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1、高一下英语动词的时态_高一下英语动词的时态高一下英语动词的时态-,asshownintheleveloftheceilingofAfixedpointAplaceAlightbar,rodbeverticalwiththeceiling.ThelowerendoftherodwithalightrollerO.AnotherthreadupperendfixedontheceilingoftheBpoint,thinlinesacrossthepulleyO,thebottomisaGweightoftheobject.BOlongthinlinewithanAngleofthetaceili
2、ng=30o.Systemremainstill,excludingallfriction.Inthefollowingstatementiscorrect()A.finelineBOtensionsizeisontheceilingB.arodontheactingforceofpulleysizeisC.a.rodandthreadforcesizeofpulleyisGDapoleontheactingforceofpulleysizeisGWithheavyGastheresearchobject,heavyGinonlybythepullofgravityandthinlineand
3、balance,sothinlineofpullingforceandthegravityofweightsintheweightback.Becausethreadwithoutgravity,subjecttoallthefriction,sothesizeofthetensilestrengthofthethinlineforGeverywhere,soAoptionsthreadBOisGtothetensionsizeoftheceiling.Withfinelinesandpulleyofthecontactpointastheresearchobject,thecontactpo
4、intbalance,hasknowntwothreadtensionforceandpulleyupwardforcetothecontactpointoflargereverse,suchasdrawingavailablepulleysizeupwardforcetothepointofcontactforG.Withrollerastheresearchobject,pulley,onlybyapressurerodandfinelinecontact,soCoptionerror,shouldbe0.Andarodforthepulleyshouldfinelinetothereac
5、tionbetweenthepulleyandthepowerofthelargereverse,suchasistheG,sotheDoptioniscorrect.-Asshown,AandBareheavy,AbybypasssmallpulleyPthreadhanging,Bintheroughleveltable;PulleyPisadiagonalshortdepartmentontheceilingoftheO;Nodesarethethreewires,pullabblevelobjects,cpullspringsalongtheverticaldirection;Spri
6、ng,finelines,smallpulleysandfinelineofgravityandthefrictionbetweenthepulleysarenegligible,thewholedeviceinstaticbalancestate.IfthesuspensionslashtheOPtensionissmallpulleys,ofthefollowingstatementsinthewrongPutOPropetensionisdecomposedintohorizontaldirectionandverticaldirection,OPlinewithanAngleof30d
7、egrees,thehorizontaldirectionishorizontalF1=Fcos30=30n,verticaldirectiononF2=Fsin303n=10),AspringofelasticwithF2areequal,shouldbe3n10),soAwrong;O3n,Ptheropepullingforceof20)crownblockandeffort,theweightofAqualityfor3kg(10),soBiswrong;DesktopfrictionwithF1equaltoB,about30n,soCiswrong;OPwithverticaldi
8、rectionAngleof60degrees,sotheDof.-AqualityformwedgestorestontheAnglefortheslope.NowputablockofcontentverticaldownwardconstantforceF,asshown.Thecontentblock高一下英语动词的时态高一下英语动词的时态A.isstillinthestationarystateBefore111pressurizedbyequilibriumconditionsMgsinethetaequalsmumgcosinethetaSolutiontomu=tantheta
9、.PressureF,thenstressanalysis(mg+F)sinetheta=mu(mg+F)cosinethetaBalancebytype,object,resultantforceiszero,accordingto1,theobjectwithoutacceleration,sostillstillYes,heisstaticRemember,aslongasthelaunchontheinclineAngleofthetaMu=tantheta.AslongasnotinthehorizontalforcefTheobjectisalwaysbalancedResulta
10、ntforceisconstant,butthefrictioninthechangeThisknowledgeiscalledfrictionAngle,thisisjustforthisconclusionisderived,andtheknowledgeofthisproblemisexamined222problemsolvingisthekeytojudgewhethertheforceofthedownalongtheslopeisgreaterthanthemaximumstaticfrictionforceMumgcosinethetaforslidingfrictionGen
11、erallythinkthatslidingfrictionvalueisthesameasthevalueofthemaximumstaticfrictionforceBytheknownavailablemgsinethetamuorlessmgcosinetheta(thatis,thedownwardforceislessthanthemaximumstaticfrictionforcealongtheslope)ThefrictionforceshouldbeequaltothesizeofthemgsinethetaSimilarlyavailable(mg+F)sinetheta
12、muorless(mg+F)cosinethetaThefrictionforcesize=(mg+F)sinethetamaximumstaticfrictionforceorlessSotheobjectisstillinthestationarystate-Asshowninfigure1-6,smallcircleofAhangingAqualityform2blockandinanothersetuprightonAbigcircle,thereisAthinline,theendtiedinAsmallcircle,Acrossingattheotherendfixedontheb
13、igcirclepointBafterAsmallpulleyhangingAqualityform1wedges.Ifthesizeofthesmallring,pulley,ropeandqualityaswellasthefrictionisnegligible,ropeandareelongation,ifthebalanceofthechordABcentralAngleforalpha,isthequalityofthetwowedgesthanm1/m2shouldbe()B.a.cos(alpha/2)sin(alpha/2)C.2sin(alpha/2)d.2sinealph
14、aStressanalysis.UpwardforceofA,Abigcirclealongthenormaldirection!Alongthetangentdirectionofresultantforceiszero!Angleofosculation=theta/2).Gravityandnormal=Angletheta高一下英语动词的时态高一下英语动词的时态So,alongthetangentdirection:M2gsintheta=m1gcos(theta/2)Sinetheta=2sin(theta/2)cos(theta/2)M1/m2=2sin(theta/2)Optio
15、ns:C-Threerootlength,asshowninthefigureareLlightropeconnectedrespectivelyintwopoints,C,DA,Bbothendshangingintheleveloftheceiling,distancebetween2L,CpointnowhangAweightofAmassofM,tokeeptheCDlinelevel,canforcefinDpointoftheminimumfor1/2mgACCDBDropearetightTheABCDisaisoscelestrapezoid高一下英语动词的时态高一下英语动词的
16、时态ThebiggertheDpullingforceOnthecirculararcastheendpointmobileDincreasedInAconstantverticalframe(youcanhaveAhorizontalline,extendingApulley/CsideofE,itiseasytoprovethatAngleAEC=D/2anddonotchangewiththeendpoint)/thesizeofthetensionandtheAngleoftherope,thegreatertheAngle,thegreaterthetension.Movingont
17、hecirculararc,Angle,tensionMovingonthestraightrod,Angleischangeless,tensionisconstantAskwhymobileoncirculararc,getbigger,Anglecanbeanalyzespecific.AnswerattheendsoftheropedependsontheAngleoftheropehorizontaldistance,youcandraw,thatswhylaterthesameAngle.Supplement:ropemustbelong,ifthestringattheendso
18、fthehorizontaldistance,thegreatertheprojectionwillbethatthestringinahorizontaldirection,thegreaterthenamelyAnglesmaller,thegreatertheAngle-Thefollowingistrue()a.allequalandopposite,respectively,thetwoforcesactingontwoobjectsmusthaveApairofactionandreactionb.allequalandopposite,thetwoforcesactingonth
19、esameobjectmustbeApairofactionandreactionc.allequalandopposite,functioninAstraightlineandthetwoforceactingbetweentwoobjectsmusthaveApairofactionandreactiond.interactionofApairofforceswhichAforceisactuallysaidreactionisarbitraryApairofparsing】【actionandreactionshouldbeproducedinApairofinteractingobje
20、cts,A,B,Cisnotcorrect.Thenamingofactionandreactionisarbitrary,D.Force:betweentwoobjectsinteractthroughdifferentformssuchasattract,relativemotion,deformationandtheforceof,exertoneselftodoSTH.Forcearepaired,theforceisthereactionforce.Reactionforce,forceandreactingforcebetweentwoobjects,alwaysappearatt
21、hesametime,andthesizeisequalintheoppositedirection,alongthesamelines,theeffectontheobjectsrespectively.Thisismechanicsfoundation.-MA,qualityisproportionaltotheforceF,andaccelerationisinverselyproportionaltoAB,theforceFisproportionaltothequalityofmandtheaccelerationofaC,theaccelerationoftheobjectdire
22、ctionisalwayswiththesamedirectionitbyforceDtheaccelerationofabody,acombinedwithsufferedexternalforceFdirectproportion,andmisinverselyproportionaltothemassofthebodyMmustnote:whenthemassoftheobject,theobjectofagreeswellwithoutsideforceFandtheaccelerationofaobjectisproportionaltotheiswrong,becauseitisd
23、ecidedtoacceleration.Butwhenmisthemassoftheobject,theobjectsaccelerationisdirectlyproportionaltotheforceFaandobjectssufferedwheniscorrect.-Inthelevelofthefrictionfactorof0.2hasaqualityofm=1kgofpellets,ballsandhorizontallightspringand45degreestotheverticalAngleofanelongatedlightropeislinkedtogether,t
24、he高一下英语动词的时态高一下英语动词的时态smallballinthestaticstate,andthelevelinthefaceofballelasticityiszero,whencutlightrope,take10g=o;Lightspringtensionsize,smallballsizeoftheaccelerationanddirection,attheinstantoftheshearspringspheresthesizeoftheacceleration?/ballbysunnyandbalance:gravity,elasticandtensile,decompo
25、sitionbyforce,to:Tsin(PI/4)=mgTcos(PI/4)=FSolution:2gmT=)F=gm,thisisthespringforce,justcutoffbecausethereisnochangedeformation,elasticityisalsooneoftheCutlinepatrol,idealsmallclubbygroundsupportrole,onlyhorizontalacceleration,andthereis:a=(F-mu)mg/mSolution:a=g-gmu,thedirectionisalongthedirectionoft
26、hespringforce/theropetension=mg/sin45oBecausethereisnofriction,springtension=theropetension*cos45o=mgTheropebroke,ballhorizontalfrictionandspringtensionMa=springtension-umg=mg-umgA=g-ug=8directionpointingtothespring-Asshowninfigure5,thequalityarembetweenAandBballwearingAregardlessofthequalityofthesp
27、ring,onAsmoothhorizontalplane,Aballclosetotheverticalwalls,todaywithAhorizontalforceFBtopushspring,afterthebalance,willsuddenlyFisremoved,atthismoment(B)1Bballvelocityiszero,theaccelerationiszero(2)thespeedofBballiszero,theaccelerationofsize(3)inthespringforthefirsttimetorestoretheoriginallongafter,
28、Adidntleavethewall(4)afterAleftwall,A,BtwogoalsaretotherighttodouniformmotionaboveistrueA.only(1)b.(2)(3)c.(1)(4)d.(2)(3)(4)/afterremovalofF,Bballrightacceleratedmotion,Aballoppressedstatic,springbackaftertheoriginallength,Aballbearingacceleratedmotiontotheright,Bballdoslowmotiontotheright./justremo
29、vedtheforceFis,becauseoftheinertiaobjecttokeeptheoriginalmotionstate,theinstantaneousspeediszero,thespringoftheelasticnottoF,accelerationthroughforceanalysis,A:horizontalelasticityandthesupportofmetope,elasticconstant,theresultantforceiszero,theaccelerationiszero.B:towithdrawforceF,Bhorizontalonlyby
30、elastic,accelerationfortheF/m./theproblemofNewtonssecondlawistheinstantaneouscorrespondingrelation.Tosolvetheproblemtobeclear,springdeformationisneedsaprocess,itselasticityisnotmutation.Aballintheelasticityandwallundertheforceofbalance,removedafterF,hasnoeffectonA,A,velocityandaccelerationinthismome
31、nt,stillis0.高一下英语动词的时态高一下英语动词的时态BballisundertheeffectofFandelasticequilibrium,namely,F=F,Fisremovednow,elasticatthismomentcannotaffordtimetochange,sotherestF=Fplay=F=ma,theaccelerationofreasonBa=F/m,atthistime,changesintherateofBisabouttobegin,or0.-Qualityof1objectsalongtheslopeAnglefor37fixedcoarse
32、cantbystaticdownwardmovement,windforcesonobjectsalongthehorizontaldirectiontotheright,anditssizeisdirectlyproportionaltothewindspeedV,expressedinKproportioncoefficient,anaccelerationaandtherelationshipbetweenthewindspeedVasshowninfigureb,please:(1)theobjectaccelerationatthefirstmovementa1,(2)between
33、theobjectandcantkineticfrictionfactor,(3)proportioncoefficientk/bysecondfiguretoknow(1)whenthefirstmovementoftheobjectaccelerationa1=4m/s2(2)whenthewindspeedV=0,isthefirstmovementobjectaccelerationa1=4m/s2Forceanalysisoftheobject:mg*sin37f=ma1.-N=mg*cos37F=mu*NHaveu=0.25N=8Nf=2N(3)whenthewindsreachm
34、aximumVwind=5,theaccelerationofabodyisa=0,namelystaticobjectsForceanalysisoftheobject:mg*sin37f=f*cos37windHavetoF=5nwindAndbecausethesizeofthewindisproportionaltothewindspeedV,FtheproportioncoefficientK=F/Vwindwind=1/(1)theinitialmoment:mgsine?-?Mgtimes?A.1=ma0Bythefigurereada0=4m/s2generationinto1
35、type,Solution:?=gsin?-ma0gcos?=0.25;(2)themomentattheendoftheaccelerationiszero:mgsine?-?N-kvcos?A.2=0N=mgtimesagain?+kvsin?Atthispointisobtainedbyfigurev=5m/sPlugin2typecansolve:k=mg(sin?-?Cos?V(sin?+cos?=0.84kg/s.-Asshown,isAheavy10nballs,inF=20nthepullforceoftheverticalupward,upwardmovementfromAp
36、ointbystatic,Ffunctionsbackto1.2,theknownkineticfrictionfactorbetweenthestemandballfor3/6),trytostarttowithdrawforceFtime,smallballafterhowlongtimewillpassawayfrompointAtopointBis2.25m.(g=10m/s2)/startontheverticaldirectionofF1=F-10G=n=G,OnthebarandtheexternalforceF2=F1sina-muF1cosa=5/2N,Ballquality
37、ism=1kg,thenstartaccelerationa1=F2/m=5/2m/s2,Whent=1.2s,mobiledisplacementforS1=1/2a1*t2=1.8m,speedv1=a1*t=3m/s,thedistancefrompointBS2=s-S1=0.45m,高一下英语动词的时态高一下英语动词的时态AfterremovedtheforceF,accelerationma2=Gsina+muGcosa=15/2n,soa2=15/2m/s2,WhentheballforthefirsttimearrivedatpointB,S2=v1t1-1/2a2*t12,t
38、hedataint2=0.2sBallcanreachthemaximumdistanceforS3=v12/2a2=0.6m,thetimet2=v1/a2=0.4s,thedistancefrompointBS4=S3-S2=0.15m,intheprocessoftheballslidema3=Gsina-muGcosa=5/2n,sothea3=5/2m/s2,TheballwhenthesecondafterpointB,S4=1/2a3*t32,andt33/5=)s,theballofthesecondafterpointBtotaltimet4t3+t2=)=3/5+0.4s.
39、-Carhangingtwoqualityofdifferentballs,thequalityislessthanthefollowing,whenthecartotherightforuniformlyacceleratedmotion(airresistance),howthecaratthistimethetwoballs,pleasedrawthegraph,thetwoballsandcarroofhangingball,3pointsisastraightline,andwhy?Solution:becauseismadeuniformlyacceleratedmotiontot
40、heright,sothetwopiecesofstringaretiltedtotheleft.Setabovetheballqualitym,mbelowtheballquality.Accelerationa.firstropetensionT1,thesecondpullT2,Angleofthefirststringinverticaldirectionintoa,thesecondisB.Stressanalysiswascarriedoutonthesecondgoal,firstbythetwoforce,pullingforceofgravityT2cosB=MgT2sinB
41、=MatgB=a/gForthefirstgoalanalysisagainT1cosA=mg+T2cosBT1sinA-T2sinB=matgA=a/gsothreepointoneline.Analysisperfectly-Asshown,thereisA,Btwowedgeblock,qualityarem,togetherinAhorizontalplane,theyandthecontactAngleistheta.NowtoblockAlevelthrustF,excludingallfriction,wanttomakeAandBtogetherwithoutrelatives
42、lidingmotion,shallnotexceedthemaximumhorizontalforceF.resolutionAB,withABthewholeastheresearchobject,byNewtonssecondlaw:maF=2WhenApushingBaccelerateforward,BofAreactionFBAwillriseupandAandBrelativeslidingoccurs.WithAastheresearchobject,itsstressdistributionasshowninthefigurebelow.ThefigureshowsthatA
43、andBtogetherwithoutrelativeslidingmotionofcriticalconditionis:N=0VerticaldirectionFBAcosq=mgHorizontaldirection-FBAsinqF=maThesimultaneousmgtanqtypeavailableF=2Thehorizontalthrustamaximumof2mgtanqtthekeytothisproblemistoforcecharacteristicaccordingtotheobjecttoanalyzewhyABpossiblerelativesliding,soa
44、stofindoutthecriticalcondition.-Asshown,thethincordendpointAinthetopleftcornerofthecar,theotherAlighterropeAdepartmentintheleftsideB,BinApointdirectlybelow,distancetoBA,B,tworopeknotsandtheotherendatCphaseisAqualityasmballs,ropelengthofACforB,ropelengthofBCforB高一下英语动词的时态高一下英语动词的时态 (notpainted).RopeA
45、Ctowithstandthemaximumtensionare4mg,ropeBCtowithstandthemaximumtensionare3mg.O:(1)theropeBCwhenjustbestraight,carsaccelerationishowold?(2)whenthecarsaccelerationis2timesthesquarerootof3g,andtworopetensionrespectivelyhowold?(3)tonotbreaklightrope,maximumaccelerationofthecartotheleftishowmuch?/(1)theballbytwoforces,theAClineofmg,tensionandgravityforcesizeofF=mgtan60=3mg)ByNewtonssecondlawto3mg)=maa=3g)(2)theballbythreeforces,mg,thegravityoftheverticaldownwardalongthedirectionoftheCAropepullingforceofF1,alongthedirectio