CHAPTER 6:UNIFORMCIRCULARM OTION ANDGRAVITATION
NGLEANDANGULARVELOCITY1.Semi- lometersshouldtheodometerread?SolutionGiven:d 1.15 m r 1.15 m2π rad 0.575 m , Δθ 200,000 rot 1.257 10 6 rad21 rotFindΔsusingΔθ Δs,sothatr()Δ s Δ θ r 1.257 10 6 rad (0.575 m)5 7.226 1 0 m 723 ersecond?Whatisthisinrev/min?SolutionGiven:r 0.420 m, v 32.0 m s .v 32.0 m sUse ω 76.2 rad s.r 0.420 mConverttorpmbyusingtheconversionfactor:49
CollegePhysicsStudentSolutionsManualChapter61 rev 2π rad ,1 rev60 s 2π rad 1 min 728 rev s 728 rpmω 76.2 rad s ion(a)Usev rωtofindthelinearvelocity:2 π rad 1 min v rω (0.100 m ) 50,000 rev/min 524 m/s 0.524 km/s1 rev60 s rad1y(b)Given:ω 2π 1.988 10 7 rad s ; r 1.496 1011 m7y 3.16 10 sUsev rωtofindthelinearvelocity:()()v rω 1.496 1011 m 1.988 10 -7 rad s 2.975 10 4 m s 29.7 km 00mradiuscurvebankedata20.0 angle?50
onUsingtan θ v2gives:rgv2tan θ v rgtan θ rg(100 m)(9.8 m s 2 )tan 20.0 18.9 m ttheNorthPoleis9.830 m/s 2 mparethiswiththeacceptedvalueof5.979 10 24 kg .Solution(a)Usingtheequationg GMgives:r22()()6371 10 3 m 9.830 m s 2GMr2gg 2 M 5.979 10 24 kg 1122Gr6.673 10 N m tdistancetoEarth,some6.29 1011 neffect,muchlessthatanunknownforcecausesit.)51
on(a)UseF GMmtocalculatetheforce:r2()GMm 6.673 10 11 N m 2 kg 2 (100 kg )(4.20 kg )Ff 2 7.01 10 7 N2r(0.200 m)(b)ThemassofJupiteris:mJ 1.90 10 27 kgFJ (6.673 10 11)()N m 2 kg 2 1.90 10 27 kg (4.20 kg )(6.29 1011m2) 1.35 10 6 NFf 7.01 10 -7 N 0.521FJ 1.35 10 -6 sactualmass.Usingr3G 2 M ,wecansolvethemassofJupiter:2T4π4π 2 r 3MJ G T23()4π 24.22 10 8 m 6.673 10 -11 N m 2 kg 2 (0.00485 y ) 3.16 10 7 s y[()]2 1.89 10 27 tisinanorbitofthesame52
thatintersectsthesatellite’sorbitatanangleof90 terthantherivet’s.)Solution(a)UseFc mac,thensubstituteusinga v2GmMandF .r2rGmM mv 2 rr2GM Ev rS(6.673 10 11)()N m 2 kg 2 5.979 10 24 kg 2.11 10 4 m s900 10 3 (a):()v tot v 2 v 2 2v 2 2 2.105 10 4 m s 2.98 10 4 m s(c)Usingkinematics:d vtott t (d)F (d3.00 10 3 m 1.01 10 7 s4vtot 2.98 10 m s)()Δp mvtot 0.500 10 3 kg 2.98 10 4 m s 1.48 108 N-7Δtt1.01 10 e’sframeofreference,vi v tot , and vf 0.So,thechangeinthekineticenergyoftherivetis:Δ KE 211122mvtot mvi 0.500 10 3 kg 2.98 10 4 m s 0 J 2.22 10 5 J222()(53)
College"Physics" Student"Solutions"Manual" Chapter"6" " 50" " 728 rev s 728 rpm 1 min 60 s 2 rad 1 rev 76.2 rad s 1 rev 2 rad , π ω π " 6.2 CENTRIPETAL ACCELERATION 18." Verify&that ntrifuge&is&about 0.50&km/s,∧&Earth&in&its& orbit is&about p;linear&speed&of&a .
Part One: Heir of Ash Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 Chapter 24 Chapter 25 Chapter 26 Chapter 27 Chapter 28 Chapter 29 Chapter 30 .
TO KILL A MOCKINGBIRD. Contents Dedication Epigraph Part One Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Part Two Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18. Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 Chapter 24 Chapter 25 Chapter 26
DEDICATION PART ONE Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 PART TWO Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 .
vec tor [6], [7], [8]. In P om plun and M atari c [6], a segm ent is rec ogn ize d w hen the roo t m ea n squ are (R M S ) value of the joint velocities falls below a ce rtain thresho ld,i.e.,assum ing that there w ill be a pause in the m otion b etw ee n m otion prim i
About the husband’s secret. Dedication Epigraph Pandora Monday Chapter One Chapter Two Chapter Three Chapter Four Chapter Five Tuesday Chapter Six Chapter Seven. Chapter Eight Chapter Nine Chapter Ten Chapter Eleven Chapter Twelve Chapter Thirteen Chapter Fourteen Chapter Fifteen Chapter Sixteen Chapter Seventeen Chapter Eighteen
18.4 35 18.5 35 I Solutions to Applying the Concepts Questions II Answers to End-of-chapter Conceptual Questions Chapter 1 37 Chapter 2 38 Chapter 3 39 Chapter 4 40 Chapter 5 43 Chapter 6 45 Chapter 7 46 Chapter 8 47 Chapter 9 50 Chapter 10 52 Chapter 11 55 Chapter 12 56 Chapter 13 57 Chapter 14 61 Chapter 15 62 Chapter 16 63 Chapter 17 65 .
HUNTER. Special thanks to Kate Cary. Contents Cover Title Page Prologue Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter
Accounting information and managerial work. Accounting, Organizations and Society, 35 (3), 301-315. ABSTRACT . Despite calls to link management accounting more closely to management (Jonsson, 1998), much is still to be learned about the role of accounting information in managerial work. This lack of progress stems partly from a failure to incorporate in research efforts the findings regarding .
