Có 40+ tài liệu thuộc chủ đề "Physics"
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b) If the mass of the light pulley may be neglected, the net force on the pulley is the vector sum of the tension in the chain and the tensions in the two parts of the rope. 5.3: a) The two sides of the rope each exert a force with vertical component T sin θ , and the sum of...
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b) Gravity is directed opposite to the direction of the bucket’s motion, so Eq. (6.2) gives the negative of the result of part (a), or 265 J . and so the work done by friction is. 80 10 6 N. 75 10 3 m ) cos 14. 62 10 9 J , or J to two...
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Note that the result is independent of the speed, and that an extra figure was kept in part (b) to avoid roundoff error.. of the result of part (a) and the horizontal displacement. (7.5), depends only on the magnitude of the velocities, not the directions, so the speed is again 24 . 7.6: a) (Denote the top of the ramp...
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8.5: The y-component of the total momentum is. This quantity is negative, so the total momentum of the system is in the y -direction.. using the value of the arctangent function in the fourth quadrant p x 0 , p y 0. 00 kg 10. 8.8: a) The magnitude of the velocity has changed by....
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(12.0 rad s 3 ) t , so at t 3.5 s, α 42 rad s 2 . is proportional to the time, so the average angular acceleration between any two times is the arithmetic average of the angular accelerations. t The angular velocity is not linear function of time, so the average angular velocity is not the...
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10.1: Equation (10.2) or Eq. (10.3) is used for all parts.. 00 m)(10.0 N) sin 90. 00 m)(10.0 N) sin 120. 00 m)(10.0 N) sin 30. 00 m)(10.00 N) sin 60. 00 m)(10.0 N) sin 180. 10.7: I 3 2 MR 2 2 mR 2 , where M 8 . 10.9: v 2 as 2...
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11.1: Take the origin to be at the center of the small ball. from the center of the small ball.. 11.2: The calculation of Exercise 11.1 becomes. 11.4: a) The force is applied at the center of mass, so the applied force must have the same magnitude as the weight of the door, or 300 N. 11.6: The other person...
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12.2: Use of Eq. (12.1) gives. (12.1) twice gives. 12.10:. 12.11:. 38 10 7 m.. 12.16: a) Using g E 9 . 80 m s 2 , Eq 12 . 12.18: M gR G 2 2 . 44 10 21 kg and. 12.19: 2 E r G mm F. 12.21: From eq. (12.1), G...
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13.3: The period is 0 440 . 13.5: This displacement is 4 1 of a period.. 13.8: Solving Eq. (13.12) for k,. 13.9: From Eq. (13.12) and Eq. (13.10), T 2 π 140 0 . 13.10: a) 2 2 sin. (13.4) if ω. (13.4) is not satisfied. (13.4) if 2. 13.11: a) x. 13.12: a) From Eq. (13.19. A...
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14.1: w mg ρVg. 14.4: The length L of a side of the cube is. 14.7: p p 0 ρgh. 14.8: The pressure difference between the top and bottom of the tube must be at least 5980 Pa in order to force fluid into the vein:. 14.9: a) ρgh. 14.11: a) ρgh. 14.12: p ρgh....
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15.1: a) The period is twice the time to go from one extreme to the other, and s. 15.4: Denoting the speed of light by c. 15.6: Comparison with Eq. (15.4) gives a) 6 . 15.9: a) sin. (15.12) with v ω k. (15.12) with v ω k . ωA cos ( kx ωt ) and a...
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16.3: From Eq. (16.5), p max BkA 2 π BA λ 2 πBA f v. 16.4: The values from Example 16.8 are B 3 . 16 10 4 Pa, f 1000 Hz, m.. 16.5: a) Using Equation (16.7), B v 2 ρ. 16.6: a) The time for the wave to travel to Caracas...
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17.1: From Eq. 17.2: From Eq. 17.5: a) From Eq. 17.8: For ( b. 17.9: Combining Eq. (17.2) and Eq. 17.11: From Eq. 17.12: From Eq. 17.13: From Eq. 76 10 4 Pa.. 17.16. 17.17. 17.18: d. 17.19: a) αD 0 T. 17.20: α T. 17.21: α. 17.22: From Eq. 17.23 β V 0 T. 75 10...
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18.1: a) n m tot M. (18.3) gives. 18.3: For constant temperature, Eq. (18.6) becomes. 18.4: a) Decreasing the pressure by a factor of one-third decreases the Kelvin. 18.6: The temperature is T 22 . 18.7: From Eq. 18.9: From Eq. 18.10: a) 5 . 18.11: V 2 V 1 ( T 2 T 1. 18.12: a)...
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33 10 3 J . 19.4: At constant pressure, W p V nR T , so. 50 10 3 J.. 19.8: a) W 13 p 1 ( V 2 V 1. 19.9: Q 254 J, W. 19.10: a) p V. 78 10 4 J.. 15 10 5 J J...
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20.1: a) 2200 J 4300 J 6500 J. 20.2: a) 9000 J 6400 J 2600 J. 20.4: a) Q 1 e Pt. 43 10 5 J.. 43 10 5 J. 180 10 3 W)(1.00 s. 20.5: a) e MW MW 0 . 20.6: Solving Eq. (20.6) for r. 50 10...
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21.1: m lead 8 . 21.2: current C s and t 100 s 10 4 s Q = It = 2.00 C. 21.3: The mass is primarily protons and neutrons of m 1 . 10 10 28. 35 10 9 C. 27 10 24 q n p 1 ....
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In part i), the paper is oriented to “capture” the most field lines whereas in ii) the area is oriented so that it “captures” no field lines.. 36 cos m) (0.1 ) C N 10 4 ( (back). 36 cos m) (0.1 ) C N 10 4 ( front). 0 90 cos m) (0.1 ) C N 10 4 (...
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23.8: From Example 23.1, the initial energy E i can be calculated:. 23.10: The work is the potential energy of the combination.. 23.11: K 1 U 1 K 2 U 2 . 23.12: Get closest distance γ . 23.13: K A U A K B U B. 23.14: Taking the origin at the center...
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24.1: Q CV. 24.3: a) 604 V. 24.5: a) Q CV 120 μC b) C ε 0 A d. 24.6: (a) 12.0 V since the plates remain charged.. If d is doubled, C 1 2 C , so V 2 V 24 . V 24.7: Estimate r 1 . 24.10: a) 1...