fisica1-youn-e-freedman-exercicios-resolvidos

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7.51: K 1 + U1 + Wother = K2 + U 2 Point 1 is where he steps off the platform and point 2 is where he is stopped by the 1 2 cord. Let y = 0 at point 2. y1 = 41.0 m. Wother = − kx , where x = 11.0 m is the amount 2 the cord is stretched at point 2.The cord does negative work. 1 2 K1 = K2 = U2 = 0, so mgy 1 − kx = 0 2 and k = 631N/m. Now apply F = kx to the test pulls: F = kx so x = F k = 0.602 m. 7.52: For the skier to be moving at no more than 30 .0 m s ; his kinetic energy at the bottom of the ramp can be no bigger than 2 2 mv (85.0 kg)(30.0 m s) = = 38,250 J 2 2 Friction does − 4000 J of work on him during his run, which means his combined PE and KE at the top of the ramp must be no more than 38 ,250 J + 4000 J = 42,250 J. His KE at the top is 2 2 mv (85.0 kg)(2.0 m s) = = 170 J 2 2 His PE at the top should thus be no more than 42 ,250 J − 170 J = 42,080 J, which gives a height above the bottom of the ramp of 42,080 J 42,080 J h = = = 50.5 m. 2 mg (85.0 kg)(9.80 m s ) 7.53: The net work done during the trip down the barrel is the sum of the energy stored in the spring, the (negative) work done by friction and the (negative) work done by 1 2 1 2 gravity. Using kx = ( F k) , the performer’s kinetic energy at the top of the barrel is 2 2 K 2 1 (4400 N) 2 3 = − (40 N)(4.0 m) − (60 kg)(9.80 m s )(2.5 m) = 7.17 × 10 2 1100 N m J, 2(7.17× 10 J) and his speed is = 15.5 m s. 60 kg 3
7.54: To be at equilibrium at the bottom, with the spring compressed a distance x 0 , the spring force must balance the component of the weight down the ramp plus the largest value of the static friction, or kx 0 = wsin θ + f . The work-energy theorem requires that the energy stored in the spring is equal to the sum of the work done by friction, the work done by gravity and the initial kinetic energy, or 1 2 1 2 kx0 = ( wsin θ − f ) L + mv , 2 2 where L is the total length traveled down the ramp and v is the speed at the top of the 1 2 3 ramp. With the given parameters, kx = 248 J and kx = 1.10 10 N. Solving for k gives k = 2440 N m. 2 0 0 × 7.55: The potential energy has decreased by 2 2 (12.0 kg)(9.80 m s )(2.00 m) − (4.0 kg) × (9.80 m s )(2.00 m) = 156.8 J. The kinetic 1 2 2 energy of the masses is then ( m1 + m2) v = (8.0 kg) v = 156.8 J, so the common speed is 2 (156.8 J) v = = 4.43 m s , or 4.4 m s to two figures. 8.0 kg
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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
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2.49: a) For a given initial upward
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2.51: a) From Eqs. (2.17) and (2.18
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2.53: a) The change in speed is the
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.0 m 2.56: a) 1.25m s. 25 = 20.0 s
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1 8 m ⎛ 365 d ⎞⎛ ⎞⎛ 2.62:
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2.66: a) The simplest way to do thi
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2 2.69: a) x = 1 2) a t , and with
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2.71: a) Travelling at 20 m s , Jua
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2.77: Let t 1 be the fall for the w
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2.82: a) 2.83: a) From Eq. (2.14),
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2 2.86: a) The helicopter accelerat
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2.90: a) Suppose that Superman fall
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2 2.93: The velocities are vA = α
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1 2.96: The time spent above y max
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2.98: a) Let the height be h and de
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3.1: a) v (5.3 m) − (1.1m) = 1.4
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3.7: a) b) r ˆ ˆ 2 v = αi − 2
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3.11: Take + y to be upward. Use Ch
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3.15: a) Solving Eq. (3.17) for v =
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v y . g 2 9.80 m s 0 16.0 m s 3.18:
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3.22: Substituting for t in terms o
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3.25: Take + y to be downward. a) U
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3.34: a) a 2 rad = ( 3 m/s) /(14 m)
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3.45: a) The a = 0 and a y = −2β
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3.48: a) The equations of motions a
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3.52: a) Setting y = −h in Eq. (3
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3.58: Equation 3.27 relates the ver
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3.60: a) This may be done by a dire
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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
Page 205 and 206: 5.124: For convenience, take the po
Page 207 and 208: 5.126: In all cases, the tension in
Page 209 and 210: 6.1: a) ( 2.40 N) (1.5 m) = 3.60 J
Page 211 and 212: 6.10: a) From Eq. (6.6), 1 2 5 ⎛
Page 213 and 214: 6.18: As the example explains, the
Page 215 and 216: 6.27: a) The friction force is µ k
Page 217 and 218: 6.35: a) The static friction force
Page 219 and 220: 6.42: The initial and final kinetic
Page 221 and 222: 6.53: a) In terms of the accelerati
Page 223 and 224: 6.58: The work per unit mass is ( W
Page 225 and 226: 6.66: Let x be the distance past P.
Page 227 and 228: 6.71: The velocity and acceleration
Page 229 and 230: 6.79: In terms of the bumper compre
Page 231 and 232: 6.85: f = .25 mg soW f = W = −(0.
Page 233 and 234: 6.94: a) The number of cars is the
Page 235 and 236: (8.00 hp)(746 W/hp) 6.99: a) = 358
Page 237 and 238: 6.101: a) Denote the position of a
Page 239 and 240: 6.104: From F = m a, Fx = max, Fy =
Page 241 and 242: 7.1: From Eq. (7.2), mgy = (800 kg)
Page 243 and 244: 7.10: (a) The flea leaves the groun
Page 245 and 246: 2 7.16: U = 1 ky , where y is the v
Page 247 and 248: 1 2 1 2 7.24: From kx = mv , the re
Page 249 and 250: dU 3 3 7.34: From Eq. (7.15), F =
Page 251 and 252: 7.39: a) At constant speed, the upw
Page 253 and 254: 1 2 7.46: a) U − U = mg( h − 2R
Page 255: 7.49: a) K 1 + U1 + Wother = K2 + U
Page 259 and 260: 7.57: The two design conditions are
Page 261 and 262: 7.62: a) The skier’s kinetic ener
Page 263 and 264: 1 2 1 2 7.69: With U = 0 , K = 0 K
Page 265 and 266: 7.74: a) From either energy or forc
Page 267 and 268: 2 2 2 2 7.78: a) a = d x dt = −ω
Page 269 and 270: 3 7.83: a) Along this line, x = y ,
Page 271 and 272: 7.86: a) The slope of the U vs. x c
Page 273 and 274: Capítulo 8
Page 275 and 276: 8.7: ∆p ∆t ( 0.0450 kg)( 25.0 m
Page 277 and 278: 8.14: The impluse imparted to the p
Page 279 and 280: 8.21: a) Taking v A and v B to be m
Page 281 and 282: 8.25: m + m )(3.00 m s) = m (4.00 m
Page 283 and 284: 8.31: Use conservation of the horiz
Page 285 and 286: 8.37: Let +y be north and +x be sou
Page 287 and 288: 8.43: a) In Eq. (8.24), let mA = m
Page 289 and 290: 8.52: It turns out to be more conve
Page 291 and 292: 8.63: The total momentum must be ze
Page 293 and 294: 8.69: (This problem involves solvin
Page 295 and 296: 8.72: a) The stuntman’s speed bef
Page 297 and 298: 8.76: Just after the collision: ∑
Page 299 and 300: 8.80: a) From the derivation in Sec
Page 301 and 302: 8.83: a) In terms of the primed coo
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Page 305 and 306: 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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