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TIPLER GENE MOSCA Abbreviations for Units A ampere Å angstrom (10 10 m) atm atmosphere Btu British thermal unit Bq becquerel C coulomb C degreeCelsius cal calorie Ci curie cm centimeter dyn dyne eV electron volt F degree Fahrenheit fm femtometer, fermi (10 15 m) ft foot Gm gigameter (109 m) G gauss Gy gray g gram H henry h hour Hz hertz in inch J joule K kelvin kg kilogram km kilometer keV kilo-electron volt lb pound L liter mmeter MeV mega-electron volt Mm megameter (106 m) mi mile min minute mm millimeter ms millisecond N newton nm nanometer (10 9 m) pt pint qt quart rev revolution R roentgen Sv seivert s second T tesla u unified mass unit V volt W watt Wb weber y year yd yard mm micrometer (10 6m) ms microsecond mC microcoulomb ohm Some Conversion Factors Length 1 m 39.37 in 3.281 ft 1.094 yd 1 m 1015 fm 1010 Å 109 nm 1 km 0.6214 mi 1 mi 5280 ft 1.609 km 1 lightyear 1 c y 9.461 1015 m 1 in 2.540 cm Volume 1 L 103 cm3 10 3 m3 1.057 qt Time 1 h 3600 s 3.6 ks 1 y 365.24 d 3.156 107 s Speed 1 km/h 0.278 m/s 0.6214 mi/h 1 ft/s 0.3048 m/s 0.6818 mi/h Angle–angular speed 1 rev 2p rad 360 1 rad 57.30 1 rev/min 0.1047 rad/s Force–pressure 1 N 105 dyn 0.2248 lb 1 lb 4.448 N 1 atm 101.3 kPa 1.013 bar 76.00 cmHg 14.70lb/in2 Mass 1 u [(10 3 mol 1)/NA] kg 1.661 10 27 kg 1 tonne 103 kg 1 Mg 1 slug 14.59 kg 1 kg weighs about 2.205 lb Energy–power 1 J 107 erg 0.7376 ft lb 9.869 10 3 L atm 1 kW h 3.6 MJ 1 cal 4.184 J 4.129 10 2 L atm 1 L atm 101.325 J 24.22 cal 1 eV 1.602 10 19 J 1 Btu 778 ftlb 252 cal 1054 J 1 horsepower 550 ft lb/s 746 W Thermal conductivity 1 W/(m K) 6.938 Btu in/(h ft2 F) Magnetic field 1 T 104 G Viscosity 1 Pa s 10 poise# #### # # # # # ## # This page intentionally left blank SIXTH EDITION WITH MODERN PHYSICS W. H. Freeman and Company New York Paul A. Tipler Gene Mosca PHYSICS FOR SCIENTISTS AND ENGINEERS PART IV ELECTRICITY AND MAGNETISM 21 The Electric Field I: Discrete Charge Distributions / 693 22 The Electric Field II: Continuous Charge Distributions / 727 23 Electric Potential / 763 24Capacitance / 801 25 Electric Current and Direct-Current Circuits / 839 26 The Magnetic Field / 887 27 Sources of the Magnetic Field / 917 28 Magnetic Induction / 959 29 Alternating-Current Circuits / 995 30 Maxwell’s Equations and Electromagnetic Waves / 1029 PART V LIGHT 31Properties of Light / 1055 32 Optical Images / 1097 33 Interference and Diffraction / 1141 PART VI MODERN PHYSICS: QUANTUM MECHANICS, RELATIVITY, AND THE STRUCTURE OF MATTER 34 Wave-Particle Duality and Quantum Physics / 1173 35 Applications of theSchrödinger Equation / 1203 36 Atoms / 1227 37 Molecules / 1261 38 Solids / 1281 39 Relativity / 1319 40 Nuclear Physics / 1357 41 Elementary Particles and the Beginning of the Universe / 1389 APPENDICES A SI Units and Conversion Factors / AP-1 B Numerical Data / AP-3 CPeriodic Table of Elements / AP-6 Math Tutorial / M-1 Answers to Odd-Numbered End-of-Chapter Problems / A-1 Index / I-1 viii Contents in Brief Preface xvii About the Authors xxxii * optional material Chapter 1 MEASUREMENT AND VECTORS / 1 1-1 The Nature of Physics 2 1-2 Units 31-3 Conversion of Units 6 1-4 Dimensions of Physical Quantities 7 1-5 Significant Figures and Order of Magnitude 8 1-6 Vectors 14 1-7 General Properties of Vectors 14 Physics Spotlight: The 2005 Leap Second / 21 Summary 22 Problems 23 PART I MECHANICS Chapter 2 MOTION INONE DIMENSION / 27 2-1 Displacement, Velocity, and Speed 28 2-2 Acceleration 35 2-3 Motion with Constant Acceleration 37 2-4 Integration 47 Physics Spotlight: Linear Accelerators / 51 Summary 52 Problems 53 Chapter 3 MOTION IN TWO AND THREE DIMENSIONS / 63 3-1Displacement, Velocity, and Acceleration 64 3-2 Special Case 1: Projectile Motion 71 3-3 Special Case 2: Circular Motion 78 Physics Spotlight: GPS: Vectors Calculated While You Move / 82 Summary 83 Problems 84 Chapter 4 NEWTON’S LAWS / 93 4-1 Newton’s First Law: The Law ofInertia 94 4-2 Force and Mass 95 4-3 Newton’s Second Law 97 4-4 The Force Due to Gravity: Weight 99 4-5 Contact Forces: Solids, Springs, and Strings 101 4-6 Problem Solving: Free-Body Diagrams 104 4-7 Newton’s Third Law 109 4-8 Problem Solving: Problems with Two or MoreObjects 111 Extended Contents ix Physics Spotlight: Roller Coasters and the Need for Speed / 114 Summary 115 Problems 116 Chapter 5 ADDITIONAL APPLICATIONS OF NEWTON’S LAWS / 127 5-1 Friction 128 5-2 Drag Forces 139 5-3 Motion Along a Curved Path 141 *5-4Numerical Integration: Euler’s Method 147 5-5 The Center of Mass 149 Physics Spotlight: Accident Reconstruction— Measurements and Forces / 158 Summary 159 Problems 160 Summary 194 Problems 195 Chapter 7 CONSERVATION OF ENERGY / 201 7-1 Potential Energy 202 7-2The Conservation of Mechanical Energy 209 7-3 The Conservation of Energy 219 7-4 Mass and Energy 228 7-5 Quantization of Energy 231 Physics Spotlight: Blowing Warmed Air / 233 Summary 234 Problems 236 Chapter 8 CONSERVATION OF LINEAR MOMENTUM / 247 8-1Conservation of Linear Momentum 248 8-2 Kinetic Energy of a System 254 8-3 Collisions 255 *8-4 Collisions in the Center-of-Mass Reference Frame 271 8-5 Continuously Varying Mass and Rocket Propulsion 273 Physics Spotlight: Pulse Detonation Engines: Faster (and Louder) / 277Summary 278 Problems 279 Chapter 9 ROTATION / 289 9-1 Rotational Kinematics: Angular Velocity and Angular Acceleration 290 9-2 Rotational Kinetic Energy 292 9-3 Calculating the Moment of Inertia 294 9-4 Newton’s Second Law for Rotation 301 9-5 Applications of Newton’s SecondLaw for Rotation 303 9-6 Rolling Objects 310 Physics Spotlight: Spindizzy—Ultracentrifuges / 316 Summary 317 Problems 318 x Contents Courtesy of Rossignol Ski Company Chapter 6 WORK AND KINETIC ENERGY / 173 6-1 Work Done by a Constant Force 174 6-2 Work Done by aVariable Force–Straight-Line Motion 179 6-3 The Scalar Product 182 6-4 Work–Kinetic-Energy Theorem—Curved Paths 188 *6-5 Center-of-Mass Work 190 Physics Spotlight: Coasters and Baggage and Work (Oh My!) / 193 Contents xiii 21-4 The Electric Field 704 21-5 Electric Field Lines711 21-6 Action of the Electric Field on Charges 714 Physics Spotlight: Powder Coating—Industrial Static / 719 Summary 720 Problems 721 Chapter 22 THE ELECTRIC FIELD II: CONTINUOUS CHARGE DISTRIBUTIONS / 727 22-1 Calculating from Coulomb’s Law 728 22-2 Gauss’s Law738 22-3 Using Symmetry to Calculate with Gauss’s Law 742 22-4 Discontinuity of 749 22-5 Charge and Field at Conductor Surfaces 750 *22-6 The Equivalence of Gauss’s Law and Coulomb’s Law in Electrostatics 753 Physics Spotlight: Charge Distribution—Hot and Cold / 754 Summary755 Problems 756 Chapter 23 ELECTRIC POTENTIAL / 763 23-1 Potential Difference 764 23-2 Potential Due to a System of Point Charges 767 23-3 Computing the Electric Field from the Potential 772 En E S E S 23-4 Calculation of V for Continuous Charge Distributions 773 23-5Equipotential Surfaces 781 23-6 Electrostatic Potential Energy 787 Physics Spotlight: Lightning—Fields of Attraction / 791 Summary 792 Problems 794 Chapter 24 CAPACITANCE / 801 24-1 Capacitance 802 24-2 The Storage of Electrical Energy 806 24-3 Capacitors, Batteries, andCircuits 810 24-4 Dielectrics 817 24-5 Molecular View of a Dielectric 824 Physics Spotlight: Changes in Capacitors— Charging Ahead / 828 Summary 829 Problems 831 Chapter 25 ELECTRIC CURRENT AND DIRECT-CURRENT CIRCUITS / 839 25-1 Current and the Motion of Charges840 25-2 Resistance and Ohm’s Law 844 25-3 Energy in Electric Circuits 849 25-4 Combinations of Resistors 854 25-5 Kirchhoff’s Rules 860 25-6 RC Circuits 868 Physics Spotlight: Vehicle Electrical Systems: Driven to Innovation / 874 Summary 875 Problems 877 Chapter 26 THEMAGNETIC FIELD / 887 26-1 The Force Exerted by a Magnetic Field 888 26-2 Motion of a Point Charge in a Magnetic Field 892 26-3 Torques on Current Loops and Magnets 900 26-4 The Hall Effect 904 Physics Spotlight: Earth and the Sun— Magnetic Changes / 908 NASA/GoddardSpace Flight Center Scientific Visualization Studio Summary 909 Problems 910 Summary 986 Problems 988 Chapter 29 ALTERNATING-CURRENT CIRCUITS / 995 29-1 Alternating Current in a Resistor 996 29-2 Alternating-Current Circuits 999 *29-3 The Transformer 1004 *29-4 LC andRLC Circuits without a Generator 1007 *29-5 Phasors 1010 *29-6 Driven RLC Circuits 1011 Physics Spotlight: The Electric Grid: Power to the People / 1019 Summary 1020 Problems 1022 Chapter 30 MAXWELL’S EQUATIONS AND ELECTROMAGNETIC WAVES / 1029 30-1 Maxwell’sDisplacement Current 1030 30-2 Maxwell’s Equations 1033 30-3 The Wave Equation for Electromagnetic Waves 1034 30-4 Electromagnetic Radiation 1040 Physics Spotlight: Wireless: Sharing the Spectrum / 1049 Summary 1050 Problems 1051 PART V LIGHT Chapter 31 PROPERTIESOF LIGHT / 1055 31-1 The Speed of Light 1056 31-2 The Propagation of Light 1059 31-3 Reflection and Refraction 1060 31-4 Polarization 1070 31-5 Derivation of the Laws of Reflection and Refraction 1077 31-6 Wave–Particle Duality 1079 31-7 Light Spectra 1080 *31-8 Sources of Light1081 Chapter 27 SOURCES OF THE MAGNETIC FIELD / 917 27-1 The Magnetic Field of Moving Point Charges 918 27-2 The Magnetic Field of Currents: The Biot–Savart Law 919 27-3 Gauss’s Law for Magnetism 932 27-4 Ampère’s Law 933 27-5 Magnetism in Matter 937 PhysicsSpotlight: Solenoids at Work / 947 Summary 948 Problems 950 Chapter 28 MAGNETIC INDUCTION / 959 28-1 Magnetic Flux 960 28-2 Induced EMF and Faraday’s Law 961 28-3 Lenz’s Law 965 28-4 Motional EMF 969 28-5 Eddy Currents 974 28-6 Inductance 974 28-7 Magnetic Energy977 *28-8 RL Circuits 979 *28-9 Magnetic Properties of Superconductors 983 Physics Spotlight: The Promise of Superconductors / 985 Atlas Photo Bank/Photo Researchers, Inc. xiv Contents Contents xv Physics Spotlight: Optical Tweezers and Vortices: Light at Work / 1088 Summary1089 Problems 1090 Chapter 32 OPTICAL IMAGES / 1097 32-1 Mirrors 1097 32-2 Lenses 1108 *32-3 Aberrations 1121 *32-4 Optical Instruments 1122 Physics Spotlight: Eye Surgery: New Lenses for Old / 1131 Summary 1132 Problems 1134 Chapter 33 INTERFERENCE ANDDIFFRACTION / 1141 33-1 Phase Difference and Coherence 1142 33-2 Interference in Thin Films 1143 33-3 Two-Slit Interference Pattern 1145 33-4 Diffraction Pattern of a Single Slit 1149 *33-5 Using Phasors to Add Harmonic Waves 1152 33-6 Fraunhofer and Fresnel Diffraction 115933-7 Diffraction and Resolution 1160 *33-8 Diffraction Gratings 1162 Physics Spotlight: Holograms:Guided Interference / 1165 Summary 1166 Problems 1167 PART VI MODERN PHYSICS: QUANTUM MECHANICS, RELATIVITY, AND THE STRUCTURE OF MATTER Chapter 34 WAVE–PARTICLE DUALITY AND QUANTUM PHYSICS / 1173 34-1 Waves and Particles 1174 34-2 Light: From Newton to Maxwell 1174 34-3 The Particle Nature of Light: Photons 1175 34-4 Energy Quantization in Atoms 1180 34-5 Electrons and Matter Waves 1181 34-6 The Interpretation ofthe Wave Function 1185 34-7 Wave–Particle Duality 1187 34-8 A Particle in a Box 1189 34-9 Expectation Values 1193 34-10 Energy Quantization in Other Systems 1196 Summary 1198 Problems 1199 Chapter 35 APPLICATIONS OF THE SCHRÖDINGER EQUATION / 1203 35-1 TheSchrödinger Equation 1204 35-2 A Particle in a Finite Square Well 1206 35-3 The Harmonic Oscillator 1208 35-4 Reflection and Transmission of Electron Waves: Barrier Penetration 1211 35-5 The Schrödinger Equation in Three Dimensions 1217 35-6 The Schrödinger Equation for TwoIdentical Particles 1220 Summary 1223 Problems 1224 Chapter 36 ATOMS / 1227 36-1 The Atom 1228 36-2 The Bohr Model of the Hydrogen Atom 1229 36-3 Quantum Theory of Atoms 1234 36-4 Quantum Theory of the Hydrogen Atom 1236 36-5 The Spin–Orbit Effect and Fine Structure1241 36-6 The Periodic Table 1244 36-7 Optical Spectra and X-Ray Spectra 1251 Summary 1255 Problems 1257 Chapter 37 MOLECULES / 1261 37-1 Bonding 1261 *37-2 Polyatomic Molecules 1269 37-3 Energy Levels and Spectra of Diatomic Molecules 1271 Summary 1278 Problems1279 Chapter 38 SOLIDS / 1281 38-1 The Structure of Solids 1282 38-2 A Microscopic Picture of Conduction 1286 Example 3-4 Rounding a Curve A car is traveling east at 60 km/h. It rounds a curve, and 5.0 s later it is traveling north at 60 km/h. Find the average acceleration of the car.PICTURE We can calculate the average acceleration from its definition, . To do this, we first calculate , which is the vector that when added to , results in .vSfv S i v S aSav v S/ t SOLVE 1. The average acceleration is the change in velocity divided by the elapsed time. To find , we firstfind the change in velocity:aSav aSav vS t 2. To find , we first specify and . Draw and (Figure 3-7a), and draw the vector addition diagram (Figure 3-7b) corresponding to :vSf v S i v S vSfv S iv S fv S i v S 3. The change in velocity is related to the initial and final velocities: N EW S vivf vf vi v (a) (b) j i F I G U R E 3 - 7 vSf v S i v S 4. Substitute these results to find the average acceleration: aSav vf S vi S t 60 km/h jn 60 km/h in 5.0 s 5. Convert 60 km/h to meters per second: 60 km/h 1 h 3600 s 1000 m 1 km 16.7 m/s CHECK The eastward component ofthe velocity decreases from 60 km/h to zero, so we expect a negative acceleration component in the x direction. The northward component of the velocity increases from zero to 60 km/h, so we expect a positive acceleration component in the y direction. Our step 6 result meets both of theseexpectations. TAKING IT FURTHER Note that the car is accelerating even though its speed remains constant. PRACTICE PROBLEM 3-1 Find the magnitude and direction of the average acceleration vector. 6. Express the acceleration in meters per second squared: 3.4 m/s2in 3.4 m/s2jn aSav vSf v S i t 16.7 m/s jn 16.7 m/s in 5.0 s xviii Preface In this edition, the problem-solving steps are again juxtaposed with the neces- sary equations so that it’s easier for students to see a problem unfold. A boxed Problem-Solving Strategy is included in almost every chapter toreinforce the Picture, Solve, and Check format for successfully solving problems. INTEGRATED MATH TUTORIAL This edition has improved mathematical support for students who are taking cal- culus concurrently with introductory physics or for students who need a math review. Thecomprehensive Math Tutorial reviews basic results of algebra, geometry, trigonometry, and calculus, links mathematical concepts to physics concepts in the text, provides Examples and Practice Problems so students may check their understanding of mathematical concepts. After eachproblem statement, students are asked to Picture the problem. Here, the problem is analyzed both conceptually and visually. In the Solve sections, each step of the solution is presented with a written statement in the left-hand column and the corresponding mathematical equations in theright-hand column. Check reminds students to make sure their results are accurate and reasonable. Taking It Further suggests a different way to approach an Example or gives additional information relevant to the Example. A Practice Problem often follows the solution of an Example,allowing students to check their understanding. Answers are included at the end of the chapter to provide immediate feedback. NE W! PROBLEM-SOLVING STRATEGY Relative Velocity PICTURE The first step in solving a relative-velocity problem is to identify and label the relevantreference frames. Here, we will call them reference frame A and reference frame B. SOLVE 1. Using (Equation 3-9), relate the velocity of the moving object (particle p) relative to frame A to the velocity of the particle relative to frame B. 2. Sketch a vector addition diagram for the equation .Use the head-to-tail method of vector addition. Include coordinate axes on the sketch. 3. Solve for the desired quantity. Use trigonometry where appropriate. CHECK Make sure that you solve for the velocity or position of the moving object relative to the proper reference frame. vSpB v S pA v S AB vSpB v S pA v S AB Preface xix Conceptual Example 8-12 Collisions with Putty Mary has two small balls of equal mass, a ball of plumber’s putty and a one of Silly Putty. She throws the ball of plumber’s putty at a block suspended by strings shown in Figure 8-20. The ball strikesthe block with a “thonk” and falls to the floor. The block subsequently swings to a maximum height h. If she had thrown the ball of Silly Putty (instead of the plumber’s putty), would the block subsequently have risen to a height greater than h? Silly Putty, unlike plumber’s putty, is elastic andwould bounce back from the block. PICTURE During impact the change in momentum of the ball–block system is zero. The greater the magnitude of the change in momentum of the ball, the greater, the magnitude of the change in momentum of the block. Does magnitude of the change inmomentum of the ball increase more if the ball bounces back than if it does not? v F I G U R E 8 - 2 0 SOLVE The ball of plumber’s putty loses a large fraction of its forward momentum. The ball of Silly Putty would lose all of its forward momentum and then gain momentum in the oppositedirection. It would undergo a larger change in momentum than did the ball of plumber’s putty. The block would swing to a greater height after being struck with the ball of Silly Putty than it did after being struck with the ball of plumbers putty. CHECK The block exerts a backward impulse onthe ball of plumber’s putty to slow the ball to a stop. The same backward impulse on the ball of Silly Putty would also bring it to a stop, and an additional backward impulse on the ball would give it momentum in the backward direction. Thus, the block exerts the larger backward impulse onthe Silly-Putty ball. In ac- cord with Newton’s third law, the impulse of a ball on the block is equal and opposite to the impulse of the block on the ball. Thus, the Silly-Putty ball exerts the larger forward impulse on the block, giving the block a larger forward change in momentum. PEDAGOGYTO ENSURE CONCEPTUAL UNDERSTANDING Student-friendly tools have been added to allow for better conceptual understanding of physics. New Conceptual Examples are introduced, where appropriate, to help students fully understand essential physics concepts. These Examplesuse the Picture, Solve, and Check strategy so that students not only gain fundamental conceptual understanding but must evaluate their answers. In addition, margin notes allow students to easily see the links between physics concepts in the text and math concepts. Example M-13Radioactive Decay of Cobalt-60 The half-life of cobalt-60 is 5.27 y. At you have a sample of that has a mass equal to 1.20 mg. At what time (in years) will 0.400 mg of the sample of have decayed? PICTURE When we derived the half-life in exponential decay, we set In this example, we areto find the time at which two-thirds of a sample remains, and so the ratio will be 0.667. SOLVE N N 0 N N 0 1 2 . 60Cot 60Cot 0(60Co) 1. Express the ratio as an exponential function:N N 0 NN 0 0.667 e lt 2. Take the reciprocal of both sides: N 0 N 1.50 elt 3. Solve for t: t ln 1.50 l 0.405 l 4. The decay constant is related to the half-life by (Equation M-70). Substitute for and evaluate the time:l(ln2) t1 2 l (ln2) t1 2 t ln 1.5ln 2 t1 2 ln 1.5 ln 2 5.27 y 3.08 y CHECK It takes 5.27 y for the mass of a sample of to decrease to 50 percent of its initial mass. Thus, we expectit to take less than 5.27 y for the sample to lose 33.3 percent of its mass. Our step-4 result of 3.08 y is less than 5.27 y, as expected. PRACTICE PROBLEMS 27. The discharge time constant of a capacitor in an circuit is the time in which the ca- pacitor discharges to (or 0.368) times itscharge at If for a capacitor, at what time (in seconds) will it have discharged to 50.0% of its initial charge? 28. If the coyote population in your state is increasing at a rate of 8.0% a decade and con- tinues increasing at the same rate indefinitely, in how many years will it reach 1.5 times itscurrent level? t t 1 st 0 .e 1 RCt 60Co M-12 INTEGRAL CALCULUS Integration can be considered the inverse of differentiation. If a function is integrated, a function is found for which is the derivative of with respect to THE INTEGRAL AS AN AREA UNDER A CURVE; DIMENSIONALANALYSIS The process of finding the area under a curve on the graph il- lustrates integration. Figure M-27 shows a function The area of the shaded element is approximately where is evaluated anywhere in the interval This approximation is highly accurate if is very small. The total areaunder some stretch of the curve is found by summing all the area elements it covers and taking the limit as each approaches zero. This limit is called the integral of over and is written M-74 The physical dimensions of an integral of a function are found by multiplying the dimensions of theintegrand (the func- tion being integrated) and the dimensions of the integration variable For example, if the integrand is a velocity functiont. f(t) f dt areai lim tiS0 a i fi ti tf ti ti ti. fifi ti, f(t) . t.F(t) f(t)F(t)f(t) f(t) fi t1 t2 t ti t1 t2 t3 F I G U R E M - 2 7 A general function The area of the shadedelement is approximately where is evaluated anywhere in the interval. fifi ti, f(t) . NE W! See Math Tutorial for more information on Differential Calculus New Concept Checks enable students to check their conceptual understanding of physics concepts while they read chapters. Answersare located at the end of chapters to provide immediate feedback. Concept Checks are placed near relevant topics so students can immediately reread any material that they do not fully understand. PHYSICS SPOTLIGHTS Physics Spotlights at the end of appropriate chapters discusscurrent applications of physics and connect applications to concepts described in chapters. These topics range from wind farms to molecular thermometers to pulse detonation engines. New Pitfall Statements, identified by exclamation points, help students avoid common misconceptions.These statements
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