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84 Pages

### Chapter 20 Problems

Course: PEP 111, Fall 2007
School: Stevens
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Word Count: 7129

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Model: 20.1. This is a wave traveling at constant speed. The pulse moves 1 m to the right every second. Visualize: Please refer to Figure Ex20.1. The snapshot graph shows the wave at all points on the x-axis at t = 0 s. You can see that nothing is happening at x = 6 m at this instant of time because the wave has not yet reached this point. The leading edge of the wave is still 1 m away from x = 6 m. Because the...

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Stevens - PEP - 111
21.1. Model: The principle of superposition comes into play whenever the waves overlap.Visualize:The graph at t = 1 s differs from the graph at t = 0 s in that the left wave has moved to the right by 1 m and the right wave has moved to the left by
Stevens - PEP - 111
22.1. Visualize: Please refer to Figure Ex22.1.Solve: (a)(b) The initial light pattern is a double-slit interference pattern. It is centered behind the midpoint of the slits. The slight decrease in intensity going outward from the middle indicates
Stevens - PEP - 111
23.1. Model: Light rays travel in straight lines.Solve: (a) The time ist=x 1.0 m = = 3.33 10 -9 s = 3.33 ns c 3 10 8 m / s(b) The refractive indices for water, glass, and zircon are 1.33, 1.50, and 1.96, respectively. In a time of 3.33 ns, l
Stevens - PEP - 111
24.1. Model: Balmer's formula predicts a series of spectral lines in the hydrogen spectrum.Solve: Substituting into the formula for the Balmer series,=91.18 nm 91.18 nm = = 410.3 nm 1 1 1 1 - 2 - 2 2 22 n 2 6where n = 3, 4, 5, 6, . and wher
Stevens - PEP - 111
ELECTROMAGNETIC AND WAVES FIELDSw.1. Model: The net magnetic flux over a closed surface is zero. Visualize: Please refer to Ex34.1. Solve: Because we can't enclose a &quot;net pole&quot; within a surface, Q, = f B . d i = 0 . Since the magnetic field isunif
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Stevens - PEP - 111
14.1. Solve: The frequency generated by a guitar string is 440 Hz. The period is the inverse of the frequency, henceT= 1 1 = = 2.27 10 -3 s = 2.27 ms f 440 Hz14.2. Solve: Your pulse or heart beat is 75 beats per minute. The frequency of your hear
Stevens - PEP - 111
1.1.Solve:1.2.Solve:Solve: (a) The basic idea of the particle model is that we will treat an object as if all its mass is concentrated into a single point. The size and shape of the object will not be considered. This is a reasonable approxim
Stevens - PEP - 111
2.1.Solve:Model: The car is represented by the particle model as a dot. (a) Time t (s) Position x (m) 0 1200 1 975 2 825 3 750 4 700 5 650 6 600 7 500 8 300 9 0(b)2.2. Solve:Diagram (a) (b) (c)Position Negative Negative PositiveVelocity
Stevens - PEP - 111
3.1. Solve: (a) If one component of the vector is zero, then the other component must not be zero (unless the whole vector is zero). Thus the magnitude of the vector will be the value of the other component. For example, if Ax = 0 m and Ay = 5 m, the
Stevens - PEP - 111
4.1. Solve: A force is basically a push or a pull on an object. There are five basic characteristics of forces. (i) A force has an agent that is the direct and immediate source of the push or pull. (ii) Most forces are contact forces that occur at a
Stevens - PEP - 111
5.1.Model: We can assume that the ring is a single massless particle in static equilibrium. Visualize:Solve:Written in component form, Newton's first law is( Fnet ) x = Fx = T1x + T2 x + T3 x = 0 NT1 x = - T1T1y = 0 N Using Newton's first l
Stevens - PEP - 111
6.1. Model: We will assume motion under constant-acceleration kinematics in a plane.Visualize:Instead of working with the components of position, velocity, and acceleration in the x and y directions, we will use the kinematic equations in vector f
Stevens - PEP - 111
7.1. Solve: (a) From t = 0 s to t = 1 s the particle rotates clockwise from the angular position +4 rad to -2 rad. Therefore, = -2 - ( +4 ) = -6 rad in one sec, or = -6 rad s . From t = 1 s to t = 2 s, = 0 rad/s. From t = 2 s to t = 4 s the partic
Stevens - PEP - 111
8.1. Visualize:Solve: Figure (i) shows a weightlifter (WL) holding a heavy barbell (BB) across his shoulders. He is standing on a rough surface (S) that is a part of the earth (E). We distinguish between the surface (S), which exerts a contact forc
Stevens - PEP - 111
Solve: (a) The momentum p = mv = (1500 kg)(10 m /s) = 1.5 10 4 kg m /s . (b) The momentum p = mv = (0.2 kg)( 40 m /s) = 8.0 kg m /s .9.1. Model: Model the car and the baseball as particles.9.2. Model: Model the bicycle and its rider as a particl
Stevens - PEP - 111
10.1. Model: We will use the particle model for the bullet (B) and the bowling ball (BB).Visualize:Solve:For the bullet,KB =For the bowling ball,1 1 2 mB vB = (0.01 kg)(500 m /s) 2 = 1250 J 2 2 1 1 2 mBB vBB = (10 kg)(10 m / s) 2 = 500 J 2
Stevens - PEP - 111
11.1. Visualize:r Please refer to Figure Ex11.1. rSolve: (b) (c)(a) A B = AB cos = ( 4)(5)cos 40 = 15.3. r r C D = CD cos = (2)( 4)cos120 = -4.0. r r E F = EF cos = (3)( 4)cos 90 = 0.11.2. Visualize:r Please refer to Figure Ex11.2. rSolve
Stevens - PEP - 111
12.1.Solve: (b)Model: Model the sun (s), the earth (e), and the moon (m) as spherical. (a)Fs on e =Gms me (6.67 10 -11 N m 2 / kg 2 )(1.99 10 30 kg)(5.98 10 24 kg) = 3.53 10 22 N = (1.50 1011 m ) 2 rs2 e -Fm on e =GMm Me (6.67 10 -1
Stevens - PEP - 111
13.1. Model: The crankshaft is a rotating rigid body.Solve: The crankshaft at t = 0 s has an angular velocity of 250 rad/s. It gradually slows down to 50 rad/s in 2 s, maintains a constant angular velocity for 2 s until t = 4 s, and then speeds up
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