Chapter 1 18. 9.19 nm/s Chapter 2 6. (a) 5.00 m/s (b) 1.25 m/s (c) -2.50 m/s (d) -3.33 m/s (e) 0 14. 0.182 mi west of the flagpole 16. (a) graph (b) 41.0 m/s, 41.0 m/s, 41.0 m/s NOTE that the three values are the same only because all three selected intervals happen to be "small enough". For an interval of 2.0 s (3.0 s to 5.0 s), you would get 41.75 m/s. (c) v_av = 17.0 m/s, much less than in (b) 22. (a) 0, 1.60 m/s², 0.80 m/s² (b) 0, 1.60 m/s², 0 34. (a) 5.51 km (b) 20.8 m/s, 41.6 m/s, 20.8 m/s; 38.7 m/s Questions from Optional Reading 3 1. (a) 1.5 m/s² (b) 3 m/s (c) 12 m (d) 3 m/s (e) 3 m 2. (a) 20 m/s (b) 5 m/s² 3. 10 s 4. (a) 26 m/s (b) 60 m Questions from Optional Reading 4 1. 10 m/s (the x velocity can never change) 2. 10 m/s E 3. 2 s (v_y changes by -19.6 m/s²) 4. 4 s 5. 40 m 6. 19.6 m (the average v_y is 9.8 m/s for the UP trip) Chapter 3 6. (a) 6.12 units at 113° counterclockwise from +x-axis (b) 14.8 units at 22.5° counterclockwise from +x-axis 20. (a) 74.6° NofE (b) 470 km 30. (a) clears the bar by 0.85 m (b) falling; v_y = -13.4 m/s 36. 61.4 s 38. (a) 10.1 m/s at 8.53° EofN (b) 45.0 m 52. 14.0 m/s Questions from Optional Reading 5 1. 31.4 rad/s 2. 62.8 m/s 3. 0.2 s 4. 94.2 rad 5. 188.4 m Question from Optional Reading 6 The forces are: F_BR, the force on the ball by the rope, UP W_B, the weight of the ball, DOWN The Third-Law partners of these forces are: F_RB, the force on the rope by the ball, DOWN F_G,EB, the force of gravity on the Earth by the ball, at the Earth's center pointing towards the ball Chapter 4 6. 7.41 min 8. 312 N 26. a = 4.43 m/s², T = 53.7 N 40. (a) 55.2 degrees (b) 167 N 50. (a) 18.5 N (b) 25.8 N 56. 21.5 N 58. (a) 50.0 N (b) 0.500 (c) 25.0 N 60. 0.814 64. (a) 9.8 N (b) 0.58 m/s² 68. 0.685 m/s² Chapter 5 32. 5.11 m/s 38. friction has magnitude 4.23 N work by friction is -16.9 J work done by F is 47.9 J 42. (a) 2.29 m/s (b) 15.6 J 50. 2.92 m/s 66. (a) 582 trips (b) 90.5 W or 0.121 hp 70. (a) 306 J (b) -147 J (c) 0 (d) 147 J Chapter 6 2. (a) 5.40 N⋅s toward the net (b) -27.0 J 14. 260 N directly away from the wall 20. (a) 0.490 m/s (b) 0.0201 m/s 22. thrower 2.48 m/s catcher 2.25 cm/s 24. (e) 2.22 m/s in the direction opposite to the velocity of B 28. 529 m/s 32. (a) 20.9 m/s East (b) 8689 J; both vehicles are warmer and both are deformed 38. (a) 2.50 m/s (b) 37500 J 44. 41.5 mi/h 72. (a) 1.07 m/s @ 29.7 degrees clockwise from the x axis (b) 31.8% 74. (a) 7.06 m/s (b) 2.54 m Chapter 7 2. 2.15 m, 123 m, 773 m 16. 2360 m/s² 26. 1380 N so he will not make it 28. (a) 24.9 kN (b) 12.1 m/s 36. (a) 5590 m/s (b) 3.98 h (c) 1470 N 42. (a) 3.77 m/s² (b) 3.26 s 50. (a) 0 (b) 1290 N (c) 2060 N 52. (a) 2.06 m/s (b) 54 degrees (c) 4.70 m/s Chapter 8 8. 0.00669 nm, 0 36. (a) 5.35 m/s² (b) 42.8 m (c) 8.91 rad/s² 38. 30.3 rev/s 44. 35.6 rad/s 50. (a) 1.91 rad/s (b) initial 2.53 J final 6.44 J 52. (a) 3.58 rad/s (b) 539 J, which is the work done by the man during the walk inwards 56. 508 N (ignoring the mass of the door) 60. (a) 46.8 N (b) 0.234 kg⋅m² (c) 40.0 rad/s Chapter 9 6. 22.1 N towards bottom of page in diagram 10. (a) 2.45 mm (b) 0.75 mm 12. 55.5 MPa; the arm should survive 18. (a) -0.0538 m³ (b) 1090 kg/m³ (c) for the enormous pressure change, the fractional change in volume is small (~5%) 20. (a) 65.1 N (b) 275 N 24. 2.31 lb (for interested students only) 25. The answer in the back of the book for this one is incorrect. The online version, however, is correct. The correct answer for the book's numbers is 26.6 N⋅m. 40. 154 in/s 48. (a) 15.1 MPa (b) 2.95 m/s (c) 4.34 kPa 64. 455 kPa 74. answers will be posted after this HW is due 82. 0.721 mm Chapter 13 6. (a) 327 N (b) 1250 N/m 8. (a) 575 N/m (b) 46.0 J 24. 2.23 Hz 36. 58.8 s 42. (a) 0.20 Hz (b) 0.25 Hz Chapter 14 10. (a) 0.05 fW (b) 0.05 mW 14. (a) 0.000132 W/m² (b) 81.2 dB 16. (a) 0.00796 W/m² (b) 109 dB (c) 2.82 m 18. answers will be posted after this HW is due 24. 41.2 kHz 26. (a) 0.0217 m/s (b) 2.000029 MHz (c) 2.000058 MHz 30. answers will be posted after this HW is due 32. answers will be posted after this HW is due 42. 9.00 kHz 50. 29.7 cm 60. 200 m/s Question from Optional Reading 19 Point A, with Δpath = 0, would now be a point of destructive interference, as would point C. Point B, with Δpath = (1/2) a wavelength would now be a point of constructive interference. In the resulting standing wave, the positions of nodes and antinodes would now be reversed.