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200 Pages

### Chapter 12

Course: PHYS 221, Spring 2012
School: Clemson
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Word Count: 11793

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A 12.1. Model: 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 13.2 m/s 26 mph for the hands is a little high, but reasonable. 12.2. Solve: Model: Assume constant angular acceleration. 2 rad min (a) The final...

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Clemson - PHYS - 221
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
Clemson - PHYS - 221
14.1. Solve: The frequency generated by a guitar string is 440 Hz. The period is the inverse of the frequency,henceT=1 1 = = 2.27 103 s = 2.27 ms f 440 Hz14.2. Model: The air-track glider oscillating on a spring is in simple harmonic motion.Solve: Th
Clemson - PHYS - 221
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 m3The 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
Clemson - PHYS - 221
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 samemass its volume must beVwater =Assess:mwater water=22,600 kg = 22.6 m3 1000 kg m3Since
Clemson - PHYS - 221
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 isW = p dV = ( area under the pV curve )
Clemson - PHYS - 221
18.1. Solve: We can use the ideal-gas 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
Clemson - PHYS - 221
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 isQC = QH Wout = 250 J
Clemson - PHYS - 221
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 thetension is doubled will be vstring = 2TS = 2vstring = 2 ( 200 m/s ) = 283 m/s20.2.
Clemson - PHYS - 221
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
Clemson - PHYS - 221
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 that22.1. Model: Two
Clemson - PHYS - 221
23.1. Model: Light rays travel in straight lines.Solve: (a) The time ist= 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
Clemson - PHYS - 221
24.1. Model: Each lens is a thin lens. The image of the first lens is the object for the second lens.Visualize: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
Clemson - PHYS - 221
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 =
Clemson - PHYS - 221
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
Clemson - PHYS - 221
27.1.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
Clemson - PHYS - 221
28.1. Visualize: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
Clemson - PHYS - 221
29.1. Model: The mechanical energy of the proton is conserved. A parallel-plate capacitor has a uniformelectric 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
Clemson - PHYS - 221
30.1. Solve: The potential difference V between two points in space isV = V ( xf ) V ( xi ) = Ex dxxi xfwhere 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 ) =
Clemson - PHYS - 221
31.1. Solve: The wires cross-sectional area is A = r 2 = (1.0 103 m ) = 3.1415 106 m 2 , and the electron2current through this wire is i = 31.3, the drift velocity isvd =Ne = 2.0 1019 s 1 . Using Table 31.1 for the electron density of iron and Equatio
Clemson - PHYS - 221
32.1. Solve: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
Clemson - PHYS - 221
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
Clemson - PHYS - 221
34.1.Visualize: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 havev=Assess:E 1.0 V = = 2.0 104 m/s lB (1.0 m ) ( 5.
Clemson - PHYS - 221
35.1. Model: Apply the Galilean transformation of velocity.Solve: (a) In the laboratory frame S, the speed of the proton isv=(1.41106m/s ) + (1.41 106 m/s ) = 2.0 106 m/s2 2The angle the velocity vector makes with the positive y-axis is = tan 1
Clemson - PHYS - 221
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 ist = 180 + 30 = 210 (b) The instantaneous value of the emf is rad180= 3.665
Clemson - PHYS - 221
37.1. Model: S and S are inertial frames that overlap at t = 0. Frame S moves with a speed v = 5.0 m/salong x-direction relative to frame S. Visualize: theThe figure shows a pictorial representation of the S and S frames at t = 1.0 s and 5.0 s. Solve: F
Clemson - PHYS - 221
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 isN 10 nA 1.0 108 C/s = = = 6.25 1010 s 1 6.3 1010 s 1 t e 1.60 1019 C
Clemson - PHYS - 221
39.1. Solve: A steady photoelectric current of 10 A is indicated in the graph. The number of electrons persecond is10 A = 10Cs= 1.0 10 5C 1 electron = 6.25 1013 electrons/s s 1.6 10 19 C39.2. Model: Light of frequency f consists of discrete quanta,
Clemson - PHYS - 221
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
Clemson - PHYS - 221
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
Clemson - PHYS - 221
42.1.Solve: (a) A 4p state corresponds to n = 4 and l = 1. From Equation 42.3, the orbital angularmomentum 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
Clemson - PHYS - 221
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
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