Sears and Zemansky's University Physics for JEE Mains and Advance Vol I

753 Pages·2019·85.047 MB·English

by  Hugh D. Young, Roger Freedman, A. Lewis Ford, Nitin Jain, Neel Kamal Sethla

THIRTEENTH EDITION
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Hugh D. Young
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Roger A. Freedman a a
UNIVERSITY PHYSICS n r
A. Lewis Ford sk s a
JEE y n
Nitin Jain for Mains and Advance ’s d
Neel Kamal Sethia
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EI THIRTEENTH EDITION
University Physics for JEE, Volume I, 13th Edition, is an Indian adaptation of the internationally-renowned bestseller V
‘University Physics with Modern Physics by Young, Freedman and Ford’. The Indian adaptation is modified as per the JEE E
E
syllabus and examination pattern, strives to meet the requirements of the aspirants. The title provides a perfect blend of Sears and Zemansky’s

M R
theoretical clarity along with optimum number of practice exercises, detailed solved examples, conceptual questions, and
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application-based questions. Along with these, new sets of question aligned with JEE structure, pattern and content extent S
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also incorporated. n I UNIVERSITY PHYSICS
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a T
SALIENT FEATURES n Y
d
 Topics thoughtfully synchronized to meet students’ requirements
A P JEE
 Numerous illustrations, exhaustive worked-out examples and application-based exercises for self-learning d for Mains and Advance
 Research-based problem solving approach used in examples as well as in problem solving strategies v H
a
 Vast collection of objective questions as per JEE pattern and syllabus n Y
c
 All question types as per JEE (Main and Advance) covered
e S
 Single/Multiple Correct Choice Type
I (cid:31)(cid:30) (cid:29)(cid:28)(cid:27) (cid:26) (cid:25) (cid:24)
 Passage Type C
 Assertion-Reason Type
S
 Integer Type
 Matrix-Match Type
(cid:31)
ABOUT THE ADAPTERS
(cid:30)
Nitin Jain, B.E. (Hons.) from BITS Pilani, has more than two decades of experience in imparting proficiency in the (cid:29) HIGHLIGHTS
(cid:25)
subject to aspirants of IIT-JEE and Physics Olympiads. His association with apex institutions at Kota enabled him to (cid:24) 
cultivate a distinct expertise to identify the peculiar problems of students, which he puts to good use to reveal to Customised as per JEE syllabus and examination pattern
them the nuances of cracking the IIT-JEE and the Physics Olympiads.
Neel Kamal Sethia, B.Tech. (Hons.) from IIT Kharagpur, has been teaching physics for the past sixteen years. His Cover Im 13e  Introduced new set of questions for effective learning
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Freedman
Ford
MRP Inclusive `750.00
of all Taxes
Jain
ini.np.peeaarsrsoonn.c.coomm Sethia Hugh D. Young I Roger A. Freedman
A. Lewis Ford
This edition is manufactured in India and is authorized for sale only Nitin Jain I Neel Kamal Sethia
in India, Bangladesh, Bhutan, Pakistan, Nepal, Sri Lanka and the Maldives.
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PhET SimulaTionS
Available in the Pearson eText and in the Study Area of MasteringPhysics
Section Page
1.6 Estimation 10
1.7 Vector Addition 13
2.4 * Forces in 1 Dimension 49
2.4 * The Moving Man 51
2.5 Lunar Lander 54
3.2 Maze Game 80
3.3 * Projectile Motion 83
3.4 Ladybug Revolution, Motion in 2D 91
5.2 Lunar Lander 156
5.3 Forces in 1 Dimension, Friction, *The Ramp 159
6.2 * The Ramp 197
6.3 Molecular Motors, Stretching DNA 204
7.3 * The Ramp 240
7.5 * Energy Skate Park 247
9.3 Ladybug Revolution 316
10.6 Torque 358
12.3 Balloons & Buoyancy 416
13.2 Lunar Lander 448
13.4 My Solar System 454
14.2 Motion in 2D 485
14.3 * Masses & Springs 488
14.5 * Pendulum Lab 495
15.8 Fourier: Making Waves, Waves on a String 539
16.6 Sound, Wave Interference 575
17.6 States of Matter 614
17.7 The Greenhouse Effect 618
18.3 Balloons & Buoyancy, Friction,
Gas Properties 651
18.6 States of Matter 664
*Indicates an associated tutorial available in the MasteringPhysics Item Library.
A P O L ™
ctiv hysics n ine
Activities
www.masteringphysics.com
1.1 Analyzing Motion Using Diagrams 7.6 Rotational Inertia 11.12 Electric Potential, Field, and Force
1.2 Analyzing Motion Using Graphs 7.7 Rotational Kinematics 11.13 Electrical Potential Energy and Potential
1.3 Predicting Motion from Graphs 7.8 Rotoride–Dynamics Approach 12.1 DC Series Circuits (Qualitative)
1.4 Predicting Motion from Equations 7.9 Falling Ladder 12.2 DC Parallel Circuits
1.5 Problem-Solving Strategies for 7.10 Woman and Flywheel Elevator–Dynamics 12.3 DC Circuit Puzzles
Kinematics Approach 12.4 Using Ammeters and Voltmeters
1.6 Skier Races Downhill 7.11 Race Between a Block and a Disk 12.5 Using Kirchhoff’s Laws
1.7 Balloonist Drops Lemonade 7.12 Woman and Flywheel Elevator–Energy 12.6 Capacitance
1.8 Seat Belts Save Lives Approach 12.7 Series and Parallel Capacitors
1.9 Screeching to a Halt 7.13 Rotoride–Energy Approach 12.8 RC Circuit Time Constants
1.10 Pole-Vaulter Lands 7.14 Ball Hits Bat 13.1 Magnetic Field of a Wire
1.11 Car Starts, Then Stops 8.1 Characteristics of a Gas 13.2 Magnetic Field of a Loop
1.12 Solving Two-Vehicle Problems 8.2 Maxwell-Boltzmann 13.3 Magnetic Field of a Solenoid
1.13 Car Catches Truck Distribution–Conceptual Analysis 13.4 Magnetic Force on a Particle
1.14 Avoiding a Rear-End Collision 8.3 Maxwell-Boltzmann 13.5 Magnetic Force on a Wire
2.1.1 Force Magnitudes Distribution–Quantitative Analysis 13.6 Magnetic Torque on a Loop
2.1.2 Skydiver 8.4 State Variables and Ideal Gas Law 13.7 Mass Spectrometer
2.1.3 Tension Change 8.5 Work Done By a Gas 13.8 Velocity Selector
2.1.4 Sliding on an Incline 8.6 Heat, Internal Energy, and First Law 13.9 Electromagnetic Induction
2.1.5 Car Race of Thermodynamics 13.10 Motional emf
2.2 Lifting a Crate 8.7 Heat Capacity 14.1 The RL Circuit
2.3 Lowering a Crate 8.8 Isochoric Process 14.2 The RLC Oscillator
2.4 Rocket Blasts Off 8.9 Isobaric Process 14.3 The Driven Oscillator
2.5 Truck Pulls Crate 8.10 Isothermal Process 15.1 Reflection and Refraction
2.6 Pushing a Crate Up a Wall 8.11 Adiabatic Process 15.2 Total Internal Reflection
2.7 Skier Goes Down a Slope 8.12 Cyclic Process–Strategies 15.3 Refraction Applications
2.8 Skier and Rope Tow 8.13 Cyclic Process–Problems 15.4 Plane Mirrors
2.9 Pole-Vaulter Vaults 8.14 Carnot Cycle 15.5 Spherical Mirrors: Ray Diagrams
2.10 Truck Pulls Two Crates 9.1 Position Graphs and Equations 15.6 Spherical Mirror: The Mirror Equation
2.11 Modified Atwood Machine 9.2 Describing Vibrational Motion 15.7 Spherical Mirror: Linear Magnification
3.1 Solving Projectile Motion Problems 9.3 Vibrational Energy 15.8 Spherical Mirror: Problems
3.2 Two Balls Falling 9.4 Two Ways to Weigh Young Tarzan 15.9 Thin-Lens Ray Diagrams
3.3 Changing the x-Velocity 9.5 Ape Drops Tarzan 15.10 Converging Lens Problems
3.4 Projectile x- and y-Accelerations 9.6 Releasing a Vibrating Skier I 15.11 Diverging Lens Problems
3.5 Initial Velocity Components 9.7 Releasing a Vibrating Skier II 15.12 Two-Lens Optical Systems
3.6 Target Practice I 9.8 One-and Two-Spring Vibrating Systems 16.1 Two-Source Interference: Introduction
3.7 Target Practice II 9.9 Vibro-Ride 16.2 Two-Source Interference: Qualitative
4.1 Magnitude of Centripetal Acceleration
9.10 Pendulum Frequency Questions
4.2 Circular Motion Problem Solving
9.11 Risky Pendulum Walk 16.3 Two-Source Interference: Problems
4.3 Cart Goes Over Circular Path
9.12 Physical Pendulum 16.4 The Grating: Introduction and Qualitative
4.4 Ball Swings on a String
10.1 Properties of Mechanical Waves Questions
4.5 Car Circles a Track
10.2 Speed of Waves on a String 16.5 The Grating: Problems
4.6 Satellites Orbit
10.3 Speed of Sound in a Gas 16.6 Single-Slit Diffraction
5.1 Work Calculations
10.4 Standing Waves on Strings 16.7 Circular Hole Diffraction
5.2 Upward-Moving Elevator Stops
10.5 Tuning a Stringed Instrument: 16.8 Resolving Power
5.3 Stopping a Downward-Moving Elevator
Standing Waves 16.9 Polarization
5.4 Inverse Bungee Jumper
5.5 Spring-Launched Bowler 10.6 String Mass and Standing Waves 17.1 Relativity of Time
5.6 Skier Speed 10.7 Beats and Beat Frequency 17.2 Relativity of Length
5.7 Modified Atwood Machine 10.8 Doppler Effect: Conceptual Introduction 17.3 Photoelectric Effect
6.1 Momentum and Energy Change 10.9 Doppler Effect: Problems 17.4 Compton Scattering
6.2 Collisions and Elasticity 10.10 Complex Waves: Fourier Analysis 17.5 Electron Interference
6.3 Momentum Conservation and Collisions 11.1 Electric Force: Coulomb’s Law 17.6 Uncertainty Principle
6.4 Collision Problems 11.2 Electric Force: Superposition Principle 17.7 Wave Packets
6.5 Car Collision: Two Dimensions 11.3 Electric Force: Superposition Principle 18.1 The Bohr Model
6.6 Saving an Astronaut (Quantitative) 18.2 Spectroscopy
6.7 Explosion Problems 11.4 Electric Field: Point Charge 18.3 The Laser
6.8 Skier and Cart 11.5 Electric Field Due to a Dipole 19.1 Particle Scattering
6.9 Pendulum Bashes Box 11.6 Electric Field: Problems 19.2 Nuclear Binding Energy
6.10 Pendulum Person-Projectile Bowling 11.7 Electric Flux 19.3 Fusion
7.1 Calculating Torques 11.8 Gauss’s Law 19.4 Radioactivity
7.2 A Tilted Beam: Torques and Equilibrium 11.9 Motion of a Charge in an Electric Field: 19.5 Particle Physics
7.3 Arm Levers Introduction
7.4 Two Painters on a Beam 11.10 Motion in an Electric Field: Problems
7.5 Lecturing from a Beam 11.11 Electric Potential: Qualitative
Introduction
SEaRS and ZEmanSkY’S
uniVERSiTY
PHYSiCS
FOR THE JEE
VOLUME I
Copyright © 2019 Pearson India Education Services Pvt. Ltd

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BRiEf ConTEnTS
MEcHanIcs WaVEs/acOUsTIcs
1 Units, Physical Quantities, and Vectors 1 15 Mechanical Waves 516
2 Motion Along a Straight Line 37 16 Sound and Hearing 555
3 Motion in Two or Three Dimensions 73
THERMOdynaMIcs
4 Newton’s Laws of Motion 112
5 Applying Newton’s Laws 144 17 Temperature and Heat 599
18 Thermal Properties of Matter 642
6 Work and Kinetic Energy 192
19 The First Law of Thermodynamics 676
7 Potential Energy and Energy
Conservation 225
8 Momentum, Impulse, and Collisions 265 appEndIcEs
9 Rotation of Rigid Bodies 308 A The International System of Units A-1
B Useful Mathematical Relations A-3
10 Dynamics of Rotational Motion 340
C The Greek Alphabet A-4
11 Equilibrium and Elasticity 380 D Periodic Table of Elements A-5
E Unit Conversion Factors A-6
12 Fluid Mechanics 409
F Numerical Constants A-7
13 Gravitation 444 G Surface Energy and Surface Tension A-9
14 Periodic Motion 479 Photo Credits C-1
140 CHAPTER5Applying Newton’s Laws
5.6Correct and incorrect free-body dia- 5.2 Using Newton’s Second Law:
grams for a falling body.
140 CHAPTER5Applying Newton’s Laws Dynamics of Particles
(a)
5.6Correct and incorrect free-body dia- 5We. a2re noUws rienadgy Nto ediwscutsos ndy’sna mSiecscporonbdlem Lsa. Iwn t:h ese problems, we apply
grams for a falling body. Newton’s second law to bodies on which the net force is notzero. These bodies
Build Skills (a) Oacntsly o tnh eth fios rfcael loinf gg rfaruviitt.y aerqeu anlo ttoi tnhD eey qmunailasisbm roifui mtch esa bn odod hfy e tnPicmaee ras rtietis ca aclcceceelselerraatitniogn. :The net force on the body is
We are now ready to discuss dynamicsproblems. In these problems, we apply
(b) Correct fOrenel-yb othdey fdoiracger aomf gravity Nareew ntootn’isn seeqcuoianlidb FSrliauw(cid:31)m t moa nSabdo dhieens(c Noe nea wrwetho aincc’hcs e stlheecerao ntniendtg lf.ao wTrc,h eve einsc etnoto rf tofozrcermero )o. nT htheese b bo(o5d.dy3i )eiss
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We most often use this relationship in component form:
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solve a broad range of physics problems. You can safely draw problems. You can solve anydynamics problem using this strategy.
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Motsot bEexa mmoprlee sc oemncpisleo ya na dd ifaogcruasmed—. htrshepisleelu c.a liTtca chlw eecel ae dsfrioaeretsui.cno tDdnio oiinnsn o’dttfh i fmitfshe eeerm exknaoitmnr iefzprteoilce mt hfi nritischc letrui eoddsneuo slwf to,an rlbhclueitl thli esdv oanp lorumeweva. i kroIeetuv stesu urorsrnneesed sy,o osauoust
often a pencil sketch that sVhiosuwasl ttuhhnaadttS etturhhspeetpa emnoxdsipne rh ueiosnsw sssiti oegwnaned f b oowerbc teaoa xmigniiseev sdteh p iettl husaesan .md tCo ewba anohs ga yftgoo aiurtn dm soahenwoa wnninsh t.iihtlilia sml? optuisohn uexpctehpet
hill. The direction of the kinetic friction force is now reversed, so
Most Exwahmapt lae ss teumdepnlot ys hao duialdg rdarmaw—. the acceleration is different from the downhill value. It turns out
often a pencil sketch that shows Rolling Friction tthhaatt tthhee emxipnruess ssiiognn fboerc aoxmise sth pel ussa.m Cea ans yfoour dsohwownh tihlli sm?otion except
what a student should draw. It’s a lot easier to move a loaded filing cabinet across a horizontal floor using a
cart with wheels than to slide it. How much easier? We can define a coefficient of
Rrfloaoltl lislnuignr fgfar ciFecr tdiiocivntiidmoernd , wbyh icthhe i su pthwea hrdo rnizoornmtaall ffoorrccee neexeedretedd f obry ctohnes tsaunrtf ascpee.e Td roann sa-
Ipto’sr taa tiloont eeansgieinr eteor sm coavlle mar ltohaed terda cfitliivneg r ecsaibsitnaentc ea.crToyspsi caa hl ovrailzuoenst aolf flmoroarr eu s0in.0g0 2a
ctoa r0t .w00it3h fwohr esetelse lt hwanh eteol ss liodne sitte. eHl orwai lms uacnhd e0a.s0i1e rt?o W0.e0 c2a fno dr erfiunbeb ear ctoireefsfi coine ncto onf-
rflceorfafieltlt ciesin.ue gTnr fthfa rtechisecae tn divo ihavniligudmheerswd , s awbhyyoh t wirctuh hoce kni suse .p trhweeaa shrodon rn irzoaorinlmrtoaaall dff ootrrrcaceien nse exaeerderet gedde fn oberyr a cltolhyne s mtsauunrctfh as cpmee.oe rTde r oafnun esa-l
portation engineers call mrthe tractive resistance.Typical values of mrare 0.002
tFol u0i.0d0 R3 efosri sstteaenl cweh eaelnsd o nT estremeli rnaaills Sanpde 0e.d01 to 0.02 for rubber tires on con-
cSrteictek.i nTgh eysoeu vr ahlauneds oshuot wth eo nwei nredaoswon o rfa ail fraosatd-m troavinins ga rcea rg ewniellr aclolyn vminucceh ymoour oef f tuheel
eefxfiisctieennct et hoafn fl huigidh wreasyi sttraunckcse.,the force that a fluid (a gas or liquid) exerts on a
body moving through it. The moving body exerts a force on the fluid to push it
Fluid Resistance and Terminal Speed
out of the way. By Newton’s third law, the fluid pushes back on the body with an
Seqtiucakli nagn dy oouprp ohsainted foourtc eth.e window of a fast-moving car will convince you of the
exiTstheen cdei roefc tfliounido fr ethseis tflaunidce r,etshiset afnocrcee f othrcaet aa cfltiunigd o(na ag abso odry liisq ualiwd)a yexs eorptsp oosni tae
bthoed yd imreocvtiionng othf rtohueg hb oidt.y T’sh ev emloocvitiyn gr ebloatdiyv ee xtoe rtths ea flfouricde. Tonh et hme aflguniidtu tdoe poufs hth iet
ofluuti do rfe tshiest awnacye. fBoyrc Ne uewsutaolnly’s i nthcirreda sleasw w, tihthe tflhuei sdp peuedsh oefs t bhea cbko doyn tthhreo buogdhy th we iflthu iadn.
equal and opposite force.
The directionof the fluid resistance force acting on a body is always opposite
the direction of the body’s velocity relative to the fluid. The magnitudeof the
fluid resistance force usually increases with the speed of the body through the fluid.
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MasteringPhysics offers an extensive set of assignable mathematics
NinE aW V!i dVeidoe Tou tTourt soorl uStoiolunt iporno vfoidre Edv ienr yth Eex Satmudpyl eArea NreEvWiew! Mtuatothrieamls—atciocvs eRrienvgie dwif fTeuretonrtiiaall sand integral calculus as well
Eoaf cMha EsxtearminpglPeh iys seicxsp®la ainnedd i na nthde s oPelvaerdso bny eaTne xint.structor Masa asltgeeribnrgaP ahnysdi ctsr iogoffenrosm ane teryx.t ensive set of assignable mathematics
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NEW! Bridging Problems
At the start of each problem set, a
332 CHAPTER10Dynamics of Rotational Motion
Bridging Problem helps students ◄
make the leap from routine BRIDGING PROBLEM Billiard Physics
Acue ball (a uniform solid sphere of mass mand radius R) is at 3.Draw two free-body diagrams for the ball in part (b): one show-
exercises to challenging problems rest on a level pool table. Using a pool cue, you give the ball a ing the forces during the hit and the other showing the forces
sharp, horizontal hit of magnitude Fat a height habove the center after the hit but before the ball is rolling without slipping.
with confidence and ease. of the ball (Fig. 10.37). The force of the hit is much greater 4.What is the angular speed of the ball in part (b) just after the
than the friction force ƒthat the table surface exerts on the ball. hit? While the ball is sliding, does vcmincrease or decrease?
The hit lasts for a short time ¢t. (a) For what value of Does vincrease or decrease? What is the relationship between
Each Bridging Problem poses a hcewntielrl 1thhe =ba0ll2 ,r othlle wbaitlhl owuitl ls slilpidpein agc?r o(sbs) tIhfe ytaobul eh ifto rth ae wbhalill ed, ebaudt vcmand vwhen the ball is finally rolling without slipping?
eventually it will roll without slipping. What will the speed of its EXECUTE
moderately diffi cult, multi-concept center of mass be then? 5.In part (a), use the impulse–momentum theorem to find the
speed of the ball’s center of mass immediately after the hit.
problem, which often draws on earlier SOLUTION GUIDE oTrheemn utos efi tnhde trhoet aatniognualal rv seprseieodn iomf mtheed iiamtepluyl saef–temr othmee hnittu. m(H tihnet:-
See MasteringPhysics®study area for a Video Tutor solution. To write down the rotational version of the impulse–momentum
chapters. In place of a full solution, IDENTIFY andSET UP theorem, remember that the relationship between torque and
1.Draw a free-body diagram for the ball for the situation in part (a), angular momentum is the same as that between force and linear
it provides a skeleton solution guide einxcelrutsd inbgo thy oaunr icmhopiucles ivoef fcoorocred ionna teth ea xbeas.l l Naontde atnh aitm tphue lsciuvee 6.mUsoem yeonuturm re.)sults from step 5 to find the value of hthat will
consisting of questions and hints. 2.tTmtoimhraqeseu s ce gua aie vrs oefpouse rnectddhe et hva epcbm pac,lle ilnae tnadend rf ootahrfn emag ucatilusmaser. e t so¢preqteugdei v vaeps. ptWlhieeh dab taf olmlr’ sut shctae tnb tesea rtm hoeef 7.cIsanean cugposueanl radtthr les(a bpwb)ea, e flldoa t grio mat hirnmoe l letf rdiwaniniadtsth eltoalhyutei to a snbfltaaiepllr lp m’tishn oegct iehionmintt.me aTrne-hoddefi a-rnmot ewtalayrsti isato efns tNeparlee etmwhdeot o tahinnoi’tdns.
A full solution is explained in rselilpaptiionngs?hip between vcmand vfor the ball to roll without oexf ptrhees siboanlsl faosr ivt cmis asnldid ivnga. s Ufusne cttihoensse oefq uthaeti oenlasp steod wtirmitee
tsince the hit.
a Video Tutor, provided in the 8.Using your results from step 7, find the time twhen vcmand v
have the correct relationship for rolling without slipping. Then
Study Area of MasteringPhysics® 10.37 find the value of vcmat this time.
EVALUATE
and in the Pearson eText. h mass m 9.I(fa )y aonud h (abv)e f oarc cyeosusr stoe laf! pool table, test out the results of parts
R 10.Csoalnid y boaul ls,h yoowu thwaot uifld y ohua vues etdo ah hito tlhloew t ocpy lionfd tehre r actyhleirn dthearn t oa
cause rolling without slipping as in part (a)?
◄ In responPrsoebl etmos professors, the PrFoorb inslterumctor- aSsseigntesd h onmoewwork, gino tco wluwwd.meas temringophryseic s.com
biomedr.e, iq.cu.,ir a.in.gl.l :c yParl cooublluersm.i seB IoOnf: inBtcieorsecdaisei nncpge dsri fpofircoubbltlyel.m eCsPm.: Cusm ul(atBiveI pOrob)le,m sm incoorproera tidng imffiat ecriual flrtom p earrolierb chlaeptmers.s C ALC: Problems
requirinDIgS CcUSaSlIcOuN lQuUsES (TICONASLC), and more cuQm10.4uAlafotuir-vwehe elp-drrivoe bcarl ies amccesler attihnga fotr ward from rest.
draw oeQnn1e0 d.e1, thWae hrcerlinti icecyallri nq ducearnh-thiteyaa disp btohtlete st oirnrq saune a(aupCtpolmiPeodb )tiol.e t ehne gbionlet sa.r We thigyh its- Sfrhicotwio nth feo drciree cdtuioe nt oth teh ec apra’sv ewmheenetl st htuart na cacnedl ehroawte st hthise ccaaurs efos ra-
the torque more important than the actual forceapplied to the ward.
wrench handle? Q10.5Serious bicyclists say that if you reduce the weight of a
About 2Q100.2%Can oa sfin gple rfoorceb aplpeliemd tos a baodry ech anngee bwoth itos trran srlae-vbiiskee, idt is. mTorhe eeffescteive rife yovu idso isoo inn thse wahereels rather than in
tional and rotational motion? Explain. the frame. Why would reducing weight in the wheels make it
driven Qb10y.3 dSupepotsae yioleu cdou lds utseu wdheeelsn of ta-nyp tyeper ifn other dmesigan onf acee adsiear otna t hge baictyhcliest rtheand re dnucaingt itohe nsaamel laym ount in the
soapbox-derby racer (an unpowered, four-wheel vehicle that frame?
coasts from rest down a hill). To conform to the rules on the total Q10.6The harder you hit the brakes while driving forward, the
throughwe igMht oaf tshet veehricilne agndP rihdery, sshoiuclds y.ou design with large mas- more the front end of your car will move down (and the rear end
sive wheels or small light wheels? Should you use solid wheels or move up). Why? What happens when cars accelerate forward?
wheels with most of the mass at the rim? Explain. Why do drag racers not use front-wheel drive only?
Problem diffi culty is now indicated by a three-dot ranking
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290 CHAPTER9Rotation of Rigid Bodies
Example 9.6 Moments of inertia for different rotation axes
Amachine part (Fig. 9.15) consists of three disks linked by light- massless. In parts (a) and (b), we’ll use Eq. (9.16)to find the
weight struts. (a) What is this body’s moment of inertia about an moments of inertia. Given the moment of inertia about axis A,
axis through the center of disk A, perpendicular to the plane of the we’ll use Eq. (9.17)in part (c) to find the rotational kinetic energy.
diagram? (b) What is its moment of inertia about an axis through
the centers of disks Band C? (c) What is the body’s kinetic energy EXECUTE:(a) The particle at point Alies onthe axis through A, so
its distance rfrom the axis is zero and it contributes nothing to the
i4f. 0i t rardotast?es about the axis through Awith angular speed v= moment of inertia. Hence only BandCcontribute, and Eq. (9.16)
gives
IDSEONLTUIF>TYI OaNndSET UP:We’ll consider the disks as massive parti- IA= ami ri 2 ==100.0.1507 kkgg2#1m0.250 m22+10.20 kg210.40 m22
cles located at the centers of the disks, and consider the struts as (b) The particles at Band Cboth lie on axis BC, so neither parti-
9.15An oddly shaped machine part. cle contributes to the moment of inertia. Hence only Acontributes:
Adixskiss tBhr aonudg hC B IBC= ami ri 2 = 0.30 kg 0.40 m2=0.048 kg#m2
mB 5 0.10 kg (c)From Eq. (9.17),1 21 2
Bring Physics to Life A(pxeirsp ethnrdoiucguhla rd itsok p Alane 0.50 m 0.30 m EVALUKAATE=:T12IhAev m2o=me12 n1t0 .o0f5 i7n ekrgtia# mab2o21u4t .a0x irsa dA > iss2 2gr=eat0e.r4 t6h aJn that
of figure) A 0.40 m C mC 5 0.20 kg ambaocuhti naex pisa rBt Cro.t aHtee nabceo uot fa xtihse B tCw.o axes it’s easier to make the
mA 5 0.30 kg
Deepen knowledge of physics by building ApplicationMoment of Inertia of a CAUTION Moment of inertia depends on the choice of axisThe results of parts (a) and (b)
Bird’s Wing of Example 9.6 show that the moment of inertia of a body depends on the location and ori-
connections to the real world. 190 CHAPTER6Work and Kinetic Energy Wwhmwuiionnhmmggemssne niumantp g ob obavifrien rdi ddnr fleahdarpaotpisdiwas ly n,si t ms(sauo rapw olt luh itnwoneg id n7sb g,0tir hsi dte b wr ecsoiaaththntaos tua mepl dsseae mrktr eh.sa eeAliltcs- te0hn.0et a4at8ixo iksng toh#frm ot2uh.ge”h Wa Bxei sah.n aIdvt ’eCs tnioso b0t e.e0 sn4po8eu ckgigfih c#t moa n2jd.u” ssta❙ ysa, y“,T “hTeh me ommoemnte notf ionfe rintiear toifa tohfis t hbiosd yb oadbyo uist
gornydp)h. uBsy) choanst rimasmt,e tnhsee A wnidnegasn t hcaotn adroer h(Vaurldtu tro In Example 9.6 we represented the body as several point masses, and we eval-
ApplicationTendons Are Nonideal Wormk–ovEe ndueer tgo yth eTirh laeroger meomm efnot rof Sinetrrtiaa.ight-Luianteed Mtheo tsiuomn ,in Eq. (9.16)directly. When the body is a continuousdistribution
Springs VaryCionngdo rFso flrapc ethseir wings at about one beat per of matter, such as a solid cylinder or plate, the sum becomes an integral, and we
Malotntuagsc,c hlse ttsihf fleeyxm ee rltatos f tobicroc nceeossl l.av igAae ttnhe nefib dteoernns d.c ooTnnhsse it sghtrasa topfh Icnia Sl eccssateosicaeoor n now df6h .oisl2entr whataoiekglde hdiontefg-fr,l tiibhnvueeeti rd ma wtt hoimnetgio sows nts ot tewirmakide–tyshe. npare ercfgeoyrn tstohtaenotrne enxemeeatdm, fW optortloce tesu .=s Woe fKe c 2ascluac-cnuh lKn uosc1 w,atlfo cop urrc lotaahvltecieo u stnlphaseat -eti nt heS emctoiomne n9t. 6o;f mineeartniwa.h Wilee, wTialbl lge iv9e. 2segvievreasl
sareh wsopawolslan bsheyo w(tao t shamen ataellp nkpdalioenndg affrrooormoce) .ts htTerhe ehtc intheden slde oignn o df oes tlheti’ss tchoenosriedmer ias ptraureti celvee tnh awt huennd etrhgeo feosr ac ed ivsaprliaecsmde wimomimetenhne tns pxitoosws nihosti.ifo l Eeni a.nb ceAehristn ib agion ad fcSoytre es cdhst ioeoovwnne n br6 ayi.ln2 a ,fTaamblieli a9r. 2sihsa upnesif oirnm t;etrhmast ios,f ththee dire nmsiatyss heas s atnhde
NEW! Applications of Physics ◄ naftoenoun tnid dedoex nahb liye b sxiitnpe rtrteihtnsgeg r l,ase stismioos nptf lohe[reE,c qswe.t or(w6arhki.g 7ilihet) t] d.-rl eoiNnleaeosx t bienhe gath hsat ahvttoai o ntrbh eeof nwcaeent dfaoipvrpcidleye w tthhiteeh w txoo-tcarokl m–deipsnopenlragecnyet mt hFeexno,trw exhmiinc, thEo q wa. e(l a6nr.og6we), n tsaouaCl lmmeAoaUwbecTe h vIrtO o asoNl evfug aesmr maCyet.oa nmalJtlul p lbssu epett gicoanmaisgnu eittsnhsne etw F stmhi ig¢teoh. m xvi6.nea.nW 1ltuth6 eeoe,f sionelirdti apaYrtosu omf athy eb eb otedmy.pted to try to compute the
Throughout the text, free-standing captioned photos wdohnil ed osterse tocnhlyin agb. oAust a9 3re%s uoltf ,t hthee w roerlakx tinhga tt ewna-s of Fxin each small segment is approximatelymm coaosmsn esantnatd no tmf. iunTlethirpetil ayc iohnfga a nt hgbeeo dtioynt a blky mi anasesstusi cmbyin tgh eth saqtu aalrl et hoef mthaes ds iisst acnocnec fernotmra ttehde acte tnhtee rc oenf tmera ossf
apply physics to real situations, with particular emphasis done to stretch it. eonf ekrignye tiinc seengemrgeyn its ¢thxea siusm eq oufa tlh teo cthhaen wgeosr kin F tahxtoe o¢f itlxnheadne, giavatxhindi sdLu. aasRnloe d sso eimsngt.am tsThseah nMte tts eit,mso atppnaitlvda o tctithoheanudn; s ag ibite soduote asnn’ at xwiso rthkr!o Fuogrh e oxnaem epnlde,, pwehrepne nad uicnuiflaorr mto tthhien rroodd,
on applications of biomedical and general interest. eWqtuotal= to¢ tKheh otoldtasl fworo vrka rydionnge foonrc eths ea sp waretilcl laes dfoutthrrrae itcn emogdno atsmhtt aeetnnh ette n oocftnie irenneste e.rdrt,ii asa p islda iIcste=amnMceenL tL2. 3S2o[fcraosme (tbh)e i na xTiasb, lwe e9 .2w]o. uIfl dw oeb ttoaoink tthhee minacsosr arse cctornecseunl-t
Here’s an alternative derivation of the workI–=enMergLy 2the2o=reMmL f2or4 .a ❙force t>hat
390 CHAPTER12mtFhlaeuy iwd voMarrekyc i hnwatnietgihcr saplo. sAitsi oan p. rIet liinmvionlavreys, wmea kniontge tah acth athnNeg oea1w coc >fteh lv2aeatr rawitaieob nlke n afo>rowofm hth oxew tp otao rv tcix>caillneculate the kinetic energy of a rotating rigid body,
can be expressed in various ways, using ax=dwvex cdatn, vaxpp=lyd txhed te,naenrdg yt hper icnhcaiipnles of Chapter 7to rotational motion. Here are
ANnFapolropr, mwlbicuaatl tbailo osnomdL aflillso dwtis eitnnu rtihbneag rnh ucufelome sr afu nocT hrau odarrsebt araui vihslea elantaimrtvtie-s: oVwniese c maaoxnuuoss=t t hfaodepdrvrpct,x elwys= iabth edgd tvtrwxhexeea d detcexnter nt=pshtroreeamv sl>tx sue tdudub vprxbeexose i a nomtt pstoph ovoefi sn bse>gat r tcafhkatie ssogt esfy lst itahd naeinnd (flg6d so,. o1twshm1oe ) te hto oeau xnktae eamretm ppt ohltesehst e f.t ruflobonewt. a Tgt horaeints’tgs.
pathology can cause the flow to become turbu- why you have to keep squeezing a tube of toothpaste or a packet of ketchup (both
Fb1yo0 rt0cee0n dexoenr (teNd) Maximullueismsntee ttf.neu Tniln uddgroi abntguo enl ebxonltsoectoneicds mi tofleanockwhe nswi qintFuhoer is.aoe ms, twe tthhhicioshs crisoe psweuh liWyst, atEotq=. (6vaaL. ti7p xst1)crThxeo2tehFseu eslfxslu r sp odrfl ruenxue stsai =dostth usfta )hrLt ethet x ot e1dhbx i2keflafm efcoteeakowrpx teo a. dntlfh x cwteeh =o eflrr eukflqLi odudxw1 ioxcr2n eomtedhm v abtixtoyn di dgsstvhxu fxoesa ut drnat x iegontrf efa oai tgtrseci rvce e otF(hnn6ax t.v1aniosi2 ntl)hueerm. aYet mofluoorws pfi hrnaegtreeir cts hp prroreousvgsuihdr aee
Now dvxdxdxis thcey clihnadnrgicea iln p vipeelo ocfi tlye ndgvtxhd Lurainngd trhaed iduiss pRlatcuernmse onut td txo, sboe ipnroportional to LR4.If
500 Tendon being stretched dEviaqnr.ig a( bx61-l.ev1 e2fl)ro>ocwmite2 i xecsat ovn 1 Evsaxxu,nebewidnsnsroe gtcvci etdw2Riu sebatebceety r y ct sed hahaav ea sffsxneaaeg cfc Retopto oro7btr rhi y1 1nooed tf of7sv l n.i0x mT1e>.9-dhih0t0xisa s.2 (9l f afdg0r, xio 1vt.m04heT%es = xhru 1iress1eq ad.uc5nuih2drcae t(nxidaog2 np5et)rs2o e i% sntthshc ueiernr e eiccan orsitenerearscges rrtseehap)ate.ois oTrneens-hq ibusy isr ei2md4 pp=lree 1srs6eul;ardteie >odcnirfsefhaeisrp--
2cthh1ea.m9rg e.hs.a.s WT mwhaoeg nns tmihtueadylle a pr0el.a 21s2t50i.c 0 N scp.mhWearhpeaast r tai,s rt eht hege ri vecpehnua lrsgpievo esoi tfnio vreeca ece hbl eesctpwtrhieeceranel 2Cq12hO.a2=r2ge+. .5q.10T0w=1 noC+ p4oi.si0 Tna0etnt 2dcnxohnC ar=erlaigxs-ei nsg0l o.3a3cr0aetTfi0 epnh daeml la liyc.an ettWfie dngx drhoaal=n to aft0 hrv.ee2x 0dtxhv0-exetna ah xdaimxmesprd irle,jaosaaudr wigst anaestinsn trsnrv ii dgatetxhWf2i su nooe)t2dc foloca. thletnonoSh =a ndaewturnh nbghaeeLssdierc v:tgh1tt1ievhiteo2ura >mintberit vnls bo mxgceo datudt2wvnhs xc eperl ereeuens.ps uspalu te hrrie ing.a nhsDdu-cu bhlesoo twtaloene strtith aelelrilom yRl ie4dtslie,de vtew ap(etweendhd ibecnlhoc oteed,n edpvsree tnsos uanr aesrm raonawldl
(a) if the two charges are equal and (b) if one sphere has four times direction of Etxhteen stioont a(ml mf)orce exerted by these two charges on >a nega-
t2h1e. 1c0ha.r.geW ofh tahte Iof tWheer? Were Not Neutral? A75-kg person holds t2i1v.e2 p3o.in.t cBhIOargBea qse3 =Pa-ir6in.0g0 innC DthNaAt ,i s Ip.lacTehde a tt wTWthuoher e bonsui rtdlWihegeetoni stsn cp?=oeeef12d m tohvfe 2a2 fl-ow12 minvg12 fluid exceeds a certain( 6c.r1i3ti)cal value, the flow is no
This is the same as Eqlo. n(6g.e6r) ,l asmo itnhaer .w Ionrskte–aedn,e trhgey flthoewo rpematt eisr nv ableidco emveens ewxittrheomutely irregular and com-
out his arms so that his hands are 1.7 m apart. Typically, a person’s DNA double helix are connecttehed asbsuym pptaioinrs t haot ft hpeb lnaeexst,e fason rdc( eait d icseh cnaoninngseetas, ncto.ntinuously with time; there is no steady-state pattern. This
hand makes up about 1.0% of his or her ◄body weight. For round thymine, cytosine, and guanine). Because of the girereogmuleart,r icch asohtiacp fleow is called turbulence.Figure 12.20shows the contrast
numbers, we shall assume that all the weight of eNacEh Whan!d P ish dEueT t oSimouf ltahetsieo nmso laecnuldes ,T uadteonriniae lsbonds with thyminbee twaenedn lacmytinoasri naned turbulent flow for smoke rising in air. Bernoulli’s equation
is notapplicable to regions where there is turbulence because the flow is not
the calcium in the bones, and we shall treat thSei xhtaenedsn a as spsoiigntnabbEoxlaenmd psPle hw6.E7ithT MgTuouatintooinn rewi. aitFlhsi ga u evraner yaEibn2g1l e.f2o 3rsctesuhodwesn tthse ttohstye ammdyin.aek–ead enine
charges. One mole of Ca contains 40.18 g, and each atom has 20 bond. Each charge shown is (cid:31)e, and the H¬NWhdetihsetar na cfleo wi sis laminar or turbulent depends in part on the fluid’s viscosity.
protons and 20 electrons. Suppose that only 1.0c%o nofn tehec tpioosnitisv ebeAt0nw. 1ai1er-0ter ancnmk g.r li(edaea)r Col-fa lmlicfauessl a0pt.1e0h t0h ekegn nisoe tamtftaocrehcened ttaoh athate t nhenyddm otfi nhae e e wxTuoehrrnket– sgde rnoeeenargt reaylr d yttheheineno rivengimes c,. osisnictye, ththee t ogtrael awteorr kt hdeo ntee nindveonlcvye sf tohre tdhies -fluid to flow in sheets
horizontal air track by a spring with force constant 20.0 Nm tance moved (our target variable). In Figs. 6.22b and 6.22c we
c(caho)au rHlgooemswb sim no afn euyan cCbhaa l ahantaocnemdd s cwdhoeaerregs ee uadncobhea shl aeananccdeh d ch oanbntyda i poncno?fhe n g(Mytbaas)ti inaivHc?se ost (we.c c)r 7h mWian6rahgg naPeyPt. hhEy(mtthhFIytTsoeehsieg rv igee .tiic l n stii6rdg sNs.ei ea 2nam®ra2tto ¬t mas1 r )fo.oa.rau5 invHcI0cnenta ltimsiiaobt¬ vditn>aoltes,le lti atyh,Niooon en cdt trhrhn o i c(eegrbnt o sher)shsi tmppg ii(frheduaa ibntt)elh. gr isieFrfPn ei iiavtnasohei dertenu pi ?atinaolhs yis errnrTtt urstmosetorrhat,noac cveexhkkadni ei me sfiddossoe uf tuaefpmrue,nc mr sTdndedto ehieis tsdtnh et h aexdogi an nntcutg c ,tae.l tehis hdldo etcetar tthoeru>hht aaliietssattthht ieeoScotasioBttOnhbfoehht rolsnetu aee ho¬ rl)t t(ensflaf wdwtae amoohgt h ioutHlwyetihriehl nd lerlfl eeeaycii¬ A’su pr osp t ’iaohl ssodmraens esiNm imqed mihtsbtiu oiavpiaotiantapderhinnti ao nlien xzarnoew g-d.e-)nd)m, h.rii oserWio enr urvece en tSi ti stshoealcteinrkcoke etgsttei ocoilxlith yndybe= e .ed 1tr Ih) t20n soea.t a 5onftfltahp,d cteos whx.t w r,eeT i =ag ghai hlelssidti d stmt(teu(oiltrnoemh ’ tsbeitve oh eiitdnne as i lcrttdiagihosiame raslpte it iucpt vnytrtoahiae sorileriiyns.-- fln(Woewehd eewnda stwo le ae mndsiisuncrauer s tashneaddt
force would the person’s arms have to exert on his hands to pre- ktiinoetnics f riacrtieo n pwaitrha clloeelffi ctioe nte makch= 0o.t4h7e.r. Remember, howheovriezFroo,n rt taahl, aflstuo i idon n olfyt h ate hgei vheonr izvoinsctaol siftoyr,c eflso wdo swpeoerkd. iNs oat ed ethteart mining factor for the
vent them from flying off? Does it seem likelyT thhaet hcios marmpsr earheenOsSOi¬vLUeTH IOli¬NbrNarsyet ,o thfe A Oc-tievxPehrtsy as ifcosrc ae ponp bleotths theEs outHqrnne. sst+(cet6hat.e ba1osl0nf;e ) dt towug iruthvbhseeueesn l t e htNahne ec cw -ewro.,i otArikrck–fla eldon owsenpr geep yeab dttyht eeitrshon err e tegmhalaic wdth eieers dn so.et neaI drbtr hlteeehg easu pwtl raloiornriwkgti deasossp n eiiente dtsh esu fldodwen lpya tbteecrno mcaens
c2a1p.1a1ble. .ofT ewxoer vtienrgy ssumchal la 8fo.5r5c-eg?spheres, 15.0 camnadp aartp frpolme tc-ebn-aseIcDaeodEnnlNes dtTcatI tnFulrYito,k tanesonow d irnwiiSs aeEte Thl c aseaUln Pohni:nsoyTtg dha urtesoh lesgf eoteo hrNcnee a¬aceotvxonesHamrttaein,¬dl ta -wabbcNyhc ieltclesheheer.a t ti.sis po( rnb0i n).fg 0oC r5imsa2 ul9nclao usnt lmatWbe bafy neert tho dhec emxeam pusf easpotcnerhtridy cnet hge boep yto ohrg noenrl oirtt dh uotefehganr ehgc.tltnoio dermesssro., vAiwnet h ftilahcorehtwh ipe srifl p towhewe i twh nsoeapuglleta, et fivdrviasec,r t ioitaofhtn ieE ostqnhe.as (nd 6i in.ws1 t0iutth)hr.eb adnecnessi tdya omfp t hoeu tfl; utihde,
ter to center, are charged by adding equal numbers of electrons to oTf hCehnap tceor m2ptao reso ltvhee thsitsr epnrogbtlhe mo. f Inthsteea db, owned’illn gu sef otrhcee offlrfi octtwiho enp .aetlteercnt riosn s tianbleand tends to maintain its laminar nature (Fig. 12.29a).
NEW! Video Tutor Demonstratieoacnh so fa nthdem T. uDtisorergiaarldsing all other forces, how many electrons hydrogen with the bonding force of the adenine–thymWihneen mtheo lcerictiuclael ss.peed is reached, however, the flow pattern becomes unstable.
The disturbances no longer damp out but grow until they destroy the entire
“Pause and predict” demonstratiwoonu ldv iydoeu ohasv oe fto k aedyd tpo heaycshi cssp here so that the two spheres will laminar-flow pattern (Fig. 12.29b).
accelerate at 25.0gwhen released? Which way will they accelerate? Figure E21.23
..
cbtsshetoeuetnf do ocSerefte nupa tdstsssesy tie egoAinn nrtgaergaaba tng lheoese ff te s uMotrut uotadhtsrceietaoenilrmrst is unbe nbg.a dPys T heeahrydsses k2hpesiigtointsataialr1carna arednitnn.stcniv1 hego et gid2wwv,anttd ,ah aiti aonttwndhoolin e osdnnfhdgeemonss a J raot mietiaahcofu lnwln see flsftvsd o htob m tta rioteuiooHhcdnhyfrolxe d eeo eotpee sh wwwnso b oe u,aP euoed beseSrb nuveteayae tdhla cmrc darsemcthao ihh.r oenl tnc iDsaaen aatoogcoannn bolnh sansedl Inttliy ecs ra boesmoot pinonettn eunhoxr dtrig aiTahonn eneuoebnge e trtfdEtye i ho1oxopad lie.neatncr u0n to. c dsott vtgtghti orroet tteAihoanhiorn cnkpyfse g er pFe ?ool rxii ottongEoehrtthxrneoocttprs en o a?r(?swcena .W csn e(tsoadhr hSue) yey uvnI olo?fpeteuc0shgb2 tropt(fhhhr.euo1yhbt1 nooalean. a ei12)eswronrxdln g04e W ogni sbwssena nr . wyno rwea.tehgtesmo )xhhverBia h.!utc ?eee-r).tIhBoo O wAnebt Bhcnnosyasea nTuumcisdhssmteiei ye nn±Pde mgoe aeb ftiaHih, nyor tnahe faiHpdtn en a tgtdcghih ry eieCesnttoH hooN bmcCeDsfyo i enNCbtnOtoeadrAl1is—Hi sesicbiCen 2n.n( ostgasHT1e. n)ha2C ( h tdilaa 92aaNseasdpsm un n eetd fwTdOiwdfi n nuoh 2 caiioeeiftngHre h Hfl1euasnto0— htoi,tnagn . lde wl03ytnmuolnheNs .0y iwha2o semyn0 i1n f H8gmetos mdiwohp0 nf i i e aom ssne2tet tpthlhe.2deaeeo Nre se,Fenc l f N dbceuicDrfgsuoycelo.tCmNeuuts ro(Hsr lc bAae,esae ) is afr snEta.dedu ueA 2oareCEc,bel1u neduoaaat.bi-eCnn2ccnlnhhg4Ndeei neNC H (a) (b)
a related problem situation. 21.13 . In an experiment in space, one proton is held fitahlxseoe O dth —aatnH tdh—esOe ,t hNr—ee Hco—mNbi,n aantido nOs— arHeO —parNal lceolm tob ienaacthio ontsh,C ear.n Cd aalNscsuulmatee
another proton is released from rest a distance of 2.50 mthme naewt faoyr.ce that cytosine exerts on guanine due to the Hpreceding three
(a) What is the initial acceleration of the proton after it is creolmeabsineadti?ons. Is this force attractive or repulsive?
Biom(vebe)dl oicScitkaye–ltlctyihm Beq gaurasalpeithadsti ovEfe n thd(en -rooe lfen-auCsmehdb paerrpso!tt)o enra’s c Pcmeroloetirobantli.oenm–tisFmigeu r◄ea End21.2214.24 .. BIO Base Pairing in DNA, II. Refer to Exercise 21.23.
21.14 . Anegative charge of -0.550 mCexerts an upward 0.200-N Figure EH21.24sho0.w29s0 the bonding of the cytosine and guanine mol-
aTftnt2a(q ohhomti13s eter .uuacx 1=sudg-e5ben en=0o+ krsia...ntnv5tu..5n+ ao5aed.db04nw0nTe d0i. - bh0unotami irr0n nnialcCeeo kdoCheccln s tcagsmiaonhircponiggw.aconiudneerni aC g )monin( lescfhtim f ?hcaatab ahctartehng gheertsgeenh adrf e rieon gts q our0eobetc1td.rusf 3eiief og i0odp tsaaih0r mnrernca ea.oemdtn t C e tboasthxhddrnsielgrai e, ira=qrn ucegnmt3)ncaheg?+kti ellsqesn(y 2d b2sz o . bb)e0tiw o=teer0Wauonnl xo?s cah-cwtemaah t3 atidai.. otl0rad.i gW 0rnno(eed ea s hnn.)ets . haCxW Ctee a rhhmitnCsaasd yar togt qgHoiines1s-sineCecaCpOHFcasraiNseus¬glCueclcue,emur Hsd1ael.eai s ¬n CTtEsaeCguhl 2 O2s mOett2Noh h1 O 2 Oe,e rt. eh 2tn¬eNah et04 ac 1nHt¬.0ttHo2mhn . f9n3mtmeo0mh0Ha s r1bH0H1eneci ¬ d e nbt ha oHt 2rNt22hnieN oa¬NOd, etni cnsCcaCNo .gyn1 mI tdds ios b ist sihNiCOdtnniauaseGn¬e Ct Nfciu eooeaoxHnrnsncei snlaer¬ yNetar Csa er tetN oote rp naatHac chcgrhteoaiu vlm0afle.eon1b lroi1 inctnr0oee ar s ntee diapmaouculneo.hl s snI ,toin govt hatetthn?ehhirdees.
21.16 In Example 21.4, suppose the point charge on the y-axis H
at y = -0.30 m has negative charge -2.0 mC, and the other H 0.290
cnheat rfgoercs er eomn aQin. Hthoew s admoees. yFoinudr athnes wmearg dniiftfuedr ef raonmd tdhiarte cinti oEnx aomf pthlee Cytosine H O2 1H nm 2O GNuanine H
222111...411?78 E..x..plIIanni nEE xtxhaaemm dppilfleefe 2r21e1n..34c,e, cswa.lhcautl aitse tthhee nneett ffoorrccee o(mn acghnairtgued qe 1a.nd H C C CN 0n.3m00H N C 1 C C NC
d2i1r.e1c9tio.n.) Tonh rcehea progien qt 1cheaxregretes da rbey a trhraen ogtehde ra ltownog cthhea rxg-easx?is. Charge N 1 C 2 0n.2m90 1 2 C N
xq1==0.+230.00 0m m.CChiasr gaet tqh3e =ori-gi8n.,0 0an md Cc.hWarghee rqe2 is= q-35l.o0c0a tmedC ifi st haet O2 1H 2N
n22e11t.. 22f01orc..e.. oRTnwe qpo1e apisto 7Ein.x0te 0cr hcNiasreign e2 st1h .ae1r 9e- flxoo-rcd aqirt3eedc=t ioon+n 8?th.0e0 y -maCxi.s as follows: Section 21.4 Electric Field and ElectrHic Forces
charge q1 = -1.50 nC at y = -0.600 m, and charge q2 = 21.25 . CP A proton is placed in a uniform electric field of
+3.20 nCat the origin y = 0 .What is the total force (magni- 2.75 * 103 N C.Calculate:(a) the magnitude of the electric force

tude and direction) exerted by these two charges on a third charge felt by the proton; (b) the proton’s acceleration; (c) the proton’s
q3 = +5.00 nClocated a1t y = 2-0.400 m ? speed after 1.0>0 msin the field, assuming it starts from rest.