College Physics (Instructor's)11th EditionSERWAY Publisher: CENGAGE LISBN: 9781305965317
College Physics (Instructor's)
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Chapter 1 - Units, Trigonometry. And VectorsChapter 1.9 - VectorsChapter 1.10 - CornoonentsofavectorChapter 2 - Motion In One DimensionChapter 2.1 - Displacement, Velocity, And AccelerationChapter 2.2 - Motion DiagramsChapter 2.4 - Freely Fall Ing ObjectsChapter 3 - Motion In Two DimensionsChapter 3.1 - Displacement, Velocity, And AccelerationChapter 4 - Newton's Laws Of Motion
Chapter 4.2 - The Laws Of MotionChapter 4.4 - Static Friction ForcesChapter 4.6 - Applications Of Newton's LawsChapter 5 - EnergyChapter 5.1 - WorkChapter 5.2 - Kinetic Energy And T He Work- Energy TheoremChapter 5.3 - Gravitational Potential EnergyChapter 5.5 - Spring Potential EnergyChapter 5.6 - Systems And Energy ConservationChapter 6 - Momentum, Impulse, And CollisionsChapter 6.1 - Momentum And ImpulseChapter 6.2 - Conservation Of MomentumChapter 6.3 - Collisions In One DimensionChapter 7 - Rotational Motion And GravitationChapter 7.1 - Angular Velocity And Angular AccelerationChapter 7.2 - Rotational Motion Under Constant Angular AccelerationChapter 7.3 - Tangential Velocity, Tangential Acceleration, And Centripetal AccelerationChapter 7.5 - Newtonian GravitationChapter 8 - Rotational Equilibrium And DynamicsChapter 8.4 - The Rotational Second Law Of MotionChapter 8.5 - Rotational Kinetic EnergyChapter 8.6 - Angular MomentumChapter 9 - Fluids And SolidsChapter 9.2 - Density And PressureChapter 9.3 - Variation Of Pressure With DepthChapter 9.4 - Pressure MeasurementsChapter 9.5 - Buoyant Forces And Archimedes' PrincipleChapter 9.6 - 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 12.3 - Thermal Processes In GasesChapter 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 - Concepts Of Oscillation Rates In Simple Harmonic MotionChapter 13.4 - Position, Velocity, And Acceleration As Functions 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 14.10 - Standing Waves In Air ColumnsChapter 14.11 - BeatsChapter 15 - Electric Forces And FieldsChapter 15.1 - Electric Charges, Insulators, And ConductorsChapter 15.2 - Coulomb's LawChapter 15.3 - Electric FieldsChapter 15.4 - Elect Ric Field LinesChapter 15.8 - Electric Flux And Gauss' LawChapter 16 - Electrical Energy And CapacitanceChapter 16.1 - Electric Potential Energy And Electric PotentialChapter 16.2 - Electric Potential And Potential Energy Of Point ChargesChapter 16.3 - Potentials, Charged Conductors, And Equipotential SurfacesChapter 16.6 - Combinations Of CapacitorsChapter 16.7 - Energy In A CapacitorChapter 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 Vo Ltage 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 Se RiesChapter 18.3 - Resistors In Pa RallelChapter 18.5 - Rc CircuitsChapter 19 - MagnetismChapter 19.3 - Magnetic FieldsChapter 19.5 - Magnetic Force On A Current -Carrying ConductorChapter 19.6 - MagnetictorqueChapter 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 Circu ItsChapter 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 Refraction BChapter 23 - Mirrors And LensesChapter 23.1 - Flat MirrorsChapter 23.3 - Images Formed By RefractionChapter 23.5 - Thin LensesChapter 24 - Wave OpticsChapter 24.2 - Young's Double·Siit ExperimentChapter 24.4 - Interference In Thin FilmsChapter 24.7 - ingle·Siit 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
Sample Solutions for this Textbook
We offer sample solutions for College Physics (Instructor's) homework problems. See examples below:
Chapter 1, Problem 1CQChapter 1, Problem 13PChapter 1, Problem 22PChapter 1, Problem 36PChapter 1, Problem 43PChapter 1, Problem 49PChapter 1, Problem 74APChapter 2, Problem 1CQChapter 2, Problem 6P
Chapter 2, Problem 20PChapter 2, Problem 36PChapter 2, Problem 48PChapter 2, Problem 49PChapter 2, Problem 50PChapter 2, Problem 52PChapter 2, Problem 53PChapter 3, Problem 1CQChapter 3, Problem 4PChapter 3, Problem 7PChapter 3, Problem 14PChapter 3, Problem 23PChapter 3, Problem 26PChapter 3, Problem 29PChapter 3, Problem 39APChapter 4, Problem 1CQChapter 4, Problem 15PChapter 4, Problem 22PChapter 4, Problem 23PChapter 4, Problem 37PChapter 4, Problem 46PChapter 4, Problem 61PChapter 4, Problem 62PChapter 4, Problem 70PChapter 4, Problem 73APChapter 4, Problem 85APChapter 4, Problem 92APChapter 5, Problem 1CQChapter 5, Problem 22PChapter 5, Problem 29PChapter 5, Problem 31PChapter 5, Problem 46PChapter 5, Problem 49PChapter 5, Problem 60PChapter 5, Problem 63APChapter 5, Problem 69APChapter 5, Problem 71APChapter 5, Problem 77APChapter 6, Problem 1CQChapter 6, Problem 33PChapter 6, Problem 46PChapter 6, Problem 58PChapter 6, Problem 62APChapter 6, Problem 63APChapter 6, Problem 68APChapter 6, Problem 70APChapter 6, Problem 71APChapter 6, Problem 75APChapter 6, Problem 83APChapter 7, Problem 1CQChapter 7, Problem 12PChapter 7, Problem 17PChapter 7, Problem 19PChapter 7, Problem 31PChapter 7, Problem 34PChapter 7, Problem 49APChapter 7, Problem 50APChapter 7, Problem 64APChapter 8, Problem 1CQChapter 8, Problem 28PExplanation: The weight of the beam acting downward from its midpoint, tension of the acting left...Chapter 8, Problem 37PChapter 8, Problem 40PChapter 8, Problem 50PChapter 8, Problem 60PChapter 8, Problem 80APChapter 8, Problem 84APExplanation Given Info: The mass of the arms is 6.87 kg , the mass of torso is 33.57 kg , the mass...Chapter 8, Problem 89APExplanation: The following figure shows the free body diagram.Chapter 9, Problem 1CQChapter 9, Problem 7PExplanation: 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 22PChapter 9, Problem 31PExplanation: Formula to calculate the time taken by the water to reach the ground is, t=2(Δy)ay t is...Chapter 9, Problem 45PChapter 9, Problem 46PChapter 9, Problem 47PChapter 9, Problem 73PChapter 9, Problem 83APChapter 10, Problem 1CQChapter 10, Problem 2CQSection 1: Explanation: To determine: The constants a and b. Answer: The constant (a) is 4.7×10−3...Chapter 10, Problem 15PChapter 10, Problem 21PChapter 10, Problem 26PChapter 10, Problem 44PChapter 10, Problem 60APChapter 10, Problem 64APChapter 11, Problem 1CQExplanation: Given Info: Expression for the compressive stress and strain is, FA=Y(ΔLL0) F is the...Explanation: Given Info: Initial temperature of water and calorimeter is 10.0°C , mass of aluminum...Given Info: Mass of ice cube is 40 g, initial temperature of ice cube is −10°C , temperature of...Explanation: 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 39PChapter 11, Problem 44PChapter 11, Problem 46PChapter 11, Problem 68APChapter 12, Problem 1CQChapter 12, Problem 5PChapter 12, Problem 14PChapter 12, Problem 22PChapter 12, Problem 25PChapter 12, Problem 26PChapter 12, Problem 27PChapter 12, Problem 30PChapter 12, Problem 71APSection1: To determine: The constant volume occupied by the gas. Answer: The volume of the gas is...Chapter 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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