Test #2 Extra Credit
1. A charge of 0.060 C moved vertically in the field of 0.090 T that is oriented 45
from the vertical. What speed must the charge have such that the force acting on
it is 10 N?
a. 3,500 m/s
b. 2,600 m/s
c. 1,500 m/s
d. 7,600 m/s
ANSWE
Extra Credit 1
1. A piece of plastic has a net charge of +6.00 C. How many more protons than electrons does this piece
of plastic have?
a. 6.781013
b. 4.501013
c. 1.601013
d. 3.751013
ANSWER: D
2. How much work is required to move an electron between two
Extra Credit #4
210
1. What is the resulting nucleus when
Pb undergoes alpha decay?
82
234
a.
U
92
206
b.
Hg
80
222
c.
Fr
87
210
d.
Pb
82
ANSWER: B
2. What is the resulting nucleus when
222
86
222
b.
73
222
c.
87
254
d.
90
a.
222
Rn undergoes beta-minus d
Extra Credit #3
1. A mirror has a focal length -15 cm. If an object is placed 10 cm in front of the mirror,
where will the image form?
a.) 7.5 cm behind the mirror
b.) 6 cm in front of the mirror
c.) 6 cm behind the mirror
d.) 7.5 cm in front of the mirro
Ch. 45
Page 1
CHAPTER 45 Astrophysics and Cosmology Note: A factor that appears in the analysis of energies is e2/4p0 = (1.60 1019 C)2/4p(8.85 1012 C2/N m2) = 2.30 1028 J m = 1.44 MeV fm. 1. Using the definition of the parsec, we find the equivalent dista
Ch. 44 Page 1
CHAPTER 44 Elementary Particles Note: A useful expression for the energy of a photon in terms of its wavelength is E = hf = hc/ = (6.63 1034 J s)(3 108 m/s)(109 nm/m)/(1.60 1019 J/eV); E = (1.24 103 eV nm)/ = (1.24 1012 MeV m)/.
1.
The total
Ch. 43 Page 1
CHAPTER 43 Nuclear Energy; Effects and Uses of Radiation Note: A factor that appears in the analysis of energies is e2/4p0 = (1.60 1019 C)2 /4p(8.85 1012 C2/N m2)= 2.30 1028 J m = 1.44 MeV fm.
1.
We find the product nucleus by balancing the
Ch. 42 Page 1
CHAPTER 42 Nuclear Physics and Radioactivity Note: A factor that appears in the analysis of energies is e2/4p0 = (1.60 1019 C)2/4p(8.85 1012 C2/N m2) = 2.30 1028 J m = 1.44 MeV fm.
1.
To find the rest mass of an particle, we subtract the res
Chapter 41
p. 1
CHAPTER 41 Molecules and Solids Note: At the atomic scale, it is most convenient to have energies in electron-volts and wavelengths in nanometers. A useful expression for the energy of a photon in terms of its wavelength is E = hf = hc/ =
Chapter 40 p. 1
CHAPTER 40 Quantum Mechanics of Atoms Note: At the atomic scale, it is most convenient to have energies in electron-volts and wavelengths in nanometers. A useful expression for the energy of a photon in terms of its wavelength is E = hf =
Chapter 39 p. 1
CHAPTER 39 Quantum Mechanics Note: At the atomic scale, it is most convenient to have energies in electron-volts and wavelengths in nanometers. A useful expression for the energy of a photon in terms of its wavelength is E = hf = hc/ = (6.
Chapter 38 p. 1
CHAPTER 38 Early Quantum Theory and Models of the Atom Note: At the atomic scale, it is most convenient to have energies in electron-volts and wavelengths in nanometers. A useful expression for the energy of a photon in terms of its wavele
Ch. 36 p. 1
CHAPTER 36 Diffraction and Polarization 1. We find the angle to the first minimum from sin 1min = m/a = (1)(680 109 m)/(0.0345 103 m) = 0.0197, so 1min = 1.13. Thus the angular width of the central diffraction peak is ?1 = 21min = 2(1.13) = 2.
Ch. 35 p. 1
CHAPTER 35 The Wave Nature of Light; Interference 1. We draw the wavelets and see that the incident wave fronts are parallel, with the angle of incidence 1 being the angle between the wave fronts and the surface. The reflecting wave fronts are
Ch. 34 p. 1
CHAPTER 34 Lenses and Optical Instruments 1. (a) From the ray diagram, the object distance is about 3& focal lengths, or 250 mm.
FI O F
(b) We find the object distance from (1/do) + (1/di) = 1/f; (1/do) + (1/88.0 mm) = 1/65.0 mm, which gives d
Ch. 33
p. 1
CHAPTER 33 Light: Reflection and Refraction 1. (a) The speed in ethyl alcohol is v = c/n = (3.00 108 m/s)/(1.36) = (b) The speed in lucite is v = c/n = (3.00 108 m/s)/(1.51) = 2.21 108 m/s. 1.99 108 m/s.
2.
We find the index of refraction from
Chapter 32 p. 1
CHAPTER 32 Maxwells Equations and Electromagnetic Waves 1. The electric field between the plates depends on the voltage: E = V/d, so dE/dt = (1/d) dV/dt = (1/1.3 103 m)(120 V/s) = 9.2 104 V/m s. The displacement current is ID = 0A (dE/dt)
Chapter 31 page 1
CHAPTER 31 AC Circuits 1. (a) The reactance of the capacitor is XC1 = 1/2pf1C = 1/2p(60 Hz)(7.2 106 F) = 3.7 102 . (b) For the new frequency we have XC2 = 1/2pf2C = 1/2p(1.0 106 Hz)(7.2 106 F) = 2.2 102 . We find the frequency from XL =
Chapter 30, p. 1
CHAPTER 30 Inductance; and Electromagnetic Oscillations 1. The magnetic field of the long solenoid is essentially zero outside the solenoid. Thus there will be the same linkage of flux with the second coil and the mutual inductance will b
Chapter 29 Page 1
CHAPTER 29 Electromagnetic Induction and Faradays Law 1. The average induced emf is = N ?B/?t = (2)[(+58 Wb) ( 80 Wb)]/(0.72 s) =
3.8 102 V.
2.
Because the plane of the coil is perpendicular to the magnetic field, the initial flux throu
Chapter 28 p.1
CHAPTER 28 Sources of Magnetic Field 1. The magnetic field of a long wire depends on the distance from the wire: B = (0/4p)2I/r = (107 T m/A)2(65 A)/(0.075 m) = 1.7 104 T. When we compare this to the Earth's field, we get B/BEarth = (1.7 10
Chapter 27 p. 1
CHAPTER 27 Magnetism 1. (a) The maximum force will be produced when the wire and the magnetic field are perpendicular, so we have Fmax = ILB, or Fmax/L = IB = (7.40 A)(0.90 T) = 6.7 N/m. (b) We find the force per unit length from F/L = IB
Chapter 26 p. 1
CHAPTER 26 DC Circuits 1. (a) For the current in the single loop, we have Ia = V/(Ra + r) = (8.50 V)/(68.0 + 0.900 ) = 0.123 A. For the terminal voltage of the battery, we have Va = Iar = 8.50 V (0.123 A)(0.900 ) = 8.39 V. (b) For the curr
Chapter 25 p. 1
CHAPTER 25 Electric Currents and Resistance 1. The rate at which electrons pass any point in the wire is the current: 9.38 1018 electron/s. I = 1.50 A = (1.50 C/s)/(1.60 1019 C/electron) = The charge that passes through the battery is ?Q =
Chapter 24 p. 1
CHAPTER 24 Capacitance, Dielectrics, Electric Energy Storage 1. From Q = CV, we have 2500 C = C(950 V), which gives C = 2.6 F.
2.
From Q = CV, we have 28.0 108 C = (12,000 1012 F)V, which gives V = From Q = CV, we have 75 pC = C(12.0 V), w
CHAPTER 23 Electric Potential 1. We find the work done by an external agent from the work-energy principle: = ?K + ?U = 0 + q(Vb Va) Wab = ( 7.0 106 C)(+ 6.00 V 0) = 4.2 105 J (done by the field). We find the work done by an external agent from the work-e
Chapter 22 p. 1
CHAPTER 22 Gauss's Law 1. Because the electric field is uniform, the flux through the circle is = ? E dA = E A = EA cos . (a) When the circle is perpendicular to the field lines, the flux is = EA cos = EA = (5.8 102 N/C)p(0.15 m)2 = 41 N m
Chapter 21 p. 1
CHAPTER 21 Electric Charge and Electric Field 1. The magnitude of the Coulomb force is F = kQ1Q2/r2 = (9.0 109 N m2/C2)(2.50 C)(2.50 C)/(3.0 m)2 =
6.3 109 N. 1.88 1014 electrons.
2.
The number of electrons is N = Q/( e) = ( 30.0 106 C)/( 1
Chapter 20
CHAPTER 20 Second Law of Thermodynamics; Heat Engines 1. For the heat input, we have QH = QL + W = 8500 J + 2700 J = 11,200 J. We find the efficiency from e = W/QH = (2700 J)/(11,200 J) = 0.24 =
24%.
2.
We find the efficiency from e = W/QH = (8