fisica1-youn-e-freedman-exercicios-resolvidos

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1.94: a) The maximum and minimum areas are (L + l) (W + w) = LW + lW + Lw, (L – l) (W – w) = LW – lW – Lw, where the common terms wl have been omitted. The area and its uncertainty are then WL ± (lW + Lw), so the uncertainty in the area is a = lW + Lw. b) The fractional uncertainty in the area is a lW + Wl l w = = + A WL L W , the sum of the fractional uncertainties in the length and width. c) The similar calculation to find the uncertainty v in the volume will involve neglecting the terms lwH, lWh and Lwh as well as lwh; the uncertainty in the volume is v = lWH + LwH + LWh, and the fractional uncertainty in the volume is v V = lWH + LwH + LW H LWh = l L + w W + h H , the sum of the fractional uncertainties in the length, width and height. 1.95: The receiver's position is ( + 1.0 + 9.0 − 6.0 + 12.0) i ˆ + ( − 5.0 + 11.0 + 4.0 + 18.0) ˆj = ( 16.0) iˆ + ( 28.0) ˆj . The vector from the quarterback to the receiver is the receiver's position minus the 16 .0 iˆ + 35.0 ˆ , a vector with magnitude quarterback's position, or ( ) ( )j 2 2 .0 o ( 16.0) + ( 35.0) = 38.5, given as being in yards. The angle is ( ) 24.6 right of downfield. 16 arctan = to the 35. 0
1.96: a) b) i) In AU, ii) In AU, iii) In AU, 2 2 ( 0.3182) + (0.9329) = 0.9857. 2 2 2 + ( −.4423) + ( .0414) = 1.3820 ( 1.3087) − 2 2 2 − (1.3087)) + (0.9329 − ( −.4423)) (0.0414) = 1.695. ( 0.3182 + c) The angle between the directions from the Earth to the Sun and to Mars is obtained from the dot product. Combining Equations (1-18) and (1.21), φ = ⎛ ( −0.3182)(1.3087 − 0.3182) + ( −0.9329)( −0.4423 − 0.9329) + arccos ⎜ ⎝ (0.9857)(1.695) d) Mars could not have been visible at midnight, because the Sun-Mars angle is less than 90 o .
Page 1 and 2: Física I - Sears, Zemansky, Young
Page 3 and 4: 5 1.1: 1mi × ( 5280ft mi) × ( 12
Page 5 and 6: 1.15: a) If a meter stick can measu
Page 7 and 8: 8 11 1.28: The moon is about 4× 10
Page 9 and 10: 1.34: r o o 1.35: A ; A x = ( 12.0
Page 11 and 12: 1.39: Using components as a check f
Page 13 and 14: 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: 1.91: A r and C r are perpendicular
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 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
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
Page 167 and 168:
5.35: First, determine the accelera
Page 169 and 170:
5.40: Differentiating Eq. (5.10) wi
Page 171 and 172:
5.49: a) Setting arad = g in Eq. (5
Page 173 and 174:
5.56: The tension in the lower chai
Page 175 and 176:
5.63: (Denote F r by F.) a) The for
Page 177 and 178:
5.67: Consider the forces on the pe
Page 179 and 180:
5.70: It’s interesting to look at
Page 181 and 182:
5.72: The key idea in solving this
Page 183 and 184:
5.77: See Exercise 5.40. a) The max
Page 185 and 186:
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
Page 191 and 192:
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
Page 195 and 196:
mg mg 5.103: Without buoyancy, kv t
Page 197 and 198:
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
Page 203 and 204:
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 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
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
Page 303 and 304:
8.86: (a) For momentum to be conser
Page 305 and 306:
8.90: a) For the x - and y - direct
Page 307 and 308:
8.96: The trick here is to realize
Page 309 and 310:
8.98: The two fragments are 3.00 kg
Page 311 and 312:
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°
Page 319 and 320:
2 9.10: a) ω = ω + α t = 1.50 ra
Page 321 and 322:
9.18: The following table gives the
Page 323 and 324:
9.24: From a rad = ω 2 r, 4 which
Page 325 and 326:
vr 5.00 m s 9.33: The angular veloc
Page 327 and 328:
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)
Page 333 and 334:
dθ 2 2 3 2 9.65: a) ω z = = 2γt
Page 335 and 336:
9.70: a) The angular acceleration w
Page 337 and 338:
9.75: I approximate my body as a ve
Page 339 and 340:
9.81: a) Consider a small strip of
Page 341 and 342:
9.84: Taking the zero of gravitatio
Page 343 and 344:
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
Page 371 and 372:
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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