34. REASONING In each case the object is in equilibrium. According to Equation 4.9b, Fy = 0, the net force acting in the y (vertical) direction must be zero. The net force is composed of the weight of
50. REASONING AND SOLUTION The period of the moon's motion (approximately the length of a month) is given by
4 2 r 3 T= = GM E
4 2 ( 3.85 108 m ) ( 6.67 1011 N m 2 / kg 2 )( 5.98 1024 kg )
3
= 2.38 10
51. REASONING AND SOLUTION given by T = 2 r/v. The speed is
The sample makes one revolution in time T as so that v = 55.3 m/s
v2 = rac = (5.00 102 m)(6.25 103)(9.80 m/s2) The period is
T = 2 (5.00 102
52. REASONING AND SOLUTION The free body diagram for the plane is shown below to the left. The figure at the right shows the forces resolved into components parallel to and perpendicular to the line o
54. REASONING AND SOLUTION a. The centripetal force the normal force exerted on the rider by the wall .
is
provided
by
b. Newton's second law applied in the horizontal direction gives FN = mv2/r = (55
56. CONCEPT QUESTIONS a. The period for the second hand is the time it takes for it to go once around the circle or Tsecond = 60 s. b. The period for the minute hand is the time it takes for it to go
56. REASONING Since the wire beneath the limb is at rest, it is in equilibrium and the net force acting on it must be zero. Three forces comprise the net force, the 151-N force from the limb, the 447-
57 SSM REASONING There are four forces that act on the chandelier; they are the forces of tension T in each of the three wires, and the downward force of gravity mg. Under the influence of these force
59. REASONING The toboggan has a constant velocity, so it has no acceleration and is in equilibrium. Therefore, the forces acting on the toboggan must balance, that is, the net force acting on the tob
61. CONCEPT QUESTIONS a. Since the speed and mass are constant and the radius is fixed, the centripetal force is the same at each point on the circle. b. When the ball is at the three oclock position,
48. REASONING At first glance there seems to be very little information given. However, it is enough. In part a of the drawing the bucket is hanging stationary and, therefore, is in equilibrium. The f
48. REASONING The centripetal force is the name given to the net force pointing toward the center of the circular path. At the lowest point the net force consists of the tension in the arm pointing up
36. REASONING According to Equation 5.3, the magnitude Fc of the centripetal force that acts on each passenger is Fc = mv 2 / r , where m and v are the mass and speed of a passenger and r is the radiu
36.
REASONING AND SOLUTION a. The apparent weight of the person is given by Equation 4.6 as FN = mg + ma = (95.0 kg)(9.80 m/s2 + 1.80 m/s2) = 1.10 103 N b. c. FN = (95.0 kg)(9.80 m/s2) = 931 N FN = (9
37. REASONING AND SOLUTION The block will move only if the applied force is greater than the maximum static frictional force acting on the block. That is, if F > sFN = smg = (0.650)(45.0 N) = 29.2 N T
38. REASONING The normal force (magnitude FN) that the pilots seat exerts on him is part of the centripetal force that keeps him on the vertical circular path. However, there is another contribution t
42. REASONING AND SOLUTION The deceleration produced by the frictional force is
a= fk m = k mg m = k g
The speed of the automobile after 1.30 s have elapsed is given by Equation 2.4 as
v = v0 + at = v
44. REASONING AND SOLUTION The normal force exerted by the wall on each astronaut is the centripetal force needed to keep him in the circular path, i.e., Fc = mv2/r. Rearranging and letting Fc = (1/2)
45. REASONING AND SOLUTION Let s represent the length of the path of the pebble after it is released. From Conceptual Example 2, we know that the pebble will fly off tangentially. Therefore, the path
46. REASONING AND SOLUTION Newtons second law applied in the vertical and horizontal directions gives L cos 21.0 W = 0 L sin 21.0 R = 0 a. Equation (1) gives
21.0
(1) (2)
L=
W 53 800 N = = 57 600 N co
47.
SSM
REASONING AND SOLUTION Since the magnitude of the centripetal
2 acceleration is given by Equation 5.2, a C = v / r , we can solve for r and find that
r=
v2 ( 98.8 m / s ) 2 = = 332 m a C 3.00(
62. REASONING The weight of the part of the washcloth off the table is moff g. At the instant just before the washcloth begins to slide, this weight is supported by a force that has magnitude equal to
63. REASONING According to Newtons second law, the acceleration has the same direction as the net force and a magnitude given by a = F/m. SOLUTION Since the two forces are perpendicular, the magnitude
68. REASONING AND SOLUTION From Newton's second law and the equation: v = v0 + at, we have v v0 F = ma = m t a. When the skier accelerates from rest (v0 = 0 m/s) to a speed of 11 m/s in 8.0 s, the req
97. REASONING According to Equation 4.4, the weights of an object of mass m on the surfaces of planet A (mass = MA, radius = R ) and planet B (mass = MB , radius = R ) are WA = GM A m R
2
and
WB =
GM
102. REASONING Since the mountain climber is at rest, she is in equilibrium and the net force acting on her must be zero. Three forces comprise the net force, her weight, and the tension forces from t
104. REASONING AND SOLUTION The figure to the left below shows the forces that act on the sports car as it accelerates up the hill. The figure to the right below shows these forces resolved into compo
is
F =m v v0 t = (73 kg) (11 m/s) 0 m/s = 1.0 102 N 8.0 s
b. When the skier lets go of the tow rope and glides to a halt (v = 0 m/s) in 21 s, the net force acting on the skier is
F =m v v0 t = (73 kg)
Math 152: Calculus II
Rutgers University
Exam #2, Spring 2017
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Write here the numbers of
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CS 112Final Exam
Please answer all questions in the space provided. There are twelve problems on this
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COMPLEX NUMBERS
Real solutions x of the equation
x2 + 1 = 0
(1)
do not exist, because there is no real number x such that x2 = 1.
If we assume the existence of a new number i (obviously not a real one
Math 152: Calculus II
Rutgers University
Exam #1, Spring 2017
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