fisica1-youn-e-freedman-exercicios-resolvidos

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2.59: (a) Denote the time for the acceleration (4.0 s) as t 1 and the time spent running at constant speed (5.1 s) as t 2 . The constant speed is then at 1 , where a is the unknown acceleration. The total l is then given in terms of a, t 1 and t 2 by and solving for a gives 1 2 l = at + at 2 1 1t2 , l a = 2 (1 2) t + t t 1 1 2 (100 m) = 2 (1 2)(4.0 s) + (4.0 s)(5.1s) = 3.5 m s 2 . (b) During the 5.1 s interval, the runner is not accelerating, so a = 0. 2 2 (c) ∆v ∆t = [(3.5 m s )(4 s)] (9.1s) = 1.54 m s . (d) The runner was moving at constant velocity for the last 5.1 s. 2.60: a) Simple subtraction and division gives average speeds during the 2-second intervals as 5.6, 7.2 and 8.8 m s . b) The average speed increased by 1.6 m s during each 2-second interval, so the acceleration is 0.8 2 m s . c) From Eq. (2.13), with v 0 = 0, v = 2(0.8 m s )(14.4 m) = 4.8 m s. Or, recognizing that for constant acceleration the average speed of 5.6 m/s is the speed one 2 second after passing the 14.4-m mark, 5.6 m s – (0.8 m s )(1.0 s) = 4.8 m s . d) With both the acceleration and the speed at the 14.4-m known, either Eq. (2.8) or Eq. (2.12) gives the time as 6.0 s. 2 e) From Eq. (2.12), x – x 0 = (4.8 m s )(1.0 s) + 21 (0.8 m s )(1.0 s) 2 = 5.2 m. This is also the average velocity (1/2)(5.6 m s + 4.8 m s ) times the time interval of 1.0 s. 2 2.61: If the driver steps on the gas, the car will travel 2 2 (20 m s)(3. 0 s) + (1 2)(2.3 m s )(3.0 s) = 70.4m. If the brake is applied, the car will travel 2 2 (20 m s)(3. 0 s) + (1 2)( −3.8 m s )(3.0 s) = 42.9 m, so the driver should apply the brake.
1 8 m ⎛ 365 d ⎞⎛ ⎞⎛ 2.62: a) ⎟ ⎞ 4 24h 3600s d = ct = ( 3.0 × 10 )(1y) ⎜ ⎟ ⎜ ⎟ ⎜ s ⎝ 1y ⎠⎝ 1d ⎠⎝ 1h ⎠ 15 = 9.5× 10 m 8 m −9 b) d = ct = (3.0× 10 )(10 s) = 0.30 m s 11 d 1.5× 10 m c) t = = = 500 s = 8. 33 min 8 c 3.0× 10 m s 8 d 2(3.84× 10 m) d) t = = = 2.6 s c 8 3.0× 10 m s 9 d 3× 10 mi e) t = = = 16,100 s = 4. 5 h c 186,000 mi s 2.63: a) v = 2π RE t = 464m s 4 b) v = 2π r t = 2.99× 10 m s ( r is the radius of the earth's orbit) c) Let c be the speed of light, then in one second light travels a distance c (1.00s). The number of times around the earth to which this corresponds is c ( 1.00s) 2 πR = E 7. 48
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Física I - Sears, Zemansky, Young
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5 1.1: 1mi × ( 5280ft mi) × ( 12
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1.15: a) If a meter stick can measu
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8 11 1.28: The moon is about 4× 10
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1.34: r o o 1.35: A ; A x = ( 12.0
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1.39: Using components as a check f
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1.43: a) The magnitude of A r + B r
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1.48: a) ˆ ˆ ˆ 2 2 2 i + j + k =
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1.53: Use of the right-hand rule to
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1.59: a) To three significant figur
Page 21 and 22: 1.65: Let D r be the fourth force.
Page 23 and 24: 1.68:a) R = A + B + C x R y = x x x
Page 25 and 26: 1.70: The third leg must have taken
Page 27 and 28: −1 200−20 o 1.73: a) Angle of f
Page 29 and 30: 1.75: Let +x be east and +y be nort
Page 31 and 32: 1.78: (a) Take the beginning of the
Page 33 and 34: 1.83: a) Parallelog ram area = 2 ×
Page 35 and 36: 1.88: a) This is a statement of the
Page 37 and 38: 1.91: A r and C r are perpendicular
Page 39 and 40: 1.96: a) b) i) In AU, ii) In AU, ii
Page 41 and 42: Capítulo 2
Page 43 and 44: 2.6: The s-waves travel slower, so
Page 45 and 46: 2.14: (a) The displacement vector i
Page 47 and 48: 2.18: a) The velocity at t = 0 is a
Page 49 and 50: 2.20: a) The bumper’s velocity an
Page 51 and 52: 2.26: a) x 0 < 0, v 0x < 0, a x < 0
Page 53 and 54: 2.28: a) Interpolating from the gra
Page 55 and 56: 2.31: a) At t = 3 s the graph is ho
Page 57 and 58: 2.35: a) b) From the graph (Fig. (2
Page 59 and 60: 2.37: a) The car and the motorcycle
Page 61 and 62: 2.43: a) Using the method of Exampl
Page 63 and 64: 2.45: a) v y = v0 y − gt = ( −6
Page 65 and 66: 2.49: a) For a given initial upward
Page 67 and 68: 2.51: a) From Eqs. (2.17) and (2.18
Page 69 and 70: 2.53: a) The change in speed is the
Page 71: .0 m 2.56: a) 1.25m s. 25 = 20.0 s
Page 75 and 76: 2.66: a) The simplest way to do thi
Page 77 and 78: 2 2.69: a) x = 1 2) a t , and with
Page 79 and 80: 2.71: a) Travelling at 20 m s , Jua
Page 81 and 82: 2.77: Let t 1 be the fall for the w
Page 83 and 84: 2.82: a) 2.83: a) From Eq. (2.14),
Page 85 and 86: 2 2.86: a) The helicopter accelerat
Page 87 and 88: 2.90: a) Suppose that Superman fall
Page 89 and 90: 2 2.93: The velocities are vA = α
Page 91 and 92: 1 2.96: The time spent above y max
Page 93 and 94: 2.98: a) Let the height be h and de
Page 95 and 96: 3.1: a) v (5.3 m) − (1.1m) = 1.4
Page 97 and 98: 3.7: a) b) r ˆ ˆ 2 v = αi − 2
Page 99 and 100: 3.11: Take + y to be upward. Use Ch
Page 101 and 102: 3.15: a) Solving Eq. (3.17) for v =
Page 103 and 104: v y . g 2 9.80 m s 0 16.0 m s 3.18:
Page 105 and 106: 3.22: Substituting for t in terms o
Page 107 and 108: 3.25: Take + y to be downward. a) U
Page 109 and 110: 3.34: a) a 2 rad = ( 3 m/s) /(14 m)
Page 111 and 112: 3.45: a) The a = 0 and a y = −2β
Page 113 and 114: 3.48: a) The equations of motions a
Page 115 and 116: 3.52: a) Setting y = −h in Eq. (3
Page 117 and 118: 3.58: Equation 3.27 relates the ver
Page 119 and 120: 3.60: a) This may be done by a dire
Page 121 and 122: 3.64: Combining equations 3.25, 3.2
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3.66: (a) v x ( runner) = vx (ball)
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3.69: Take + y to be upward. a) The
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3.73: a) The acceleration is given
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3.76: a) dv dt = = d dt (1/ 2) vxax
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3.85: The three relative velocities
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3.90: As in the previous problem, t
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3.92: The x-position of the plane i
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4.1: a) For the magnitude of the su
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2 4.11: a) During the first 2.00 s,
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4.24: (a) Each crate can be conside
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4.27: a) b) For the chair, a y = 0
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4.33: a) The resultant must have no
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2 4.39: a) Both crates moves togeth
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4.43: a) The engine is pulling four
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4.48: a) The net force on a point o
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4.51: a) L is the lift force b) ∑
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2 4.54: The velocity as a function
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4.57: In this situation, the x-comp
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5.1: a) The tension in the rope mus
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5.14: a) In level flight, the thrus
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2L 5.20: Similar to Exercise 5.16,
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5.28: a) The stopping distance is 2
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5.35: First, determine the accelera
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5.40: Differentiating Eq. (5.10) wi
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5.49: a) Setting arad = g in Eq. (5
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5.56: The tension in the lower chai
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5.63: (Denote F r by F.) a) The for
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5.67: Consider the forces on the pe
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5.70: It’s interesting to look at
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5.72: The key idea in solving this
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5.77: See Exercise 5.40. a) The max
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5.82: We take the upward direction
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5.85: Denote the magnitude of the a
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5.92: (a) There is a contact force
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5.96: (a) 80 mi h is 35 .7 m s in S
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5.101: a) The rock is released from
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mg mg 5.103: Without buoyancy, kv t
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5.106: (a) To find find the maximum
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5.109: a) For the same rotation rat
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5.116: a) Differentiating twice, a
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5.120: a) There are many ways to do
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5.124: For convenience, take the po
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5.126: In all cases, the tension in
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6.1: a) ( 2.40 N) (1.5 m) = 3.60 J
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6.10: a) From Eq. (6.6), 1 2 5 ⎛
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6.18: As the example explains, the
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6.27: a) The friction force is µ k
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6.35: a) The static friction force
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6.42: The initial and final kinetic
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6.53: a) In terms of the accelerati
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6.58: The work per unit mass is ( W
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6.66: Let x be the distance past P.
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6.71: The velocity and acceleration
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6.79: In terms of the bumper compre
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6.85: f = .25 mg soW f = W = −(0.
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6.94: a) The number of cars is the
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(8.00 hp)(746 W/hp) 6.99: a) = 358
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6.101: a) Denote the position of a
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6.104: From F = m a, Fx = max, Fy =
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7.1: From Eq. (7.2), mgy = (800 kg)
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7.10: (a) The flea leaves the groun
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2 7.16: U = 1 ky , where y is the v
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1 2 1 2 7.24: From kx = mv , the re
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dU 3 3 7.34: From Eq. (7.15), F =
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7.39: a) At constant speed, the upw
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1 2 7.46: a) U − U = mg( h − 2R
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7.49: a) K 1 + U1 + Wother = K2 + U
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7.54: To be at equilibrium at the b
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7.57: The two design conditions are
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7.62: a) The skier’s kinetic ener
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1 2 1 2 7.69: With U = 0 , K = 0 K
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7.74: a) From either energy or forc
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2 2 2 2 7.78: a) a = d x dt = −ω
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3 7.83: a) Along this line, x = y ,
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7.86: a) The slope of the U vs. x c
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Capítulo 8
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8.7: ∆p ∆t ( 0.0450 kg)( 25.0 m
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8.14: The impluse imparted to the p
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8.21: a) Taking v A and v B to be m
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8.25: m + m )(3.00 m s) = m (4.00 m
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8.31: Use conservation of the horiz
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8.37: Let +y be north and +x be sou
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8.43: a) In Eq. (8.24), let mA = m
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8.52: It turns out to be more conve
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8.63: The total momentum must be ze
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8.69: (This problem involves solvin
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8.72: a) The stuntman’s speed bef
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8.76: Just after the collision: ∑
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8.80: a) From the derivation in Sec
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8.83: a) In terms of the primed coo
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8.86: (a) For momentum to be conser
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8.90: a) For the x - and y - direct
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8.96: The trick here is to realize
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8.98: The two fragments are 3.00 kg
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8.101: a) If the objects stick toge
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8.107: a) For t < 0 the rocket is a
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8.111: a) The tension in the rope a
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9.1: a) 1.50 m = 0.60 rad = 34.4°
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2 9.10: a) ω = ω + α t = 1.50 ra
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9.18: The following table gives the
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9.24: From a rad = ω 2 r, 4 which
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vr 5.00 m s 9.33: The angular veloc
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9.40: a) In the expression of Eq. (
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2π 2 2 9.49: a) ω = T , so Eq. (9
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9.60: For this case, dm = γ dx. a)
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dθ 2 2 3 2 9.65: a) ω z = = 2γt
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9.70: a) The angular acceleration w
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9.75: I approximate my body as a ve
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9.81: a) Consider a small strip of
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9.84: Taking the zero of gravitatio
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9.90: a) In the case that no energy
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9.94: a) From the parallel-axis the
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9.99: a) Following the hint, the mo
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Capítulo 10
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10.6: (a) τ A = (50 N)(sin 60° )(
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1 2 2 = m 2 8.25 kg 0.0750 m = 0.02
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10.23: n = mg cos α mg sin θ −
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10.26: a) The angular speed of the
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10.32: I = mL 1 2 1 2 2 = ( 117 kg)
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10.40: The skater’s initial momen
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10.46: (a) Conservation of angular
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2 2π rad 10.51: a) Solving Eq. (10
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2 2θ 2θ 2θ I 10.57: t = = = . α
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10.63: The net torque on the pulley
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2 10.67: For the disk, K = (3 4) Mv
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10.72: (a) Στ = Iα and a = Rα T
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10.75: a) Use conservation of energ
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10.79: a) v = 10K 7m = ( 10)( 0.800
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10.84: (a) Στ = Iα 1 1 2 mg cos
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10.90: Angular momentum is conserve
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10.95: a) The initial angular momen
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10.100: a) The distance from the ce
Page 387 and 388:
10.102: Denoting the upward forces
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11.1: Take the origin to be at the
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11.12: a) b) x = 6 .25 m when FA =
Page 393 and 394:
11.18: a) Denote the length of the
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11.30: a) The volume would increase
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11.43: For the airplane to remain i
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11.49: The horizontal component of
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11.56: (a) and (b) Lower rod: Στ
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11.59: a) ∑τ = 0, axis at lower
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−4 −4 2 11.64: a) ∆w = −σ
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11.68: (a) ΣτElbow = 0 F (3.80 cm
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11.71: a) Consider the forces on th
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11.74: a) The center of gravity of
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11.79: a) Take torques about the po
Page 415 and 416:
11.87: Use subscripts 1 to denote t
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11.93: a) Taking torques about the
Page 419 and 420:
11.95: Assume that the center of gr
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