fisica1-youn-e-freedman-exercicios-resolvidos

Recommendations

Info

9.15: From Eq. (9.11), ω 2 θ − θ0 0 z = = t αzt 60.0rad (2.25rad s )(4.00s) − = − 2 4.00s 2 10.5rad s. 2 0z 0 z t 6.00 s s av = (70rad s)(6.00s) = 420rad. z ωz 140 rad s 9.16: From Eq. (9.7), with ω = , α = = = 23.33rad . easily found from θ = ω t − The angle is most 9.17: From Eq. (9.12), with ωz = 0, the number of revolutions is proportional to the square of the initial angular velocity, so tripling the initial angular velocity increases the number of revolutions by 9, to 9.0 rev.
9.18: The following table gives the revolutions and the angle θ through which the wheel has rotated for each instant in time and each of the three situations: t (a) rev's θ (b) rev's θ (c) rev's θ The θ and a) 0.05 0.50 180 0.03 11.3 0.44 158 0.10 1.00 360 0.13 45 0.75 270 0.15 1.50 540 0.28 101 0.94 338 0.20 2.00 720 0.50 180 1.00 360 –––––––––––––––––––––––––––––––––––––– ωz graphs are as follows: b) c)
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 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 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: 2 9.10: a) ω = ω + α t = 1.50 ra
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
show all

fisica1-youn-e-freedman-exercicios-resolvidos

You also want an ePaper? Increase the reach of your titles

Delete template?

Save as template?