cheat cheat - 5A system mnsists at three particles with...

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Unformatted text preview: 5A system mnsists at three particles with these masses and velocities: mass 3.0 ltg, moving north at 3.0 nits; mas 4.0 kg, moving south at 5.0 hits; and mass 7.0 kg, movi'ig north at 2.0 nits. What is the total momentum of the system? Given: m.=3.[lig,ir i=3.0mis north; n12=4.0kg,v;=5.0mis south; m3=T.0kg,\r3=2.0mfs riorlh Q'le‘fion Pm? fin, = fi, +37!2 +fi, = iiilfil + $5, + raj, = iiili'liiori‘li + ltlz‘lt‘lsottfll +m3l‘3llflft‘0 = (mll'l — iiilirI +m3‘l‘3)!l0l'flt' p‘m :3.0}.3film's-4.0kgic5.0niii+7.0kgx2.0iiiis:Skgimisiian‘ii T! A ball of mass 5.0 kg moving with a speed of 2.0 rnl's in the + air—direction hits a we] said bounces back with the same speed in the —ii-direction. What is the clia'ige of mornentu'n ofthe hall? Given: m=5.0lig ii“ =+2_l]mi's , 'il'u=—2.EH1‘I.I'B (hiestim: flip ? Remaiti : The vector quantities are written eiltier with an arrow orwith a bold diameter. The change in morrientum is 5p = p,— p.= mvf— mv.= mtvf — v.) = {5.Dkg} [—2.0rrits—Zfll1'iis} = 20kg ”Us in the —ii direction 144' ASH-kg body isinitialy moving northward at 15 mis. Then a force ot 15 N, toward the east, acts on it for 4.0 s. {a} At the end of the 4.0 a, what is the bodfs final velocity? [b] What is the change in moniean duiing the 4.0 s? The change of niorrientum i1 each direction is: flnmm =0: Asm= Farm = mflvfl=mvfl The magnitude ofv in ttie that direction is Vt = Betti-hum” Val—2} = Sqrtth2+ {Fmbtv'nfl’i = settt15mt321+ (15NH {4.053(30kg) )2}: 25 rni's T = tan'1 [tin-Iv“) = tan"{15rn.’s .-' 20rni's] = 3? ” North of East 20! Asubrna'i'ieofinesslfix 1tJ'i ligand initidlyat resttiresatorpedo ofiness250 kg. The torpedo has an initial speed of 100.0 nits What isthe initid recoil speed of the submar'ne? Neglect ttie drag force ot the water. Given: m.n=2.5x1n‘i.g ; mm: 25cm, vi“ = 1DD.IJm.'s Question: triennial the submarine '? 9 Pi =pr U=lTIm-V|up+mu'Vmu1i—)flhrp-Vup=-flhn-anj -) Vmujl = ll'liu-p - [-Vueri'rnm =250|iJg ‘ 100.[I'n.is I 25‘105=0.IJ1 nits = ‘lci'ni's opposite to the torpedo 27IParticle A is at the oriljn aid has a mass at 30.0 g. Particle B has a mass at 10.0 g. Where must particle B be located if the coontiiates ofthe center of mass are (it, y} = (2.0 cm, 5.0 cm}? Given: mA=30.Dg., xA.=D, YAFD, mB=10.0g, xc.= 2.0cm, ym=5.0crri Question: (Kay's) '? I = mnxfl + Maxi = Maxi J? = may.) " may: = Maya ‘" in +rri, :iii +JIiIiJ m N's-‘5'; Mn+m& _{—lflg)xl_ J’ _(—IDE}J’& _ :10g+30g H 2—10gi-30g xh —x,,,, "1. +i't1. = 2.06m 30.03 +10.03 n _y& no +iiiira = 5.04am 30.03 +10.03 Ill], 10.0g m, 10.03 Jr, =3.0c:|:i1 y, =20cm It0ilI Atoyuau'wilhamassot 12Dgrnovesto the right with aspeedoffljfi rnis Asmal child drops a 3011-9 piece of claiiI onto the can The claiiI sticks tothe ca" aid the mar continues to the right. What is the change i1 speed of the car? Consider the fiictional force between the car and the ground to be negligible. Given: Mm=1ZDg,V|=fl_?5l“lrS, M=3CLUQ Csiestion: L'iir ? The momentum being conserved : p.- = p. = {mir+mi:m}vr pi: mu. v. étmcm-mmlw = mm in = mu, vlf [mm+mm) Av = w—v. = V: t t muttmurI-rriuan i {iv = (0.?5rnil's it: (1209112Dg-k3tlflgD—1) = -I].15mis 444' A DEED-kg bullet is shot honziontalir and collides with a 2.00-lig block of wood. The bullet embeds in the block and the blocli slides along a l'iorizontal surface for 1.5!] m. It the coefficient of kinetic fiiction between the blocli and surfam is 0.400, what was the origi'ial speed of the bullet? Mil—Ag in...” + mm)_ 1 0—H: 2(0 Aw?“ 90”“ 5 )0 50'" = 350m is him a. 020i; 491' A projectieot 1.0—kg mass approaches a staiionarybodyot 5.0 kg at 10.0 niismd, alter colliding, rebounds in the reverse di'ection along the sairie ine with a speed of 5.0 inis. What is the speed of the 5.0-kg body after the collision? Given :m1=1_Dkg, m2=5.[lig, V1i=1fl_m“llr5, v2i=0mfs, vf=5_flmts in the opposite direction(velo-city vir changes sign} Question: the speed v”? we have miv“ + I'I'lz'il': = mivirl'l'l'lztl'zf 1.0lig‘1flfl rnts-HD = 1 .Dltg {-5.0W3]+5_EI£Q'V3 -) 10 leg in +5kg n1 =5.l]log"iia 9v; = +3n‘iis and Va is firms in the +1: dieclion 52! Two identicd puclis are on a1 a: taile. Pucli A has an inilid velocity of 2.0 rnis in the +x-direction. Flick E is at rest. Flick A collides elastically with puck B and A moves oil it 1.0 inis at an ange of ED" move the x—exis. What is the speed and direction of pucli El alter the collision? There is conservation of inear mcmentum, pi.=m\r.u =pu=mviu+mvEm -)an= Vin—Visit p.,.=m\i,,.,,,,=0l=p..,,=millm.+mii“I —>vw=-vm Thefinalspeatofpucliflis m: earths-1 + m2} = sqrtt {Vu— miz +t-mi2i um: sqrt[ i 2mg—[1_I]mi|'s}cosfifl It)2 + (—(1.l]mi's si'i so ”n2; = 1 .i' rnis The direction oi puck B is tan"B mun“:- = inn" ((4.0sz Js'n so ”12.0w: —(1.0mi's}cos so “ i = so“ so 30ubelowthe : direction 1 ‘iii' hi w- lWhalistliemtztizmzlinecfiz-iofa solidimmdis}.ofmass491g,wiflaafllickmsof5_00emmdmdins of20.Iun.aboutanaxis‘fli:wghitscmHmdperpendicifluboit? I: o.5m*=o.5(4dflo.2oom) =09 m2 3.Abowlingballmadefma hashzlfflnndilis mad'ultbqwlimgball'fheymmihocflhe seine nlahrial (and therefme have It: same masspw multinational By what fiche is the (3)1335: and (b) rotational inertia offline child's hsflredilceii compared In the adult ball? The'irohime ofa sphere is V=(Itit3)p Rs andM=pV Given: REMQ, same ihnsity material, Quit-005i Mia '-' [mama Mmmufi pvaiipviw p temp Rafi) memos“ ’) MmecliM=Rm1mm3= wPWflf=B I=(3.|'53MRI Lmiflmu= (35) “wilffifimlMJ! Lam =WSXU4)=32 12.111epifllconiofalmmmengiueiswomdamundadmmofndius 6.00m.WIiiilefl1ecoildis pulledwithafmoeofTSNbostmttheeagine,whatmagnimdebucquedoesthecoudapplytnfimdmm? T he hocq'uel t |= F I'M: 'J"SI\I‘(6.CHIl'irl0'I in)= 4.5N_m 19. Ahapdnm,ofhngtlizmlwidthlfifim._isli.e1dopenflmmgleoif65.0'wflhmpedioflnflom. Aropefiflhfiedmflnnisededgeofflndoormdfismmflnwflbdfindflmdnminsui‘ha posifiomlhalfliempepnflsperpendiculaflybeflnhspdom.If‘fliemassoffliehapdomislfi.8kg, whatisthetm'queemdonthatnpdom'b'y‘fliempe? Thethdmrisineqmflnium$themrqnembedmtbednmbythewmtmqummd ExemiU-Thmmbiem'ityis t = mg MWM) =lfi.flg*930mfsz'cos(65deg)*(l.65mi'2}= 5?.4N—m 13. Adfldnfmzs540.01gissitfingonaholimtalseazwatadistanoeocflflmfmmthemppmbmg extimfifiemagnmdeoftBeMmieabuttheafismiemflnwdghtofthedfld? |t|=er=rwmg=2flm‘ 40.[Icg’9.31 W52=730an 23.15.5tmeu5edlngrindwheatintoflmristurnedflnxonghlimohfiomsbyacomslszmu of 2|.0Napp1iedtio‘fliefimofa 10.0cinmdi'u5 shaficmecbedmflnwheelHowmudiu-udiisdomeon thestonedim'mgflnu revolutions? 'I'hewulidombylhestmeis W = pig = r, FAG =(c.1mm)(2i1cm(12m)(2wdim)=15c-J 25. A flywheelnfinzss lfllghasanefiisctis’emdiuocf0fi2 m(ammeflaemassisoeneemted alongaMohamedatflnefliectivenfliusncfflmflywlnel}.(a}Whattmqueismgnicedto tilingfluswhnlErommsttnaspeedoleIrpminafimeintenral-ocffl-UD s?(b)Hawmm:hwuliis domedm'inglhefiflfl s? Th tmque. “=12” ill: q=fi1+cfli=l+aai=aai mdr=Fri= mgl=ll(rifl)rl= tl'lf_l agi®wn 51} w: [1821a)(0.62m)2 [2E13d][limn]_ f=lfl1 N =—30I (DUI-pm) rev 05 =_9N-m. 28.AweighloleIflNrfitsonahveratapointfljflmfiomasupport.0ntli.esamesideofthe support,atadislanoeocffl.0m&omit,anupwacdfioroewifi1ngnitnde?isapplieiNngecttJie weighlofflmbouditselflffliesysbmisinequflibfimwhatisn Er=0=—F(3.0 mi+uzcomm5| m), so F=— NW: 20014. 3.0111 33.Aboiisepe.i.litmslsnd53.0mabm'elhegmundona5.0-m—1ongladderflutleensagajmslfliewallat apolint4.?mahau‘elln gromminjnMweighsfisIN mdlhehdhweiglis 120N_Assumingm fiicfionbetweenfluhousemdflieuppamdofflnhdder,fimifln Eoreeoffiic‘finmlhztthedlivewa’y exubmllnhothmoftheladdx. 4.1" III. 4-? 50. Alawnspcrihklu’hastlnee5pmtsthat5pnywatu,each15.flmleng.A5fllewatErisspua3-ei the spcinklutums nmndiuaeixde.TlnspdnlleIhasatutalmumenlofinelfiauf9JUX lfl'algrmzlf thesprinl'lu'staxts Ermiestahdtahesfl.205turea€hilsfinalspeeduf2.2 mfguhatfiwcedovaeadi modem-tonthesprinlhr? St = 3FR= In -)F=( IaBRJ, a=&mf&t=22‘2‘3.144'3.20 F=[E'QBE-EIZ‘GQ‘PSJM.205))i(3‘15.0e-02)=0.88N 52..Fommasma.reanangedasshown.ney memectdhyrigimasshssmdsofhngflas 0351:: andfl5Im. Wha‘ttocrq'ueumsth appliadtocauseanangularacoelualinnuifl'fi iatlI's2 aboutflieaxis shown? . -111 f:N =fl flinm_5m}+rfisum[15_m]]=isni~r r: Elli51=flll+ma+mc+m13k1=sincealliiummasmatet’flflimflfiomtheuis. 2r Jill = (41 lg-t- 3.0 kg + 5.0 kg +2.. mums mifzfmqs ladli'fsl) = 3.WhinishawwmemfiewksoffiefiaofamflthDD-kgpamdmwmmnm wimmeflmEaclifioothasusmficemurfflflEOml. Given; 111.40.3ng memmmsfi-nmm’ W 1),? Thmgaprammmefimhisfiurfiwflflflzm’ P... :51: [no.rsgixrstsm'shr summed: =22m anmappingamathmmmaram.mmhunmmsmx104m“. Whutisfllapmssmeimmsemflnflnidwiflfinthesyfinge? Give: :F={Nthfionappfledbythenmhéflklfl'kfllemthefiameisappfied Mung! Mmmhhfluidis .112: Fax: 4.1omrsnoxiur‘m2) = 33.011): 1-1. Hmhighmymsmkwmwamflmpmsmmfirewmhemwsbmwm helowuunnqun'icpm Given :anm— p= lfl.tl:PsflIeprEsuremerhctim (harden; theheight=|1=d? p=p_+pgd, h=a=cpflypg =1n.csi>ar[(1momg.rm‘)x[9_somish] =1om (ymuflineadahciggtassun) whmh=aesuhaguuuwammhemm 4041:6202anwww.mmnmafmmemom‘mmm grswityis5.D.thtisitsiuilislaocelemliun? mmfimmmmmmmwfimaammmy. hqmusiledimcfinu.WewmdmwflnFBDmmNewtun‘ssecmdkwalmgy 5F,=mg—Fa=ma -) Fg—Fg’m; Fg= fluweiglitoftzlredisplmedunits!= '5'pr Themassoffliepiaceofmtalism=vlpfurthemlal find F 5 [HIV PwNPl] = stl -[ WP.“ = Bil-[1511)] =5 H'S-DF 0-50 swmis 309’- offlieamelafimofgrm'flygmmaflmi 2. Thepmmehfldeahofieofchmpagmhdjmfighummmpmsmmideihmd efflieln‘flhhasuuinnermrlilseflflcm.Whutistlnfiicfimalfiurmenflnmfldmtothemrjefflie m? Fan Ffi-fld-qaj _ SF, sigma...— r.= 0 —) r.= FfimrFu: {P —P.m:l*(pf} Maison-Hum} laun*1.[l13el35 PsisthBJ-ll‘mfllflm —1:'-1JZ}N 5.!1 lD-kghabysinunatiuee—hggedsmollhedimelarufmchucfflumml’srmdfiaetislflmAtto— tgsrhltsi‘tsousfiOuI-Wmu'filnhasfiwdlmlxfiet,enchwithadimeuruffiflun'firhoqmfls fliegteuterpmmmtnflnfloormdhyhrwmmh’! Gixui :mh=1[tg,r=f1M)un=lcm=fl.fllm; m,=flkg,r‘ =(6M)un=3cm=ll.t]|3m Momllbtk? PiFlI't-E'GIJ‘I2 Pil=rn..g;i'-'-l|::-r’I P, _ limp-'1 _ 4x10kgx(fl.l)3m}1 _2 n :1, Small?! SKEDirEXKDDl): llamhssshrgecfiindficslhmpenwflhahng‘mjncfljndfinlnfl.Thehwuputuf‘the counimrholdsllfimaufwmmfllemficemfitflbummefflmouminerisifl]m’.'fln Migttoffluluwupmufflucoutaiinisljflmundttemckcmlaiussmlnmofmfijflm high'flieIalslwlumeuffliecnlumnofmiut'heneckisflltllm‘.‘itfhatismemagnirudeofflie fiamemamdhyflemmhehommufflutouninefl Gm.fl=5.t)thn1,d=tl.5tlan-2.5Dm.gp =pu... QnEfiomF? P=Pstrn+pgd; P=FiA -) F=P*A= (Psun*l.013e—5+ pgd}.i'i= F =(1.013E05Pa+(1e03kgfm3)‘93‘1d§.G.SDm+2.SDm))*5.mm3=1.UE06 N 34Acyliudlicaldiskhaswhmll.wulfl'amauniimsssfilfikgmdiskisfluafiugunthemflaceof sucruewutzrwiflliisflatsmfineshrimmalTheateuufmhflatsmfimeisfl.fltlm1.(a]fifhatisflle specificgravityeflhedist'.‘(hJHwflrhdmlhemlmelisnshommsmioefficJHowfirsbme flmwuterlwelisiistwsmface? Given: V=8.9?xlfl"m3, p = W“; Question: pigm— ‘i’ . . . p _ :m _ 8.16hg flmmmmmfl ,9, Pp, 8.9TE—03m’xflflED3trg'im3) = 0.910 b)Tl1e object floats, it is in equilibrium withpert ofthe object 9.1me tlieweight ofvolume of fluid displswd is equal to the object weight V Ad 0.910xlr' 0.9‘1l2'xRETIRE—03m3 P. r . V = V —=—= =0.910 d'= =— 13; 13 Pa 93 " p] I", V“ _’ 21 0-640”: c)Huwfsrabovethewatierlevelisitstop surfsoe =1.28m Ad_ _ d _ _ ‘ Afi—UfllO—Hi d_—0.910 d—0.091.28m=0.13cm 27.1112 periudncl'oscfllalinnuifa spring-and—mass system is 0.5I 5 and the amplitude is S.I cm. What is the magnitude of“! acceleration al lhepnisit 11me extension ofthe wiring? Given: A=S.lli:m=5.[lxllram, T=fl.505.l Qufitinn: a atmaxim'um extension 3' At thcmaximum magnitude ofthe acccluation x=A the amplitude, vm = (El-A1 am=sz But m = 21:; T —) am = Aapri)‘ =(S.If.‘i:lr1[lamprfll132.Ir (0505):] = 19 ms.“ 43.A.nideal spring with a spring constant oflS Him is suspuded ventrally Abody ofmass 0.60 kg is atlauliedtolheumshetdndspimgzndrdeased. (a)Whatisthemen:inuoftliesyuingwhmtlLespeed isamaximum?(b)Whatistl1e ' speed? Given: m=fl.60kg Question: 3) 1311:;- ‘i' ll) 11-: '3 Itismflieerniifibrhmprfintthatmespwdismaxmhmauditismininmmwhmme AmpliuideisreachedThefonceaareappliedinflieydimctiou Writing newton’s law at equilibrium, SFY=D a) SF,= loi—mg=0 -) x=mgfk=(0_6(lkgx9_30mlsl)flfl~l'=0_39m b) vm=wA=vam)=0.39mv[15NiO.60kg)=1.95ml‘s 48.4!Mytsnsmdedva‘ficaflyfiummithalspdngofsplingcnnstantZ.5me.Thes‘piri.ngisinifially inflsdamedposifimnehcdyistlmreleasedandesefllamabomitsequiliiuiumpusfim'flie mo'linnistbscribedhy y=(4.l] cm)5i.u[(l].'i"fl Easy] Whatistliemaximnmliheficmel'gyoifmebmtfl Given: k=2.§ Nina, QuEfiom:EEmax'i Themaxinmmkinflicmergyoccmswhmthevelmfitgismasjnnmuamely, Vm=Am and KE=smvm =sm(Am)2=1;amA2m =smA3mm), themasscamel 9 KE= ££A2k= s (4.0x102m)%42_s~rim) =2x10" J 62A pendulum clock has a period of 0.550 s on Earth. It is taken to another planet and found to have a period of 0.652 s. The change in the pendulum‘s length is negligible. (a) Isthe gravitational field strength on the other planet greater than or III: than that on Earth? {b} Find the gravitational field strength on the other planet 3] T=2pv(Lr‘g) therefore T is inverselyr proportional to the square root of acceleration of Gravity, the bigger the period the smaller the acceleration ofgrav'rty on the planet b)Name1y hm. = V g}; v31 -> g; =ng (rim)2 g1=9.80mi'52 x (ti-55031362)2 =5_5imis‘ Rhyming ofahmiugforkmovesbeckandfimthwhenitis ad Manhattan "Iladistamette prongmresbetweuiitsmepmifionsislildmm Ifttlefiequeucyofthe hmiugfixrkis mflfkwhatmthemfiumwlmityandthemfimnmhafimofflleprmg? Assume SI-Itt-L Given; f=440.0I-Iz._ A=(224i2)1m=(2.24xlfl'3i2} III, Questions: Fm?amn '? -) u...= [as = 2pl A= 2p(4dflHz)((2.24x10'3f2)=3.10mts —) a.,...= m” A= 4p11’A = 4p2{44(lI-lz)2{224x10-3f2) =ssaomrs’ 32 A ITO—goljectoua sprihgosciflateslefitofigldoua fiictiomhsssmfseewitha frequezcyof 3.00Hzandanampnmdearlzo cm. (flmatisthespuingcomtmi? [b)Ifthe otg'ect starts atx=12.l]m1att=0mdtheequflflnfmnpofintisatx=Qwhatequafiondeeuibeeits position as s fmctionoftime‘? Given: f=3.DflHz._ iflit=l2.0cn1=12ir10'2n:i= at t=0 Question: k? Kantian '? -) m =i:(ldm) -> H’m; m=2p ezpemnzrémln radis a) k=4p2(3_oon)’(1ioxm"kg)= GUNim b) the eqintionis m) = (12.0:(10'2111) oos[(5.00p radish] 4?. The displacementol an object in SHM is given byytt} = [3.0 cm) sin [(1.53'r radish]. What is the frequency of the oscillations? Given m=1.5?radi's, Question :l’? Wehavew = 2M: 2pt—) l=ui2p= 1.5radl'si2‘114: 0.2SIJHz 3.11ei11tensityoffl1esmmdwavefiomaje‘tairplmeasilistakingofl'is1.0):lfl‘me‘atadistanee ofSBm.Whatisfieintensitynfthesuundwmfllatreanhesflieemnfapmmdingata distanoeuflm mfiomflmrunway? Assume flntlllesmdwavenfliatfi fiumfln airplane equalein Answer. Theinlne‘nsityufflu smdasafumrlinm uftltedislznceisgivmby 12m = may}: r warm”: = nzrn’ —) 12 = {14:12:12)} = (13:11:1 wrmfixrs .rimmumf=1m 310-3 Win12 13. What is Illa wavelength ufa mewlmse weed and periodase 75.01115 and imms rEp-ectivel'y Given: FTSflmJ's ;T=5.Dllm5 Question.- 3. '3 T=5.0l}mF Silflxlfl" set: 9 f=1iT = [USBlelfl'a] Ha; 1.=i'.. f9 Ella-Ff: (75.0sz )(S.IO;10'3}=0.3?5m 21.. Awmunasiringhasequafim JLY. r)=(4_0 mm)si.n(mt—h) wherew=fij x lfllndiszndk= 6.Ind.|‘1n (zJW'IIatislheamplihlth-uflhzme? [h)WhaI‘i5tILe wm'elsngth? (E)thlisfieperi.ud? (d)Wha1isthewavuspeaiT (ejhwhidldimcflnndmfllewm travel? 33 )‘(x._t)=A5iJ1[m‘t—h}mdthesinm5nfanyamglei5a‘lbojtequal+l,y(x,l) maximum is at best Audfiatisfiedafinfionuflhe amplitude b) l=2pf1= 2p!6.0mdfm=2x3.14!6.0md1h=1.0m c} m=2prr -)I=2p.l'w =2x3.14!6.03102mdl's=0.0152c d) v=3dT=LOmIDfllsec= lIOmfs Thesignnkamdmtmuppnsitesnthewavehnukimfimfldimcfiunfinamlymtbgfigh) 25A sinewu'eishavefimghflnfigblmamximhghmhminnfigurempmmen‘lslheshapeof flmcwdattime r=0;flndaxkulhnrepamttheshapeoffllecurdatfiml=.105.(Nuhefl:atfl:.e hmiamlzlandwnficalicalgmdifih’ent.)mm(a)fl:eamplitudeand(b)&ewwehngfiaofthe wawe?(c)Matislhespeedofflmwaw?mm[d)lhefiequenqmd[e)‘fl1eperiuducfthewave? Tbmnpljtndeisym=2.6mfl1fiaforeflleanlphh11kurhichisymisA=2_fiun l=Mlthbe1weenZliheyJ=lfim—Zm= 14m. 1.'= fix Edi =(ij— 5.5 mynlflsec = ZQDM’S f=vfl=29_flm'sfl4m=l_4Hz T = lff= 1.31.4 Hz = 0.73 49Acmdufhngfll 1.5misfixaiatbafimdsllsmasspumitlmgthisljghnmdflnmis 12 N. (3)thfisfllefinqnmcynfthefimdzmmhluaciflzfiun? [b)Wh21tmsiunisraquiredifthen=31mdehas afiequncyufOSImz? a)AcmdfimedatboihmdshasanuchaHmfl1enfl5,flnmflumflnfiequencyuffllamssl'blewmfi Is givmhy fn=nvf2L; u=l,2,3,___ whereLis flatleugfllofflleoord forn=1 itistheflmdamunalosciflaflon v= vfl'fp), T= 12N, p= 1.2 gxm—)V=v(12m1_2 x10"kg1m)=1oumrs f1 =VQL= 100:r 211.5n1= 33.3 Hz isthe Wufmrfimmwmloscmafim h) f; = 3‘;sz = Bmmrg) -) T= 4L2f31pa‘9 = 4(15m) x(o_sox10’Hz)‘(1_2xln'3kg/m)r9= 3mm ...
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    Jill Tulane University ‘16, Course Hero Intern