fisica1-youn-e-freedman-exercicios-resolvidos

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3.62: a) Using the same algebra as in Problem 3.58(a), v 0 = 138 . m/s . b) Again, the algebra is the same as that used in Problem 3.58; v = 8.4 m/s , at an angle of 9 .1° , this time above the horizontal. c) The graph of v x (t) is a horizontal line. A graph of y (t) vs. x (t) shows the trajectory of Mary Belle as viewed from the side: d) In this situation it’s convenient to use Eq. (3.27), which becomes − 1 2 y = ( 1.327) x − (0.071115 m ) x . Use of the quadratic formula gives x = 23.8 m . 3.63: a) The algebra is the same as that for Problem 3.58, 2 2 gx v0 = . 2 2cos α0( x tanα0 − y) In this case, the value for y is − 15.0 m , the change in height. Substitution of numerical values gives 17.8 m/s. b) 28.4 m from the near bank (i.e., in the water!).
3.64: Combining equations 3.25, 3.22 and 3.23 gives 2 2 2 v = v cos α + ( v sinα − gt) 0 = v = v 2 0 2 0 2 0 0 − 2g( v − 2gy, 0 0 0 2 2 = v (sin α + cos α ) − 2v sinα gt + ( gt) 0 0 1 sinα 0t − gt 2 where Eq. (3.21) has been used to eliminate t in favor of y. This result, which will be seen in the chapter dealing with conservation of energy (Chapter 7), is valid for any y, positive, negative or zero, as long asv 2 > 0 . For the case of a rock thrown from the roof of a building of height h, the speed at the ground is found by substituting y = −h into the above expression, yielding v = 2 v + 0 2gh , which is independent of α 0 . 2 ) 0 2 0 2
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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
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: 3.60: a) This may be done by a dire
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
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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
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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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