ¯ (a) n → p + − + (b) ) 0.783 MeV 39 1.82 MeV 41 (a) 4.274 MeV (b) 1.927×10−5 Answer Key (c) Since U-238 is a slowly decaying substance, only a very small number of nuclei decay on human timescales; therefore, although those nuclei that decay lose a noticeable fraction of their mass, the change in the total mass of the sample is not detectable for a macroscopic sample. 43 (a) 8 15 O7 + − → 7 15 N8 + (b) 2.754 MeV 44 57,300 y 46 (a) 0.988 Ci (b
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) The half-life of 226 Ra is now better known. 48 1.22×103 Bq 50 (a) 16.0 mg (b) 0.0114% 52 1.48×1017 y 54 5.6×104 y 56 2.71 y 58 (a) 1.56 mg (b) 11.3 Ci 60 (a) 1.23×10−3 (b) Only part of the emitted radiation goes in the direction of the detector. Only a fraction of that causes a response in the detector. Some of the emitted radiation (mostly particles) is observed within the source. Some is absorbed within the source, some is absorbed by the detector, and some does not penetrate the detector. 62 (a) 1.68×10 – 5 Ci (b) 8.65×1010 J (c) $ 2.9×103 64 (a) 6.97×1015 Bq This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key (b) 6.24 kW (c) 5.67 kW 68 (a) 84.5 Ci 1651 (b) An extremely large activity, many orders of magnitude greater than permitted for home use. (c) The assumption of 1.00 μA is unreasonably large. Other methods can detect much smaller decay rates. 69 1.112 MeV, consistent with graph 71 7.848 MeV, consistent with graph 73 (a) 7.680 MeV, consistent with graph (b) 7.520 MeV, consistent with graph. Not significantly different from value for 12 C, but sufficiently lower to allow decay into another nuclide that is more tightly bound. 75 (a) 1.46×10−8 u vs. 1.007825 u for 1 H (b) 0.000549 u (c) 2.66×10−5 76 (a) –9.315 MeV (b) The negative binding energy implies an unbound system. (c) This assumption that it is two bound neutrons is incorrect. 78 22.8 cm 79 (a) 92 235 U143 → 90 231 Th141 + 2 4 He2 (b) 4.679 MeV (c) 4.599 MeV 81 a) 2.4×108 u (b) The greatest known atomic masses are about 260. This result found in (a) is extremely large. (c
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) The assumed radius is much too large to be reasonable. 82 (a) –1.805 MeV (b) Negative energy implies energy input is necessary and the reaction cannot be spontaneous. (c) Although all conversation laws are obeyed, energy must be supplied, so the assumption of spontaneous decay is incorrect. Test Prep for AP® Courses 1 (c) 3 (a) 5 When 95 7 241 Am undergoes α decay, it loses 2 neutrons and 2 protons. The resulting nucleus is therefore 93 237 Np. 1652 Answer Key During this process, the nucleus emits a particle with -1 charge. In order for the overall charge of the system to remain constant, the charge of the nucleus must therefore increase by +1. 9 a. No. Nucleon number is conserved (238 = 234 + 4), but the atomic number or charge is NOT conserved (92 ≠ 88+2). b. Yes. Nucleon number is conserved (223 = 209 + 14), and atomic number is conserved (88 = 82 + 6). c. Yes. Nucleon number is conserved (14 = 14), and charge is conserved if the electron’s charge is properly counted (6 = 7 + (-1)). d. No. Nucleon number is not conserved (24 ≠ 23). The positron released counts as a charge to conserve charge, but it doesn’t count as a nucleon. 11 This must be alpha decay since 4 nucleons (2 positive charges) are lost from the parent nucleus. The number remaining is found from: −0.693 N() = N0 1 2 = 3.4×1017 −(0.693)(0.035) 0.00173 N() = 4.1×1011 nuclei Chapter 32 Problems & Exercises 1 5.701 MeV 3 99 Mo57 → 43 42 5 1.43×10−9 g 7 ¯ 99 Tc56 + − + (a) 6.958 MeV (b) 5.7×10−10 g 8 (a) 100 mSv (b) 80 mSv (c) ~30 mSv 10 ~2 Gy 12 1.69 mm 14 1.24 MeV 16 7.44×108 18 4.92×10–4 Sv 20 4.43 g 22 0.010 g 24 95% 26 This content is available for free at http://cnx.
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org/content/col11844/1.13 Answer Key 1653 (a) =1+1=2, =1+1=1+1, efn = 0 = −1 + 1 (b) =1+2=3, =1+1=2, efn=0=0 (c) =3+3=4+1+1, =2+2=2+1+1, efn=0=0 28 = (i − f)2 4 − 4(1.007825) − 4.002603 4 He = 1 H 2 = = 26.73 MeV (931.5 MeV) 30 3.12×105 kg (about 200 tons) 32 = (i − f)2 1 = (1.008665 + 3.016030 − 4.002603)(931.5 MeV) = 20.58 MeV 2 = (1.008665 + 1.007825 − 2.014102)(931.5 MeV) = 2.224 MeV 4 He is more tightly bound, since this reaction gives off more energy per nucleon. 34 1.19×104 kg 36 2− + 41 H → 4 He + 7γ + 2 38 (a) =12+1=13, =6+1=7, efn = 0 = 0 (b) =13=13, =7=6+1, efn = 0 = −1 + 1 (c) =13 + 1=14, =6+1=7, efn = 0 = 0 (d) =14 + 1=15, =7+1=8, efn = 0 = 0 (e) =15=15, =8=7+1, efn = 0 = −1 + 1 (f) =15 + 1=12 + 4, =7+1=6 + 2, efn = 0 = 0 40 = 20.6 MeV 4 He = 5.68×10-2 MeV 42 (a) 3×109 y (b) This is approximately half the lifetime of the Earth. 43 (a) 177.1 MeV (b) Because the gain of an external neutron yields about 6 MeV, which is the average BE/ for heavy nuclei. (c) = 1 + 238 = 96 + 140 + 1 + 1 + 1, = 92 = 38 + 53 efn
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= 0 = 0 45 (a) 180.6 MeV 1654 Answer Key (b) = 1 + 239 = 96 + 140 + 1 + 1 + 1 + 1, = 94 = 38 + 56 efn = 0 = 0 47 238 U + → 239 U + 4.81 MeV 239 U → 239 Np + − + 0.753 MeV Np → Pu + − + 0.211 MeV 49 (a) 2.57×103 MW (b) 8.03×1019 fission/s (c) 991 kg 51 0.56 g 53 4.781 MeV 55 (a) Blast yields 2.1×1012 J to 8.4×1011 J, or 2.5 to 1, conventional to radiation enhanced. (b) Prompt radiation yields 6.3×1011 J to 2.1×1011 J, or 3 to 1, radiation enhanced to conventional. 57 (a) 1.1×1025 fissions, 4.4 kg (b) 3.2×1026 fusions, 2.7 kg (c) The nuclear fuel totals only 6 kg, so it is quite reasonable that some missiles carry 10 overheads. The mass of the fuel would only be 60 kg and therefore the mass of the 10 warheads, weighing about 10 times the nuclear fuel, would be only 1500 lbs. If the fuel for the missiles weighs 5 times the total weight of the warheads, the missile would weigh about 9000 lbs or 4.5 tons. This is not an unreasonable weight for a missile. 59 7×104 g 61 (a) 4.86×109 W (b) 11.0 y Test Prep for AP® Courses 1 (b) 3 (c) 5 (d) 7 (d) 9 (b) Chapter 33 Problems & Exercises 1 3×10−39 s 3 This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1655 1.99×10−16 m (0.2 fm) 4 (a) 10−11 to 1, weak to EM (b) 1 to 1 6 (a) 2.09×10−5 s (b) 4.77×104 Hz 8 78.0 cm 10 1.40×106 12 100 GeV 13 67.5 MeV 15 (a) 1×1014 (b) 2×1017 17 (a) 1671
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MeV (b) = 1. = − 1; ′ = − 1; = 0; ′ = − 1 + 1 = 0 ¯ − → −+ + ¯ ⇒ − antiparticle of +; of ¯ ; of (c) 19 (a) 3.9 eV (b) 2.9×10−8 21 (a) The composition is the same as for a proton. (b) 3.3×10−24 s (c) Strong (short lifetime) 23 a) Δ++() = ) 1656 Answer Key Figure 33.20. 25 (a) +1 (b) = 1 = 1 + 0, = = 0 + ( − 1), all lepton numbers are 0 before and after (c) () → () + ( ) 27 (ab) 277.9 MeV (c) 547.9 MeV 29 No. Charge = −1 is conserved. i 31 = 0 ≠ f = 2 is not conserved. = 1 is conserved. (a)Yes. = −1 = 0 + ( − 1), = 1 = 1 + 0, all lepton family numbers are 0 before and after, spontaneous since mass greater before reaction. (b) → + 33 (a) 216 (b) There are more baryons observed because we have the 6 antiquarks and various mixtures of quarks (as for the π-meson) as well. = −1, − 1 3 35 Ω. 37 = 1, (a)803 MeV This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key (b) 938.8 MeV (c) The annihilation energy of an extra electron is included in the total energy. 1657 39 ¯ 41 a)The antiproton - → 0 + − b) 43 (a) 5×1010 (b) 5×104 particles/m2 45 2.5×10−17 m 47 (a) 33.9 MeV (b) Muon antineutrino 29.8 MeV, muon 4.1 MeV (kinetic energy) 49 (a) 7.2×105 kg (b) 7.2×102 m3 (c) 100 months Test Prep for AP® Courses 1 (d) 3 (d) 5 (b) 7 (a) 9 (c), though this comes
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from Einstein's special relativity 11 (a) 13 (d) 15 (b) 17 (b) Chapter 34 Problems & Exercises 1 3×1041 kg 3 (a) 3×1052 kg (b) 2×1079 1658 (c) 4×1088 5 0.30 Gly 7 (a) 2.0×105 km/s (b) 0.67 Answer Key 9 2.7×105 m/s 11 6×10−11 (an overestimate, since some of the light from Andromeda is blocked by gas and dust within that galaxy) 13 (a) 2×10−8 kg (b) 1×1019 15 (a) 30km/s ⋅ Mly (b) 15km/s ⋅ Mly 17 960 rev/s 19 89.999773º (many digits are used to show the difference between 90º ) 22 23.6 km 24 (a) 2.95×1012 m (b) 3.12×10−4 ly 26 (a) 1×1020 (b) 10 times greater 27 29 31 1.5×1015 0.6 m−3 0.30 Ω (34.6) (34.7) (34.8) This content is available for free at http://cnx.org/content/col11844/1.13 Index Index Symbols (peak) emf, 1030 RC circuit, 945 A aberration, 1176 aberrations, 1175 absolute pressure, 454, 478 absolute zero, 532, 565 AC current, 887, 901 AC voltage, 887, 901 acceleration, 43, 82, 146, 178 acceleration due to gravity, 68, 82 accommodation, 1155, 1176 Accuracy, 22 accuracy, 29 acoustic impedance, 755, 761 active transport, 518, 519 activity, 1400, 1410 adaptive optics, 1174, 1176 adhesive forces, 465, 478 adiabatic, 632 adiabatic process, 661 air resistance, 113, 128 alpha, 1379 alpha decay, 1391, 1410 alpha rays, 1410 Alternating current, 886 alternating current, 901 ammeter, 950 ammeters, 938 ampere, 869, 901 Ampere’s law, 991, 999 amplitude, 681, 711, 1074, 1090, 1092 amplitude
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modulation, 1080 amplitude modulation (AM), 1092 analog meter, 950 Analog meters, 939 analytical method, 128 Analytical methods, 107 Anger camera, 1427, 1453 angular acceleration, 391, 425 angular magnification, 1172, 1176 angular momentum, 413, 425 angular momentum quantum number, 1353, 1364 angular velocity, 222, 249 antielectron, 1395, 1410 antimatter, 1394, 1410 antinode, 706, 711, 743, 761 approximation, 29 approximations, 27 arc length, 221, 249 Archimedes' principle, 459, 478 astigmatism, 1161, 1176 atom, 1318, 1364 atomic de-excitation, 1340, 1364 atomic excitation, 1340, 1364 atomic mass, 1386, 1410 atomic number, 1358, 1364, 1386, 1410 atomic spectra, 1279, 1306 Average Acceleration, 43 average acceleration, 84, 82 Average speed, 41 average speed, 82 Average velocity, 40 average velocity, 82 Avogadro’s number, 546, 565 axions, 1518, 1524 axis of a polarizing filter, 1213, 1224 B B-field, 973, 999 back emf, 1031, 1056 banked curve, 249 banked curves, 230 barrier penetration, 1408, 1410 baryon number, 1478, 1492 Baryons, 1478 baryons, 1492 basal metabolic rate, 296, 301 beat frequency, 707, 711 becquerel, 1400, 1410 Bernoulli's equation, 496, 519 Bernoulli's principle, 497, 519 beta, 1379 beta decay, 1393, 1410 beta rays, 1410 Big Bang, 1504, 1524 binding energy, 1281, 1306, 1403, 1410 binding energy per nucleon, 1405, 1410 bioelectricity, 895, 901 Biot-Savart law, 991, 999 birefringent, 1221, 1224 Black holes, 1511 black holes, 1524 blackbodies, 1278 blackbody, 1306 blackbody radiation, 1278, 1306 Bohr radius, 1331, 1364 Boltzmann constant, 544, 565 boson, 1476, 1492 bottom, 1486, 1492 bow wake,
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739, 761 break-even, 1442, 1453 breeder reactors, 1448, 1454 breeding, 1448, 1454 bremsstrahlung, 1287, 1306 Brewster’s angle, 1216, 1224 Brewster’s law, 1216, 1224 bridge device, 950 bridge devices, 944 Brownian motion, 1319, 1364 buoyant force, 458, 478 C capacitance, 843, 855, 945, 950 capacitive reactance, 1049, 1056 capacitor, 841, 855, 945, 950 capillary action, 470, 478 carbon-14 dating, 1398, 1410 Carnot cycle, 640, 661 Carnot efficiency, 640, 661 Carnot engine, 640, 661 carrier particle, 178 carrier particles, 177 carrier wave, 1080, 1092 cathode-ray tube, 1364 cathode-ray tubes, 1320 Celsius, 531 1659 Celsius scale, 565 center of gravity, 364, 380 center of mass, 237, 249 centrifugal force, 233, 249 centrifuge, 226 centripetal acceleration, 225, 249 centripetal force, 228, 249 change in angular velocity, 391, 425 change in entropy, 650, 661 change in momentum, 319, 343 Chaos, 1520 chaos, 1524 characteristic time constant, 1045, 1057 characteristic x rays, 1287, 1306 charm, 1486, 1492 chart of the nuclides, 1389, 1410 chemical energy, 287, 301 classical physics, 12, 29 Classical relativity, 125 classical relativity, 128 classical velocity addition, 1267 coefficient of linear expansion, 537, 565 coefficient of performance, 648, 661 coefficient of volume expansion, 539, 565 coherent, 1191, 1224 cohesive forces, 465, 478 Colliding beams, 1473 colliding beams, 1492 color, 1487, 1492 color constancy, 1164, 1176 commutative, 104, 129, 128 complexity, 1519, 1524 component (of a 2-d vector), 128 components, 106 compound microscope, 1165, 1176 Compton effect, 1291, 1306 Conduction, 590 conduction, 606 conductor, 782, 805 Conductors, 786 confocal microscopes, 1223, 1224 conservation laws, 933, 950 conservation of mechanical energy, 279
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, 301 conservation of momentum principle, 324, 343 Conservation of total, 1478 conservation of total baryon number, 1478, 1492 conservation of total electron family number, 1492 conservation of total, 1478 conservation of total muon family number, 1492 conservative force, 277, 301 constructive interference, 704, 711 constructive interference for a diffraction grating, 1197, 1224 constructive interference for a double slit, 1193, 1224 contact angle, 470, 478 Contrast, 1222 contrast, 1224 Convection, 590 convection, 607 1660 Index converging (or convex) lens, 1122 converging lens, 1142 converging mirror, 1142 conversion factor, 19, 29 Coriolis force, 234, 249 corner reflector, 1116, 1142 correspondence principle, 1277, 1306 cosmic microwave background, 1505, 1524 cosmological constant, 1516, 1525 cosmological red shift, 1504, 1525 Cosmology, 1502 cosmology, 1525 Coulomb force, 792, 805 Coulomb forces, 791 Coulomb interaction, 799, 805 Coulomb's law, 790, 805 critical angle, 1112, 1142 Critical damping, 694 critical damping, 711 critical density, 1516, 1525 critical mass, 1446, 1454 critical point, 557, 565 Critical pressure, 557 critical pressure, 565 critical temperature, 557, 565, 1521, 1525 criticality, 1447, 1454 curie, 1400, 1411 Curie temperature, 970, 999 current, 915, 950 Current sensitivity, 939 current sensitivity, 950 cyclical process, 636, 661 cyclotron, 1472, 1492 D Dalton’s law of partial pressures, 560, 565 dark matter, 1515, 1525 daughter, 1411 daughters, 1390 de Broglie wavelength, 1296, 1306 decay, 1379, 1390, 1411 decay constant, 1398, 1411 decay equation, 1392, 1396, 1411 decay series, 1390, 1411 deceleration, 82 defibrillator, 853, 855 deformation, 203, 212, 675, 711 degree Celsius, 531, 565 degree Fahrenheit, 531, 565 Density, 441 density, 478 dependent variable, 75, 82 derived
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units, 16, 29 destructive interference, 704, 711 destructive interference for a double slit, 1193, 1225 destructive interference for a single slit, 1202, 1225 dew point, 561, 565 dialysis, 518, 519 diastolic pressure, 455, 478 Diastolic pressure, 474 dielectric, 846, 855 dielectric strength, 855 dielectric strengths, 846 diffraction, 1190, 1225 diffraction grating, 1196, 1225 Diffusion, 515 diffusion, 519 digital meter, 950 digital meters, 939 dipole, 799, 805 Direct current, 886 direct current, 901 direction, 102 direction (of a vector), 128 direction of magnetic field lines, 973, 999 direction of polarization, 1213, 1225 Dispersion, 1118 dispersion, 1142 displacement, 34, 82 Distance, 36 distance, 82 Distance traveled, 36 distance traveled, 82 diverging lens, 1124, 1142 diverging mirror, 1142 domains, 969, 999 Doppler effect, 736, 762 Doppler shift, 736, 762 Doppler-shifted ultrasound, 759, 762 Double-slit interference, 1329 double-slit interference, 1364 down, 1481, 1492 drag force, 198, 212 drift velocity, 872, 901 dynamic equilibrium, 358, 380 Dynamics, 141, 144, 179 dynamics, 178 E eddy current, 1024, 1057 efficiency, 290, 301 Elapsed time, 39 elapsed time, 83 elastic collision, 328, 343 elastic potential energy, 677, 711 electric and magnetic fields, 1090 electric charge, 777, 805 electric current, 869, 901 electric field, 786, 805, 1074, 1092 electric field lines, 805, 1092 Electric field lines, 1071 electric field strength, 793, 1092 electric fields, 794 electric generator, 1057 Electric generators, 1028 electric potential, 824, 855 electric power, 883, 901 Electrical energy, 287 electrical energy, 301 electrocardiogram (ECG), 900, 901 electromagnet, 999 electromagnetic force, 806 electromagnetic induction, 1018, 1057 electromagnetic spectrum, 1092 electromagnetic waves, 1073, 1090, 1092 Electromagnetism, 970 electromagnetism, 999 electromagnets
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, 970 electromotive force, 924 This content is available for free at http://cnx.org/content/col11844/1.13 electromotive force (emf), 950, 1092 electron, 806 Electron capture, 1396 electron capture, 1411 electron capture equation, 1396, 1411 electron family number, 1478, 1492 electron volt, 828, 856 electrons, 777 electron’s antineutrino, 1394, 1411 electron’s neutrino, 1396, 1411 electrostatic equilibrium, 786, 806 electrostatic force, 790, 806 electrostatic precipitators, 803, 806 Electrostatic repulsion, 783 electrostatic repulsion, 806 electrostatics, 800, 806 electroweak epoch, 1508, 1525 electroweak theory, 1488, 1493 emf, 934 emf induced in a generator coil, 1029, 1057 emissivity, 603, 607 endoscope, 1114 energies of hydrogen-like atoms, 1332, 1364 energy, 301 energy stored in an inductor, 1044, 1057 energy-level diagram, 1330, 1364 English units, 15, 29 entropy, 649, 661 equipotential line, 856 equipotential lines, 838 escape velocity, 1511, 1525 event horizon, 1511, 1525 external force, 146, 178 external forces, 144 External forces, 179 Extremely low frequency (ELF), 1079 extremely low frequency (ELF), 1092 eyepiece, 1166, 1176 F Fahrenheit, 531 Fahrenheit scale, 565 far point, 1158, 1176 Faraday cage, 788, 806 Faraday’s law of induction, 1019, 1057 Farsightedness, 1158 farsightedness, 1176 fermion, 1476, 1493 ferromagnetic, 969, 999 Feynman diagram, 1470, 1493 Fiber optics, 1114 fiber optics, 1142 fictitious force, 233, 249 field, 806 fine structure, 1351, 1364 first law of thermodynamics, 621, 661 first postulate of special relativity, 1239, 1267 fission fragments, 1445, 1454 flat (zero curvature) universe, 1516, 1525 flavors, 1481, 1493 Flow rate, 490 flow rate, 519 fluid dynamics,
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519 Index 1661 fluids, 440, 478 Fluorescence, 1340 fluorescence, 1364 focal length, 1122, 1142 focal point, 1122, 1142 Food irradiation, 1437 food irradiation, 1454 force, 144, 178 Force, 179 force constant, 676, 711 force field, 175, 176, 178, 792 fossil fuels, 298, 301 free charge, 806 free charges, 786 free electron, 806 free electrons, 782 free radicals, 1438, 1454 free-body diagram, 144, 179, 167, 178 free-fall, 68, 83, 149, 178 Frequency, 680 frequency, 712, 1074, 1092 frequency modulation, 1080 frequency modulation (FM), 1092 friction, 147, 178, 212, 301 Friction, 192, 282 full-scale deflection, 939, 950 fundamental, 744, 762 fundamental frequency, 706, 712 fundamental particle, 1480, 1493 fundamental units, 16, 29 G galvanometer, 939, 950 gamma, 1379 gamma camera, 1427, 1454 Gamma decay, 1397 gamma decay, 1411 gamma ray, 1088, 1092, 1306 Gamma rays, 1284 gamma rays, 1411 gauge boson, 1493 gauge bosons, 1476 gauge pressure, 454, 478 gauss, 975, 999 Geiger tube, 1383, 1411 general relativity, 1509, 1525 geometric optics, 1103, 1142 glaucoma, 475, 478 gluons, 1470, 1493 Gluons, 1489 grand unified theory, 1493 Grand Unified Theory (GUT), 1488 gravitational constant, 237 gravitational constant, G, 249 gravitational potential energy, 272, 301 Gravitational waves, 1512 gravitational waves, 1525 gray (Gy), 1429, 1454 greenhouse effect, 605, 607 grounded, 801, 806 grounding, 839, 856 GUT epoch, 1508, 1525 H Hadrons, 1476 hadrons, 1493 half-life, 1397, 1411 Hall effect, 982, 999 Hall emf, 982, 999 harmonics, 744, 762 head, 100 head (of a vector), 128 head-to-tail method, 100, 129, 128 Hearing, 724, 749 hearing, 762 heat, 576, 607 heat engine, 627, 6
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61 heat of sublimation, 589, 607 heat pump, 661 heat pump's coefficient of performance, 646 Heisenberg uncertainty principle, 1302 Heisenberg’s uncertainty principle, 1303, 1306 henry, 1041, 1057 hertz, 1092 Higgs boson, 1490, 1493 high dose, 1430, 1454 hologram, 1347, 1364 Holography, 1347 holography, 1364 Hooke's law, 203, 212 horizontally polarized, 1213, 1225 Hormesis, 1431 hormesis, 1454 horsepower, 293, 301 Hubble constant, 1504, 1525 hues, 1162, 1176 Human metabolism, 625 human metabolism, 661 Huygens’s principle, 1188, 1225 Hydrogen spectrum wavelength, 1329 hydrogen spectrum wavelengths, 1364 hydrogen-like atom, 1331, 1364 hydrogen-spectrum wavelengths, 1328 hyperopia, 1158, 1176 I ideal angle, 249 ideal banking, 230, 249 ideal gas law, 544, 565 ideal speed, 249 Ignition, 1442 ignition, 1454 Image distance, 1128 impedance, 1051, 1057 impulse, 319, 343 Incoherent, 1191 incoherent, 1225 independent variable, 75, 83 index of refraction, 1108, 1142 inductance, 1040, 1057 induction, 783, 806, 1057 inductive reactance, 1047, 1057 inductor, 1042, 1057 inelastic collision, 332, 343 inertia, 146, 179 Inertia, 179 inertial confinement, 1442, 1454 inertial frame of reference, 165, 179, 1239, 1267 inflationary scenario, 1509, 1525 Infrared radiation, 1083 infrared radiation, 1306 infrared radiation (IR), 1092 Infrared radiation (IR), 1289 infrasound, 749, 762 ink jet printer, 802 ink-jet printer, 806 Instantaneous acceleration, 49 instantaneous acceleration, 83 Instantaneous speed, 41 instantaneous speed, 83 Instantaneous velocity, 40 instantaneous velocity, 83 insulator, 806 insulators, 782 intensity, 709, 712, 732, 762, 1091, 1092 intensity reflection coefficient, 756, 762 Interference microscopes, 1222 interference microscopes, 1225 internal energy, 622, 661 Internal kinetic energy,
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328 internal kinetic energy, 343 internal resistance, 924, 950 intraocular pressure, 475, 478 intrinsic magnetic field, 1351, 1364 intrinsic spin, 1351, 1364 ionizing radiation, 1284, 1306, 1381, 1411 ionosphere, 787, 806 irreversible process, 635, 661 isobaric process, 628, 661 isochoric, 630 isochoric process, 661 isolated system, 325, 343 isothermal, 632 isothermal process, 661 isotopes, 1387, 1411 J joule, 265, 301 Joule’s law, 916, 950 junction rule, 933, 950 K Kelvin, 532 Kelvin scale, 565 kilocalorie, 576, 607 kilogram, 16, 29 kilowatt-hour, 301 kilowatt-hours, 295 kinematics, 83, 114, 128 kinematics of rotational motion, 395, 425 kinetic energy, 268, 301 kinetic friction, 192, 212 Kirchhoff’s rules, 932, 950 L Laminar, 504 laminar, 519 laser, 1343, 1364 laser printer, 806 Laser printers, 802 Laser vision correction, 1161 laser vision correction, 1176 latent heat coefficient, 607 latent heat coefficients, 585 law, 11, 29 law of conservation of angular momentum, 416, 425 law of conservation of charge, 780, 806 1662 Index law of conservation of energy, 287, 301 law of inertia, 146, 179, 179 law of reflection, 1142 law of refraction, 1110 Length contraction, 1248 length contraction, 1267 Lenz’s law, 1019, 1057 leptons, 1476, 1493 linear accelerator, 1474, 1493 linear hypothesis, 1431, 1454 Linear momentum, 316 linear momentum, 343 liquid drop model, 1445, 1454 liter, 490, 519 longitudinal wave, 702, 712 loop rule, 934, 950 Lorentz force, 975, 999 loudness, 749, 762 low dose, 1430, 1454 M MACHOs, 1518, 1525 macrostate, 656, 661 magic numbers, 1389, 1411 magnetic confinement, 1442, 1454 magnetic damping, 1024, 1057 magnetic field, 973, 999, 1074, 1092 magnetic field lines, 973, 1000, 1093
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Magnetic field lines, 1072 magnetic field strength, 1093 magnetic field strength (magnitude) produced by a long straight currentcarrying wire, 990, 1000 magnetic field strength at the center of a circular loop, 991, 1000 magnetic field strength inside a solenoid, 992, 1000 magnetic flux, 1018, 1057 magnetic force, 975, 1000 magnetic monopoles, 972, 1000 Magnetic resonance imaging (MRI), 998 magnetic resonance imaging (MRI), 1000 magnetized, 969, 1000 magnetocardiogram (MCG), 999, 1000 magnetoencephalogram (MEG), 999, 1000 magnification, 1128, 1143 magnitude, 102 magnitude (of a vector), 128 magnitude of kinetic friction, 212 magnitude of kinetic friction fk, 193 magnitude of static friction, 212 magnitude of static friction fs, 193 magnitude of the intrinsic (internal) spin angular momentum, 1355, 1364 mass, 146, 179 Mass, 179 mass number, 1386, 1411 massive compact halo objects, 1518 maximum field strength, 1090, 1093 Maxwell’s equations, 991, 1000, 1071, 1093 mechanical advantage, 371, 380 mechanical energy, 279, 301, 856 Mechanical energy, 829 mechanical equivalent of heat, 577, 607 meson, 1468, 1493 Mesons, 1478 metabolic rate, 296, 301 metastable, 1342, 1364 meter, 16, 29, 1000 Meters, 988 method of adding percents, 24, 29 metric system, 17, 29 Michelson-Morley experiment, 1240, 1267 Microgravity, 242 microgravity, 249 microlensing, 1518, 1525 microshock sensitive, 894, 901 microstate, 656, 661 Microwaves, 1082, 1289 microwaves, 1093, 1306 micturition reflex, 477, 478 mirror, 1143 model, 11, 29, 41, 83 moderate dose, 1430, 1454 Modern physics, 14 modern physics, 29 mole, 546, 565 moment of inertia, 401, 401, 425 motion, 113, 128 motor, 1000 Motors, 986 muon family number, 1478, 1493 Mutual inductance, 1040 mutual inductance, 1057 myopia, 1158, 1176 N natural frequency, 698, 712 near point, 1158, 1176 Nearsightedness, 1158 near
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sightedness, 1176 negatively curved, 1516, 1525 Nerve conduction, 895 nerve conduction, 901 net external force, 147, 179 net rate of heat transfer by radiation, 604, 607 net work, 267, 301 neutral equilibrium, 366, 380 neutralinos, 1518, 1525 neutrino, 1393, 1411 neutrino oscillations, 1518, 1525 neutron, 1386, 1411 Neutron stars, 1512 neutron stars, 1525 Neutron-induced fission, 1445 neutron-induced fission, 1454 newton, 149 Newton's universal law of gravitation, 236, 249 newton-meters, 265 Newton’s first law of motion, 145, 179, 179 Newton’s second law of motion, 146, 179 Newton’s third law of motion, 153, 180, 179 node, 743, 762 Nodes, 705 This content is available for free at http://cnx.org/content/col11844/1.13 nodes, 712 non-inertial frame of reference, 233, 249 nonconservative force, 282, 301 normal force, 158, 179 north magnetic pole, 967, 1000 note, 762 notes, 749 Nuclear energy, 287 nuclear energy, 301 Nuclear fission, 1444 nuclear fission, 1454 Nuclear fusion, 1439 nuclear fusion, 1454 nuclear magnetic resonance (NMR), 998, 1000 nuclear radiation, 1379, 1411 nuclear reaction energy, 1392, 1411 nucleons, 1386, 1411 nucleus, 1411 nuclide, 1386, 1411 Null measurements, 942 null measurements, 951 numerical aperture, 1176 numerical aperture, 1167 O objective lens, 1166, 1177 ohm, 875, 901 Ohm's law, 874, 901 ohmic, 875, 901 ohmmeter, 951 ohmmeters, 943 Ohm’s law, 915, 951 optically active, 1220, 1225 orbital angular momentum, 1350, 1364 orbital magnetic field, 1350, 1364 order, 1193, 1225 order of magnitude, 17, 29 oscillate, 712, 1093 Osmosis, 518 osmosis, 519 osmotic pressure, 518, 519 Otto cycle, 638, 661 over damping, 712 overdamped,
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695 overtones, 706, 712, 744, 762 P parallel, 917, 951 parallel plate capacitor, 842, 856 parent, 1390, 1411 Partial pressure, 560 partial pressure, 565 particle physics, 1493 particle-wave duality, 1295, 1306 Particle-wave duality, 1304 Pascal's principle, 451 Pascal's Principle, 478 Pauli exclusion principle, 1358, 1365 peak emf, 1057 percent relative humidity, 563, 565 percent uncertainty, 24, 29 perfectly inelastic collision, 332, 343 period, 680, 712 periodic motion, 680, 712 permeability of free space, 990, 1000 perpendicular lever arm, 360, 380 Index 1663 phase angle, 1054, 1057 phase diagram, 565 phase diagrams, 558 phase-contrast microscope, 1223, 1225 phon, 750, 762 Phosphorescence, 1342 phosphorescence, 1365 photoconductor, 801, 806 photoelectric effect, 1280, 1306 photomultiplier, 1384, 1411 photon, 1280, 1290, 1306 photon energy, 1280, 1306 photon momentum, 1291, 1307 physical quantity, 15, 29 Physics, 8 physics, 29 pion, 1467, 1493 pit, 221, 249 pitch, 726, 749, 762 Planck’s constant, 1278, 1307 planetary model of the atom, 1326, 1365 point charge, 792, 806 point masses, 336, 343 Poiseuille's law, 507, 519 Poiseuille's law for resistance, 506, 519 polar molecule, 799, 806, 847, 856 polarization, 783, 806, 1225 Polarization, 1213 polarization microscope, 1223, 1225 polarized, 786, 806, 1213, 1225 population inversion, 1343, 1365 position, 34, 83 positively curved, 1516, 1525 positron, 1396, 1411 positron decay, 1395, 1411 positron emission tomography (PET), 1427, 1454 potential difference, 824, 924, 951 potential difference (or voltage), 856 potential energy, 277, 279, 301 potential energy of a spring, 278, 301 potentiometer, 943, 951 power, 292, 301, 1123, 1143
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power factor, 1054, 1057 precision, 23, 29 presbyopia, 1158, 1177 pressure, 444, 447, 478 Pressure, 451 probability distribution, 1300, 1307 projectile, 113, 128 Projectile motion, 113 projectile motion, 128 Proper length, 1248 proper length, 1267 Proper time, 1243 proper time, 1267 proton, 806 proton-proton cycle, 1440, 1454 protons, 777, 1386, 1411 PV diagram, 557, 565 Q quality factor, 1429, 1454 quantized, 1277, 1307 quantum chromodynamics, 1487, 1490, 1493 quantum electrodynamics, 1470, 1493 Quantum gravity, 1509, 1525 quantum mechanical tunneling, 1408, 1411 Quantum mechanics, 14 quantum mechanics, 29, 1277, 1307 quantum numbers, 1353, 1365 quark, 343, 1493 quarks, 327, 1480 quasars, 1512, 1525 R R factor, 607 factor, 594 rad, 1429, 1454 Radar, 1082 radar, 1093 radians, 221, 250 radiant energy, 287, 302 radiation, 590, 602, 607 radiation detector, 1383, 1411 radio waves, 1070, 1079, 1093 radioactive, 1379, 1411 Radioactive dating, 1398 radioactive dating, 1411 radioactivity, 1379, 1412 radiolytic products, 1438, 1454 radiopharmaceutical, 1425, 1454 radiotherapy, 1435, 1454 radius of a nucleus, 1387, 1412 radius of curvature, 221, 250 rainbow, 1143 range, 119, 128 range of radiation, 1381, 1412 rate of conductive heat transfer, 593, 607 rate of decay, 1400, 1412 ray, 1102, 1143 Ray tracing, 1125 Rayleigh criterion, 1204, 1225 RC circuit, 951 real image, 1127, 1143 reflected light is completely polarized, 1216 reflected light that is completely polarized, 1225 refraction, 1106, 1143 relative biological effectiveness, 1429 relative biological effectiveness (RBE), 1454 relative humidity, 561, 565 relative osmotic pressure, 518, 519 relative velocities, 125 relative velocity, 128 relativistic Doppler effects, 1269, 1267 Relativistic kinetic energy, 12
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63 relativistic kinetic energy, 1267 Relativistic momentum, 1257 relativistic momentum, 1267 relativistic velocity addition, 1253, 1267 Relativity, 14 relativity, 29, 125, 128, 1239, 1267 Renewable forms of energy, 298 renewable forms of energy, 302 resistance, 874, 901, 914, 951 resistivity, 878, 902 resistor, 914, 945, 951 resonance, 698, 712 resonant, 1075, 1093 resonant frequency, 1053, 1057 resonate, 698, 712 Rest energy, 1259 rest energy, 1267 rest mass, 1257, 1267 restoring force, 675, 712 resultant, 101, 128 resultant vector, 101, 128 retinex, 1177 retinex theory of color vision, 1165, 1177 retinexes, 1165 reverse dialysis, 518, 519 Reverse osmosis, 518 reverse osmosis, 519 reversible process, 633, 661 Reynolds number, 512, 519 right hand rule 1, 975 right hand rule 1 (RHR-1), 1000 right hand rule 2, 989 right hand rule 2 (RHR-2), 1000 right-hand rule, 423, 425 RLC circuit, 1093 rms current, 888, 902 rms voltage, 888, 902 rods and cones, 1162, 1177 roentgen equivalent man, 1429 roentgen equivalent man (rem), 1454 rotation angle, 221, 250 rotational inertia, 401, 425 rotational kinetic energy, 405, 425 Rydberg constant, 1328, 1333, 1365 S saturation, 561, 565 scalar, 37, 83, 106, 128, 830, 856 Schwarzschild radius, 1511, 1525 scientific method, 12, 29 scintillators, 1384, 1412 screening, 799, 806 second, 16, 29 second law of motion, 317, 344 second law of thermodynamics, 635, 636, 640, 661 second law of thermodynamics stated in terms of entropy, 661 second postulate of special relativity, 1240, 1268 Self-inductance, 1042 self-inductance, 1057 semipermeable, 517, 519, 895, 902 series, 915, 951 shear deformation, 209, 212
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shell, 1359, 1365 shielding, 1432, 1454 shock hazard, 890, 902, 1036, 1057 short circuit, 891, 902 shunt resistance, 940, 951 SI unit of torque, 361 SI units, 15, 29 SI units of torque, 380 sievert, 1430, 1454 significant figures, 25, 29 simple circuit, 875, 902 Simple Harmonic Motion, 681 simple harmonic motion, 712 simple harmonic oscillator, 681, 712 1664 Index simple pendulum, 686, 712 simplified theory of color vision, 1163, 1177 single-photon-emission computed tomography (SPECT), 1454 single-photon-emission computed tomography(SPECT), 1427 slope, 75, 83 solenoid, 992, 1000 Solid-state radiation detectors, 1385 solid-state radiation detectors, 1412 sonic boom, 739, 762 sound, 724, 762 sound intensity level, 733, 762 sound pressure level, 735, 762 south magnetic pole, 967, 1000 space quantization, 1351, 1365 special relativity, 1268 special relativity., 1239 specific gravity, 461, 478 specific heat, 579, 607 speed of light, 1093 spin projection quantum number, 1355, 1365 spin quantum number, 1355, 1365 spontaneous symmetry breaking, 1509, 1525 stable equilibrium, 365, 380 Standard Model, 1490 standard model, 1493 standing wave, 705, 1075, 1093 static electricity, 806 static equilibrium, 358, 368, 380 static friction, 192, 212 statistical analysis, 658, 661 Stefan-Boltzmann law of radiation, 603, 607 step-down transformer, 1034, 1057 step-up transformer, 1034, 1057 Stimulated emission, 1343 stimulated emission, 1365 Stokes' law, 202, 212 strain, 208, 212 strange, 1481, 1493 strangeness, 1478, 1493 stress, 208, 212 sublimation, 558, 566, 607 Sublimation, 589 subshell, 1359, 1365 Superconductors, 1521, 1525 supercriticality, 1447, 1454 superforce, 1508, 1525 superposition, 704, 712 Superstring theory, 1491, 1515, 1525 superstring theory
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, 1493 surface tension, 465, 478 synchrotron, 1472, 1493 synchrotron radiation, 1472, 1493 system, 146, 179 systolic pressure, 455, 478 Systolic pressure, 474 T tagged, 1425, 1455 tail, 100, 128 tangential acceleration, 392, 425 tau family number, 1493 Television, 1081 Temperature, 530 temperature, 566 temperature coefficient of resistivity, 880, 902 tensile strength, 205, 212 tension, 161, 179 terminal speed, 514, 519 terminal voltage, 926, 951 tesla, 975, 1000 test charge, 792, 806 the second law of thermodynamics stated in terms of entropy, 652 theory, 11, 29 theory of quark confinement, 1487, 1493 therapeutic ratio, 1435, 1455 thermal agitation, 1082, 1093 thermal conductivity, 593, 607 thermal energy, 282, 287, 302, 552, 566 thermal equilibrium, 536, 566 thermal expansion, 537, 566 thermal hazard, 890, 902, 1036, 1057 Thermal stress, 541 thermal stress, 566 thin film interference, 1208, 1225 thin lens, 1125 thin lens equations, 1128 thought experiment, 1509, 1525 three-wire system, 1036, 1057 thrust, 154, 180, 179 timbre, 750, 762 time, 39, 83 Time dilation, 1242 time dilation, 1268 TOE epoch, 1508, 1525 tone, 750, 762 top, 1486, 1493 Torque, 360 torque, 380, 400, 425 Total energy, 1259 total energy, 1268 total internal reflection, 1112 trajectory, 113, 128 transformer, 1057 transformer equation, 1034, 1058 Transformers, 1032 transverse wave, 702, 712, 1075, 1093 triple point, 559, 566 Tunneling, 1409 tunneling, 1412 turbulence, 504, 519 TV, 1093 twin paradox, 1268 U ultra high frequency, 1081 ultra-high frequency (UHF), 1093 ultracentrifuge, 227, 250 ultrasound, 749, 762 Ultraviolet (UV) microscopes, 1222 ultraviolet (UV) microscopes, 1225 Ultraviolet radiation, 1287 ultraviolet radiation, 1307 ultraviolet radiation (UV), 1085,
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1093 uncertainty, 23, 29 uncertainty in energy, 1303, 1307 uncertainty in momentum, 1301, 1307 This content is available for free at http://cnx.org/content/col11844/1.13 uncertainty in position, 1301, 1307 uncertainty in time, 1303, 1307 under damping, 712 underdamped, 695 uniform circular motion, 250 units, 15, 29 unpolarized, 1213, 1225 unstable equilibrium, 365, 380 up, 1481, 1493 useful work, 296, 302 V Van de Graaff, 1472, 1493 Van de Graaff generator, 806 Van de Graaff generators, 801 vapor, 559, 566 Vapor pressure, 560 vapor pressure, 566 vector, 37, 83, 99, 128, 806, 830, 856 vector addition, 122, 128, 796, 806 vectors, 97, 794 velocity, 122, 128 vertically polarized, 1213, 1225 very high frequency, 1081 very high frequency (VHF), 1093 virtual image, 1130, 1143 virtual particles, 1467, 1494 viscosity, 506, 519 viscous drag, 513, 519 Visible light, 1084 visible light, 1093, 1288, 1307 voltage, 824, 915, 951 voltage drop, 915, 951 voltmeter, 951 Voltmeters, 938 W watt, 292, 302 wave, 700, 712 wave velocity, 700, 712 wavelength, 701, 712, 1074, 1093 wavelength in a medium, 1187, 1225 weakly interacting massive particles, 1518 weight, 149, 179 Weight, 157 Wheatstone bridge, 944, 951 WIMPs, 1518, 1525 work, 263, 302 work-energy theorem, 268, 302, 406, 427, 425 X x ray, 1307 x rays, 1285, 1365 X rays, 1334 X-ray, 1088, 1093 x-ray diffraction, 1338, 1365 xerography, 801, 806 Y y-intercept, 75, 83 Z z-component of spin angular momentum, 1365 z-component of the angular momentum, 1365 Index 1665 -component of spin angular momentum, 1355 -component of the angular momentum, 1353 Zeeman effect, 1350, 1365 zeroth law of therm
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odynamics, 536, 566 zircon, 1143ent with greater accuracy than is possible with the means at our disposal. The reasons will be apparent after studying later paragraphs (page 15). 28.7 (c) Approximate Numbers In all measurement, you must first decide what will be the limits of accuracy of your work and then realize that a possible error exists in the result. In order to make this small, the possible error is not permitted to exceed one-half of the smallest unit employed in the operaThus, in the previous examples, tion. if 1/8 inch were the smallest unit on the measuring instrument used, a possible error of 1/16 inch more or less is found in the answer. Accordingly the length of the room in part (a) would be between 14 feet, 15/16 inch and 15 feet, 1/16 inch and the width of the door would be between 29 15/16 and 30 1/16 Similarly, if the present room inches. temperature is given as 21.5°C., and the thermometer is marked off into 0.1 degree units, the possible error will be.05 centigrade degrees and the actual reading may be between 21.45°C. and 21.55°C. The reading 21.5°C. represents the temperature as accurately as the means of measurement at our disposal will permit. Other examples follow: Measurement 55.3 cm. 719 ft. 19.0 ft. 0.0032 in. From 55.25 cm. Limits of Accuracy up to but not including 55.35 cm. a “ “ 718.5 ft. 18.95 ft. 0.00315 in. “ “ “ “ “ “ “ 719.5 ft. 19.05 ft. 0.00325 in. To allay the fear that naturally arises over the presence of this possible error, let us calculate what per cent of the whole it represents. The percentage error is possible error divided by the accepted number times Using the first measurement in the preceding table we get: 100. The percentage error =: From this, we see that approximate numbers contain an extremely small error. The more precise the measuring device is, the smaller the unit of measurement will be. The smaller the unit of measurement is, the smaller the possible error will be. (d) How to Deal with Approximate Numbers 55.3 - 55.25 55.3 - 55.35 X 100
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or 55.3 = +_X100or- — X.oo 55.3.05.05 = -f.09% or -.09% X 100 1. Significant Digits When performing a measurement, only those numbers in the measurement that are certain, like the 15 feet, 30 inches, 21.5°C. are recorded. The digits in these numbers paragraphs, previous of 13 Chap. 2 MECHANICS are called the significant digits in the result, there being two in the first two numbers above, and three in the third. Mathematicians have laid down certain rules to guide us in working with them. They are as follows: 1. All the digits from 1 to 9 including any zeros between them or after them are significant digits. 2. The position of the decimal is disregarded in determining the number of significant digits. 3. Rounding off Approximate Numbers A number like the value of pi (tt), 3.14159, is correct to 6 significant digits. It may be made consistent with measurements having fewer than six significant digits by the process of rounding-off. This means dropping necessary number of digits off the end of a number and adding one to the last remaining digit if the next digit was five or more. the 4. Examples 3. The zeros preceding the first digit Rounding-off Accuracy are not significant. Significant Digits IN Numbers Number 604 60.4 6.04 0.604 0.0604 0.06004 604.0 60.40 Significant Digits 3 3 3 3 3 4 4 4 2. Representing Significant Digits in Large Numbers digit, Where a number like 600 is correct it should be to one significant written as 6 X 10^. To express it as 600 indicates digits. Similarly, the distance from the earth to the sun is correct to only two signiwritten ficant 93 X 10® rather than 93,000,000 miles. digits and should significant three be 3.14159 3.142 3.1 6 4 2 significant digits a a Calculations With Significant Digits How to carry out mathematical operations with approximate numbers is indicated by the following rules: 1. Addition and Subtraction In operations involving addition and subtraction of approximate numbers of which the least precise has N places of decimals, round off the other numbers where possible to N + 1 places and the answer to N places. 2. Multiplication and Division In operations involving multiplication and division of approximate numbers of which the least
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accurate has N significant digits, round off the others where possible to N + 1 digits and the answer to N digits. Examples 1. Add 2.0149 3.02864 1.239 1.97 14 Method Round off all to three decimals where possible. Round off answer to 2 decimals. Result 2.015 3.029 1.239 1.97 Total = 8.253 Proper answer = 8.25 MEASUREMENT Method Result 2. Subtract 21.347 As for addition from 32.5 Sec. 1:3 32.5 21.35 3. Multiply 2.1 by 2.56 and by 9.547 Round oflF all numbers to 3 significant digits. Keep 3 significant digits in all partial products. Round off answer to 2 significant digits. 4. Divide 96.568 As for multiplication by 7.02 Difference = 11.15 Proper answer = 11.2 2.1 X 2.56 = 5.376 Proper product = 5.38 5.38 X 9.55 = 51.3790 Proper answer = 51. 96.57 = 13.75 7.02 = 13.75 Quotient Proper answer =13.8 On page 13 we multiplied 7.54 by getting 28.7274 for an answer. 3.81, However, as the multiplier and multiplicand are correct to three significant digits only, 28.7 is the proper answer. You may wonder about the usefulness of the above material, but be assured that the method is used daily by scientists and mathematicians whose work has contributed so much to each modern development and invention. (e) Applying Approximate Numbers in Elementary Physics The diameter of a cylindrical solid is 2.50 cm. and it is 10.04 cm. long. (7r = 3.1416). Its mass is 280.76 gm. Calculate its density. Radius = 1.25 cm. Length = 10.04 cm. =3.1416 = 3.142 TT Area of the end of the cylinder = tt R^ = 3.142 X 1.25 X 1.25 = 4.910 sq. cm. Volume of the cylinder = area of end X length = 4.910 X 10.04 = 49.30 c.c. The mass given = 280.76 gm. The proper mass = 280.8 gm. 49.30 c.c. of solid weigh 280.8 gm. •.
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. 1 1 1-^ c.c. of solid weighs r • u 280.8 49.30 = 5.695 gm. The density = 5.70 gm. per c.c. 15 Chap. 2 : 4 I MECHANICS QUESTIONS 1. 2. (a) Why Is measurement necessary? (b) Why was it necessary to establish standard units? (a) Name the two systems of measurement and name the fundamental units for each. (b) What advantages has the metric system over the British system of measurement? (c) Why is the British system still used in a few countries? 3. (a) What apparatus is commonly used in the laboratory for the measurement of (i) length (ii) mass (ili) time? (b) Suggest other pieces of apparatus that could be used to measure these fundamental units more ac- curately. 4. 5. 6. 7. 8. decimetres (a) State the number of millimetres, centimetres and metre. How many metres are there In 1 kilometre? (b) Using the above units construct (i) square measure (ii) tables of in 1 cubic measure. (c) Define: litre, millilitre. (a) Distinguish between the mass and weight of an object. (b) State the number of milligrams, centigrams, decigrams, and grams in 1 kilogram. (a) Why was the apparent motion of the sun adopted as the basis for reckoning time? (b) Define: mean solar day, second. (a) Distinguish between exact and approximate numbers. Give an example of each. (b) What gives rise to approximate numbers? Why? (a) What do you mean by possible error? 16 9. (b) How do you calculate percentage error? (a) What is meant by significant digits? (b) State the number of significant digits in each of: 32060, 36.060, 0.32060,.032060, 3 X 10^, 3.56 X 105. 10. (a) What is meant by rounding off a number? (b) Round off the following to two significant digits: 36.7, 34.32, 37.495. 11. (a) State the rules for carrying out the following mathematical operations with approximate numbers: (i) addition and subtraction, (ii) multiplication and division. (b) Do the
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following: (i) 10.3575 + 9.75-8.65248. (ii) 7.935 X 2.4248 2.3. B 1. (a) Express in cm.: 15.2 m., 38 mm., 6 m. 5 cm. 4 mm. (b) Express in sq. cm.: 3 sq. m., 236 sq. mm., 6 sq. m. 5 sq. cm. 44 sq. mm. (c) Express in c.c.: 2.5 cu. m., 2300 cu. mm., 6 cu. m. 50 c.c. 465 cu. mm. 2. (a) Determine the number of inches In 950 mm., 40 cm., 10 dm. (b) Determine the number of (i) cm. in 1 foot, (ii) km. in 1 mile. (c) Which is the greater distance, 100 yd. or 100 metres? Express the difference In (i) (d) An object is at the rate of 40 miles per hour. Calculate the rate in (i) ft. per sec. (ii) metres per sec. (iii) kilometres per hour. 3. A tank is 50 cm. long, 3 dm. wide, ft. (ii) cm. travelling and 150 mm. high. (a) Calculate the area of a cover MEASUREMENT. Sec. 1:4 in sq. cm. (ii) in sq. for the tank (i) dm. (b) Calculate the volume of the tank, and express In (i) c.c. (ii) cu. dm. (iii) litres. 4. (a) What mass of water will the above tank hold (1 c.c. of water weighs 1 gm.)? Express this mass in gm., mg., eg., dg., kg. (b) Calculate grams In an ounce (ii) kilograms in a number the of (i) ton? (c) Calculate your own weight in kilograms. 5. A beaker is 20 cm. high and has a diameter of 1 4 cm. Calculate its volume in (i) litres (ii) millilitres (iii) pints. 6 Using proper "rounding-off” tech- niques, make the following calculations: (a) Add 9.75+10.357+76.92 + 5.674. (b
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) Subtract (i) 10.357 (ii) 5.674 (iii) 9.75 from 76.92. (c) Multiply (i) 2.6X7.93X1.732. (ii) 77.5X1.4142X.0032. (iii) 46X23.55X0.25. (d) Divide (i) 154 by.1 1. (ii) 9.5 by 19.03. (iii) 134.5 by 15. 17 CHAPTER 3 DENSITY AND SPECIFIC GRAVITY 1 gm. per c.c. G.G.S, system of units is In the F.P.S. system, however, a cu. ft. of water is found to weigh 62.5 ib. (approx.), and hence the density of water in this system of units will be expressed as 62.5 lb. per cu. ft. Densiof various subties stances are given in the table on page 21. (in gm. per c.c.) Fig. 3:1 Relative Densities of A— Solids, B— Liquids. To find the corresponding densities in lb. per cu. ft. the numerical values must be multiplied by 62.5. I is e.g., iron “heavier” ; 5 MEANING OF DENSITY In ordinary conversation we often say that one substance is “heavier” than than another, aluminum. Obviously we cannot mean that any given piece of iron is heavier than every piece of aluminum, but rather that for pieces of equal size, the iron would be the heavier. In science we use the term density to express the physical difference implied in the above everyday statement, that is, we say that iron has a greater density than aluminum. Density is defined as the mass of a unit volume of a substance. The method of determining density is described in chapter 5, experiments 1, 2 and 3. numerical densities measure must always be accompanied by suitable units, e.g., gm. per c.c., lb. per cu. ft., etc., according to the units; in which the mass and volume of the substance have been measured. It should further be noted that the numerical value of the density of any given substance will depend on the system of Thus, since one gram of units used. water (at its maximum density)
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occupies a volume of one cubic centimetre (Sec. 1:2), the density of water in the stating the In 18 DENSITY AND SPECIFIC GRAVITY Sec. 1:6 Research Scientist Determining the Density of a Substance by Comparison with Standard Density Floats Suspended in a Solution of Known Density. Canadian Industries Ltd. We have previously defined mass as the quantity of matter in a body. According to modern theory, matter is comprised of molecules, the molecules of any given substance being identical Hence, to each other (Sec. differences in density between various substances (Fig. 3:1), are due to the relative masses of their molecules as spatial arrangement. well as Variations in the density of a given substance are due to changes which vary the closeness of packing of the molecules. Ill: 2). their to I of water is : 6 DENSITY OF WATER It is frequently said that the density 1 gm. per c.c. However, solids and gases, expand liquids, when heated and contract when cooled with no change whatever in mass. They therefore at different densities have like different temperatures. Careful experiments to show changes in density of water with changes in temperature can be carried out with the aid of the dilatometer shown in Fig. 3:2. If this instrument, filled with water. 19 Chap. 3 MECHANICS is placed in a water bath with a thermometer, and ice slowly added, the volume change can be observed on the Fig. 3:2 A Dilatometer. scale, for changes of temperature down to 0°C, The water contracts, i.e., density increases, until a temperature of its scale) to (not Volume is smallest which the volume is the temperature at which the density is the greatest, namely 4°C. We say that water has its maximum density at 4°C. It is at this temperature that 1 cubic centimetre of water has the mass of 1 gram. The changes mentioned above are shown graphically 3:3. (Most liquids show a gradual increase in goes down.) temperature density Fig. the in as at 0°G. This fact, together with the fact that there is a sudden expansion as water (evidenced by those freezes burst pipes in winter!) is of profound importance in nature. With wintry conditions the coldest layers of a pond or lake are those at the surface, and when these layers freeze the ice so formed remains on the surface because its
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density is less than that of the water beneath it. Without this unusual behaviour of water the pond would freeze solid from the bottom upwards, greatly to the detriment of all life and to aquatic life in particular. : 7 MEANING OF SPECIFIC GRAVITY I For many purposes, instead of density, it is found more convenient to use the density of a substance relative to that of water. The relationship so obtained is called the specific gravity of It may be calculated in the substance. the following way: g ^ _ density of the substance density of water Temperature Fig. 3:3 Graph to Show the Effect of Changes of Temperature on the Density of Water. o.o. — - „ mass of unit volume of the substance mass of unit volume of water g g _ mass of any volume of the substance mass of an equal volume of water reached. 4°G. is further cooling makes it expand, i.e., its density The temperature at decreases After that, again. Thus specific gravity is a ratio and no It is simply a numunits are required. 20 DENSITY AND SPECIFIC GRAVITY Sec. 1:8 7 ber stating how many times as heavy as water, bulk for bulk, the substance is. Further, it must be evident that the number giving the specific gravity of a substance will be the same whatever the units in which the masses are measured, (Chap, 5, Exp. 4). In the C.G.S. system the density of water is 1 gm. per mb, and thus in this system, density and specific gravity are numerically equal. The two tenns are not interchangeable, however, although frequently, but wrongly, so used. : 8 DENSITY OR SPECIFIC GRAVITY I OF VARIOUS SUBSTANCES or specific gravity The density of solids, liquids and gases is a property that helps us to identify them. Some of these you will have determined experimentally while others will be found in the accompanying table of specific gravi- ties. Table of Approximate Specific Gravities 10.5 11.4 13.6 19.3 21.5 8.4-8. 8. 7-8.9 2.6 0.6 0.8 1.74 2.70 7.15 7.30 7.85 8.90 Silver Lead Mercury Gold Platinum 7.0-7.7 7.1-7.7 Brass Bronze Sand Pine Oak 0.24 0.92 2
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.2 0.70 0.79 0.87 Metals Magnesium Aluminum Zinc Tin Iron (pure) Copper Alloys Steel Iron (cast)^ Miscellaneous Solids Cork Ice (0°C.) Salt Liquids Gasoline Alcohol Turpentine Gases (at S.T.P.) Hydrogen Helium Air Carbon tetrachloride Sea- water Cone, sulphuric acid 1.60 1.01-1.05 1.83 0.00009 0.00018 0.00129 Oxygen Carbon dioxide Chlorine 0.00143 0.00198 0.00322 Gases, being so light, generally have their densities expressed in grams Also, air or hydrogen is used as the standard, rather than water, for per litre, instead of grams per cubic centimetre. (preferably the latter) purposes of comparison. 21 Chap. 3 i:9 MECHANICS QUESTIONS A 1. (a) Define density and state what two measurements must be made in order to calculate the density of an 5. Find the mass of 20 cu. ft. of material whose density is 3.2 Ib./cu. ft. 6. Find the volume of an object whose mass Is 42.7 lb. and whose density is 2.10 object. Ib./cu. ft. 2. (b) Calculate the density of a piece of aluminum whose volume is 150 c.c. and whose mass is 405 gm. (a) What effect does an increase in temperature have on the density of most substances? Why? (b) In what respect may water be said to be an unusual liquid? (c) Explain why this behaviour of water is important to life. (d) Why does it a much longer and more severe period of cold weather to cause a layer of ice to form on deep bodies of water than to form on shallow bodies? unusual require 3. (a) Define specific gravity and state clearly what is needed to find the specific gravity of an object. (b) Calculate the specific gravity of a substance whose volume is 20 c.c. and whose mass is 1 60 gm. 4. (a) Distinguish between density and specific gravity. (b) What relationship exists between the specific gravity of a substance and its density? (c) What is the specific gravity of the aluminum in 1. (b)? 7. Find the specific gravity of a substance whose mass is 148.5 gm.
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and whose volume is 30.5 c.c. 8. The specific gravity of a substance is 1.85. What volume of It weighs 1 00 gm.? Find the mass of 0.5 litres of a liquid 9. whose specific gravity is 1.6. 10. Find the specific gravity of a substance whose mass Is 3.2 lb. and whose volume is.75 cu. ft. (Density of water is 62.5 lb. per cu. ft.) 11. The specific gravity of a substance is 2.7. What volume of It will weigh 1 00 lb.? 12. Find the mass of 3.2 cu. ft. of material whose specific gravity is 8.9. 13. An irregular object has a mass of 72.6 gm. On placing it in water in a graduated cylinder the level rises from 12.0 ml. to 21.5 ml. Calculate its density. 14. A specific gravity bottle weighed 24.20 gm. when empty, 67.81 gm. when filled with turpentine and 74.20 gm. when with distilled water. What is the filled specific gravity of the turpentine? 15. A flask weighs 8.8 gm. when empty, 33.6 gm. when filled with water and 28.6 gm. when filled the with specific gravity of the alcohol. alcohol. Find B 1. Calculate the density of a rectangular solid 25 cm. long, 15 cm. wide, 3 cm. thick, whose mass is 5625 gm. 2. Find the mass of 75 ml. of a liquid whose density is 0.70 gm./ml. 3. Find the volume of an object whose mass is 750 gm. and whose density is 2.80 gm./c.c. 4. Find the density of water if 1 5 cu. ft. weigh 937.5 lb. 16. The density of a salt solution is 1.20 gm. per ml. If a flask weighs 1 2.6 gm. when empty and 62.8 gm. when filled with water, how much will it weigh when filled with the solution? 17. The composition of brass is 75% copper and 25% zinc by volume. Calculate its density. 1 8. An alloy of tin and lead has a specific gravity of 10.6. Calculate the proportion of tin present in the alloy (a) by volume (b) by weight. 22 2 DENSITY
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AND SPECIFIC GRAVITY Sec. 1:9 19. A piece of wax whose real specific gravity is 0.96 has an apparent specific gravity of 0.92 owing to a bubble of air it. The volume of the being enclosed in whole is 10.0 c.c. Find the volume of the air enclosed, assuming the weight of the air to be negligible, 20. A piece of metal 1 1.2 cm. long, 4.5 cm. wide and 1 mm. thick, has a mass of 30.2 gm. Find its density. 21. Calculate the density of a cylinder 22. Calculate the density of a sphere whose radius is 1.4 cm. and whose mass is 100 gm. 23. A cube of ice whose side is 4 cm. is allowed to melt. The volume of the water is found to be 58.2 ml. Find the density of ice. 24. Four solutions of salt, of densities 1.12, 1.17, 1.19 and 1.20 gm. per c.c., proportion are 1:2:3:4 by volume. Find the density of the together mixed the in whose length is 5.2 cm., diameter is mm., and whose mass fs 58 gm. 1 mixture. 23 CHAPTER 4 BUOYANCY force will be apparent after a consideration of the forces that water exerts on an object immersed in it (Fig. 4 : 1). The water exerts a downward pressure upon the top surface of the object, and an upward pressure upon the bottom. Because the bottom of the object is deeper in the liquid than is the top, and because pressure increases with depth, the upward pressure upon the bottom surface will exceed the downward pressure upon the top surface. There will, therefore, be a net upward pressure upon It is this upward force that the block. accounts for the apparent lightness of an object when immersed in the water. The same principle applies to all bodies immersed in any liquid or gas. Archimedes (287-212 b.g.), a Greek mathematician and inventor, was the first to study the buoyancy of liquids and to enunciate an important principle connected therewith. The story is told that Hiero, king of Syracuse, had sent his jeweller a known mass of gold to be made into a new crown. When the crown was delivered and tested it was found to have the right weight, but there was a suspicion that some silver had been substituted for
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gold in the interior of the Consequently Archimedes was crown. commissioned to determine whether or not the crown was pure gold, at the same time being instructed not to mar the crown in any way. Archimedes puzzled over this problem at great length. All that he had to work I : 10 INTRODUCTION TO BUOYANCY We are all familiar with the fact that things seem lighter under water. Those of us who have helped to build a dock know that a large stone which can only be raised with difficulty when out of the water may be raised quite easily when Similarly, we find that under water. Fig. 4:1 Explanation of Buoyancy. an anchor becomes heavier on emerging from the water. In general then, bodies immersed in water (or in any fluid) appear to lose some of their weight. This is, of course, due to the fluid exerting a buoyant force or lift on them. The reason why fluids exert a buoyant 24 BUOYANCY Sec. 1:11 with was the well-known fact that an object is easier to lift when immersed in water than when on land. medes’ Principle (Chap. 5, Exp. 5, 6) should be studied carefully in conjunction with the following statements: One day as he hopped into his bath, which had been filled brimful of water, he caused a considerable quantity to overflow onto the brick floor, and he was suddenly struck by the idea that the weight he lost on submerging was equal to that of the water which overflowed. He was so excited by this discovery, so the story goes, that he ran home “Eureka! Eureka!” (“I have found it! I have found it!”) As you that Archimedes enunciated, see if ypu can suggest the experiment that Archimedes probably performed to determine whether* or not the crown was pure gold. unclothed shouting, principle study the 1:11 ARCHIMEDES' PRINCIPLE The experiment to demonstrate Archi- 1. When a body is immersed in a liquid it displaces some of the liquid to make room for itself. Note that the volume of the liquid displaced is equal to the volume of the object. 2. The body is buoyed up by the liquid and therefore seems to weigh less than it does in air. 3. This loss in weight is exactly equal the weight of the liquid dis- to placed (Fig. 4:
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2). Archimedes’ Principle, therefore, may be stated as follows: An object, when immersed in a liquid, loses in apparent weight an amount equal to the weight of the liquid displaced. or an alternative statement is: The buoyant force of a fluid (liquid or gas) upon an object immersed in it, equal to the weight of the fluid displaced. is Examples 1. A rectangular piece of metal, 15 cm. long, 6 cm. wide and 3 cm. thick, weighs 2700 gm. in air. Find its weight when immersed in water. = 2700 gm. Weight of object in air = 15 X 6 X 3 = 270 c.c. Volume of object Volume of water displaced = 270 c.c. Weight of water displaced = 270 gm. (Density of water = 1 gm./cc.) Weight of object in water = 2700 — 270 (Archimedes’ Principle) = 2430 gm. 25 Chap. 4 MECHANICS. 2. Find the weight of the object in example 1, if it were immersed in carbon tetrachloride (S.G. = 1.60). Weight of object in air = 2700 gm. Volume of object = 15 X 6 X 3 = 270 c.c. Volume of carbon tetrachloride displaced = 270 c.c. 1 c.c. of carbon tetrachloride weighs 1.60 gm. (S.G. = 1.60) Weight of carbon tetrachloride displaced = 270 X 1.60 = 432 gm. Weight of object in carbon tetrachloride = 2700 — 432 Archimedes’ Principle is used for the accurate determination of the specific gravity of solids and liquids (Chap. 5, Exp. 7-9) (Archimedes’ Principle) = 2268 gm. : 12 PRINCIPLE OF FLOTATION I Archimedes’ Principle applies to floating bodies as well as to those which are submerged in a fluid. When a body is floating in a liquid it appears to lose all its weight, as can be shown by lowering into water a block of wood, suspended by a thread from the hook of a spring balance (Fig. 4:3). As the wood settles more deeply into the water the reading of the spring balance decreases until, when the wood is floating, it records zero weight. The entire weight of the wood is now being supported
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by the upthrust of the water, this being equal to the weight of water displaced. We thus have, for floating bodies, the Principle of Flotation, which states that the weight of a floating object is equal to the weight of the fluid (liquid or gas) it displaces when floating. A fairly heavy object will of course sink lower in the liquid in which it is floating than will a lighter In either case the submerged object. displaced an amount of portion has liquid equal to the weight of the floating object. This statement is embraced in Archimedes’ Principle, i.e., a body which floats has lost its whole weight. (Remember loss is an apparent one, not that this real. The pull of the earth on the floating body is still the same.) Experimental proof for this Principle is supplied in Chap. 5, Exp. 10. Fig. 4:3 Principle of Flotation. 26 BUOYANCY Examples Sec. 1:13 1. An object loaded onto a flat barge 18 feet long and 10 feet wide, causes it to settle 2 inches deeper into the water. Calculate the weight of the object. Volume of water displaced =z 18 X 10 X 2/12 = 30 cu. ft. Weight of water displaced = 30 X 62.5 = 1875 lb. (Density of water = 62.5 Ib./cu. ft.).'. weight of object = 1875 lb. (Principle of Flotation) 2. A plastic tray 25 cm. long, 15 cm. wide is floating on water. A lead weight whose mass is 300 gm. when placed in it causes it to sink deeper into the water. Calculate the depth to which it sinks. Let the depth to which it sinks be x cm. Volume of water displaced = 25 X 15 X x = 375x c.c. Density of water = 1 gm./c.c. Weight of water displaced = 375 x gm. Weight of water displaced = weight of floating object. (Principle of Flotation) Weight of water displaced = 300 gm. 375 x = 300 - = ^“ =.8 375 Depth to which the tray sinks is.8 cm. : 13 HYDROMETERS AND THEIR USES I (a) Structure displaced its own weight of the liquid, the lighter the liquid, the deeper the 5, for the Exp. (Chap. instrument The Principle of Flotation finds application in the hydrometer, which is a convenient rapid
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determination of the specific gravity of 12). A liquids common type of hydrometer consists of a cylindrical stem, graduated or containing a paper scale, an expansion of the stem called the float, and a bulb weighted with mercury or lead shot, the ballast, to make it float upright (Fig. 4:4). The float increases the buoyancy of the hydrometer. 11, 0.80 0.85 * 0.90 Stem 0.95 Scale 1.00N / Float The liquid whose specific gravity is to be determined is poured into a tall jar. The hydrometer is gently lowered into the liquid until it floats freely. The specific gravity of the liquid is indicated by the number on the scale which is even with the surface liquid. Since the hydrometer sinks until it has of the Weighted Bulb Fig. 4:4 Structure of a Hydrometer. hydrometer will sink in it. the largest specific gravity readings are Therefore, 27 Chap. 4 MECHANICS long and having a cross-sectional area of 1 sq. cm. It is weighted at one end with a plug of lead. On one face is marked at the bottom of the scale, and the smallest at the top. Hydrometers used lighter than water have a for liquids large float, and scale gradations starting with a specific gravity of 1.000 at the bottom. Those used for liquids heavier than water have a small float, and scale gradations starting with 1.000 at the top. A universal hydrometer has 1.000 in the centre so that it may be used for Fig. 4:6 Principle of the Hydrometer. a centimetre scale, and the rod has been rendered impervious to water by dipping in hot paraffin. Float the rod in water and note the depth to which it sinks, for example, 15 cm. Hence it displaces 15 c.c. or 15 gm. of water, and thus the weight of the hydrometer is 15 gm. Now float it in some liquid whose specific gravity is to be determined, and note the level to which it sinks, for example, 20 cm. Hence the volume of the liquid displaced is 20 c.c. and the mass of an equal volume of water is 20 gm. But note, the weight of liquid displaced is equal to the weight of the hydrometer, namely, 15 gm. That is, the mass of liquid displaced is 15 gm. and the mass of
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A ship like the Queen Elizabeth, of 85,000 tons displacement, displaces 85,000 tons of sea-water when afloat. To ensure that it does not sink more deeply than is safe in the water, each ship has a safe-loading line, known as the Plimsoll line (Fig. 4:8) painted on its hull. Actually there are several such lines, to allow for regional These marks and seasonal variations. indicate the depth to which the ship may be safely loaded under the different conditions. (b) Submarines A submarine is a vessel with a cylindrically shaped enclosed hull fitted with ballast tanks at bow and stern and with smaller tanks on either side amidships. In order to submerge, the submarine allows water to enter these tanks until the total weight of the boat and ballast is nearly as great as that of the water it can displace. The submarine is now in “diving trim” and, by a proper use of horizontal or diving rudders, it can submerge completely and maintain its depth below the surface. In order to resurface, it forces the water out of its ballast tanks with compressed air. This action reduces its total weight until, when the weight is less than the upthrust, the ship rises. (c) Floating Docks When the hull of a great liner is to be serviced and reconditioned, a floating dock is used to lift it out of the water. 30 This dock consists, essentially, of a large flat tank divided into several compartments which can be filled with water to Fig. 4:9 Floating Dry-dock. sink the dock to a sufficient depth for the ship to be drawn into it by tugs. The water is then blown out of the -tanks by compressed air, and the dock rises, lifting the ship with it (Fig. 4:9). (d) Balloons When we apply Archimedes’ Principle to air, we see that a body in air will experience a force of buoyancy equal to the weight of air it displaces. A balloon will rise if the weight of the air displaced is greater than the weight of the balloon envelope and its attachments. It will continue to rise until it reaches a level of more rarified air where the weight of the air displaced is equal to that of the balloon. The first balloons, built in 1783, employed hot air for upthrust, having open bottoms with burning braziers sl
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ung underneath them, but by August of the same year the French scientist Charles had sent up the first balloon to be filled with the newly discovered gas, hydrogen. Two years later a trip was made from France to England in such a balloon, motive power being provided by oars. Modern balloons are made of a gastight silk fabric fitted with valves. Increased height is obtained by releasing water or sand ballast, while the descent BUOYANCY Sec. 1:14 is brought about by slowly releasing the gas from the envelope. Then, since atmospheric pressure decreases with height, the gas inside the balloon expands as it rises (in accordance with Boyle’s law) and thus there is danger that the balloon For this will burst at great heights. reason, in balloon ascents to high altitudes, the envelope is not filled to capacity at ground level. The greatest height reached by a manned balloon, 72,395 feet (over I 3/2 miles) was made by Stevens and Anderson, two United States Army officers, at Rapid City, South Dakota, in 1935. Their trip was made in a tightly-sealed hollow metal sphere attached to a huge helium-filled balloon. (e) Weather or Sounding Balloons The main use of hydrogen balloons today is to collect information about the upper atmosphere for meteorological pur- poses. These sounding balloons, as they are called, expand as they rise, until they spring a leak or burst, when the meteorological instruments which they carry are parachuted to earth. Temperature, pressure and humidity reports at various altitudes are automatically sent to weather stations by radio transmitters. Analysis of the readings have shown that sounding balloons have reached heights of 25 miles or more. (f) Airships An airship is a balloon built on a light rigid framework, propelled by air-screws and steered by rudders. The design of this type of craft developed rapidly in the early decades of this century. Although long flights were successfully made in these ships, a series of disasters resulting from the ease with which they caught fire, has caused their further development to be abandoned. 31 Chap. 4 : 15 I A MECHANICS QUESTIONS 1. (a) State Archimedes’ Principle. 7. air a piece of iron whose (b) In 1 00 c.c. has a mass of volume Is 890 gm. When this is immersed in water, calculate: (i) the buoyant force of the water on it. (ii) the weight of the
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iron in water. 2. Describe briefly how Archimedes’ Principle is used to determine the specific gravity of (a) a solid denser than water (b) a liquid. 3. (a) State the Principle of Flotation. (b) Show that it is a modification of Archimedes’ Principle. (c) When a piece of wood is floated in water in a graduated cylinder the level rises from 1 5.7 ml. to 1 8.3 ml. (i) Calculate the mass of the wood. (ii) If its specific gravity is 0.60, what is its volume? 4. Explain how Archimedes could have determined whether King Hiero’s crown was pure gold or a mixture of gold and silver. 5. Explain how Archimedes’ Principle or the Principle of Flotation applies to each of the following statements: (a) In landing a fish, you find that it seems to weigh more when it is pulled out of the water than it does beneath the water surface. (b) As a ship in harbour is being unloaded, it slowly rises higher in the water. In order to make a submarine (c) submerge, large tanks aboard it are filled with water. (d) The same ship with the same cargo will ride higher on the Atlantic Ocean than on the Great Lakes. (e) Plimsoll lines are used on ships. (a) What is hydrometer? (b) Why are the smaller numbers of purpose the of a 6. 32 the hydrometer scale near the top? (c) Compare the size of float required for denser and less dense liquids. Explain the difference. for hydrometers used State three methods used for finding the specific gravity of a liquid. Which of the three do you think Is the most accurate? Why? B 1. 15 c.c. material weigh of (a) If 45 gm. in air, find the weight when immersed in water. 2. (b) If 3 cu. ft. of a substance weigh 350 lb. in air, find the weight when immersed in water. (a) Iron has a density of 7.8 gm. per c.c. Find the weight of 10 c.c. of it when immersed in water. (b) A substance has a density of 1 87.5 lb. per cu. ft. Find the weight when 5 cu. ft. of it are immersed in water. 3. (
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a) If an object weighs 140 gm. in air and 1 1 5 gm. in water, what is the volume of the water displaced? (b) If an object weighs 140 lb. in air and 115 lb. in water, what is the volume of the water displaced? 4. A piece of silver weighs 65.1 gm. in air and 58.9 gm. in water. Find its specific gravity. 5. A piece of metal weighs 500 gm. in air and 430 gm. in water, (i) What is specific gravity? (Ii) What is its volume? Its 6. A 15 lb. weight weighs only 9 lb. in (ii) Find water, (i) Find its specific gravity, Its volume, (iii) Find its density. 7. A piece of metal weighing 1 20.4 gm. has a volume of 14.5 c.c. (i) What will it weigh In water? (Ii) Find also the density of the metal. 8. An object weighs 42.2 gm. in air, 29.4 gm. in water and 25.6 gm. in a liquid.. BUOYANCY Sec. 1 : 15 9 Calculate the specific gravity of the liquid. A metal weighs 56.3 gm. in air, 45.8 gm. when immersed in water and 48.6 gm. when immersed in a liquid. Calculate the specific gravity of (i) the metal (ii) the liquid. 10. A piece of metal weighs 138.8 gm. in air, 123.2 gm. in water and 125.7 gm. in a liquid. Find the specific gravity of the metal and of the liquid. 11. An object weighs 42.2 gm. in air and 29.4 gm. in water. How much will it weigh in a liquid of density 0.80 gm. per c.c.? 12. Was King Hiero’s crown made of pure gold if in air it weighed 1500 gm., and when immersed in water it weighed 1 400 gm.? 13. A rectangular block of soap 8 cm. long and 6 cm. wide floats in water with 2.5 cm. of its thickness submerged. Calculate the mass of the soap. 14. A cube of wood, side 50 cm., floats in water with its base horizontal and 6 cm. of its height above the surface. Find
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