fisica1-youn-e-freedman-exercicios-resolvidos

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2.36: a) The truck’s position as a function of time is given by x T = v T t, with v T being the truck’s constant speed, and the car’s position is given by x C = (1/2) a C t 2 . Equating the two expressions and dividing by a factor of t (this reflects the fact that the car and the truck are at the same place at t = 0) and solving for t yields 2vT 2(20.0 m s) t = = = 12.5 s a 2 C 3.20 m s and at this time x T = x C = 250 m. b) a C t = (3.20 m/s 2 )(12.5 s) = 40.0 m/s (See Exercise 2.37 for a discussion of why the car’s speed at this time is twice the truck’s speed.) c) d)
2.37: a) The car and the motorcycle have gone the same distance during the same time, so their average speeds are the same. The car's average speed is its constant speed v C , and for constant acceleration from rest, the motorcycle's speed is always twice its average, or 2v C . b) From the above, the motorcyle's speed will be v C after half the time needed to catch the car. For motion from rest with constant acceleration, the distance traveled is proportional to the square of the time, so for half the time one-fourth of the total distance has been covered, or d 4. 2.38: a) An initial height of 200 m gives a speed of 60 m s when rounded to one significant figure. This is approximately 200 km/hr or approximately 150 mi hr . (Different values of the approximate height will give different answers; the above may be interpreted as slightly better than order of magnitude answers.) b) Personal experience will vary, but speeds on the order of one or two meters per second are reasonable. c) Air resistance may certainly not be neglected. 2.39: a) From Eq. (2.13), with v = 0 and = −g , y a y v0 y = 2g( y − y0) = 2(9.80 m s )(0.440 m) = 2.94 m s, which is probably too precise for the speed of a flea; rounding down, the speed is about 2 .9 m s . b) The time the flea is rising is the above speed divided by g, and the total time is twice this; symbolically, 2 2g( y − y0 ) 2( y − y0 ) 2(0.440 m) t = 2 = 2 = 2 2 g g (9.80 m/s ) or about 0.60 s. = 0.599 s, 2.40: Using Eq. (2.13), with downward velocities and accelerations being positive, v = (0.8 m s ) 2 + 2(1.6 = 4.1 m s . 2 m s )(5.0 m) = 16.64 m 2 s 2 (keeping extra significant figures), so v y 2 y
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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
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: 2.35: a) b) From the graph (Fig. (2
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
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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.
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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
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6.104: From F = m a, Fx = max, Fy =
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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
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7.49: a) K 1 + U1 + Wother = K2 + U
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7.54: To be at equilibrium at the b
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7.57: The two design conditions are
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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
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Capítulo 8
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8.7: ∆p ∆t ( 0.0450 kg)( 25.0 m
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8.14: The impluse imparted to the p
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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
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8.31: Use conservation of the horiz
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8.37: Let +y be north and +x be sou
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8.43: a) In Eq. (8.24), let mA = m
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8.52: It turns out to be more conve
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8.63: The total momentum must be ze
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8.69: (This problem involves solvin
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8.72: a) The stuntman’s speed bef
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8.76: Just after the collision: ∑
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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
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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
Page 387 and 388:
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
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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