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Intro Physics I

by: Kylie Bartoletti DVM

Intro Physics I PHYS 2211

Kylie Bartoletti DVM

GPA 3.67

Edwin Greco

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Edwin Greco
Class Notes
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This 0 page Class Notes was uploaded by Kylie Bartoletti DVM on Monday November 2, 2015. The Class Notes belongs to PHYS 2211 at Georgia Institute of Technology - Main Campus taught by Edwin Greco in Fall. Since its upload, it has received 12 views. For similar materials see /class/234275/phys-2211-georgia-institute-of-technology-main-campus in Physics 2 at Georgia Institute of Technology - Main Campus.


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Date Created: 11/02/15
JENNi bit 39Som I Wevl lbol 3M 3531 900 39 p 5mm 1m emv6 rollisms Art WaxWMS 391 7M 14 Wt We 1 5quot iquot so Mfmreot rmOMW 029a When a ping pong ball collides with a bowling ball why is the effecton the ping pong ball more noticeable than the effect on the bowling ball 1 The momentum of the bowling ball does notchange 39 2 The change in the bowling ball39s momentum is less than the change in the g pong ball s momentum The change in the bowling ball s elocity is less than the change in the ping pong hall s velocity QZ9d A bullet of mass m traveling horizontally at a very high speed v embeds itself in a block of mass M that is sitting at rest on a very slippery sheet of ice What is the speed of the block just after the bullet embeds itself in the block Eb 39 E Pr Vr m QW MNJT J 9 V 029e A space satellite of mass 500 kg has 1 lt 5100 0 400 gt ms velocity lt 12 0 8 gt ms just before F 39 being struck by a rock of mass 3 kg with K2lt 10239 039 0398 gt ms velocity lt 3000 o 900 gt ms 3 lt 102 039 03 gt quotV5 4 lt 3688 0 1191 gt ms After the collision the rock s velocity is 5 lt 3533 0 1192 gt ms lt 700 0 300 gt ms Now what is the velocity of the space satellite I p 397 rm0 I IS 47 BMWquot awees ahar56 aloorJ W6 44 054V iclfs vfo 1k oW WAC4 432 1 will must 1 Jaw to shvf 11 WW WW BA cw Suvuwdq nlb39 nv39flwhj ISMf 1H Mast71 4 enf y b f f l t5 AFQJVM 1 b y fwig amp 39 OT A V39so quot39 m g G L 6 x E 9 9 3 PM PM K 39 39 I I I l I Mmmw f MCth v y r i F Ar 1 F M 0 s P QM W1 39 quotH A A BMGMI mV x M mwx 4 x IQ m ML 4 3 L V AW EZM 39 M4 w b MM 0 MW 1 5 M to Manet n PICS 135 1 gym assHelm from 5 gm 5 mink FrWm S E Eu 1 39 53900 K Ania 47 3 1134 3000 O fwm 01 I 2 3 J 47ooof3007 soot y g 5 e3 39 0 lt9gtlugglvggt 00 40 ltquot quotv quot 3939Wgt 5 00 lm Cmer we aux4134 you ever W64 fo Macaw w 46M 7 whiff WT wen 740va 50 39 ASK a Pol175 5w A39ll awakhc M2 Marts 3 So 79 M M looM H 1 spa 04 MK Cows5M4 M m z m plyW mavmeIvw I u J fdiv z l wl Mr 0 POW391 7k MGH af 139 WQ 9 earle Novwt 6 SM M M new FMS 4 3exoooa A4 Pm cvlw MM 3 4quot M0H007m I 39 m 5 NM I 5 fix Poirt w 71 CW Vf lt3sz I 3 movrtlns 7 IH 3 am45 s mi EVWH 7 59V M smw s a 330 246060 5 79 xm wmlg Po tMuPaeM VWJM I J we Aft7 An JIMrt f 3947 a bx 1 42 mowan Ov WJFR 1 3 O S F f a I 4 39 9 39 67 lo lt013x ojomS 39I39 4 36 1030 07V73XI063 S 2 x10 In ylo c ldgt 39MS gt T P39 39 VN 5 PI 5quot 4 4 23 M07 swaga7 MS 3 Wch v 4 LMS Mou olch 130 Hma IXWOGS P0FHM 3 quot39 W U Cyquot Aaz 0 13xlol7n j v 15 km m1fku n w4w MT LH S Q31a Why was our prediction of the Earth s motion so poor 1The Momentum Principle does not apply to gravitational forces 2 We neglected air resistance 3 We did not use enough significant figures 4 We assumed the force on the Earth was constant in magnitude and direction over three months H 11 ffetFVIM Jibi 4 W 05W 39li39dVW AHA W v Arc stMt II W 66 WIS MW Wre WM VH Tn resan 14 0 6H 5 5 AM A we 3lt 7M3 ITEKATE 39 vaK Min13960 smf V Mst 3 A7quot A f A39fs 7 47quot 3 oralI46 FM f eta 01 Cowq TO496 thfs gumr3 va 14w p 4 4 M Mw 5 4n W574 XFW5SM 1W 1quot 6 Vm39afda 41 fwa lc tS 5HT 52 k 3 HOOKC MW 3737 7 Ro e v f Hcm39f no lthYe ml K515i Wquot 3 5 L0 1quotquoter 07 V4ch yu nj L 3 0 yM ft er y 115 239 slam 54754 Lo L S auxV4134 1L itS fWES 4 791 1 404 L0 v f tcmf A Cumpdssof 42m I m 3W 14W 1 L0 L L L 032b Which diagram shows the directions of momentum and net force during step 1 2 3 4 None of these wlifahamptg Cd kulm lye Posmm all A 5ch MW 0 7 to A 30ij W a 1m Mfewa i 0 03 setamlS a synC Ms afum waves W1 Powhatan lc fs iwa 14 caulk Poceem W 3 30 rteIs A39f figs 015 3 L C39f 3 I LO 7th Mt 006 391 L Mm syfm Half 31 o uwovnoukls 3 SFYI j L 3 a biavaqu PICS A f 3 01 s 1 V uPozm AF Fm arr n Ft 3 0quotquot o x quotf quot39 0305 M4q7 0 03 N A 01x 1vod s37 0490112 Kyla437 l 3 a e P57 3 t FA None of these 032c Which diagram shows the directions of momentum and net force during step 2 t 2 L 40 t 575Rquot BBGK 33 mm L s o IK Sunom I SfM Aj DMUVM 3 W5 4 018 S 335 anm 0 34 quot Own u MS 3 m a 010707 3 W lm m a 7quot 9i 4 a M s omzu rig15 0070 0005 quot quotquot 3 A W m J a 2 gt 1 A F rl 439 VM H a 7quot on 0 0quotIW5 57 0 06 42W 7 032d Which diagram shows the directions of mamentum and net force during step 3 4 None of these Site 3 3L MLOS 0 1 P z 7 5115 quot 5W quotquotquot Wm 39 amt oLfOtJIMquot PM L 7 0quot S em 3st397 oa M W quot quot quotquotgt19 5 lt0 03915910gtM 395 FWH 1 90le via A W wh onsj A 6 013 hum7 3 mavh lawn now A A M 00 40 0J9 0m 3 mch M 7 O1 quot 0 OCL a mm 3 W x we AM cwro 147 s M by s yaw quot rm LUI H39 a 02 018 016quot 01439 012 positiony O 008 39 006 39 004 39 002 39 O 02 04 06 08 1 time a AS we 1amp6me A f ch Armc 13914 ow sobhim Hbo MCNAW WW M 5 03S39 vvrf7 A 44wc M 54 aria1145c SoftHon 14 01 2 AW 60 fdeA ch lass 7L U96 SCH C Lk Vozo ohm Mm q sts 1W lm 39 9 t 39 g 5M6quot fly 17 0 tc 4 wr m I s Monr7 FOEJ 3003 3 In 4 0 46 exmo 39s M mm WIAUan we AM ken 467 1 0 0446 04 1k 57 811407 MC 1 1R CtPevj atW15 7n Sovn v67 I Mow MIf louoCcSSCS 1M jJthC 01146 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2 r q WK 1 5 1 fl f 50m b 7 S 5 07 0 WW Awass 1W5 scmenfs EF meMSMIfM l rVK4W 67 M quot 7 9 gt 2 F Ar Z 1 l in 41 lfmf f ILVH A quot5 or IA ymun M W NA rg Aue w do Ang miequ MAYHWf M W39H 11M to V A cw ov m 057a You push a crate out of a carpeted room and along a tiled hallway While on the carpet you exert a force of 30 N and the crate moves 2 m While on the tile you exert a force of 12 N and the crate moves 8 m How much work do you do 1210 2180 3 156 J 4 105 5 42 j 6 We need to know the mass of the crate 29 IM I3 6amp0 a i 1 2 2 5 3 gfm LM Levem 39 61m 39ie f SmY 3 ed a M fdrCm 1 the MT 4316 femH s m mr m call Jae Conkw of rm68 0 am 0801 bk Ss 9 5 mi A o39 4 N V8 CfAI FVaWMS39 bowm lemwf M l m f lq39l wf quotto AK CM Av kW 43 Wm M m mt waded m ceMev of Wm w w M m xeuf 1 sj 39w 23103 c loafH PM ci cC Cvblquot Hf bJIOIC o x fobff OCW M Tl VQ cmfev of mhg 39 v33 45gtVM I H WCquotS om Lc 5795M H 4646 comv OVL M BS quot 3 BoM f 445 no bfZC w JMetIsfmg39 no ro tMI MJ l ma a1 I 40 VHN IWIW 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Qua quot9 Iquot pol 7 0t WSKfvowz 4M s roc 77 A 517qu h erpPeJ armym disf 071M45st HM KUUOS K 54AHh1CJW 5634 W la 74 3ww pwk4cww wwmF b7qwmmmg suqm H nthe insM Mam tlIC Cemoeva V4433 Am movel 00844066 01 a Lewde L39 0 5 mj Ms WNWM 1 01M 1 39 AW 5 39 anM l I M Lid A l l i F5 SYMJ I ch How Much HaMAJ KMHK emu 01015 the 01sz RM in 4 81le fieWC 0 FA CL FL 37 FL43 Of 0 Solu m Pull n o 043 rm Lm 377 5 4M 3 UOK P Y P v M Sb IMMIS o lml 1 0 V V 5 L M 939 mm 4 014798 3W EAMDI Pow Pmiclt Pm M7513 AB 5 W 39 533 L 7 i I p a M CW I y J M V6quot d FA War ZakS a 7 AH 40rd I ch 3 2 F0 or AK 3 M465 7 emf W lsf 7 WMS 0f be M4m s 04 AE SW 3 AK 1quotILKWJr FLa mws H WWW of 46 VetJ f5 46 Vl ff AC C WS Vm diamwe H Ms M plmc Am FCLd AKWMS I summm n Joke 501mm 1044 Le Powrtvfamfce Ame1796 O AKVA 5 4 392 quotquot KW FL PICS vat xtiap We cm of hj CLAW r A CLIh39 014435 M 13 ansM op WI 01 f aLM 61439 cm ox vatfmltxx 4th WMwe out F evtmuyly 1M c w hh 514914 st Ocf to H3 ll 145444 L 446 you Kfvpphh ww You AM Pullt 7M C AW 0 1 4f CHMl c4 JIM JIVMKJ YW Pull on 4 L1H Ain va 0 WWW h I A a 39o a DVQ 0 rivf fm fiin 17 MKS WLIC Lam imrmyes M MP VW 0 H wwwis mm m a at M4 kl w oJmH 3 39quotf JayN176 5M Adan1amp1 Elna37 0 FA L m o 3 FA 0 F MILz m FLZ 6 FL 1 gillVi bf 6mm 22m s mm j Biaw MMM36m Zgt A Bellet 5mm A wooden APFN u AMA is to mta Waf Hg 6M9 0 M4433 cu so A quot1914 speed BUMH min jug Mm 6H m memes M 4M h AWL Lion wrt d A oumcan Wed am 5 a W IL C q A A A L N s3 x NWon Itqu A A I 2 KMVi MW 1 H 3 AL o 534 svbs tif f 7 z 39z I MK m 1mK M9 3 K1 l7 RM 1 KMVi a m I w EMMYvi iwxviavk 3m K1 m f a f af xw QM o mwm mg 056 UZ fS i A 3 Mrr39aslcss S Cv j 53 W4 Aim WI 1Q 6 A IO39Yo 1473 M A M Ak if Hf new urzi ou 70V quotI 939 fr M 1W 31 ij de r W a 3 3 9 II F A MSW1 vrt F r I A4 1 k A Q LM CW of 1H aLaSK MW39M A AIV MCC A I a burn of WM w39 MS unwound 75mm 16 am am Rs Vcw M Ms MW 3X lmv s FJ if B M 4 3mm um 35 idlef 1M elm QlO2b The spokes of a bicycle wheel have low mass so almostall of the mass of the wheel is concentrated in the rim What is the moment of inertia of a bicycle wheel of radius R and mass M 1MR2 2 2 7t MR2 V HZRRM 4 12MR2 t 511MR2 QlO3a A yoyo is in the xy plane You pull up on the string with a force of magnitude 06 N What is the direction of the torque you exert on the yoyo r 0005 rn R 0035 m 1 x 2 x 3 3 4 y 5 2 6 z 7 zero magnitude zaxis points outofpage QlO3b A yoeyo is in the xy plane You pull up on the string with a force of magnitude 06 N What is the magnitude of the torque you exert 1 0005 N m r0005 m R0035 m 2 0003 N m 3 0021 N m 4 0035 N m z 5 06 N m 6 ca n not be determined witho ut knowing the length of the string M no quotWWw QlO7a Child runs and jumps on playground merry g goround For the system of the child disk excluding the axle and the Earth which 7 statement is true from just before to just after impact K total kinetic energy P total 1 K P and i do notchange linear momentum Iquot total angular 2 P and 139 do notchange momentum about the axle 3 I does not change 4 K and P do notchange 5 K and Z do notchange 0107b What is the initial angular momentum 1 lt0 00 gt ofthe child disk about the axle 2 lt 0 Rmv 0 gt 3 lt0 Rmv 0 gt 4 lt0 0 Rmv gt 5 lt 0 0 Rmv gt 0107c The disk has moment of inertia l and afterthe 1 lt0 0 0 gt collision it is rotating with angular speed to The 2 lt 0 o O gt rotational angular momentum of the disk alone not 3 lt0 loo 0 gt counting the child is 4 lt0 0 oo gt QlO7d After the collision what is the speed in ms pf the child 1 0R 2 o 3 0R2 4 mR 5 02R QlO7e After the collision what is the translational angular 1 lt 0 0 0 gt momentum of the child about the axle 2 lt 0 Rmm O gt 3 lt O Rmoo 0 gt 4 lt0 RmooR 0 gt 5 lt0 RmmR 0 gt QlO7f 10P51 What principle should we use to find 1 The momentum principle 17 s 2 The energy principle 3 The angular momentum principle Wrf 4W 209 lwvxar cm W 5 W Cd sWC L alds 3K Mt i Qua4 4be buys 39t o Jsn v EweHA 0 EM 3n OSC yrws hasquot 1 06c 39 3o OM MM t w oswi v ws 1 r 3vd w m 54 CKCN4 w y o J 1 w we M4 4 mama 0 M73 eo MrMIC Z39 gotMM 0 W677 M12 V oscfr7WS 1NI EL N4 a 51 td lieu A Uiicvo i 3 6146 possi ir f 393 0 1 U Emh Mica061W 44mmch 46 rm7 sw 6 MWUgMR ewe7 Zquot Lt l3 gulf vulqu 03 M in MY of HS 32quot mkvow f Q 11 E 0 23045 M quot1 WVAff n 730614 f39 14 Wt 300 oscil fw5 loo Mms in 80601 6 3914 M 200 oscrHMws 66 23 MM nr WJ W is 00 0M4 0 may in alum 7 14y M 5427 3410 O e snrl396I30W 393 t 39 IQ mo hut7 PlHKV 6VJAM of W 1quot If A 393 323 Mfr 58m quot A 395 5 l 05 l I we 300 1 300 43 1 3100 f A 439 6 6 3004 09 6 acrl a Ewe 1 its 5 amnnm BZOUI 5 45 loo396 wtvq quot 7fquot A 5 jam TLA Eo 55300 361M a j humb v of M7quot 3 I YUM If WK MW771 a rt6 g J td rquot RK Cd391u03 6603 5 quot a I I MIN a j a I 47 39IL VH WW u 1 m W l quot L I J A o 1 i x I Aw i 13 5M RA660351365W 3 30 10 So 6070 30 9o 1100 3 MM MGM 3 in f H lamwvs QKWIC oz bv fw vh 14 mWf JOBK Sle di dfjw m 7L W77 W13 4 00 70 Sarf 1M 0 33 ruff mam 01L div SM was wenl be QRMJ W s Mfg Mcmvc x fo numms mem of Wardquot 03 va MC Mf Iv ve 14 9u F JLQJM M ll Rawt More PM 5149 14 V5 50M g AV4 In lam 3 73 p 3 Mv A 4 547 A2 3Ho 1x a 4 M Postal 5thj a 01 quottt MVl 6040 wow J 69 30 Swiftl 1k kUh1 nave LL MaoW1 gigIN Dr 6mm Mmat xvl ci2994206 39 7584 3 review I H5 a 6H WWW quotquotquotquotquot J1 va tf K U W AU WWW L F valU K J K V I x 4mm Wl ankv em 0 M M 3 UB 0 39r gt an Be ablt 1 0 yum Gama5739 3M A413 Know hm 1T0 WH C 39 046 0 do anewy pvoMOmS A e A6515 Q 3 3 1 41 Ema31 CwMSW WWW39P 0 WW ape 0 fqn szcgz o a empevW 0W 13 I I as l 5 WM V Q3 Em 1 Cf 1473 v X t 1w aw fowzs WWIM 1quot quotcm 395 KtvMJ TCH S fooHr quotquot gt 55 Ceame o M468 39 min 9 quotquot245quot 5 397 quot39 ng39 V 4m 4 nc cwrfv of quotMRS fwmvlk m7 M le Vega6 I Yaw Mbwew F5 8014 pMMC CW 144mg 2 E rto 04 ClanMIC of iMASIMI39M lampwc AWwa A v 3 W is H6 net Somg 0amp5th 7 4 WW all MASS E54 N ia I 39ak Afo on o m ar of 39fcmeI enemyI A msmw w r W It W Q 1 59 u1M WW 395 77 5M 7139 4mm 3 i 4r WK ealw fwrc Jag 3 W 556 3 how Vv dads vhc i106 mwe W 5 KM 5 K twms 4 le g1 X n 5 lt gt wa Kv a 4 30Hem areK 9 394 548441 fm q WMMJ P nccflle Drkw Av Ptctw Atom K 5754M Pkyf 4 M MRW yonct m M m M W cm s 013 3 IaM ltt39ewe 1 0 5 A 57gtanz 0k A AEW S O 5th AH at a 4h 3wa gig250 Ag 3 O 4 Eelq ig tuna37 elf AvQ o jc Seh w quot f 434 0x mum invulWS jquot coliis m yrcaM pf 1 Pa wkW W0 l PM3 Bfwa C olxfew B32 Qwv ws A 3 KVttqcvkcS jaw 02 algaW mm The W l 0 W 777 TMUWXSM amde 6er Plum Qua0144 mod6 of t 4 WV NJ q ymfoM ePNWS om OICHWM V3 fuJJMJ LPNsW143 misva div Wffwv 4644477 10 9053 ch cam 71 slam M Uv wa 3 HM EVIWW f f mm 50L M o JUPme 0 A S QVWVR M 2quot va ms M A 301 e xPenMew f a M0435 4 q OldtLC 3 7 YIO 716 314415 0 an GLOWm 3 Q1 3 019pr exfw Ms suml 4amp47 064 low2k o Hp Thy0v 1 6 W 0x butkd Lu07k Cx pfece of chmix a it Iquot w 60 in f Some colnet M11633 30M0 6W4 W M 06quot RM 47 W7C ring8 0 6t vOev V VotHw lwf qupowi 5 4 ex 3W3 er 1 V 3 HOWquot quot 3 waitm3 13915 23 How JM flu Ma of Maw be my a My 0714 M4351 i bm 39 o jc Ub aj 14 14C MW 0 c Jquot K 1 ABM LOZZISI WIS 6 J z 191 39 Mia49 a i 2 50M nucleus 4 I 9 z 7 fquot M quot gt J J APZO 3955 quotLLPJ39P Mm P3 5 MJP39 z a 2 a At do 339 LIMAquot 1M i lmoL Q92b A squishy clay ball collides in midair with a baseball and sticks to the baseball which keeps going Initial momenta p1CLAY and plBALL Fina momentum of claybal p2 1 Which equation 0 describes this ollision i a 1 p2 p1CLAY P1BALL 2 p2 gt p1CLAY p1BALL 3 p2 lt p1CLAY p1BALL ENC Q92c A squishy clay ball collides in midair with a baseball and s cks to the baseball which keeps going nitiel kinetic energies 39 chlayp Klbaseball Fnal kinetic energy of clayball K2 l Which equation correctly describes this collision 1 K2 chlay Klbaseball 2 K2 gt chlay Klbaseball 3 K2 lt chlay Klbaseball 0929 What is the energy equation for his collision written in the form Ef Ei Wext Q for the system of Eoulet hook I 1 2 1 1 2 1 Mm v2 mv 2 vaz mv AEthermal mviz AEthermal mviz Mmv2 mviz AE thermal A pingpong ball bounces elastically off a bowling ball which is ini 39ally at rest After the collision the pingpong all39s kine39 energy is Kp What is the kine 39c energy of the bowling ball 1 Kp 2 Kp 3 much greater than Kp 4 negligibly small nearly zero PHYS 2211N Dr Greco Wednesday March 4th 2009 Open and Closed Systems When Wand Q are zero we say the system is closed otherwise the system is open An open system is like a bank that can accept deposits and gives withdraws money is moving in and out of the system A closed system is like a bank under lockdown no money comes in our gets out How you define your system will determine if it is open or closed PHYS 2211N Dr Greco Wednesday March 4th 2009 Example The Weightlifter Lifts 3 Weight H above the Earth System EarthWeightsLifter System EarthWeights Surroundings Nothing Surroundings Lifter AEsySZO AEsySZWZi er AKweightA UgmvAEli erZO A KweightA U Fi erH gmv PHYS 2211N Dr Greco Wednesday March 4th 2009 Multiparticle Systems A block pulled by a wire will have innumerable particles interacting with their neighbors However modeling of this complex process can be simplified Forces internal to the system cancel due to reciprocity The force due to interactions with the surroundings acts at the center of mass of the system mW where PmMWTm vltlt c PHYS 2211N Dr Greco Wednesday March 4th 2009 I The Center Of Mass discovered by Archimedes of Syracuse We can obtain a formula for the center of mass from the definition of momentum and Newtons 3rd law Razz mIVIZMsysvcm 39 Assume constant mass d d ROFEQ quotWIFEUVI rm 7m1r1m2r2m3r3m 3 J m J quotquot4 rm m1m2m3m For solid objects with constant density the center of mass coincides with the geometric center PHYS 2211N Dr Greco Wednesday March 4th 2009 The Center of Mass for several objects m11711m12712m21721m22722m31731m32732 m11m12m21mum21mu M1R1M27 2M3723 r quotquot M1M2M3 41 o Rj ui This is not a superposition but a type of average 72 r For momentum and force we found quot i ggl that the total was a simple sum 3 Here we are computing a weighted average and the rulefor finding the total is slightly more complicated M r22 1712 PHYS 2211N Dr Greco Wednesday March 4th 2009 Example Stealing the T What is the center of mass of the T Assume that both parts have the same mass and dimensions R1ltOLW2Ogt R2 ltOL20gt Then using our forumla M1R1M2R2M3R3 r 0 quot M1M2M3 7 MltOLW20gtMltOL20gt Cm MM 7 lt093LW0gt cm 4 PHYS 2211N Dr Greco Wednesday March 4th 2009 The Center of Mass is a powerful tool because it allows us to calculate the motion of a complex system if we know the net force and the center of mass Consider two twin Olympic athletes trying to jump over a bar both leave the ground with the same velocity One clears the bar and the other does not A judge observed that the center of mass doesn39t clear the bar for the athlete who did not make it over Question Did the center of mass of the other athlete who cleared the bar make it over If v raj1 What forces act on both jumpers l 3 y 1 What path will their center of mass follow 1 39 How did the winning jumper make it over quot center of mass PHYS 2211N Dr Greco Wednesday March 4th 2009 PRS Two Hockey Pucks Two pucks lie on ice and can slide with little friction A string is attached to each puck and the string is pulled with a constant force F The string is wound around the outer edge of puck 1 but attached to the center of puck 2 They both start from rest Try to imagine what you would see as they move What do you think will happen in the next 3 seconds 1 1 will move farther than 2 2 2 will move farther than 1 3 1 and 2 will move the same distance 4 Not enough information Solution See VPvthon Demo PHYS 2211N Dr Greco Wednesday March 4th 2009 Energy in a Multiparticle System Gravitational Energy Ug m nearthe Earth39s surface Ugml gylingyzingyKF Ufgmy1rHZyzgMyym Multi article Kinetic Ener mm Knamlmtoml szlmth 2 K K Pro 1M V K nanslmtonal2M 2 y cm w K I Kvomnomzl WIth wbmnom Ltr r L l 01 39397r PHYS 2211N with Dr Greco 0905am 0955am Wednesday January 5th 2009 Scalars Quantities which can be represented by a single number eg Mass Speed Temperature Vectors represented by arrows Quantities which contain two types of information Magnitude and Direction Velocity Force Magnetic Field Scalars and Vectors make up the language of Mechanics we use them quantify physical phenomena PHYS 2211N with Dr Greco 0905am 0955am Wednesday January 5th 2009 What are vectors made off Not sugar and spice We use a right handed coordinate system The magnitude of a vector is a scalar I ririr length The symbol r arrow is used to denote position PHYS 2211N with Dr Greco 0905am 0955am Wednesday January 5th 2009 The direction of a vector is determined by dividing the vector by its magnitude I d 1l ltrxryrZgt 2 2 2 rxryrZ fltcos9x cosey cos9Zgt tl This vector is called the unit vector and always has a length of one Designated with a hat instead of an arrow PHYS 2211N with Dr Greco 0905am 0955am Wednesday January 5th 2009 Vector addition and subtraction is carried out component by component Z ltAXBXAyByAZBZgt This can also be represented graphically PHYS 2211N with Dr Greco 0905am 0955am Wednesday January 5th 2009 PRS Question 1 Which of these arrows represents the vector lt 420gt 1 21 2 E t 3 39c 3739quot 4 a 5 396 6 None of the above PHYS 2211N with Dr Greco 0905am 0955am Wednesday January 5th 2009 PRS Question 2 What is the unit vector in the direction of the vector lt 3 5 2 gt 1 lt 3 5 2 gt 2 lt 1 1 1 gt 3 lt 049 081 032 gt 4 lt 049 081 032 gt 5 lt 03 05 02 gt PHYS 2211N with Dr Greco 0905am 0955am Wednesday January 5th 2009 PRS Question 3 What is the magnitude of the vector A 2 lt1 2 3gt lt 1 2 4gt 1lt 1 4 2gt 2lt0 4 1 gt 3 15 4 17 5 I35 PHYS 2211N with Dr Greco 0905am 0955am Wednesday January 5th 2009 Velocity SI units of meterssecond Is a vector quantity that defines the speed magnitude and direction of Snapshots motion for an object The velocity vector is instantaneously tangent to an objects traj ecto Changes in velocity indicate interaction has taken place PHYS 2211N with Dr Greco 0905am 0955am Wednesday January 5th 2009 I PRS Question 4 Moving objects left the traces shown at left The dots were laid down at equal time intervals Which objects did NOT interact with another object somewhere A E B I DA 2B 3C 4D 5A and D 6A and B 7A B and D PHYS 2211N with Dr Greco 0905am 0955am Wednesday January 5th 2009 How are the velocity and position vector related 2 Dquot 9 CD NJ J 1 2 3 4 5 m Wednaso g U 20 39 a2V a Aw 4mg gm 4 1 if vgb i PM Q g 91 if 6 MS 5 imp QQ L w m wg w QZ ampm v 1 7 quot39 3 H if fnv 4v 1 3va nHN V M 06 1k W M6 in 30c osc 30611 IX hl cJot 413 femream ILIFM Wt 200 05 Won t 3 N W 3 53w was M was t 5 V39C 7L S w the H wJw l LM 9 5 A 35 V rtSWh s iV 16w W I 7 m as f 7 is An 9 an winHM s MP 9 39 quot7 S empu 39wt 0th 43 4k chMc M wdzv ABM Co If JM IW ivnVf it s MVFM 0 exclMC 6 vm gt7lw WW 57 L V39l K In 381 U 3016 7 T noise marq m m H ve Log MU NSA31 5 MS39V U NS WU Lima 7 9 PHYS 2211N Dr Greco Wednesday Feb 18th 2009 A puzzle Consider a ball initially at rest that begins to fall towards the Earth System Ball System Ball Earth ENEquot anal System r W i 0 Ball I I 13m 0 x mm W o In both cases there is an increase in kinetic energy PHYS 2211N Dr Greco Wednesday Feb 18th 2009 A puzzle cont How to resolve this violation of the energy principle There is a type of energy that is associated with pairs of interacting objects inside a system We will call this Potential Energy It is a potential energy because it has the ability to be converted into other types of energy Common types of potential energy Compressed or stretched springs galaxies of stars interacting gravitationally atoms in which the protons and electrons interact electrically TnT etc Changes in interaction energy are associated with changes in shape of a multiparticle system PHYS 2211N Dr Greco Wednesday Feb 18th 2009 Potential Energy Apply the energy principle Consider a system of to baquot 1 three balls all attracted to each other by some AE1f1zf1339Ar1 F Arl interaction Q Q Wrml W1sm 1qu Adding up the changes in particle energies AE1E2E3 Wm W5m AUziWint PHYS 2211N Dr Greco Wednesday Feb 18th 2009 Energy of a Multiparticle System The energy of a multiparicle system is the sum of the rest energies and the kinetic energies of each particle in the system plus the sum of the potential energies due to the interactions of all pairs of particles in the system ESySEm1czmzczK1K2U132 You are free to define any system you wish to analyze but be consistant about what energy flows in our out of your system ie Work and what energy changes occur inside your system Do not double count PHYS 2211N Dr Greco Wednesday Feb 18th 2009 PRS Question Earth Moon and Spacecraft A spacecraft travels from near the Earth toward the Moon How many gravitational potential energy terms are there in the Energy Principle 1 1 2 2 3 3 4 6 5 0 PHYS 2211N Dr Greco Wednesday Feb 18th 2009 Properties of Potential Energy Potential energy depends on the seperation between pairs of particles not on their individual positions i 2w3712mA72gt fiv 21 i AU fvA A1 A779 A727A772J77277V77VA7 A Ufm39A7 Any change of potential energy is associated with a change of shape Polenlial energy is conslanl for rigid systems Demo PHYS 2211N Dr Greco Wednesday Feb 18th 2009 Properties of Potential Energy Potential energy must approach zero as the separation between pairs of particles becomes very large For an attractive interaction potential energy becomes s negative as the seperation distance decrease For a repulsive interaction potential energy becomes positive as the seperation distance decreases 4 mll m212vklr39zronslam 1M m lhm hkw mum 30 K m also commm so In 395 also ronsmm chulsmn mu protum Aumttiou mo mm l uergy PHYS 2211N Dr Greco Wednesday Feb 18th 2009 PRS Question Cornet and Star A comet orbits a star in a strongly elliptical orbit The comet and star are far from other massive objects As the comet travels away from the star how does the kinetic energy and potential energy of the system change 1 K increases and U decreases 2 K increases and U increases 3 K decreases and U increases 4 K decreases and U decreases 5 K and U do not change PHYS 2211N Dr Greco Wednesday Feb 18th 2009 Gradient of Potential Energy In the limit that the separation distance becomes infinitesimal Slope is negative and decreasing d U U r Slope 0 A i 7 Mzixunum slope Force is positive I and decreasing Fr Force 2 l l i Muximuml orcc X 257503 quot Gweco Pyl WeentSdm H 271 xvi51y 3002 Lemm 7 ve om b 1 V 33 735 r iquot A Avj Z 0 397 EXAMPk PCs 601K 04 H mwfn o BQKGQHgt Vb a A A as A v v if M quot3 N f I 14 M Pr gt8 Fn Y n foxn 6 09 gt VQUC 67 quot 5 The JLC HM of cm 39Stc fs m M Es tmjenHo em PMH of r e m3 Smknw time MWvi k y5 0 MM QCLUHlt C df m 0 Momerrfum 3 NeMms 37 LAN 3 A 0336 Mavis m 0 V S tmgHP MC 4 0amp 0 Cmm f r Spu except 60 Mm exfewf fkrrf Emewm es a M other 0 8 3 6de 0 2 3 ww mM0 H WW 0 chidvyz 5W ctfnS veler Box un US LinchuMer 3 A 03517043 MAS M valoc ef defQMXJK 50w 5W0 uf A39sl IH CIQCquotI M 33 NagNJ fd I t S VClOC 7 p n 3 39 WMQMM JPJYMV PO 39FC M Up tov f39e 1A fewMS of MOmeI Pvm A gt 3 a lom 4 5 T 3quot V V 4 3 4 quot9 rm MC 2 V EXMPR I W rcLWvath C ragMm m gt A fwcricm with CcMYMH WOMva 40 a 271 Xco l9gtr g 539 Lends 1 H 011714 0 SCCMOIS ct v M expoW w akf 5 47k OCMIFM o2 quotOW39OS CCW39oS H 3 539 KW igt 45 1 lmecY 4quot 0 quotW W MS A 39 39 2 V39m 4 401007m 1 40 0 xmS2 O75 3 luLHXIO a m 230 9 n 7 1 v 339 1 i727410 t Z S 6 X 16 W 17 xm ntfj 3 d 395 r8 f t39 WA1 3 00107 m 4 0 0 Home gt quot141ij mm vi7139va quotV1473 a AP f 93 0 0 4er MM AtM39 H ohmic mkjn u m 6 Mr WC HM 3 A 3 3 APF 3 gx A Z incurpwofa 393 WP 3 4 c WM7U Jar Th6 mo menfunI Prm off 6 pl 397 3 Mom nlt C AM72 7n mamaHum 710 5Y 5WI fo CR MWH 7 Iln fecac ifmd A 1 10 I TS 3 UVIOvnatl js ME of mw c 03366195 msJ39Tdf f q 573 o EVA7 Cm a nut VI LC I JVWUUAJ45 mew on Z quot 5 NedaIs S Ccuw Lo lvv 5577 AP Em M l 1M Ckv l 1M MWI CVWI V 0 573W 1395 f w l f D re A fune MW WI 31 s f 1W WWW Wm 39139 dWW W of 4m fMwqcv H 4M MUWIMW dds nrf We 60 Ta 0 EWCAWMJ39Q 1W mAjn udlt I H NM 6t 39CLQ d VH M WSc 47W Q jWC C 38 4quot Svm tn erecs mm m M 57W some Hm 12W 5 imp 93 ff 3 Kym5 r A1 g we UWHS 0 wcc 51 M N w4cMS 1 N SC N 001 LyAv t V If NITe 4 VM Mlaw 5 A quottquot o is 1L390 00v l CC ms 5 quot 4 WM ca ntf tv 0 04 fro Th m M em va PrOnc A 390 fr ows Us 0 gumH E 3 rota1 I M S S a 3 7V 2 r 3 PHYS 2211N Dr Greco Wednesda Jan 28 2009 PRS Question You hold a tennis ball at rest above your head then open your hand and release the ball which begins to fall At this moment which statement about the gravitational forces between the Earth and ball is correct 1 The force on the ball by the Earth is larger than the force on the Earth by the ball 2 The force on the Earth by the ball is larger than the force on the ball by the Earth 3 The forces are equal in magnitude 4 There is not enough information to determine this PHYS 2211N Dr Greco Wednesda Jan 28 2009 PRS Question The tennis ball falls for 1 second During this time the change in the y component of the ball s momentum is Apy 06 kg ms What is the change in the y component of the Earth39s momentum 1 06 kg ms 2 06 kg ms 3 zero because the ball does not exert a force on the Eanh 4 zero because the Earth s momentum can t change 5 There is not enough information to determine this PHYS 2211N Dr Greco Wednesda Jan 28 2009 From Last Lecture Coulombs Law 17805 CharlesAugustin de Coulomb 1 11421 1il 2 1 4112 77 7 13921 F 7 47m in irgi 415 r2 It ain39t fiction it39s a natural fact opposites attract Paula Abdul Obeys Newtons 339d law of Reciprocity PHYS 2211N Dr Greco Wednesda Jan 28 2009 Electrostatic Forces cont Coulombs law describes the forces that hold together matter on the atomic level How does this force compare to the force of gravity on eanh Example Tension Consider an iron ball hanging from the end ofwire What interatomic force is required to keep the wire from falling toward earth J PHYS 2211N Dr Greco Wednesda Jan 28 2009 Example Tension cont System Ball Surroundings Earth ampWire Force Diagram Fundamental Principle Pff7FneuyyAt 00lFW sliml Al F Substitution ii lmgl we Amazingly the sum of Coulomb forces between electrons and protons in the wire equals the gravitational attraction between the metal ball and the Earth PHYS 2211N Dr Greco Wednesda Jan 28 2009 Model of a Solid Aquot matter consist of atoms uncuttable Democritus 460 BC 370 BC Atoms attract each other when they are close but not too close Atoms repel one another when they get too close to each other Atoms in solids liquids and gases keep moving even at very low temperatures PHYS 2211N Dr Greco Wednesda Jan 28 2009 Model of a Solid cont A chemical bond between two atoms acts like a spring BallSQring Model Eguilibrium chemical bond mum clcclrunyhmd V l y I NJ N m r r muvxxmb uu w e y 7 Too Close nunlam innel elecumm gamma PHYS 2211N Dr Greco Wednesda Jan 28 2009 Model of a Solid cont Everyday solids are made up of millions and millions of atoms balls and springs At room temperature the atoms are in motion in the term at lattice vibrations The cubic lattice of a solid is not very stable but gives good approximations Ball and S rin Cubic Lattice PHYS 2211N Dr Greco Wednesda Jan 28 2009 Model of a Solid cont From our previous analysis of Hanging Mass tension we determined that the V larger the mass the great the O V tension in the wire 139 y w lenslmgl V L Does this fit with our model of a solid Replace the wire with a long thin chain of atoms Assume massless springs The larger the mass the bigger the stretch PHYS 2211N Dr Greco Wednesda Jan 28 2009 PRS Question How does the diameter of one atom in a solid compare to the length of an atomic bond in our model 1 The bond length is greater than the atomic diameter 2 The bond length is less than the atomic diameter 3 They are the same PHYS 2211N Dr Greco Wednesda Jan 28 2009 Model of a Solid cont To apply Hooke39s law we need to know something about the length and Sim 399 CUbiC 3 Stem stiffness of our model39s springs and V V 7 7 how they pack we each NEWS wave FlGURED an WY muoszams 7 meme Nam PHYS 2211N Dr Greco Wednesda Jan 28 2009 PRS Question The mass of one atom is m kgatom The density of solid copper is 0 rho kgmA3 What is the volume occupied by one atom of copper in a solid 1 mo 2 quot70 3 m0 quot3 4 0 m PHYS 2211N Dr Greco Frida Jan16 2009 Things you must memorize gt gt a gt gt 1 pfzpl39i I7116tAlL pzymv y 2 Nvfi5i 1M rfrivanglL VanN 2 C Study tips Due the reading Stop and aswer the exercises scattered throughout the chapters designated by X eg 2X8 Show all of your work it can help you and the grader Work extra problems at the end of the chapter Ask for help when you get stuck PHYS 2211N Dr Greco Frida Jan16 2009 PRS quastion 1 The Hockey Puck A hockey puck is sliding along the ice with nearly constant momenum lt 10 05 gt kg ms when it is suddenly struck by a hockey stick with a force lt 0 0 2000 gt N that lasts for only 3 ms 3e3 s What is the new vector momentum of the puck 1lt10011gt kg ms 2lt0 0 6gt kg ms 31486 kg ms 4lt16011gt kg ms 5lt0 0 30gt kg ms PHYS 2211N Dr Greco Frida Jan16 2009 PRS quastion 2 The Textbook You push a book across a table In 1 A net force is necassary order to keep the book moving with to keep an object moving constant momentum you have to keep pushing with a constant force 2 To make the net force on the book zero you must push with a force equal and opposite ot the friction force on the book Which statement explains this 3 The force you exert must be slightly larger than the friction force PHYS 2211N Dr Greco Frida Jan16 2009 PRS quastion 3 The Kick A ball is initially on the ground and you kick it so that it has an initial velocity lt 370gt ms At this speed air resistance is negligible Assume the usual coordinate system 2 is out of the page Which components of the ball39s momentum will change in the next half second 1Px 2Py 3Pz 4 Px amp Py 5 Py amp Pz 5 Pz amp Px 7an Pya amp Pz PHYS 2211N Dr Greco l7 l y Frida Jan 16 2 009 PRS guastion 4 The Kick Which graph correctly shows py for the ball during flight 1 l 2 y 3 6 f PHYS 2211N Dr Greco Frida Jan16 2009 Multiparticle Systems Newtons 3rd law For forces due to the interaction between two objects eg Gravitational Electrical the forces of two bodies on each other are always equal and are directed in opposite directions Not true for all forces eg Magnetic OPERATION onerAP 9 In this course we call this reciprocity or why you can t pick yourself by your bootstraps Frida Jan 16 2009 Multiparticle Systems cont The momentum principle PtumlfPtumlzFmtswvA l Prum1f71f72f73 FmmyF1smF2mvvF3Wy Reciprocity tells us that the internal forces can only deform the system but can not change its total momentum PHYS 2211N Dr Greco Frida Jan 16 2009 Multiparticle Systems oont Suppose all the people of the Earth go to the North Pole and on a signal all jump straight up What is the recoil speed of the earth We can solve this problem using Newtons second and third law Define our system to be the Earth and people ldealize and ignore all other interactions eg Sun moon mars eto PHYS 2211N Dr Greco Frida Jan16 2009 Multiparticle Systems cont Draw a force diagram I I Not to scale Apply the momentum prInCIpIe people Let motion takes place in the ydirection r gt gt a ptotalf ptotalJ i I7netsurrAlL a a f pearthfppeoplef0 39 p 6 itquot I 619 mearthvearth f y mpeople vpeoplef y 0 mpeople vpeoplef Vearthf earth m ea rth PHYS 2211N Dr Greco Frida Jan16 2009 Multiparticle Systems cont Use the momentum amp position update 39 Not to scale to determine the speed of the people Change of system 99032 gt gt pPeOPlerfpP60PleiFnetsurrAt I f O mpeople Vpeople i mpeople g A t l p e I fep I vpeopleJZgAt Now all that IS needed IS a tIme Interval PHYS 2211N Dr Greco Frida Jan16 2009 Multiparticle Systems cont Use the position update formula Vpeople z39 gt r Atzvpeoplej V gt gt p 0plel ri vangt gt h At g f 2 Solve for velocity in terms of h 2gh Substitute this value into the previous result m 27gh 298ms02m Vearth f peZle 669 Vpeople i 14e 13 ms earth PHYS 2211N Dr Greco Frida Jan16 2009 PRS quastion 5 The Bullet A bullet of mass m traveling horizontally at a very high speed v embeds itself in a block of mass M that is sitting at rest on a very slippery sheet of ice You want to find the speed of the block just after the bullet embeds itself in the block What should you Choose as the system 1The bullet 2 The block 3 The bullet and the block 4 I39m lost PHYS 2211N Dr Greco Frida Jan 16 2009 Conservation of Momentum In the previous example and PBS questions we chose our system such that there were no external forces Newton W ptotalfpmm1i 39 More generally we write A psystem A p surroundings 0 Momentum Conservation is very powerful kungfu J PHYS 2211N Dr Greco Frida Jan 16 2009 Collisions Interactions that take place over a short time and have a comparatively large effect on the momentum of the system It is not necessary for the object to come into contact 5 Interactions forces from the surroundings are neglected System 0 Small gt Everyday gt Galactic GROS MORNE NATIONAL PARK of XVIVi Dr 6mm quot34 de rSE vuay 1715 9009 39 L 2 3 0 m Levewc u fw3 A K33 1s kw 57 is 1 awe cola as is RamsM iL M wt M wit to 6quotle l Jets nu f 1L6 wot 3 LJC HM u eJ NJ to moo k Cal 5 8 PW Nev Nmtc powls U4 U I 3 5U I I 214 I I I 1 5 C T 3 M E 2 i 0 quot WW ital spam I ioLMFeQ opjfuM3 M MW dCr f biz U EM 1 c J 3M Rae 1244 sfmv j v NC Ideal339le 8W J w ru spam w m Mch MWMWM is A 3001 W s mj 38 1N osmHMfms M Snht 3 EMWW 3h 0 OSLFHMMJ SPVMJ W MS sysfm M MA 6M1 Cum 5 1 sfw x3 l Helene WM 250 l l E K u WW 1 L L 3 I I If LMM LO LIAM M2 phenvaQSJLW BMW I r F 0 nd V l l E Kt u Z J 1 14 YMV o as Lulu Lb LMT A36 3 y l l CWPHCJ GHQch 0 i 1 59mm 3 mmwm7 wt V50 I 1 L La kw O iws LoLN 9 s 5 W emij as he 3M in m Mam ch63 we cm w Ms 4m Jame A ewaessx m 4w 1 mw M 1 1 Posw m mv EKJLO m e I t L s 0 m s 391 7k 3 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Idealth svaj I S 09 Idepued gafwm rmekh 0 kvu tdS M 3 3ch A good 71 42ch rod 4 H Energy m Gun cgtuqrtM3 SPI MJ mss L skm V l t F5 29 G ll F5 I a 1 R W m L In LO LivIN Lo Q CWPI C the 3PM Mb Icicawi wHL i7 0 quotZ mg u m i a E K 3quot U 3 m w Q 394 quot 3 LO g when M fth is M M It LamaHa 315 a 3 m M 2 3 2 m E K t U 3 m Emvx t 0 g g Qua fancier M 510 M3 onLC 3am Inks 230 m M t as K39WJ 2 a 1 E K hawbmm quot3933 l M encrfh 35 MC skint 39an H WN C39C MS Wxs Pdwa us A dce wnMe M velodty GrecoLecture16 Page 1 061a Which of the following 1 U S kS C expressions satis es the 2 U k C requirement that S S Fx arUs dx ksx where x is measured from the equilibrium 4 U z k x2 C position C is a constant S S 061b A horizontal spring has a mass attached which can move with negligible friction You stretch the spring and release the mass from rest For the resulting motion which of the following statements is TRUE 1 When the spring is momentarily fully compressed K has its largest value 2 When the spring momentarily has its relaxed length U has its largest value 3 When the spring momentarily has its relaxed length K has its smallest value 4 When K is large U is small and vice versa 5 When K is large U is large and vice versa wxmi n Qf b 339 g W vajs MM M ta was e U hkt 3 GR sPrht Lang 0 one um wa emsewvmkm 8 20 Consth urea37 1 mm 3 3 39quot 333 Of 5 6 y HKJH 079 M43 2 3amp3 Mjf r mBam xeiksfgam wags m s awg 2 A a mgmm K3 wow 16 s matf a O 3 qujfszD m3 v Ks 00 a3 KS 13 M32 3 awx ygmgwi i m s is m KS 37 PKS QuE Schms 756 me an HH nof ma 5 M kc asullaHms will eveM39qu cl AM Pow af 41M axij is 391er 0 4 5M3 59 Pkr t of wk enewaz acts info incrtASM 4 L Mann mo39um w ln M So I 04 39 A sis LT C N S39C is 57 M KWftf 961601564 Malay of 8N7 Mm 313 60015 gt 13 wd un t 39to 6 Mlt we cm 1n4 vfalw mmuf M cxc is ant7w Eun wxs of Mums an even The 5M 39uM turn aquot mwcr 39InS CCaul of Mmsw h R W 39gach MUM wt MKSWQ 4 644 WC 0 M CM 57 A lacMWW Mums 44a y mod eyPM sa39m all Mew m0 Corclex 114 to amp dmkv39je ln tCMPori wC 3 In K i390 SouC MJVc rtl 1 54 WHO wheel exPuM n1 Mm can A 11 1 3 wcrjbrl unfit 7 I GrecoLecture16 Page 3 061c A horizontal spring with stiffness 3 Wm has a relaxed length of 25 cm 025 m A mass of 50 grams 0050 kg is attached and you stretch the spring to a total length of 29 cm 029 m The mass is then released from rest and moves with negligible friction What is the kinetic energy of the mass at the moment when the spring returns through the position where its length is its relaxed length of 25 cm 1 24e3 2 48e3 J 3 6e2 4 126e1 5 15


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Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.