fisica1-youn-e-freedman-exercicios-resolvidos

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2 3.55: a) With α 0 = 45° , Eq. (3.27) is solved for v gx 0 x− y . In this case, y = −0.9 m is the = 2 change in height. Substitution of numerical values gives above algebraic expression for v 0 in Eq. (3.27) gives v = 42.8 0 m/s . b) Using the ⎛ x ⎞ y = x − ⎜ (188.9 m) 188 m ⎟ ⎝ ⎠ Using x = 116 m gives y = 44.1m above the initial height, or 45.0 m above the ground, which is 42.0 m above the fence. 2 2 3.56: The equations of motions are y = ( v0 sin α) t −1/ 2gt and x = ( v 0 cos α) t , assuming the match starts out at x = 0 and y = 0 . When the match goes in the wastebasket for the minimum velocity, y = 2D and x = 6D . When the match goes in the wastebasket for the maximum velocity, y = 2D and x = 7D . In both cases, sin α = cos α = 2 / 2. . 2 To reach the minimum distance: 6 = v t , and 2D gt equation for t gives t = 6D 2 ( ) 2 1 2 6 − 2 v0 6D 2 v0 D 2 0 2 1 2 = v 2 0t − . Solving the first 2 . Substituting this into the second equation gives D = D g . Solving this for v 0 gives v = 0 3 gD . 2 To reach the maximum distance: 7 = v t , and 2D gt equation for t gives t = 7D 2 v0 1 7D 2 gives D 7D − g( ) 2 D 2 0 . Substituting this into the second equation 2 = 2 v . Solving this for v 0 0 gives as expected, is larger than the previous result. 2 1 2 = v 2 0t − 2 . Solving the first v = 49gD /5 3. 13 gD , which, 0 = 3.57: The range for a projectile that lands at the same height from which it was launched is 2 v0 sin 2α g v0 sin α H 2 2 2 R = , and the maximum height that it reaches is = g . We must find R when 2 sin α D 2 H = D and v = 0 6gD . Solving the height equation for 6gD sin α , = g 2 6 gD sin(70.72° ) α . Then, g sin = (1/3) 1/ R = , or R = 5 .6569D = 4 2D . , or
3.58: Equation 3.27 relates the vertical and horizontal components of position for a given set of initial values. a) Solving for v 0 gives v 2 0 2 2 gx / 2 cos α0 = . x tan α − y Insertion of numerical values gives v = 16.6 0 m/s . b) Eliminating t between Equations 3.20 and 3.23 gives v y as a function of x , gx v y = v0 sin α0 − . v cosα 0 0 x = v0 cos 0 = 8.28 m/s, v y = − Using the given values yields v α 6.98 m/s, so 2 2 −6 .98 v = (8.28 m/s) + ( −6.98 m/s) = 10.8 m/s, at an angle of arctan ( ) = −40. 1° 8. 24 , with the negative sign indicating a direction below the horizontal. c) The graph of v x (t) is a horizontal line. 0
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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
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1.65: Let D r be the fourth force.
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1.68:a) R = A + B + C x R y = x x x
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1.70: The third leg must have taken
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−1 200−20 o 1.73: a) Angle of f
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1.75: Let +x be east and +y be nort
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1.78: (a) Take the beginning of the
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1.83: a) Parallelog ram area = 2 ×
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1.88: a) This is a statement of the
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1.91: A r and C r are perpendicular
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1.96: a) b) i) In AU, ii) In AU, ii
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Capítulo 2
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2.6: The s-waves travel slower, so
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2.14: (a) The displacement vector i
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2.18: a) The velocity at t = 0 is a
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2.20: a) The bumper’s velocity an
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2.26: a) x 0 < 0, v 0x < 0, a x < 0
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2.28: a) Interpolating from the gra
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2.31: a) At t = 3 s the graph is ho
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2.35: a) b) From the graph (Fig. (2
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2.37: a) The car and the motorcycle
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2.43: a) Using the method of Exampl
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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 and 72: .0 m 2.56: a) 1.25m s. 25 = 20.0 s
Page 73 and 74: 1 8 m ⎛ 365 d ⎞⎛ ⎞⎛ 2.62:
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: 3.52: a) Setting y = −h in Eq. (3
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
Page 123 and 124: 3.66: (a) v x ( runner) = vx (ball)
Page 125 and 126: 3.69: Take + y to be upward. a) The
Page 127 and 128: 3.73: a) The acceleration is given
Page 129 and 130: 3.76: a) dv dt = = d dt (1/ 2) vxax
Page 131 and 132: 3.85: The three relative velocities
Page 133 and 134: 3.90: As in the previous problem, t
Page 135 and 136: 3.92: The x-position of the plane i
Page 137 and 138: 4.1: a) For the magnitude of the su
Page 139 and 140: 2 4.11: a) During the first 2.00 s,
Page 141 and 142: 4.24: (a) Each crate can be conside
Page 143 and 144: 4.27: a) b) For the chair, a y = 0
Page 145 and 146: 4.33: a) The resultant must have no
Page 147 and 148: 2 4.39: a) Both crates moves togeth
Page 149 and 150: 4.43: a) The engine is pulling four
Page 151 and 152: 4.48: a) The net force on a point o
Page 153 and 154: 4.51: a) L is the lift force b) ∑
Page 155 and 156: 2 4.54: The velocity as a function
Page 157 and 158: 4.57: In this situation, the x-comp
Page 159 and 160: 5.1: a) The tension in the rope mus
Page 161 and 162: 5.14: a) In level flight, the thrus
Page 163 and 164: 2L 5.20: Similar to Exercise 5.16,
Page 165 and 166: 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
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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 =
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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
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11.87: Use subscripts 1 to denote t
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11.93: a) Taking torques about the
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11.95: Assume that the center of gr
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