Bundle: College Physics, Loose-Leaf Ver...10th EditionRaymond A. Serway, Chris Vuille Publisher: Cengage LearningISBN: 9781305367395
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Chapter 1 - IntroductionChapter 2 - Motion In One DimensionChapter 2.2 - VelocityChapter 2.3 - AccelerationChapter 2.4 - Motion DiagramsChapter 2.6 - Freely Falling ObjectsChapter 3 - Vectors And Two-Dimensional MotionChapter 3.1 - Vectors And Their PropertiesChapter 3.2 - Components Of A VectorChapter 3.3 - Displacement, Velocity, And Accleration In Two Dimensions
Chapter 3.4 - Motion In Two DimensionsChapter 4 - The Laws Of MotionChapter 4.3 - Newton's Second LawChapter 4.4 - Newton's Third LawChapter 4.5 - Applications Of Newton's LawsChapter 4.6 - Forces Of FrictionChapter 5 - EnergyChapter 5.1 - WorkChapter 5.3 - Gravitational Potential EnergyChapter 5.5 - Spring Potential EnergyChapter 6 - Momentum And CollisionsChapter 6.1 - Momentum And ImpulseChapter 6.2 - Conservation Of MomentumChapter 6.3 - CollisionsChapter 7 - Rotational Motion And The Law Of GravityChapter 7.1 - Angular Speed And Angular AccelerationChapter 7.2 - Rotational Motion Under Constant Angular AccelerationChapter 7.3 - Relation Between Angular And Linear QuantitiesChapter 7.4 - Centripetal AccelerationChapter 7.5 - Newtonian GravitationChapter 7.6 - Kepler's LawsChapter 8 - Rotational Equilibrium And Rotational DynamicsChapter 8.5 - Relationship Between Torque And Angular AccelerationChapter 8.6 - Rotational Kinetic EnergyChapter 8.7 - Angular MomentumChapter 9 - Solids And FluidsChapter 9.2 - Density And PressureChapter 9.4 - Variation Of Pressure With DepthChapter 9.5 - Pressure MeasurementsChapter 9.6 - Buoyant Forces And Archimedes' PrincipleChapter 9.7 - Fluids In MotionChapter 10 - Thermal PhysicsChapter 10.1 - Temperature And The Zeroth Law Of ThermodynamicsChapter 10.3 - Thermal Expansion Of Solids And LiquidsChapter 10.5 - The Kinetic Theory Of GasesChapter 11 - Energy In Thermal ProcessesChapter 11.2 - Specific HeatChapter 11.4 - Latent Heat And Phase ChangeChapter 11.5 - Energy TransferChapter 12 - The Laws Of ThermodynamicsChapter 12.1 - Work In Thermodynamic ProcessesChapter 12.4 - Heat Engines An D The Second Law OfthermodynamicsChapter 12.5 - EntropyChapter 13 - Vibrations And WavesChapter 13.1 - Hooke's LawChapter 13.2 - Elastic Potential EnergyChapter 13.3 - Comparing Simple Harmonic Motion With Uniform Circular MotionChapter 13.4 - Position, Velocity, And Acceleration As A Function Of TimeChapter 13.5 - Motion Of A PendulumChapter 14 - SoundChapter 14.3 - The Speed Of SoundChapter 14.6 - The Doppler EffectChapter 14.8 - Standing WavesChapter 14.9 - Forced Vibrations And ResonanceChapter 14.10 - Standing Waves In Air ColumnsChapter 14.11 - BeatsChapter 15 - Electric Forces And Electric FieldsChapter 15.1 - Properties Of Electric ChargesChapter 15.2 - Insulators And ConductorsChapter 15.3 - Coulomb's LawChapter 15.4 - Electric FieldsChapter 15.8 - The Van De Graaff GeneratorChapter 16 - Electrical Energy And CapacitanceChapter 16.1 - Electric Potential Energy And Electric PotentialChapter 16.2 - Electric Potential And Potential Energy Due To Point ChargesChapter 16.3 - Potentials And Charged ConductorsChapter 16.6 - CapacitanceChapter 16.7 - Combinations Of CapacitorsChapter 16.8 - Capacitors With DielectricsChapter 17 - Current And ResistanceChapter 17.1 - Electric CurrentChapter 17.2 - A Microscopic View: Current And Drift SpeedChapter 17.3 - Current And Voltage Measurements In CircuitsChapter 17.4 - Resistance, Resistivity, And Ohm's LawChapter 17.6 - Electrical Energy And PowerChapter 18 - Direct-Current CircuitsChapter 18.1 - Sources Of EmfChapter 18.2 - Resistors In SeriesChapter 18.3 - Resistors In ParallelChapter 18.5 - Rc CircuitsChapter 19 - MagnetismChapter 19.3 - Magnetic FieldsChapter 19.4 - Magnetic Force On A Current -Carrying ConductorChapter 19.6 - Motion Of A Charged Particle In A Magnetic FieldChapter 19.8 - Magnetic Force Between Two Parallel ConductorsChapter 20 - Induced Voltages And InductanceChapter 20.2 - Faraday's Law Of Induction And Lenz's LawChapter 20.3 - Motional EmfChapter 20.6 - Rl CircuitsChapter 21 - Alternating-Current Circuits And Electromagnetic WavesChapter 21.1 - Resistors In An Ac CircuitChapter 21.4 - The Rlc Series CircuitChapter 21.11 - Properties Of Electromagnetic WavesChapter 21.12 - The Spectrum Of Electromagnetic WavesChapter 22 - Reflection And Refraction Of LightChapter 22.2 - Reflection And RefractionChapter 22.3 - The Law Of RefractionChapter 23 - Mirrors And LensesChapter 23.1 - Flat MirrorsChapter 23.3 - Convex Mirrors And Sign ConventionsChapter 23.5 - Atmospheric RefractionChapter 24 - Wave OpticsChapter 24.2 - Young's Double-Slit ExperimentChapter 24.4 - Interference In Thin FilmsChapter 24.7 - Single Slit DiffractionChapter 24.8 - Diffraction GratingsChapter 25 - Optical InstrumentsChapter 25.2 - The EyeChapter 25.6 - Resolution Of Single-Slit And Circular AperturesChapter 26 - RelativityChapter 26.3 - Einstein's Principle Of RelativityChapter 26.4 - Consequences Of Special RelativityChapter 26.7 - Relativistic Energy And The Equivalence Of Mass And EnergyChapter 27 - Quantum PhysicsChapter 27.5 - The Compton EffectChapter 27.6 - The Dual Nature Of Light And MatterChapter 28 - Atomic PhysicsChapter 28.3 - The Bohr ModelChapter 28.4 - Quantum Mechanics And The Hydrogen AtomChapter 28.5 - The Exclusion Principle And The Periodic TableChapter 29 - Nuclear PhysicsChapter 29.3 - RadioactivityChapter 29.6 - Nuclear ReactionsChapter 30 - Nuclear Energy And Elementary ParticlesChapter 30.6 - Conservation Laws
Book Details
While physics can seem challenging, its true quality is the sheer simplicity of fundamental physical theories--theories and concepts that can enrich your view of the world around you. COLLEGE PHYSICS, Tenth Edition, provides a clear strategy for connecting those theories to a consistent problem-solving approach, carefully reinforcing this methodology throughout the text and connecting it to real-world examples. For students planning to take the MCAT exam, the text includes exclusive test prep and review tools to help you prepare.
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Chapter 1, Problem 1WUEChapter 1, Problem 18PChapter 1, Problem 32PChapter 1, Problem 39PChapter 1, Problem 45PSolution: The figure representing the relationship between the height of mountain and the elevation...Chapter 1, Problem 56APChapter 2, Problem 1WUEChapter 2, Problem 6P
Chapter 2, Problem 20PChapter 2, Problem 36PChapter 2, Problem 48PChapter 2, Problem 50PChapter 2, Problem 52PChapter 2, Problem 53PChapter 2, Problem 66APChapter 2, Problem 71APChapter 3, Problem 1WUEChapter 3, Problem 14PChapter 3, Problem 19PChapter 3, Problem 23PChapter 3, Problem 28PChapter 3, Problem 35PChapter 3, Problem 41PChapter 3, Problem 51APChapter 3, Problem 60APChapter 3, Problem 72APChapter 4, Problem 1WUEChapter 4, Problem 7WUEChapter 4, Problem 32PChapter 4, Problem 35PChapter 4, Problem 37PChapter 4, Problem 41PChapter 4, Problem 51PChapter 4, Problem 57APChapter 4, Problem 68APChapter 4, Problem 69APChapter 4, Problem 79APChapter 4, Problem 84APChapter 5, Problem 1WUEChapter 5, Problem 10WUEChapter 5, Problem 22PChapter 5, Problem 29PChapter 5, Problem 31PChapter 5, Problem 46PChapter 5, Problem 49PChapter 5, Problem 60PChapter 5, Problem 65APChapter 5, Problem 71APChapter 5, Problem 73APChapter 5, Problem 81APExplanation: Given Info: The force constant of the spring is 3.85 N/m , the length the spring is...Chapter 6, Problem 1WUEChapter 6, Problem 31PChapter 6, Problem 44PChapter 6, Problem 54APChapter 6, Problem 55APChapter 6, Problem 60APChapter 6, Problem 62APChapter 6, Problem 63APChapter 6, Problem 67APChapter 6, Problem 69APChapter 6, Problem 77APChapter 7, Problem 1WUEChapter 7, Problem 12PChapter 7, Problem 17PChapter 7, Problem 19PChapter 7, Problem 26PChapter 7, Problem 31PGiven info: The mass of the dwarf is 1.991×1030 kg and the radius of the dwarf is 6.38×106 m ....Chapter 7, Problem 49APChapter 7, Problem 50APExplanation: Given info: The mass of the block is 0.50 kg . The initial speed of the block is 4.0...Chapter 7, Problem 66APChapter 8, Problem 1WUEChapter 8, Problem 15PExplanation: Given info: The mass of the segments in kg (Mass), the lengths of the segments...Chapter 8, Problem 22PExplanation: The weight of the beam acting downward from its midpoint, tension of the acting left...Chapter 8, Problem 34PChapter 8, Problem 44PChapter 8, Problem 53PChapter 8, Problem 54PChapter 8, Problem 59PChapter 8, Problem 76APExplanation: The following figure shows the free body diagram.Chapter 9, Problem 1WUEExplanation: Given info: Mass of the lead is 20.0 kg and density of the lead is 11.3×103 kg/m3 . The...Chapter 9, Problem 7PChapter 9, Problem 16PExplanation: Given info: The diameter of the steel cable 1.27 cm and it is 5.75 m long. The...Chapter 9, Problem 34PChapter 9, Problem 43PExplanation: Formula to calculate the time taken by the water to reach the ground is, t=2(Δy)ay t is...Chapter 9, Problem 55PChapter 9, Problem 56PChapter 9, Problem 57PChapter 9, Problem 83APExplanation: Given info: The initial radius of the balloon is 0.200 m . The final radius of the...Section 1: Explanation: To determine: The constants a and b. Answer: The constant (a) is 4.7×10−3...Chapter 10, Problem 15PChapter 10, Problem 18PChapter 10, Problem 21PChapter 10, Problem 26PChapter 10, Problem 30PChapter 10, Problem 38PChapter 10, Problem 60APExplanation: Given info: M=4m , c=4186 J/kg , and Lf=3.33×105J/kg Write the expression given....Explanation: Given Info: Expression for the compressive stress and strain is, FA=Y(ΔLL0) F is the...Chapter 11, Problem 18PExplanation: Given Info: Initial temperature of water and calorimeter is 10.0°C , mass of aluminum...Chapter 11, Problem 23PExplanation: Aluminum container at 20°C contains ethyl alcohol at 30°C and ice at 0°C . Heat...Explanation: Given Info: Mass of ice block is 40 g, initial temperature of ice block is −78°C , mass...Chapter 11, Problem 35PChapter 11, Problem 40PChapter 11, Problem 66APExplanation: Humans do work and give off energy by heat. This makes the First law of thermodynamics...Chapter 12, Problem 5PChapter 12, Problem 12PChapter 12, Problem 20PChapter 12, Problem 23PChapter 12, Problem 26PChapter 12, Problem 67APSection1: To determine: The constant volume occupied by the gas. Answer: The volume of the gas is...Chapter 12, Problem 72APChapter 13, Problem 1CQChapter 13, Problem 15PChapter 13, Problem 16PChapter 13, Problem 17PChapter 13, Problem 18PChapter 13, Problem 28PChapter 13, Problem 32PChapter 13, Problem 40PChapter 13, Problem 68APChapter 13, Problem 76APChapter 14, Problem 1CQChapter 14, Problem 5CQChapter 14, Problem 18PChapter 14, Problem 32PChapter 14, Problem 36PExplanation: Given Info: Standing wave is setup in a string with both ends fixed. Formula to...Chapter 14, Problem 50PChapter 14, Problem 67PChapter 14, Problem 72APChapter 14, Problem 82APChapter 15, Problem 1CQChapter 15, Problem 5CQChapter 15, Problem 4PChapter 15, Problem 10PChapter 15, Problem 16PChapter 15, Problem 18PChapter 15, Problem 24PChapter 15, Problem 32PChapter 15, Problem 69APChapter 16, Problem 1CQChapter 16, Problem 19PChapter 16, Problem 39PChapter 16, Problem 41PChapter 16, Problem 43PChapter 16, Problem 44PChapter 16, Problem 48PChapter 16, Problem 50PChapter 16, Problem 60APChapter 17, Problem 1CQChapter 17, Problem 9PChapter 17, Problem 32PChapter 17, Problem 41PChapter 17, Problem 48PGiven Info: The tungsten wire has length 15.0 cm and radius is 1.00 mm Explanation: Formula to...Chapter 17, Problem 66APChapter 17, Problem 68APChapter 18, Problem 1CQChapter 18, Problem 10PChapter 18, Problem 12PChapter 18, Problem 15PChapter 18, Problem 23PChapter 18, Problem 30PChapter 18, Problem 31PChapter 18, Problem 43PChapter 18, Problem 61APExplanation: The magnetic force on a moving charged particle is given by, F→=q(v→×B→) The velocity...Chapter 19, Problem 42PChapter 19, Problem 48PChapter 19, Problem 50PGiven Info: The wire is lying on a horizontal table in the xy-plane. The wire carries a current of...Chapter 19, Problem 53PGiven info: The wire along the x-axis carries current of 5.00 A . The current along the y-axis...Given info: The three conductors are parallel and carry equal currents of magnitude I=2.0 A which is...Chapter 19, Problem 73APChapter 19, Problem 76APChapter 20, Problem 1CQExplanation: Given Info: current in the coil is 2A, number of turns is 300, length of the coil is 20...Explanation: Given Info: Resistance R is 6.00 Ω , length l is 1.20 m , current I is 1.00 A and...Explanation: Given Info: Number of turns in the coil N is 475 turns , Cross sectional area A is...Explanation: Given Info: Voltage ε is 9.00 V , t is 0.100 s , maximum value of I(t) is 2.00 A Since...Explanation: Given Info: Electromotive force ε is 6.0 V , Resistance R is 8.0 MΩ , Inductance L is...Explanation: Given Info: The length of insulated copper wire is 60.0 m and radius of the copper wire...Chapter 20, Problem 57APChapter 20, Problem 63APChapter 21, Problem 1CQChapter 21, Problem 6PChapter 21, Problem 16PChapter 21, Problem 22PChapter 21, Problem 24PChapter 21, Problem 29PChapter 21, Problem 30PExplanation: Given info: The AC voltage has a form Δv=(90.0 V)sin(350t) . The resistance is 50.0 Ω ....Chapter 21, Problem 42PExplanation: Given Info: Distance r is 2.0 inch , Power of radiation emitted from the cell phoneis...The dispersion of light into a spectrum is most clearly visible in a rainbow. The phenomena behind...Explanation: The equation for time is, Δt=dc d is the distance c is the speed of light Substitute...Chapter 22, Problem 11PChapter 22, Problem 18PExplanation: The ray diagram for the minimum incident angle is given below: Formula to calculate the...Given Info: The refractive index of diamond is 2.419 . Explanation: Formula to calculate the...Explanation: The following ray diagram show path of the light. Formula to calculate the angle θ3 is,...Explanation: The ray diagram for the beam through the prism is Formula to calculate the refraction...Chapter 23, Problem 1CQChapter 23, Problem 46PGiven info: The radius if left side is −15.0 cm . The radius if right side is 10.0 cm . The...Chapter 23, Problem 51APChapter 23, Problem 52APChapter 23, Problem 54APChapter 23, Problem 57APChapter 23, Problem 59APChapter 23, Problem 61APChapter 24, Problem 1CQChapter 24, Problem 5PChapter 24, Problem 14PChapter 24, Problem 39PChapter 24, Problem 41PExplanation: Given Info: The velocity of light is 343 m/s , the frequency is 37.2 kHz , the slit...Chapter 24, Problem 61PChapter 24, Problem 64APChapter 24, Problem 66APChapter 24, Problem 70APChapter 25, Problem 1CQChapter 25, Problem 10PChapter 25, Problem 30PExplanation: The focal length of the lens for the left eye is, fL=pLqLpL+qL pL is the object...Chapter 26, Problem 1CQChapter 26, Problem 9PChapter 26, Problem 34PExplanation: The formula used to calculate the initial rest energy is, ERi=mic2 ERi is the initial...Explanation: Given Info: The rest energy of electron is 938.3 MeV . The rest energy of proton is...Chapter 26, Problem 43APChapter 26, Problem 45APChapter 26, Problem 56APExplanation: When the object is inside a hot kiln, the object gets heated and thus emit thermal...Explanation: The energy of the photon is, EPeak=hcλmax h is the Plank’s constant λmax is the maximum...Explanation: The work function in joules is, ϕJ=ϕeV(1.60×10−19 J1 eV) ϕJ is the work function in...Explanation: The figure shows the situation before and after the scattering process. The Compton...Explanation: The Compton shift formula is, λ'−λ=hmec(1−cosθ) (1) h is the Plank’s constant c is the...Explanation: The equation for de Broglie wavelength of the electron in terms of kinetic energy is,...Explanation: Johnny’s speed just before impact is, v2=v02+2ay(Δy)v=v02+2ay(Δy) v0 is the initial...Explanation: Expressing the relation between energy of the spectrum of the hydrogen atom and the...Explanation: Given Info: The Bohr radius is 0.0529 nm . Formula to calculate the radius is, r=n2ao r...Explanation: Formula to calculate the energy of the photon of the longest wavelength is,...Formula to calculate the energy level is, En=−Z2(13.6 eV)n2 En is the nth energy level, n is nth...Explanation: Formula to calculate the wavelength is, λ=1μZ2(36hc5k) μ is the reduced mass, Z is...Section1: To determine: The ionization energy of the L-shell. Answer: The ionization energy of the...Explanation: Expression the angular momentum associated with the orbital motion of Earth to satisfy...Explanation: Alpha emission is the radioactive decay process where the atomic nucleus emits an alpha...Explanation: Section 1: To determine: The orbital radii of C12 . Answer: The orbital radii of C12 is...Explanation: Formula to calculate the binding energy per nucleon is, BEA=[Zmp−Nmn−m(H12)]c2A mp is...Section 1: To determine: The binding energy per nucleon of F2656e . Answer: The binding energy per...Explanation: Given info: Half-life of I131 is 8.04 days. 1 day=(24)(3600 s) The half life is,...Explanation: Given info: Time elapsed is 1730 days. Half-life of C137 is 1.10×104days . Half-life of...Explanation: The mass of the particles on the left side is, mi=m(H11)+m(L37i) m(H11) is the mass of...Given Info: The sample contains 3.5 μg of C11. Half-life of C11 is 20.4 min. Formula to calculate...The reaction is, K+→π++π0 The following table gives the conservation law violated in the reaction....Explanation: Given Info: The percentage of mass reserved is 0.70% . The amount of uranium reserved...Explanation: Given Info: The density is 3×10−3 g/m3 . The deapth is 4×103 m . The radius is 6.38×106...The reaction is, 11H+612C→713A+γ⇒11H+612C→713N+γ Thus, the nucleus A is 713N. Conclusion: The...Explanation: The reaction is, Λ0→p+π+ A strangeness of Λ0 is -1, the strangeness of p is 0 and the...Explanation: Given info: Mass of water is 1.00 L . Avogadro number is 6.02×1023 molecules/mol . The...Explanation: Given info: The mass of ocean is 1.32×1021 kg The mass per mol is 18.0×10−3 kg/mol The...
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