29.1. Model: The mechanical energy of the proton is conserved. A parallel-plate capacitor has a uniform
electric field. Visualize:
The figure shows the before-and-after pictorial representation. The proton has an initial speed vi = 0 m/s and a final speed
As discussed in Section 28.1, the symmetry of the electric field must match the symmetry of the charge distribution. In particular, the electric field of a cylindrically symmetric charge distribution cannot have a component parallel to th
Model: The electric field is that of the two charges placed on the y-axis. Visualize: Please refer to Figure EX27.1. We denote the upper charge by q1 and the lower charge by q2. Because both the charges are positive, their electric fields at P are d
26.1. Model: Use the charge model.
Solve: (a) In the process of charging by rubbing, electrons are removed from one material and transferred to the other because they are relatively free to move. Protons, on the other hand, are tightly bound in nuclei. So
25.1. Model: Balmers 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 11 1 1 2 2 2 22 6 2 n
where n = 3, 4, 5, 6, and where we have used n =
24.1. Model: Each lens is a thin lens. The image of the first lens is the object for the second lens.
The figure shows the two lenses and a ray-tracing diagram. The ray-tracing shows that the lens combination will produce a real, inverted image
23.1. Model: Light rays travel in straight lines.
Solve: (a) The time is
t= x 1.0 m = = 3.3 109 s = 3.3 ns c 3 108 m/s
(b) The refractive indices for water, glass, and cubic zirconia are 1.33, 1.50, and 1.96, respectively. In a time of 3.33 ns, light will
Visualize: The interference pattern looks like the photograph of Figure 22.3(b). It is symmetrical with the m = 2 fringes on both sides of and equally distant from the central maximum. Solve: The bright fringes occur at angles m such that
22.1. Model: Two
21.1. Model: The principle of superposition comes into play whenever the waves overlap.
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
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
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 ideal-gas law in the form pV = NkBT to determine the Loschmidt number
(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
22,600 kg = 22.6 m3 1000 kg m3
15.1. Solve: The density of the liquid is
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,
1 1 = = 2.27 103 s = 2.27 ms f 440 Hz
14.2. Model: The air-track glider oscillating on a spring is in simple harmonic motion.
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
Model: A spinning skater, whose arms are outstretched, is a rigid rotating body.
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.
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).
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.
(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
Model: The model rocket and the target will be treated as particles. The kinematics equations in two dimensions apply. Visualize:
For the rocket, Newtons second law along the y-direction is
( Fnet ) y = FR mg = maR
aR = 1 1 15 N ( 0.8 kg ) (
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
We can assume that the ring is a single massless particle in static equilibrium.
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
Note that the climber does not touch the sides of the crevasse so there are no forces from the crevasse
5.4. Model: Assume friction is negligible compared to other forces.
(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.
(b) A car drives up
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.
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
1.4. 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
43.1. Model: The nucleus is composed of Z protons and neutrons. Solve: (a) 3H has Z = 1 proton and 3 1 = 2 neutrons. (b) 40Ar has Z = 18 protons and 40 18 = 22 neutrons. (c) 40Ca has Z = 20 protons and 40 20 = 20 neutrons. (d) 239Pu has Z = 94 protons and