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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, th... |
− 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 ... |
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. ... |
://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:... |
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... |
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, th... |
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. 2... |
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 tha... |
) 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/c... |
/ 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 ... |
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 swim... |
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... |
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 ... |
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 ... |
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 = Δ... |
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 = – ne... |
= 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 c... |
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 ... |
= 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... |
� 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, becaus... |
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 ... |
) 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 ra... |
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 = ... |
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 sm... |
.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 /... |
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) 82... |
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 w... |
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 be... |
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... |
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 2... |
/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 ... |
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... |
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 availa... |
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 ... |
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%... |
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 voltag... |
, 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 ass... |
.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 ... |
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 f... |
+ 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 ... |
(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 ... |
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... |
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 sma... |
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 k... |
(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... |
) 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 ca... |
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 ... |
−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 (... |
.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) 0.752... |
–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 (... |
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... |
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) 1.0... |
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 e... |
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 kine... |
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 consis... |
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... |
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... |
). 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 ... |
.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... |
) 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 ... |
) 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 ... |
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) =... |
= 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, o... |
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 lep... |
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 tha... |
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, 24... |
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, 1... |
, 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 co... |
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, ... |
, 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 elect... |
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 electro... |
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... |
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 i... |
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... |
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 Ne... |
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, ... |
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, 14... |
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, 94... |
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 ... |
, 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 coef... |
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 un... |
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... |
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 significa... |
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 MEASURE... |
. 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 advant... |
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 95... |
) 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. f... |
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 previ... |
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... |
.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 Ca... |
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 ... |
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 ... |
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 fa... |
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 flu... |
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 low... |
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 bu... |
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 al... |
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 ... |
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 cylinde... |
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 met... |
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