TEACHER UPPLEMENT 2011 - drsphysics.com · the publisher to photocopy answers as needed from this...

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T T EACHER EACHER S S UPPLEMENT UPPLEMENT © Copyright 2011 N & N Publishing Company, Inc. 18 Montgomery Street Middletown, New York 10940 www.NN4TEXT.com (800) NN 4 TEXT email: [email protected] 2011 2011 Important Information about the Important Information about the Practice Tests in the Student Practice Tests in the Student STAReview STAReview texts. Please see page 2. texts. Please see page 2.

Transcript of TEACHER UPPLEMENT 2011 - drsphysics.com · the publisher to photocopy answers as needed from this...

TTEACHEREACHER SSUPPLEMENTUPPLEMENT

© Copyright 2011N & N Publishing Company, Inc.

18 Montgomery Street Middletown, New York 10940www.NN4TEXT.com (800) NN 4 TEXT email: [email protected]

20112011 Important Information about theImportant Information about thePractice Tests in the StudentPractice Tests in the StudentSTAReviewSTAReview texts. Please see page 2.texts. Please see page 2.

THIS BOOK IS THE PROPERTY OF:

STATEPROVINCECOUNTYPARISHSCHOOL DISTRICTOTHER

BOOK NO.

ENTER INFORMATIONIN THE SPACESTO THE LEFT ASINSTRUCTED.

YEAR CONDITIONISSUED TO USED ISSUED RETURNED

STUDENTS to whom this textbook is issued must not write on any pageor mark any part of it in any way, consumable textbooks excepted.

1 Teachers should see that the student’s name is clearly written in inkin the spaces above in every book issued.2 The following terms should be used in recording the condition ofthe book: NEW, GOOD, FAIR, POOR, and BAD.

TEACHER SUPPLEMENT: PHYSICAL SETTING: PHYSICS STAREVIEW N&N© Page 1

N O P E R M I S S I O N H A S B E E N G R A N T E D B Y N & N P U B L I S H I N G C O M P A N Y , I N C T O R E P R O D U C E A N Y P A R T O F T H I S B O O K

Author

Nancy A. Moreau

EditorsWayne Garnsey & Paul StichFran Harrison, Associate Editor

Artwork & GraphicsEugene B. Fairbanks & Wayne Garnsey

N&N Publishing Company, Inc.18 Montgomery Street Middletown, NY 10940-5116

For Ordering & Information1-800-NN 4 TEXT

Website: www.nn4text.com email: [email protected]

Commencement-LevelCommencement-Level

LIMITED PHOTOCOPY PERMISSION

The professional teacher purchaser of PHYSICAL SETTING: PHYSICS STAReview, forwhom this Teacher Supplement has been developed, has limited permission granted bythe publisher to photocopy answers as needed from this supplement.

This permission does not grant or imply approval for copying in any form any otherpages of the text PHYSICAL SETTING: PHYSICS STAReview without written priorpermission from the publisher.

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Important Information about the Practice Tests in the student STAReview texts:N&N Publishing is no longer publishing a new edition every year with the most recentstate tests. Instead, the Practice Tests at the back of the STAReviews will remain thesame until a new revision of the book’s contents is updated due to changes in statecurriculum, updates or corrections to information, and/or new information should beincluded.

The Practice Tests in the student STAReview texts do not identify the actual test used;however, the Teacher Supplement identifies the source of the test at the beginning ofeach Practice Test.

For teacher use only, this CD has a Test Archive folder with printable PDFs of themost recent tests. Also, an archive of tests can be found on the web at:

http://www.nysedregents.org/

This change has become necessary for two reasons: (1) more of the state tests arebecoming restricted, and (2) it is a significant cost savings that is helping N&NPublishing maintain our low prices in a sky-rocketing environment of production costincreases for energy, paper, printing, environmentally safe inks, and freight.

NOTES FOR THEPHYSICAL SETTING: PHYSICS INSTRUCTOR

Helping students to meet the new learning standards in Science demands new teaching and assessment strategies. Studentshave to be instructed in new writing skills for more rigorous, on-demand testing requirements. PHYSICAL SETTING: PHYSICS –STAReview (Strategic Test Achievement Review) is designed to focus students’ attention on the key themes and skills needed forthis transition.

One thing is very clear from historical performance on standard Science examinations, students consistently lose a largerpercentage of credit on the written segment than on the multiple choice segment. A good, solid review plan takes this intoaccount. The greater emphasis in PHYSICAL SETTING: PHYSICS – STAReview is on the skills and strategies for success onthe written segment.

CCHAPTERHAPTER 1 – P1 – PHYSICSHYSICS: N: NEXTEXT GGENERATIONENERATION ..1111Einstein’s E = mc2 ..........................................12The Standard Model ......................................14Anti-Matter.....................................................15Fundamental Forces ......................................15Radioactive Decay ..........................................17Quarks ............................................................18The Very Large and the Very Small..............19Lab 1 – Indirect Measurements ....................23Lab 2 – Psyching Out The System................24Part A Multiple Choice Questions.................25Part B Constructed Response........................27Part C Extended-Constructed Response.......28

CCHAPTERHAPTER 2 – V2 – VECTORSECTORS ..............................................................2929Role of Units in Problem Solving..................30Scalars and Vectors ........................................31Distance and Displacement...........................32Solving Vector Problems Graphically ...........32Vector Addition by Components ....................38Lab 3 – Displacement Vector.........................41Part A Multiple Choice Questions.................43Part B Constructed Response........................44Part C Extended-Constructed Response.......48

CCHAPTERHAPTER 3 – K3 – KINEMATICSINEMATICS ....................................................4949Motion of Objects ...........................................50Describing Motion with Diagrams................51Vector Diagrams.............................................52Describing Motion with velocity vs. time

Graphs......................................................52Determining the Slope on a p-t Graph .........53Getting Information from a v-t Graph..........58Freely Falling Objects....................................61Kinematic Equations .....................................63Lab 4 – Creating Motion Graphs ..................68Part A Multiple Choice Questions.................70Part B Constructed Response........................74Part C Extended-Constructed Response.......79

CCHAPTERHAPTER 4 – F4 – FORCESORCES ANDAND FFRICTIONRICTION ....................8181Forces ..............................................................82Newton’s Three Laws of Motion ....................84Universal Gravitation, Law of.......................87Gravitational Fields .......................................88Static and Kinetic Friction ............................92Object on an Incline .......................................94Lab 5 – Forces that Accelerate ......................96Lab 6 – Coefficient of Friction.......................98Part A Multiple Choice Questions...............100Part B Constructed Response......................104Part C Extended-Constructed Response.....107

CCHAPTERHAPTER 5 – M5 – MOTIONOTION ININ AA PPLANELANE ......................109109Section I – Two Dimensional Motion and

Trajectories ............................................110Projectile Fired at an Angle ........................114Projectile Motion ..........................................114Part A Multiple Choice Questions...............118Part B Constructed Response......................121Part C Extended-Constructed Response.....123

Section II – Uniform Circular Motion ........123Centripetal Force .........................................124Lab 7 – Centripetal Force............................129Part A Multiple Choice Questions...............132Part B Constructed Response......................134Part C Extended-Constructed Response.....135

CCHAPTERHAPTER 6 – M6 – MOMENTUMOMENTUM ................................................137137Momentum....................................................138Impulse .........................................................141Lab 8 – Conservation of Momentum ..........144Part A Multiple Choice Questions...............146Part B Constructed Response......................147Part C Extended-Constructed Response.....148

CCHAPTERHAPTER 7 – S7 – SWINGSWINGS ANDAND SSPRINGSPRINGS ..................149149Things that Swing........................................150Lab 9 – Pendulum........................................151Hooke’s Law ..................................................152Lab 10 – Vibrating Spring...........................153Lab 11 – Hooke’s Law ..................................154Lab 12 – Determine g from a Simple

Pendulum...............................................155Part A Multiple Choice Questions...............155Part B Constructed Response......................156Part C Extended-Constructed Response.....158

CCHAPTERHAPTER 8 – W8 – WORKORK, P, POWEROWER,, ANDAND EENERGYNERGY159159

Work – Definition and Mathematics...........160Work – How to Measure ..............................161Force vs. Displacement Graph.....................163Power and Horsepower Rating....................165Potential Energy ..........................................169Kinetic Energy .............................................171Lab 13 – Are You a Horse? ..........................168Lab 14 – Force, Work, and Energy ..............172Part A Multiple Choice Questions...............173Part B Constructed Response......................175Part C Extended-Constructed Response.....177

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TTABLEABLE OFOF CCONTENTSONTENTS

CCONTEXTONTEXT OFOF THETHE PPHYSICALHYSICAL SSETTINGETTING: P: PHYSICSHYSICS SSTUDENTTUDENT TTEXTEXTNote: Process Skills [most often assessed in Parts B and C of the Test] and Lab Skills [assessed in Part D on the Test]appear, not as separate areas, but in direct association with Key Ideas and Major Understandings throughout thechapters. For specific information see Appendix B.

CCHAPTERHAPTER 9 – C9 – CONSERVATIONONSERVATION OFOF EENERGYNERGY ..179179Pendulum and Roller Coaster .....................180Internal v. External Forces ..........................182Work Energy Theorem.................................183Lab 15 – A Potpourri of Energy ..................186Lab 16 – Mechanical Energy Conservation189Part A Multiple Choice Questions...............189Part B Constructed Response......................191Part C Extended-Constructed Response.....193

CCHAPTERHAPTER 10 – E10 – ELECTRICLECTRIC FFIELDSIELDS ANDAND FFORCESORCES195195

Static Electricity ..........................................196Microstructure of Matter .............................196Charge Detection and Types .......................197Coulomb’s Law .............................................199The Electric Field.........................................201Electric Potential..........................................203Potential Difference or Drop .......................203Lab 17 – Static Charges ..............................199Part A Multiple Choice Questions...............206Part B Constructed Response......................210Part C Extended-Constructed Response.....213

CCHAPTERHAPTER 11 –11 – OOHMHM’’SS LLAWAW ANDAND RRESISTIVITYESISTIVITY215215

All About Circuits ........................................216Conductivity in Solids, Liquids, and Gases 216Voltage and Ohm’s Law ...............................217Resistance.....................................................218Energy and Power........................................220Circuits .........................................................223Lab 18 – Ohm, Ohm on the Range .............222Lab 19 – Resist Me ......................................224Part A Multiple Choice Questions...............225Part B Constructed Response......................227Part C Extended-Constructed Response.....229

CCHAPTERHAPTER 12 – S12 – SERIESERIES ANDAND PPARALLELARALLELCCIRCUITSIRCUITS ......................................................................................................231231

Series Circuits ..............................................232Parallel Circuits ...........................................234Kirchhoff’s Laws...........................................235Lab 20 – Series and Parallel Circuits.........237Part A Multiple Choice Questions...............238Part B Constructed Response......................239Part C Extended-Constructed Response.....241

CCHAPTERHAPTER 13 – M13 – MAGNETISMAGNETISM ............................................243243Magnetic Force .............................................244Magnetic Fields and the “Hand Rules”.......244Magnetic Field..............................................248Electromagnetic Induction ..........................250Generator Principle......................................250Electromagnet Radiation.............................251Poles – Unlike and Like...............................253Comparing Earth’s Magnetic Field.............253Lab 21 – Magnetic Fields ............................252Part A Multiple Choice Questions...............254Part B Constructed Response......................257Part C Extended-Constructed Response.....258

CCHAPTERHAPTER 14 – W14 – WAVEAVE PPROPERTIESROPERTIES ..........................259259Waves ............................................................260Wave Characteristics ...................................262The Doppler Effect .......................................262Periodic Wave Phenomena...........................265Refraction .....................................................269Diffraction and Interference........................270Lab 22 – Ripple Tanks .................................268Lab 23 – Super Position...............................271Lab 24 – Transverse Waves.........................272Part A Multiple Choice Questions...............274Part B Constructed Response......................276Part C Extended-Constructed Response.....278

CCHAPTERHAPTER 15 – S15 – SOUNDOUND ANDAND LLIGHTIGHT ........................279279Sound as a Mechanical Wave ......................280Vibrations and Resonance ...........................281The Doppler Effect .......................................282Shock Waves .................................................283Boundary Phenomena and Echoes..............285Lab 25 – Speed of Sound in Air...................286Part A Multiple Choice Questions...............287Part B Constructed Response......................289Part C Extended-Constructed Response.....290

CCHAPTERHAPTER 16 – R16 – REFRACTIONEFRACTION ANDAND RREFLECTIONEFLECTION291291

Light Energy ................................................292Law of Reflection ..........................................293Refraction .....................................................294Photons through Media ...............................296Light Wave Bending.....................................297Snell’s Law and the Absolute Index of

Refraction...............................................298Lab 26 – Reflection ......................................302Lab 27 – Refraction and the Index of

Refraction...............................................303Part A Multiple Choice Questions...............305Part B Constructed Response......................307Part C Extended-Constructed Response.....308

CCHAPTERHAPTER 17 – D17 – DIFFRACTIONIFFRACTION ANDANDIINTERFERENCENTERFERENCE ........................................................................................311311

Light – Wave or Stream of Particles? .........312Two Point Source Interference....................313Polarization ..................................................315Electromagnetic Spectrum ..........................317Lab 28 – Laser Penny ..................................312Lab 29 – Polarized Art.................................318Part A Multiple Choice Questions...............319Part B Constructed Response......................320Part C Extended-Constructed Response.....321

CCHAPTERHAPTER 18 – M18 – MODERNODERN PPHYSICSHYSICS ..........................323323Dual Nature of Light ...................................324Photoelectric Effect [enhancement material] .........324Photon-Particle Collisions – Compton Effect

327Matter Waves – de Broglie ..........................327Atomic Models – Rutherford and Bohr.......328Atomic Spectra .............................................330Uncertainty Principle – Heisenberg ...........335Lab 30 – Electron Jump ..............................334

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Part A Multiple Choice Questions...............336Part B Constructed Response......................338Part C Extended-Constructed Response.....339

AAPPENDIXPPENDIX A –A – SSPECIFICPECIFIC RRATINGATING CCRITERIARITERIA341341

AAPPENDIXPPENDIX B –B – RREFERENCEEFERENCE TTABLESABLES::PPHYSICALHYSICAL SSETTINGETTING/P/PHYSICSHYSICS ....................345345

APPENDIX C – PROCESS SKILLS,PERFORMANCEINDICATORS ...................................351

APPENDIX D – GLOSSARY AND INDEX....355

PRACTICE TESTSSETTING/PHYSICS ..........345

APPENDIX C – PROCESS SKILLS,PERFORMANCEINDICATORS ...................................351

APPENDIX D – GLOSSARY AND INDEX....355

TEACHER SUPPLEMENT: PHYSICAL SETTING: PHYSICS STAREVIEW N&N© Page 5

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RRUBRICUBRIC FORFOR PPHYSICSHYSICS LLABSABS

QUALITY WORK POINTS POINTSPOSSIBLE EARNED

Format NameLab Title

Organization Statement of the problemMaterials listSummary of ProceduresData table drawn with rulerGraphs - axes drawn with ruler

(if needed)Data analysisa Sample calculation shown

for each problem typeb Calculations accuratec Units labeled in all steps

and answerError analysisConclusion

Mechanics Correct spelling, capitalizationand punctuation

Neatness Typed or written in pen(except graphs)

All sections in order

Total Points

PPRACTICERACTICE TTESTSESTS AANSWERSNSWERSPractice Examination Answers are in folders

identified with the date of the test and are PDF docu-ments which can be opened and read with any version ofAdobe Acrobat Reader 6.0 or higher. The free AdobeAcrobat Reader can be downloaded at the follow URL:

http://get.adobe.com/reader/

PAGE 6 TEACHER SUPPLEMENT: PHYSICAL SETTING: PHYSICS STAREVIEW N&N©

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TTOO THETHE SSTUDENTTUDENT & T& TEACHEREACHERPHYSICAL SETTING: PHYSICS STAREVIEW is written based on the new standards and assessments for physics. It is a

comprehensive review of the Key Ideas, Major Understandings, Performance Indicators, Process Skills, and Real WorldConnections as set forth in the University of the State of New York Education Department: Physical Setting: PhysicsCore Curriculum.

“O“OPENPEN FFIRSTIRST””The student should upon receiving this STAReview begin by reading this section: “To the Student.”

• Start by reviewing the Table of Contents (previous 6 pages). This will give an overview of the major topicsreviewed in this book.

• Now, become familiar with Appendix C – Index & Glossary. This section is an extensive listing of the key physi-cal terms that one needs to know in order to understand the material. A brief definition or explanation of theterm is given together with cross-referenced pages to direct the student to additional material directly relatedto the term.

OORGANIZATIONRGANIZATIONThe book is organized “conceptually” through Major Understandings, but the review is linked through the following

organizational parts.

• Standards are the overall, general goals that apply to all scientific and indeed most general learning. Forexample, each Standard contains several goals, such as “Analysis, Scientific Inquiry, and Engineering Design inorder to pose questions, seek answers, and develop solutions.”

• Within each Standard, Key Ideas are used to further define the generalized objectives to be reached. Forexample, Standard 1 has several Key Ideas such as Key Idea 1 within the Scientific Inquiry part of Standard1, that is, “to develop explanations of natural phenomena in a continuing, creative process.”

• For each Key Idea there are several related Process Skills which specifically identify what processes the stu-dent must learn in order to demonstrate the particular Key Ideas of a general Standard. These Skills are iden-tified and found in all chapters followed by explanations of the Skill and questions to test the student’s abili-ties in preparation for the final, year-end test.

• Associated with both Standards and their Key Ideas are the Performance Indicators. These tell the studentspecifically what he/she is expected to know in order to answer correctly the questions on the final, year-endtest. In other words, the Performance Indicators are the testing objectives. These are identified at the begin-ning of each chapter and again at the end of each unit with the Part A, B, and C questions.

• Finally, there are the Major Understandings. Each Performance Indicator has specific concepts and physicalunderstandings to learn. This is the “meat and potatoes” of Physical Setting: Physics STAReview. These MajorUnderstandings are first listed at the beginning of each chapter, are further developed in the text, examples,sample problems, and illustrations that follow, and are tested throughout the Unit in the Skills and at the endof each chapter in Parts A, B, and C.

MMEANINGEANING OFOF SSYMBOLSYMBOLSSymbols are critical in all science. So, the author has developed a mini-help system. Stars are used to help

navigate the student through the more complex Major Understandings in physics.

Stars indicate two important things: Some starred material may not be specifically referred to in theCore Curriculum, but this text is needed for better understanding of major concepts. Also, stars may note

special material that further explains Major Understandings, Skills, and Real World Connections. The dash-dot line ( ) in the left margin identifies the extent of this specially identified material.

In addition, the! identifies for the student “Try It” problems that should be solved in order to betterunderstand the Performance and Lab Skills. The solid gray line ( ) in the left margin identifies thethe extent of the Skill and/or Lab and any associated “Try It” problems being identified.

!!

SSKILLKILL M 1.1M 1.1 !! TTRYRY II TT

TEACHER SUPPLEMENT: PHYSICAL SETTING: PHYSICS STAREVIEW N&N© Page 7

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CHAPTER 1PHYSICS FOR NEXT GENERATIONKEY IDEA 5PERFORMANCE INDICATOR 5.3TRY IT: SKILL M 1.1, PG 0131 1.1 x 10–17 kg2 13034 MeV or 2.09 x 10–9 J3 6.3 x 1018 J4 1.1 x 10–12 kg5 5.13 MeV

TRY IT: PG 0186 The energy imparted produces mesons

before the quark is isolated.

TRY IT: SKILL 3.2, PG 0197 6.5 x 10–7 m and 6.5 x 10–5 cm8 3.0 x 1010 cm/s and 3.0 x 105 km/s9a 1.0 x 10–10 m9b 0.1 x 10–6 mm9c 0.1 x 10–3 µm

TRY IT: SKILL 3.2, PG 02110 6.5 x 10–1 m 13 1.45 x 1014 volts11 2.300 x 1012 watts 14 1.55 x 10–9 m12 4.73 x 10–2 m

TRY IT: PG 02215 103 jelly beans 20 104 hairs16 105 kg 21 1012 kg17 102 golf balls 22 106 balloons18 109 gallons/year 23 108 hot dogs19 105 m 24 108 pizzas

TRY IT: SKILL M 2.1, PG 02425 Determining the size of the atom26 Increase the number of drops, and

increasing the number of trials andaveraging the results.

CHAPTER 1 ASSESSMENTSPART A – QUESTIONS, PG 0251 3 10 42 2 11 33 3 12 14 1 13 35 1 14 46 2 15 37 3 16 38 1 17 39 2 18 3

PART B – QUESTIONS, PG 02719 320 421 222 223 124 325 426 c2

27 quark and anti-quark28 up, up, down29 0 quarks30 three31a gravity31b weak nuclear31c electromagnetic31d strong32 nuclear, weak33 1.66 x 10–27 kg

PART C – QUESTIONS, PG 02834 E = mc2

Ep = (1.67 x 10–27 kg) (3.0 x 108 m/s)2

= 1.503 x 10–10 J

Ee = mc2

= (9.11 x 10–31 kg) (3.0 x 108 m/s)= 8.2 x 10–14 J

The proton would yield 1.502 x 10–10 J moreenergy than an electron.

35 E = mc2

E = (1.67 x 10–27 kg) (3 x 108 m/s)2

E = 1.503 x 10–10 JE = 1.503 x 10–10 J (1 MeV/1.6 x 10–13 J)

= 938.9 MeV = 9.39 x 102 MeV

CHAPTER 2VECTORSKEY IDEA 5PERFORMANCE INDICATOR 5.1TRY IT: SKILL M 1.1, PG 031

m 3600s 1 ft 1 mi1 10 (–– )( ––––––)(––––––––)(–––––––)s 1 h 0.305 m 5280 ft

= 22.4 mi/hr

AANSWERSNSWERS TOTO QQUESTIONSUESTIONSPPHYSICALHYSICAL SSETTINGETTING:: PPHYSICSHYSICS STARSTAREVIEWEVIEW

Updated 07/21/04

TRY IT: SKILL 5.1 IV, PG 0332a R = 5 cm = 5 m, 37° N of W

2b R = 10 m,37° E of S

2c R = 18 m, 26° N of E

2d R = 50 m, N

2e R = 1200 m, 34° E of S

2f R = 875 cm, 65° S of W

TRY IT: SKILL 5.1, M 1.1, PG 0343a 5 m @ 37° N of W

3b 10 m @ 37° E of S

3c 17.9 m @ 26.6° N of E

3d 50 m, N

3e 1198 m @ 56.6° S of E

3f 877 cm @ 64° S of W

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5 m

4 m

3 m

37°

R

R = 5 cm = 5 cm, 37° N of W

10 m6 m

8 m

37°R

R = 10 m, 37° E of S

18 m

8 m

16 m

R26°

R = 18 m, 26° N of E

R = 50 m, N

N

50 m

suggested scale: 1 cm = 50 m

500

m, N

250 m, S

R = 1200 m, 34° E of S

1000

m

1200 m

660 msuggested scale: 1 cm = 100 m

34°

65°

R = 875 cm, 65° S of W

380 m

790

m

TEACHER SUPPLEMENT: PHYSICAL SETTING: PHYSICS STAREVIEW N&N© Page 9

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TRY IT: SKILL 5.1 IV, PG 0354a 1 cm = 1 m

4b 1 cm = 2 m

4c 1 cm = 2 m

4d 1 cm = 50 m

4e scale 1 cm = 100 mR = 1.9 cm = 190 m, 34° S of E

TRY IT: SKILL M 1.1 & M 3.1, PG 0375

TRY IT: SKILL 5.1 IV, M 3.1, PG 0396 67.4 m @ 67° S of E7a 362.8 m @ 49.5° N of E7b 398.8 m @ 5.4° W of N8 One satellite will provide your distance

from the satellite, two signals will give youtwo possible locations, the third is needed tonarrow it down to a single location.

CHAPTER 2 ASSESSMENTSPART A – QUESTIONS, PG 0431 2 8 42 1 9 33 4 10 24 2 11 15 1 12 16 3 13 47 1 14 1

PART B – QUESTIONS, PG 04415 416 317 118 419 420 221 14.4 km22a 50 N22b 86.6 N23 13.2 m24 800 m25a 192.8 N25b 229.8 N26 Displacement occurs in straight line

motion, distance does not have to, andtherefore is greater than the displacement.

Velocity

(m/s)

100100100100100

x-components(m/s)

98.5 m/s86.6 m/s70.7 m/s50.0 m/s0.00 m/s

y-components(m/s)

17.4 m/s50.0 m/s70.7 m/s86.7 m/s100 m/s

Angledegrees

1030456090

5 m

6 m

4 m

3 m

N

suggested scale:1 cm = 1 m

8 m

14 m

11.4

m

6 msuggested scale:1 cm = 2 m

44°

59°

N

suggested scale:1 cm = 2 m

10 mN

10 m

16 m

8 m

53°

suggested scale:1 cm = 50 m

155m

250

m

300

m

150m

18°

R = 1.9 cm, = 190 m, 34° S of E

R = 5 cm, 59° W of S

R = 11.4 m, 44° S of W

R = 10 m, 53° N of E

R = 155 m, 18° N of W

R

27 4 blocks, west28 30 m, north29 4 km30 2 blocks, east31 50 N32 500 km33 8 N

PART C – QUESTIONS, PG 048F opp

34 sin ø = ––––– (or F vertical)F

F opp = F sin ø= 100 N sin 60°= 86.6 N

F horizontalcos ø = –––––––––––

FF horiz = F cos ø

= 100 N cos 60°= 50.0 N

35 scale 1 cm = 2 km

35a Displacement = 16.6 km35b Distance = 10 km + 5 km + 6 km = 21 km

CHAPTER 3KINEMATICSKEY IDEA 5PERFORMANCE INDICATOR 5.1TRY IT: PG 0511

TRY IT: PG 0522 B3 A4

TRY IT: SKILL 5.1I, 5.1II, PG 0555 graph 1: m = 4.17 m/s

6 graph 2: m = 0.1 m/s

TRY IT: SKILL M 3.1, 5.1I, 5.1II, PG 056[Note: Graphs below are of a general shape.]

7

8

9

10

11

12

13

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N 5 km

10 k

m

6 km

16.6

km

55°

Velo

city

(m

/s)

Time (s)

Velo

city

(m

/s)

Velo

city

(m

/s)

Velo

city

(m

/s)

Velo

city

(m

/s)

Time (s)

Time (s)

Time (s)

Time (s)

Time (s)

Time (s)

Velo

city

(m

/s)

Velo

city

(m

/s)

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TRY IT: SKILL M 3.1, 5.1I, 5.1II, PG 05814 t = 0 s, t = 3.5 s, t = 7 s15 FG and CE16 AB17 FG18 No19a 89.6 m/s 19b 322.6 km/hr20 5 sec

21a 12 mph––––s 21c 9.1 mph––––s

21b 12 mph––––s 21d 6.5 mph––––s

22 The acceleration is decreasing in magnitude,as time and velocity increase.

TRY IT: SKILL 5.1I, 5.1II, PG 06023

24 mph/sec25 The car tops out in its acceleration and the

rate decreases from there.

TRY IT: PG 06326 You would feel weightless when an elevator

car descends rapidly.

TRY IT: SKILL 5.1, M 1.1, PG 06527 Time Speed Distance Table:

TRY IT: SKILL M 3.1, M 5.1 D, PG 06728 Students’ answers should mirror the trip to

school, except for segment BC that would beomitted.

29 Students’ answers will vary.

CHAPTER 3 ASSESSMENTSPART A – QUESTIONS, PG 0701 4 12 42 1 13 23 1 14 14 4 15 15 1 16 46 3 17 37 1 18 38 3 19 39 3 20 210 4 21 311 2

PART B – QUESTIONS, PG 07422 32 m 28 7.3 m23 0.43 s 29 432 m24 3 s 30 45 m25 0 31 8 s26 2 m/s 32 22 m27 2 m/s2 33 DE

34 1 m/s35 BC, CD, and FG36 AB, GH, and EF

37 2 43 0.2 s38 3 44 44 m39 the same 45 3 s to 4 s40 49 m/s 46 1 m/s41 10 s 47 2 s to 3 s42 12 m 48 2.9 s

49 20.5 m/s50 4 m/s2

51 157 m52 t = 5 s53 between t = 0 and t = 5 s54 24 m/s55 – 5.0 m/s2

PART C – QUESTIONS, PG 07956 - 58 graph below

9 m/s – 20 m/sm = –––––––––––––– = – 5 m/s2

3 s – 0.8 s

Time

Speed

Distance

0.00

0.00

0.00

1.00s

9.81

4.91

6.00s

58.9

176.8

5.00s

49.1

122.8

4.00s

39.2

78.6

3.00s

29.4

44.1

2.00s

19.6

19.6

Ac

ce

lera

tion

Time

59 Dist = speed x time= (40 m/s) (20s)= 800 m

60!v 40 m/s – 0 m/sa = –––; a = –––––––––––––; a = 2 m/s2!t 20 s – 0 s

61 30°30 m

62 4.6 cm ( ––––– ) = 138 m1 cm63 238 m64 d = vi t + 1–2 a t2

d = 1–2 gt2

2dt = ––g

2(238 m)t = ––––––––9.81 m/s2

t = 7.0 s

CHAPTER 4FORCES AND FRICTIONKEY IDEA 5PERFORMANCE INDICATOR 5.1TRY IT: SKILL M 2.1, PG 0831a 10 N1b the ring would accelerate in the direction of

the resultant2a min = 0 N

max = 20 Nvector 3 = 10 N (is between 0 N and 20 N)yes

2b no2c no

TRY IT: SKILL M1.1, PG 0853 2 m/s2

4 Yes, the block can be moving in dynamicequilibrium but there is no accelerationbecause there are no unbalanced forces.

5 0.2 m/s2; West

TRY IT: SKILL PG 0866 547 N

TRY IT: SKILL PG 0897 9.83 m/s2

TRY IT: SKILL5.1X, PG 0948a 96 N8b the skier will accelerate9a 28.0 N9b 0.57

9c wet, the coefficient is closer to that given forwet concrete

10a lubricating the surface10b polishing the surface

TRY IT: SKILL5.1IX, PG 09811 graph at right

Note: If force is constantand friction is eliminated.

12 inverse13 ma should be constant14 F = ma

CHAPTER 4 ASSESSMENTSPART A – QUESTIONS, PG 1001 1 13 22 2 14 43 2 15 34 1 16 15 3 17 16 1 18 17 3 19 18 2 20 49 4 21 410 1 22 111 2 23 412 2

PART B – QUESTIONS, PG 10424 48 N25 6000 N26 the ground pushing forward27 328 10 N29 , parallel to ramp, upward

30 50 N31 10 N32 1.6 m/s2

33 0.49 m/s2

34 6 m/s2

35 4 m/s2

36 1.99 x 1020 N37 1.01 x 10–47 N38 96 N

PART C – QUESTIONS, PG 10439 graph at right:40 F = ma

20 N = 5.0 kg (a)20 Na = –––––5.0 kg

a = 4 m/s2 ; to the right

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Ac

ce

lera

tion

mass

• 49 N

20 N•

• 49 N

C

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41 1 cm = 8.0 N42 graph

at right

43 45. N44 27° N of E45 F = ma

45.4 N = 10 kg aa = 4.54 m/s2 @ 27° N of E

CHAPTER 5MOTION IN A PLANEKEY IDEA 5PERFORMANCE INDICATOR 5.1TRY IT: PG 1121 The first would be half a parabola, the

second nearly linear. Air resistance, andhorizontal forces on the parachute causinghorizontal acceleration.

TRY IT: SKILL 5.1VII, PG 1142 14.3 s for 2000 m: 20.2 s

1300 m 1800 m

TRY IT: SKILL M1.1, 5.1VII, PG 1163a vH = 0.46 m/s

vV = 0.47 m/st = .097 srange = .044 m

3b vH = 121.6 m/s vV = 79.0 m/stime = 16.1 srange = 1960 m

3c vH = 146.8 km/h vV = 209.7 km/htime = 43.5 srange = 6280 m

TRY IT: PG 1174a 36°4b 7 iron

CHAPTER 5 (SECTION 1) ASSESSMENTSPART A – QUESTIONS, PG 1181 4 7 12 3 8 23 3 9 44 1 10 35 4 11 46 4

PART B – QUESTIONS, PG 12112 2.43 m 15 13. m/s13 23.1 m/s 16 17. m14 300 m/s 17 decreases18 The range decreases with decreasing angle of

elevation because time in flight decreases.19 34.6 m/s20 graph at right

PART C – QUESTIONS, PG 12321 20, 7022 sin 20 = cos 7023 Correction: Chart in Review Book, pg. 123

Quantity Column, change:“Initial horizontal vertical velocity (Vy)”“Maximum vertical distance”

CHAPTER 5 (SECTION 2)UNIFORM CIRCULAR MOTIONTRY IT: PG 1295 Students’ answers will vary

TRY IT: PG 1316 Plot force versus velocity squared, the slope

will be the mass divided by the radius.

TRY IT: PG 1317 Plot radius versus velocity squared, the

slope will be the mass divided by G force.

8 Students’ answers will vary (should bewithin 5%).

CHAPTER 5 (SECTION 2) ASSESSMENTSPART A – QUESTIONS, PG 1321 1 6 12 1 7 23 3 8 14 2 9 25 4 10 4

20 N

P 40 N

Ho

rizo

nta

l sp

ee

d

Time

Quantity Increase,decrease,

remain the same

Explanation

Initial horizontalvelocity (Vx)

Decrease Vx = Vi Cos "as " increases, Cos " decreases

Initial verticalvelocity (Vy)

Increase Vx = Vi Sin "as " increases, Cos " increases

Maximumvertical height

Increase a longer Vy means greaterheight

Maximumdistance

Increase even with less Vx , there ismore time in the air, hencegreater distance

Time in air Increase a largerVy means longer flighttime

PART B – QUESTIONS, PG 13411 East 13 8 m/s2

12 3 m/s2 14 3900 N15 B

PART C – QUESTIONS, PG 13516 d = vit + 1–2 at2

d = 1–2 gt2

2dt = ––g

2(60.0 m)t = ––––––––9.8 m/s2

t = 3.50 s

17 vf = vi + atvf = + (9.8 m/s2) (3.5 s)vf = 34.3 m/s

18a t = 3.50 s; no change18b vf = 34.3 m/s; no change

vv19 sin Ø = ––v

vv = v sin Ø= 40.0 m/s sin 40° = 25.7 m/s

vh20 cos Ø = ––– v

vh = v cos Øvh = 40.0 m/s cos 40ºvh = 30.6 m/s

dh = vhtdh = (30.6 m/s)(2.5 s)dh = 76.5 m

21 An increase in initial velocity and/or anincrease in angle to forty-five degrees wouldincrease the range.

22 graphic at right

mv223 Fc = –––r60 kg (5 m/s)2

Fc = –––––––––––10 m

Fc = 150 N

CHAPTER 6MOMENTUMKEY IDEA 5PERFORMANCE INDICATOR 5.1TRY IT: PG 1431a 18.3 m/s 2b 166. N*sec1b 2900 N 2c 22960 N2a 2.03 m/s 2d 1660 N

CHAPTER 6 ASSESSMENTSPART A – QUESTIONS, PG 1461 3 5 32 1 6 33 3 7 24 3

PART B – QUESTIONS, PG 1478 75 N 13 128 kg m/s9 2400 N 14 6600 N10 3.0 x 104 N•S 15 3 m/s11 2.0 x 103 kg 16 0.2 m/s12 4 m/s

PART C – QUESTIONS, PG 14817 vf

2 = vi2 + 2 a s

vf2 – vi

2s = ––––––––

2a

(0 m/s)2 – (20 m/s)2= ––––––––––––––––

2 (–5 m/s)= 40 m

18 F!t = m!vJ = m!v

= 1000 kg (20 m/s)= 2.0 x 104 kg m/s

CHAPTER 7SWINGS AND SPRINGSKEY IDEA 4PERFORMANCE INDICATOR 5.1, 4.1TRY IT: SKILLS 2.2, 2.3, 2.4, 4.1IV, PG 1511 increase the length2 decrease the length

3 T = 2# l–g

4 The acceleration due to gravity, and thelength of the pendulum.

5 6.9 s

TRY IT: SKILLS 2.3, 2.4, PG 153

6 T = m––k

7 the mass and spring constant8 They are analogous with

m––k replacing l–g

TRY IT: SKILLS 5.1III, PG 1559 Answers will vary, but should be

reasonable.

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C

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CHAPTER 7 ASSESSMENTSPART A – QUESTIONS, PG 1551 12 33 44 2

PART B – QUESTIONS, PG 1565 2 N/m6 40 N/m7 10 N/m8 50 N/m9 100 N/m10 2.5 N/m11 R, U, Y12 W, X, Z

PART C – QUESTIONS, PG 15813, 14, 15, 16

18.5 N – 13.5 Nm = ––––––––––––––0.66 m – 0.48 m

5 N= ––––––0.18 m

= 27.8 N/m

CHAPTER 8WORK, POWER, AND ENERGYKEY IDEA 4PERFORMANCE INDICATOR 4.1TRY IT: PG 1601a no1b yes1c no (if we neglect friction)1d yes

TRY IT: PG 1632 130 J3 100 J4 4700 J5 5 push–ups6 7.5 J7 2400 J8 table below

TRY IT: PG 166Dodge Viper 0.14 hp/lbsFerrari 0.13 hp/lbsShelby 0.12 hp/lbsLotus 0.11 hp/lbsCorvette 0.11 hp/lbsPorche 0.10 hp/lbs3000GT 0.09 hp/lbsEscort 0.04 hp/lbs

TRY IT: SKILLS 4.1VII, PG 1689 Students’ answers may vary.10 Speed is constant. No work done against

friction. Mass is constant.

TRY IT: PG 17011 check for understanding (using 60 J)

A = 60 JB = 60 JC = 40 JD = 20 JE = 0 J

TRY IT: SKILLS 4.1I, 4.1V, PG 17312 The energy comes from the work done on

the can.13 The can imparts and impulse to the nail.14 Student results should reflect a

proportional response.15 no

Trial

123

Angle

32º35º39º

Force

1.10 x 104N1.44 x 104N1.64 x 104N

Displace-ment

106 m84.4 m75.3 m

Work(Joules)

1.166 x 106 J1.215 x 106 J1.235 x 106 J

Potential Energy(Joules)

1.166 x 106 J1.215 x 106 J1.235 x 106 J

8 16 24 32 40 48

Elongation (10–2 m)

Forc

e (N

)

56 64 72 80

CHAPTER 8 ASSESSMENTSPART A – QUESTIONS, PG 1731 4 6 22 4 7 13 1 8 14 3 9 15 4 10 2

PART B – QUESTIONS, PG 17511 9.8 x 103 J 19 112 700 J 20 500 N/m13 250 J 21 0.6 J14 480 J 22 2 x 10–3 J15 5 m/s 23 2.4 J16 213 W 24 12 J17 250 W 25 4.0 x 106 J18 2.7 x 104 N

PART C – QUESTIONS, PG 17726 W = Fd F = W = mg

= (680 N) (3.5 m)= 2380 J

W27 P = ––t

2380 JP = ––––––

11.4 s

P = 209 W

28 The student generates 71.2 W more power inthe second trial.

29 The slope represents the object’s velocity.

30 W = F·d= (300 N) (6.6 m)= 1980 J

31 graph at right

CHAPTER 9CONSERVATION OF ENERGYKEY IDEA 4PERFORMANCE INDICATOR 4.1TRY IT: PG 1801a 1.00 m 1d 0 m1b .200 m 1e 4.43 m/s1c 3.96 m/s 1f 1.00 m

TRY IT: PG 1832a 0 J2b 16 J

TRY IT: SKILLS 4.1II, PG 1883 v = .271 m/s4 d = 2.4 m5 µ = 0.10

CHAPTER 9 ASSESSMENTSPART A – QUESTIONS, PG 1891 4 6 12 2 7 43 4 8 34 1 9 35 3

PART B – QUESTIONS, PG 19110 .20 J 14 5.1 m11 2 15 490 J12 4 16 decreases13 4 17 4

18 remains the same19 981 J20 14 m/s

PART C – QUESTIONS, PG 19321 At position A, the energy is kinetic and at

position B, the energy is elastic potentialenergy.

22 mgh = 1–2 kx2 (PEg = PEspring)2mghk = –––––

x2

23l T = 2 # ––g

0.65 mT = 2 (3.14) ––––––––9.8 m/s2

T = 1.60 s

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24 d = 5.8 cm3 md = 5.8 cm ( ––– )1 m

d = 17.4

25 PEG = mgh= (650kg)(9.8 m/s2)(17.4 m)= 1.11 x 105 J

26 The kinetic energy of the car on the thirdhill is 1.75 times greater than when the caris on the second hill.

CHAPTER 10ELECTRIC FIELDS AND FORCESKEY IDEA 4,5PERFORMANCE INDICATOR 4.1TRY IT: SKILLS M 2.1, M 3.1, S2.1, PG 2011 The foil is attracted.2 The foil is repelled.3 The foil is attracted.4 First there is attraction, then after touching

no reaction.5 The foil is repelled.

CHAPTER 10 ASSESSMENTSPART A – QUESTIONS, PG 2061 3 10 32 1 11 43 1 12 24 1 13 35 2 14 26 4 15 27 4 16 28 4 17 49 2 18 2

PART B – QUESTIONS, PG 21019 168 V 25 420 5.0 V 26 250 N/C21 3.2 x 10–19 C 27 4 V22 8 x 10–19 C 28 2.0 x 107 V23 4.8 x 10–19 C 29 2.7 N24 3.6 N

30 graphic at right

31 332 333 334 3

PART C – QUESTIONS, PG 21335 graphic below

36 k q1 q2Fe = –––––––r2

9 x 109 N m2/c2 (2.4 x 10–6 C)(–2.4 x 10–6 C)Fe = ––––––––––––––––––––––––––––––––––––

(0.50 m)2

= 2.07 x 10–1 N

37 graph at the right

38

39 k q1 q2Fe = –––––––r 2

(9 x 109 N m2/c2)(1.6 x 10–19 C)(–1.6 x10–19 C)Fe = –––––––––––––––––––––––––––––––––––––

(1 x 10–10 m)2

= 2.304 x 10–8 Norder of magnitude is 10–8

40 G m1m2Fg = –––––––r2

6.67 x 1011N m2/kg2 (9.11 x 10–31kg)(1.67 x 10–27kg)Fg = ––––––––––––––––––––––––––––––––––––

(1 x 10–10 m) 2

= 1.01 x 10–47 Norder of magnitude is 10–47

41 The electrostatic force is on the order ofmagnitude 1039 N larger than thegravitational force.

42 VE = ––d

1.0 x 102 VE = –––––––––––2.0 x 10–2 m

E = 5.0 x 103 V/m

SphereRTU

Charge0++

+ + + + +

– – – – –

A B• •

CHAPTER 11OHM’S LAW AND RESISTIVITYKEY IDEA 4PERFORMANCE INDICATOR 4.1TRY IT: PG 2211 table below

CHAPTER 11 ASSESSMENTSPART A – QUESTIONS, PG 2251 4 7 32 4 8 23 2 9 44 1 10 35 4 11 16 2 12 3

PART B – QUESTIONS, PG 22713 qI = ––t

q = Itq = (3.0 A)q = 90 C

14 1 V15 5.0 C16 3 x 103 A17 10 $18 8 C19 4 V20 3.6 x 104 W21 36 $22 60 W23 15 A24 2.7 x 105 J25 20 J26 1.01 x 105 J

PART C – QUESTIONS, PG 22927 graphic below

28 1 1 1––– = –– + ––Req R1 R2

R1 R2Req = ––––––R1 + R2

(5 $) (20 $)Req = –––––––––––

5 $ + 20 $100 $2

Req = ––––––25 $

Req = 4 $

29 V = I Req

VI = ––Req

24vI = –––

4 $I = 6.0 A

30 qI = ––t

qt = ––I

25 Ct = –––––––––

2.0 x 104 At = 1.25 x 10–3 s

31 W = VIt W = (1.8 x 106 V)(2.0 x 104 A)(1.25 x 10–3 s)W = 4.5 x 107 J

32a graph below

32b 10.5 V – 4.0 Vm = ––––––––––––––––––––––4.25 x 10–2 A – 1.5 x 10–2 A

m = 236.4 $

32c Slope represents resistance in the resistor.

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Hair Dryer

Microwave

CurlingIron

Vacuum

5000

900

1200

1500

120

120

120

120

41.7

Appliance Power(W)

Voltage(V)

Current (A)

7.5

10.0

12.5

2.9

Resistance ($)

16

12

9.6

5 $$ 20.0 $$24 V

A

••

1 2 3 4 5 6

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CHAPTER 12SERIES AND PARALLEL CIRCUITSKEY IDEA 4PERFORMANCE INDICATOR 4.1TRY IT: SKILLS 4.1XIII, 4.1VIII, 4.1XII, PG 2381 circuit below

2 V = IReq

V Req = ––I

12 vReq = –––– 6 A

Req = 2 $

3 Req = 2 $1 1 1––– = –– + ––

Req R1 R2

Req R2R1 = –––––––Req + R2

(2 $) (3 $)R1 = ––––––––––

–2 $ + 3 $

6 $2R1 = ––––

1 $

R1 = 6 $

CHAPTER 12 ASSESSMENTSPART A – QUESTIONS, PG 2381 42 43 4

PART B – QUESTIONS, PG 2394 1 A 8 11 A5 2 $ 9 3 A6 5 V 10 0.75 $7 12 $ 11 1/3 the value

PART C – QUESTIONS, PG 24112 V = IR

V = (0.50 A)(5 $)V = 2.5 V

13 W = VItW = (15 V)(0.5 A)(600 s)W = 4500 J

14a circuit below

V14b I1 = –––R

12 VI1 = –––18 $

I1 = 0.67 A

15 table below

Series Parallel

SeriesVT = 195 VReq = R1 + R2 +R3

= 13 $ + 15 $ +11 $= 39 $

V 195 V I = –– ; –––– = 5 AR 39 $

V1 = IR1V1 = (5 A)(13 $)

= 65 VV2 = IR2

= (5 A)(15 $)= 75 V

V2 = IR3= (5 A)(11 $)= 55 V

P1 = V1I; P1 = (65 V)(5 A) ; P1 = 325 W

P2 = V2I; P2 = (75 V)(5 A) ; P2 = 375 W

P3 = V3I; P3 = (55 V)(5 A) ; P3 = 275 W

Parallel

V V 360 VI = ––; I4 = –– ; I4 = ––––– ; I4 = 8 AR R4 45 $

360 VI5 = ––––– ; I5 = 2 A180 $

360 VI6 = ––––– ; I6 = 10 A36 $

12 v R2 = 3 $$R1

A

V

Resistor

Resistance $

Current A

Voltage V

Power W

R1

13

5

65

325

R5

180

2

360

720

R6

36

10

360

3,600

R4

45

8

360

2880

R3

11

5

55

275

R2

15

5

75

375

12 v R2

R1

AA

V

CHAPTER 13MAGNETISMKEY IDEA 4PERFORMANCE INDICATOR 4.1TRY IT: SKILLS 4.1XV, PG 2531 The magnet has been polarized with two

sets of poles.

CHAPTER 13 ASSESSMENTSPART A – QUESTIONS, PG 2541 4 9 32 4 10 23 4 11 34 3 12 15 3 13 16 1 14 37 3 15 28 2 16 3

PART B – QUESTIONS, PG 25717 graphic of magnets below

18 419 420 graphic of magnetic fields

21 Magnetism is caused by the movement ofelectrons spinning on their axis.

22 When in the influence of a magnetic field, a nailwill lose its induced magnetic properties whenthe external field is removed.

PART C – QUESTIONS, PG 25823 Students’ answers will vary.24 graphic of bar magnets below

25a The magnet causes the protons in thehydrogen of water molecules to align.

25b When the radio wave pulses arediscontinued, the protons return to theiroriginal orientation, emitting a signal whenthey do.

25c Diseased tissue contains more water thanhealthy tissue, thereby producing a strongersignal.

26a Power loss goes as the square of both thepotential and the current. However, it goesdirectly proportional with the resistance athigh current, but inversely proportionalwith resistance with high voltage.

26b A step up transformer increases the voltagein the secondary.However, the current isreduced. This is due to the conservation ofenergy.

26c A varying magnetic field is required toinduce a voltage in a conductor. Themagnetic field about a conductor with directcurrent is constant.

CHAPTER 14WAVE PROPERTIESKEY IDEA 4PERFORMANCE INDICATOR 4.3TRY IT: SKILLS 4.3IV, PG 2611 32 33 longitudinal

TRY IT: SKILLS 4.3I, PG 2644 A, B, D, C5 B, A, C, D or B, C, A, D A = C6 D, A, C, B or D, C, A, B7 B, A, C, D or B, C, A, D8 A and C9 - 11 graphic of waves below

TRY IT: PG 27112 graphic of waves below

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S SNN

N SS N

A

B

C

D

A

B

A+B

S N N S

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CHAPTER 14 ASSESSMENTSPART A – QUESTIONS, PG 2741 1 8 32 3 9 23 2 10 44 3 11 15 2 12 46 4 13 17 3 14 1

PART B – QUESTIONS, PG 27615 2 m16 0.5 m17 5 Hz18 20 Hz19 90º20 4.0 x 10–3 s21 2 Hz22 0.5 s23 5.0 x 10–3 s24 0.1 s25 5 m26 200 Hz27 0.1 m and 0.6 m28 1.529 90º

PART C – QUESTIONS, PG 27829 1.2 cm ±0.2 cm

30 14.6 cm ±0.2 cm

31 the wavelength would decrease

32 4

33 .4 cm

34 v = f %= (12 Hz)(.004 m)= .048 m/s

35 The angle of incidence equals th angle ofreflection.

36 Ø1 must equal Ø2

CHAPTER 15SOUND AND LIGHTKEY IDEA 4PERFORMANCE INDICATOR 4.3TRY IT: PG 2851 7 s2 t = time for stone to fall and time for echo

to returnd = 1/2 gt2

1200 m = 1/2(9.81 m/s2) t2

2400 mt2 = –––––––– = 15.6 s (to hit well bottom)9.81 m/s2

t = 15.6 s + 3.63 s = 19.2 s

CHAPTER 15 ASSESSMENTSPART A – QUESTIONS, PG 2871 1 7 32 4 8 43 1 9 14 4 10 45 1 11 26 4

PART B – QUESTIONS, PG 28912 2 x 10–15 s 16 213 Blue 17 858 m14 30 cm 18 resonance15 1655 m 19 C

PART C – QUESTIONS, PG 29020 v = f % ; v = c ; c = f % ; f = c / %;

f = (3.0 x 108 m/s) / (0.12m)f = 2.5 x 109 Hz

21 The Doppler Effect is the apparent change infrequency due to the relative velocitybetween source and observer.

22 A radar signal is sent toward the storm. Thereflected wave is recorded from the storm.The comparison of many of these reflectedwaves would reveal a shortening time ofreturn, indicating the storm is approaching.

23 Behind plate: d d 127 m – 18.3 mv = – ; t = – ; t = ––––––––––––– ; t = 0.315 st v 345 m/s

24 As the motorcycle approaches the front (1st)jogger, the fronts of the sound waves will beforeshortened because of the relativemotion of the cycle-jogger. The wave frontsreaching the jogger are interpreted by thejogger to have a frequency higher than theactual frequency of the cycle sound. Thejogger behind (2nd) the motorcycle willmake the opposite observation due to thelengthening of the wavelength due to theincreasing relative distance.

Ø1

Ø2

30°

30°

CHAPTER 16REFRACTION AND REFLECTIONKEY IDEA 4PERFORMANCE INDICATOR 4.3TRY IT: PG 2931 graphic below

2 33 30º

TRY IT: SKILLS 4.3VIII,4.3VIX, PG 2994 Øi = 30º n1 sin Ø1 = sin Ø2 (n2)

n1 sin Ø1n1 = 1 sin Ø2 = ––––––––n2

1 sin 30ºn2 = 1.61 sin Ø2 = ––––––– ; Ø2 = 18.1º1.61

Ø2 = 22.1ºØ3 = 11.9ºØ4 = 13.3ºØ5 = 29.9º

TRY IT: PG 3015 Students’ answers will vary.

CHAPTER 16 ASSESSMENTSPART A – QUESTIONS, PG 3051 2 7 32 2 8 13 4 9 14 1 10 15 4 11 46 3

PART B – QUESTIONS, PG 30712 1m 16 30º13 70º 17 49º14 2.04 x 108 m/s 18 2.0 x 108 m/s15 30.3º 19 42º

PART C – QUESTIONS, PG 30820 37º

21 n1 sin Ø1 = n2 sin Ø2

n1 sin Ø1n2 = ––––––––sin Ø2

1.0 (sin 60º)n2 = ––––––––––(sin 37º)

1.15n2 = ––––0.60

n2 = 1.4422 n1 sin Ø1 = n2 sin Ø2

n1 sin Ø1sin Ø2 = –––––n2

n1 sin Ø1Ø2 = arc sin ( ––––––––)n2

1.50 sin 30ºØ2 = arc sin ( ––––––––––)1

Ø2 = arc sin (0.75)

Ø2 = 48.6º

23 graphic below

24 1 1sin Øc = (––) ; Øc = arc sin (––) ; m m1Øc = arc sin( ––– )

1.50

Øc = 41.8º

25 Øi = 33º26 Ør = Øi

Ør = 33º

PAGE 22 TEACHER SUPPLEMENT: PHYSICAL SETTING: PHYSICS STAREVIEW N&N©

N O P E R M I S S I O N H A S B E E N G R A N T E D B Y N & N P U B L I S H I N G C O M P A N Y , I N C T O R E P R O D U C E A N Y P A R T O F T H I S B O O K

45°45°

Air

48.6°

30°

Lucite

TEACHER SUPPLEMENT: PHYSICAL SETTING: PHYSICS STAREVIEW N&N© Page 23

N O P E R M I S S I O N H A S B E E N G R A N T E D B Y N & N P U B L I S H I N G C O M P A N Y , I N C T O R E P R O D U C E A N Y P A R T O F T H I S B O O K

27 graphic:

CHAPTER 17DIFFRACTION AND INTERFERENCEKEY IDEA 4PERFORMANCE INDICATOR 4.3

CHAPTER 17 ASSESSMENTSPART A – QUESTIONS, PG 3191 1 4 42 2 5 13 1 6 3

PART B – QUESTIONS, PG 3207 180° 8 90°

PART C – QUESTIONS, PG 3219-119 graphic below

10 amplitude = 0.3 m

11 % = 0.4 cm

CHAPTER 18MODERN PHYSICSKEY IDEA 5PERFORMANCE INDICATOR 5.3TRY IT: SKILLS 5.3I, PG 3341 d - c = 0.57 eV c - b = 0.22 eV

d - b = 0.79 eV c - a = 4.86 eVd - a = 5.43 eV b - a = 4.64 eV

2 d - c & infrared (continues 2nd column)

E .57 eV (1.6 x 10–19 J/eVf = –– = ––––––––––––––––––––h 6.63 x 10–34 J•s

= 1.38 x 1014 Hz

d - b & infrared 1.90 x 1014 Hzd - a & ultraviolet 1.31 x 1015 Hzc - b & infrared 5.28 x 1013 Hzc - a & ultraviolet 1.17 x 1015 Hzb - a & ultraviolet 1.12 x 1015 Hz

3 The energy range in visible light is from3.19 eV to 1.59 eV. To determine the rangemultiply the highest visible light frequencyby Planck’s constant. Then, divide by 1.6 x10–19 J/eV to convert to eV. Repeat with thelowest visible light frequency to determinethe range.

Jump eV Frequency Colore & b 2.03 4.92 x 1014 orangef & b 3.03 7.33 x 1014 violete & c 1.81 2.86 x 1014 redf & c 2.84 6.87 x 1014 violetg & c 3.04 7.36 x 1014 violetf & d 2.27 5.49 x 1014 greeng & d 2.47 5.98 x 1014 greenh & e 2.14 5.18 x 1014 yellowi & e 2.15 5.20 x 1014 green

CHAPTER 18 ASSESSMENTSPART A – QUESTIONS, PG 3361 4 5 1 9 2 13 22 3 6 3 10 2 14 33 4 7 4 11 2 15 44 4 8 1 12 1 16 4

PART B – QUESTIONS, PG 33817 4.64 eV 24 1.12 x 10–27 kg m/s18 1.63 x 10–18 J 25 4.42 x 10–36 m19 1.89 eV 26 2.0 x 10–20 J20 7.37 x 10–35 m 27 2.2 x 10–19 J21 3.68 x 10–12 m 28 3.3 x 10–34 m22 3.98 x 10–19 J 29 0.31 eV23 3.32 x 10–18 J

33°

33°

0.3 –

0.3 –

0.3 –

0.50 1.0 1.5 2.0

met

ers

PART C – QUESTIONS, PG 33930 KE = 1/2 mv2

= 1/2(9.11 x 10–31 kg)(2 x 106 m/s)2

= 1.82 x 10–18 J

h h 31 % = –– ; % = ––– ; p mv

6.63 x 10–34 J•s% = –––––––––––––––––––––––––(9.11 x 10–31 kg)(2.0 x 106 m/s)

% = 3.64 x 10–10 m32 E = 2.55 eV

1.6 x 10–19 JE = 2.55 ev (–––––––––––)1 eV

E = 4.08 x 10–19 Hz

E 4.08 x 10–19 J33 E = hf ; f = –– ; f = –––––––––––––––h 6.63 x 10–34 J•s

= 6.15 x 10–14 Hz

34 Ep = 2.55 eVEp = Ei – Ef

2.55 eV = Ei – Ef

2.55 eV = [–0.54 – (–3.40)] eV ;n = 4 to n = 2

= 2.55 eV*****************************************

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