fisica1-youn-e-freedman-exercicios-resolvidos

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2.52: a) Slope = a = 0 for t ≥ 1.3 ms b) h = Area underv − tgraph max ≈ A Triangle + A Rectangle 1 ⎛ cm ⎞ ≈ (1.3 ms) ⎜133 ⎟ + (2.5 ms −1.3 ms)(133 cm s) 2 ⎝ s ⎠ ≈ 0.25 cm c) a = slope of v– t graph 133 cm s 5 a ( 0.5 ms) ≈ a(1.0 ms) ≈ = 1.0× 10 cm s 1.3 ms a ( 1.5 ms) = 0because the slope is zero. 2 d) h = area under v– t graph h(0.5 ms) ≈ h(1.0 ms) ≈ A A Triangle Triangle 1 ⎛ cm ⎞ = (0.5 ms) ⎜33 ⎟ = 8.3× 10 2 ⎝ s ⎠ = 1 (1.0 ms)(100 2 −3 cm s) = 5.0× 10 cm −2 cm 1 ⎛ cm ⎞ h ( 1.5 ms) ≈ ATriangle + ARectangle = (1.3 ms) ⎜133 ⎟ + (0.2 ms)(1.33) 2 ⎝ s ⎠ = 0.11cm
2.53: a) The change in speed is the area under the a x versus t curve between vertical lines at t = 2.5s and t = 7.5s. This area is 2 2 1 (4.00 cm s + 8.00cm s )(7.5 s − 2. 5 s) = 30.0 cm s 2 This acceleration is positive so the change in velocity is positive. b) Slope of v x versus t is positive and increasing with t. 2.54: a) To average 4 mi hr , the total time for the twenty-mile ride must be five hours, so the second ten miles must be covered in 3.75 hours, for an average of 2.7 mi hr. b) To average 12 mi hr , the second ten miles must be covered in 25 minutes and the average speed must be 24 mi hr. c) After the first hour, only ten of the twenty miles have been covered, and 16 mi hr is not possible as the average speed.
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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
Page 15 and 16:
1.48: a) ˆ ˆ ˆ 2 2 2 i + j + k =
Page 17 and 18: 1.53: Use of the right-hand rule to
Page 19 and 20: 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: 2.51: a) From Eqs. (2.17) and (2.18
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 and 116: 3.52: a) Setting y = −h in Eq. (3
Page 117 and 118: 3.58: Equation 3.27 relates the ver
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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
Page 131 and 132:
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
Page 169 and 170:
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
Page 187 and 188:
5.85: Denote the magnitude of the a
Page 189 and 190:
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
Page 193 and 194:
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
Page 199 and 200:
5.109: a) For the same rotation rat
Page 201 and 202:
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
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 ⎛
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6.18: As the example explains, the
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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
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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
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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
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)
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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
Page 255 and 256:
7.49: a) K 1 + U1 + Wother = K2 + U
Page 257 and 258:
7.54: To be at equilibrium at the b
Page 259 and 260:
7.57: The two design conditions are
Page 261 and 262:
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
Page 273 and 274:
Capítulo 8
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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
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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
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8.69: (This problem involves solvin
Page 295 and 296:
8.72: a) The stuntman’s speed bef
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8.76: Just after the collision: ∑
Page 299 and 300:
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
Page 307 and 308:
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
Page 313 and 314:
8.107: a) For t < 0 the rocket is a
Page 315 and 316:
8.111: a) The tension in the rope a
Page 317 and 318:
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
Page 323 and 324:
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. (
Page 329 and 330:
2π 2 2 9.49: a) ω = T , so Eq. (9
Page 331 and 332:
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
Page 339 and 340:
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
Page 345 and 346:
9.94: a) From the parallel-axis the
Page 347 and 348:
9.99: a) Following the hint, the mo
Page 349 and 350:
Capítulo 10
Page 351 and 352:
10.6: (a) τ A = (50 N)(sin 60° )(
Page 353 and 354:
1 2 2 = m 2 8.25 kg 0.0750 m = 0.02
Page 355 and 356:
10.23: n = mg cos α mg sin θ −
Page 357 and 358:
10.26: a) The angular speed of the
Page 359 and 360:
10.32: I = mL 1 2 1 2 2 = ( 117 kg)
Page 361 and 362:
10.40: The skater’s initial momen
Page 363 and 364:
10.46: (a) Conservation of angular
Page 365 and 366:
2 2π rad 10.51: a) Solving Eq. (10
Page 367 and 368:
2 2θ 2θ 2θ I 10.57: t = = = . α
Page 369 and 370:
10.63: The net torque on the pulley
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2 10.67: For the disk, K = (3 4) Mv
Page 373 and 374:
10.72: (a) Στ = Iα and a = Rα T
Page 375 and 376:
10.75: a) Use conservation of energ
Page 377 and 378:
10.79: a) v = 10K 7m = ( 10)( 0.800
Page 379 and 380:
10.84: (a) Στ = Iα 1 1 2 mg cos
Page 381 and 382:
10.90: Angular momentum is conserve
Page 383 and 384:
10.95: a) The initial angular momen
Page 385 and 386:
10.100: a) The distance from the ce
Page 387 and 388:
10.102: Denoting the upward forces
Page 389 and 390:
11.1: Take the origin to be at the
Page 391 and 392:
11.12: a) b) x = 6 .25 m when FA =
Page 393 and 394:
11.18: a) Denote the length of the
Page 395 and 396:
11.30: a) The volume would increase
Page 397 and 398:
11.43: For the airplane to remain i
Page 399 and 400:
11.49: The horizontal component of
Page 401 and 402:
11.56: (a) and (b) Lower rod: Στ
Page 403 and 404:
11.59: a) ∑τ = 0, axis at lower
Page 405 and 406:
−4 −4 2 11.64: a) ∆w = −σ
Page 407 and 408:
11.68: (a) ΣτElbow = 0 F (3.80 cm
Page 409 and 410:
11.71: a) Consider the forces on th
Page 411 and 412:
11.74: a) The center of gravity of
Page 413 and 414:
11.79: a) Take torques about the po
Page 415 and 416:
11.87: Use subscripts 1 to denote t
Page 417 and 418:
11.93: a) Taking torques about the
Page 419 and 420:
11.95: Assume that the center of gr
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