fisica1-youn-e-freedman-exercicios-resolvidos

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8.17: The change in velocity is the negative of the change in Gretzky’s momentum, divided by the defender’s mass, or v m A B2 = vB 1 − A2 A mB = −5.00 m s − = 4.66 m s. ( v − v ) 1 756 N (1.50m s 900 N −13.0 m s) Positive velocities are in Gretzky’s original direction of motion, so the defender has changed direction. b) 1 2 2 1 2 K ( ) ( 2 2 − K1 = mA v A2 − v A1 + mB vB2 − vB 1) 2 2 2 2 1 ⎡(756 N)((1.50 m/s) − (13.0 m/s) ) = 2 ⎢ 2 2(9.80 m/s ) ⎣ + (900 N)((4.66 m/s) − ( −5.00 m/s) = −6.58 kJ. 2 ⎤ ⎥ ) ⎦ 8.18: Take the direction of the bullet’s motion to be the positive direction. The total momentum of the bullet, rifle, and gas must be zero, so and gas ( 0.00720 kg)(601m/s −1.85 m/s) + (2.80 kg)( −1.85 m/s) + p gas = 0, p = 0.866 kg ⋅ m s. Note that the speed of the bullet is found by subtracting the speed of the rifle from the speed of the bullet relative to the rifle. v = 3.00 kg 8.19: a) See Exercise 8.21; ( )(0.800 m s) 3.60 m s. A 1.00 kg 2 2 b) (1 2) (1.00 kg) (3.60 m/s) + (1/ 2)(3.00 kg)(1.200 m/s) = 8.64 J. = 8.20: In the absence of friction, the horizontal component of the hat-plus-adversary system is conserved, and the recoil speed is (4.50 kg)(22.0 m s) cos 36.9° = 0.66 m s. (120 kg)
8.21: a) Taking v A and v B to be magnitudes, conservation of momentum is mA expressed as m AvA = mBvB , so v B = v A. m 2 2 K A (1/ 2) mAv A mAv A mB b) = = = . 2 2 K (1/ 2) m v m (( m / m ) v ) m B (This result may be obtained using the result of Exercise 8.3.) B B B B A B A A 8.22: 214 Po decay : 214 4 Po → α+ 210 X Setv v α α : KE = = α 1 = mαv 2 2KE m α α 2 α 2(1.23× 10 − 6.65× 10 27 J) kg −12 = 1.92× 10 7 m/s Momentum conservation: 0 = m v v x α α mαv = m x α − m v −27 7 (6.65× 10 kg)(1.92× 10 m/s) = −27 (210)(1.67 × 10 kg) = 3.65× 10 mαvα = 210m 5 x x m/s p
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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
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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.
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: 8.14: The impluse imparted to the p
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
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. (
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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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