21.1. Model: The principle of superposition comes into play whenever the waves overlap.
Visualize:
The graph at t = 1.0 s differs from the graph at t = 0.0 s in that the left wave has moved to the right by 1.0 m and the right wave has moved to the left by
20.1.
Model: The wave is a traveling wave on a stretched string. Solve: The wave speed on a stretched string with linear density is vstring = TS / . The wave speed if the
tension is doubled will be
vstring = 2TS = 2vstring = 2 ( 200 m/s ) = 283 m/s
20.2.
19.1.
Solve: (a) The engine has a thermal efficiency of = 40% = 0.40 and a work output of 100 J per cycle. The heat input is calculated as follows:
=
Wout 100 J 0.40 = QH = 250 J QH QH
(b) Because Wout = QH QC , the heat exhausted is
QC = QH Wout = 250 J
18.1. Solve: We can use the idealgas law in the form pV = NkBT to determine the Loschmidt number
(N/V):
(1.013 105 Pa ) = 2.69 1025 m3 N p = = V kBT (1.38 1023 J K ) ( 273 K )
18.2. Solve: The volume of the nitrogen gas is 1.0 m3 and its temperature is 2
17.1. Model: Assume the gas is ideal. The work done on a gas is the negative of the area under the pV curve.
Visualize: The gas is compressing, so we expect the work to be positive. Solve: The work done on the gas is
W = p dV = ( area under the pV curve )
16.1. Model: Recall the density of water is 1000 kg/m3. Solve: The mass of lead mPb = PbVPb = (11,300 kg m3 ) ( 2.0 m3 ) = 22,600 kg . For water to have the same
mass its volume must be
Vwater =
Assess:
mwater
water
=
22,600 kg = 22.6 m3 1000 kg m3
Since
15.1. Solve: The density of the liquid is
=
Assess:
0.240 kg m 0.240 kg = = = 960 kg m3 V 250 mL 250 103 103 m3
The liquids density is near that of water (1000 kg/m3 ) and is a reasonable number.
15.2. Solve: The volume of the helium gas in container A is
14.1. Solve: The frequency generated by a guitar string is 440 Hz. The period is the inverse of the frequency,
hence
T=
1 1 = = 2.27 103 s = 2.27 ms f 440 Hz
14.2. Model: The airtrack glider oscillating on a spring is in simple harmonic motion.
Solve: Th
13.1.
Model: Model the sun (s) and the earth (e) as spherical masses. Due to the large difference between your size and mass and that of either the sun or the earth, a human body can be treated as a particle. GM s M y GM e M y and Fe on you = Solve: Fs on
12.1.
Model: A spinning skater, whose arms are outstretched, is a rigid rotating body.
Visualize:
Solve: The speed v = r , where r = 140 cm/2 = 0.70 m. Also, 180 rpm = (180)2 /60 rad/s = 6 rad/s. Thus, v = (0.70 m)(6 rad/s) = 13.2 m/s. Assess: A speed of
11.1. Visualize: Please refer to Figure EX11.1. Solve: (a) A B = AB cos = (4)(5)cos 40 = 15.3.
(b) (c)
C D = CD cos = (2)(4)cos120 = 4.0. E F = EF cos = (3)(4)cos90 = 0.
11.2. Visualize: Please refer to Figure EX11.2. Solve: (a) A B = AB cos = (3)(4)cos11
10.1. Model: We will use the particle model for the bullet (B) and the running student (S).
Visualize:
Solve:
For the bullet,
1 1 2 K B = mBvB = (0.010 kg)(500 m/s) 2 = 1250 J 2 2 For the running student, 1 1 2 KS = mSvS = (75 kg)(5.5 m/s) 2 = 206 J 2 2 T
9.1. Model: Model the car and the baseball as particles.
Solve:
(a) The momentum p = mv = (1500 kg ) (10 m/s ) = 1.5 104 kg m/s.
(b) The momentum p = mv = ( 0.2 kg )( 40 m/s ) = 8.0 kg m/s.
9.2. Model: Model the bicycle and its rider as a particle. Also m
8.1.
Model: The model rocket and the target will be treated as particles. The kinematics equations in two dimensions apply. Visualize:
Solve:
For the rocket, Newtons second law along the ydirection is
( Fnet ) y = FR mg = maR
aR = 1 1 15 N ( 0.8 kg ) (
7.1.
Visualize:
Solve: (a) The weight lifter is holding the barbell in dynamic equilibrium as he stands up, so the net force on the barbell and on the weight lifter must be zero. The barbells have an upward contact force from the weight lifter and the gra
6.1.
Model: Visualize:
We can assume that the ring is a single massless particle in static equilibrium.
Solve:
Written in component form, Newtons first law is
( Fnet ) x = Fx = T1x + T2 x + T3 x = 0 N ( Fnet ) y = Fy = T1 y + T2 y + T3 y = 0 N
Evaluating
5.1. Visualize:
Assess: walls.
Note that the climber does not touch the sides of the crevasse so there are no forces from the crevasse
5.2. Visualize:
5.3. Visualize:
5.4. Model: Assume friction is negligible compared to other forces.
Visualize:
5.5. Visu
4.1.
Solve:
(a)
(b) A race car slows from an initial speed of 100 mph to 50 mph in order to negotiate a tight turn. After making the 90 turn the car accelerates back up to 100 mph in the same time it took to slow down.
4.2.
Solve:
(a)
(b) A car drives up
3.1.
Visualize:
Solve: (a) To find A + B , we place the tail of vector B on the tip of vector A and connect the tail of vector A with the tip of vector B. (b) Since A B = A + ( B) , we place the tail of the vector ( B ) on the tip of vector A and then con
2.1. Model: We will consider the car to be a particle that occupies a single point in space.
Visualize:
Solve:
Since the velocity is constant, we have xf = xi + vx t. Using the above values, we get
x1 = 0 m + (10 m/s)(45 s) = 450 m
Assess: 10 m/s 22 mph a
1.1.
Solve:
1.2.
Solve:
1.3.
Solve:
1.4. 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
20.1 SOLUTIONS
1339
CHAPTER TWENTY
Solutions for Section 20.1
Exercises
1. The first vector field appears to be diverging more at the origin, since both fields are zero at the origin and the vectors near the origin are larger in field (I) than they are in
19.1 SOLUTIONS
1307
CHAPTER NINETEEN
Solutions for Section 19.1
Exercises
1. (a) The flux is positive, since F points in direction of positive xaxis, the same direction as the normal vector. (b) The flux is negative, since below the xyplane F points tow
18.1 SOLUTIONS
1257
CHAPTER EIGHTEEN
Solutions for Section 18.1
Exercises
1. Positive, because the vectors are longer on the portion of the path that goes in the same direction as the vector field. 2. Negative because the vector field points in the opposi
17.1 SOLUTIONS
1197
CHAPTER SEVENTEEN
Solutions for Section 17.1
Exercises
1. One possible parameterization is x = 3 + t, 2. One possible parameterization is x = 1 + 3t, 3. One possible parameterization is x = 3 + 2t, 4. One possible parameterization is
16.1 SOLUTIONS
1105
CHAPTER SIXTEEN
Solutions for Section 16.1
Exercises
1. Mark the values of the function on the plane, as shown in Figure 16.1, so that you can guess respectively at the smallest and largest values the function takes on each small recta
15.1 SOLUTIONS
1039
CHAPTER FIFTEEN
Solutions for Section 15.1
Exercises
1. The point A is not a critical point and the contour lines look like parallel lines. The point B is a critical point and is a local maximum; the point C is a saddle point. 2. To fi
14.1 SOLUTIONS
951
CHAPTER FOURTEEN
Solutions for Section 14.1
Exercises
1. If h is small, then With h = 0.01, we find fx (3, 2) f (3 + h, 2)  f (3, 2) . h
3.012 (2+1)
f (3.01, 2)  f (3, 2) = fx (3, 2) 0.01 With h = 0.0001, we get fx (3, 2)
0.01
2 2

12.1 SOLUTIONS
861
CHAPTER TWELVE
Solutions for Section 12.1
Exercises
1. The distance of a point P = (x, y, z) from the yzplane is x, from the xzplane is y, and from the xyplane is z. So, B is closest to the yzplane, since it has the smallest x