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 ) (
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
Solve: (a) A 4p state corresponds to n = 4 and l = 1. From Equation 42.3, the orbital angular
momentum is L = 1(1 + 1) = 2 . (b) In the case of a 5f state, n = 5 and l = 3. So, L = 3 ( 3 + 1) = 12 .
42.2. Solve: (a) Excluding spin, a state is descri
41.1. Model: Model the electron as a particle in a rigid one-dimensional box of length L.
Solve: Absorption occurs from the ground state n = 1. Its reasonable to assume that the transition is from n = 1 to n = 2. The energy levels of an electron in a rigi
40.1. Model: The sum of the probabilities of all possible outcomes must equal 1 (100%).
Solve: The sum of the probabilities is PA + PB + PC + PD = 1. Hence, 0.40 + 0.30 + PC + PD = 1 PC + PD = 0.30 Because PC = 2PD, 2PD + PD = 0.30. This means PD = 0.10 a
39.1. Solve: A steady photoelectric current of 10 A is indicated in the graph. The number of electrons per
10 A = 10
= 1.0 10 5
C 1 electron = 6.25 1013 electrons/s s 1.6 10 19 C
39.2. Model: Light of frequency f consists of discrete quanta,
38.1. Model: Current is defined as the rate at which charge flows across an area of cross section.
Solve: Since the current is Q / t and Q = N / e , the number of electrons per second is
N 10 nA 1.0 108 C/s = = = 6.25 1010 s 1 6.3 1010 s 1 t e 1.60 1019 C
37.1. Model: S and S are inertial frames that overlap at t = 0. Frame S moves with a speed v = 5.0 m/s
along x-direction relative to frame S. Visualize: the
The figure shows a pictorial representation of the S and S frames at t = 1.0 s and 5.0 s. Solve: F
36.1. Model: A phasor is a vector that rotates counterclockwise around the origin at angular frequency .
Solve: (a) Referring to the phasor in Figure EX36.1, the phase angle is
t = 180 + 30 = 210
(b) The instantaneous value of the emf is
35.1. Model: Apply the Galilean transformation of velocity.
Solve: (a) In the laboratory frame S, the speed of the proton is
m/s ) + (1.41 106 m/s ) = 2.0 106 m/s
The angle the velocity vector makes with the positive y-axis is
= tan 1
To develop a motional emf the magnetic field needs to be perpendicular to both, so lets say its direction is into the page. Solve: This is a straightforward use of Equation 34.3. We have
E 1.0 V = = 2.0 104 m/s lB (1.0 m ) ( 5.
33.1. Model: A magnetic field is caused by an electric current.
Visualize: Please refer to Figure EX33.1. Solve: The magnitude of the magnetic field at point 1 is 2.0 mT and its direction can be determined by using the right-hand rule. Grab the current ca
From the circuit in Figure EX32.1, we see that 50 and 100 resistors are connected in series across the battery. Another resistor of 75 is also connected across the battery.
32.2. Solve: In Figure EX32.2, the positive terminal of the battery i
31.1. Solve: The wires cross-sectional area is A = r 2 = (1.0 103 m ) = 3.1415 106 m 2 , and the electron
current through this wire is i = 31.3, the drift velocity is
Ne = 2.0 1019 s 1 . Using Table 31.1 for the electron density of iron and Equatio
30.1. Solve: The potential difference V between two points in space is
V = V ( xf ) V ( xi ) = Ex dx
where x is the position along a line from point i to point f. When the electric field is uniform,
V = Ex dx = Ex x = (1000 V/m )( 0.30 m 0.10 m ) =
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