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# Introductory Physics I PHY 231

MSU

GPA 3.67

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This 58 page Class Notes was uploaded by Quinn Larkin on Saturday September 19, 2015. The Class Notes belongs to PHY 231 at Michigan State University taught by Staff in Fall. Since its upload, it has received 41 views. For similar materials see /class/207627/phy-231-michigan-state-university in Physics 2 at Michigan State University.

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Date Created: 09/19/15

Review Session for Midtermz Chapter 6 7 8 and 9 PHY231 Momentum impluse Momen rum SI uni r kgms p mv NewTon r39evisi red A F ma p At Impulse PHY231 Conservation of momentum Total momentum of an isolated system is conserved Isolated system means no external force acts on the system Total momentum sum of momentum for each constituents of the system Apply conservation of momentum for problems with collisions Momentum before collision Momentum after collision PHY231 Elasticinelastic collisions Two main Types of collisions Inelosl39ic Elas ric To ral momen rum is conserved for39 bo rh elas ric and inelas ric collisions To ral kine ric energy is also conser39ved for39 elas ric collisions PHY231 4 Summary collisions To ral momen rum conserved for39 an isola red sys rem ptotalf ptotalj m1V1 f 39l39 mzvz f mlvli 39l39 m2v2i Collisions of Two objec rs 1D Perfectly inelastic m1 m2vf mlvli mzvzl f fn1fn2 2vf v1i V21 ElasTic KEF KEf mlvlf 39l39 m2v2f mlvli 39l39 m2v2i v 12 2f 11 39f fn1fn2 PHY231 Vlf Vzi Angular displacement Difference be rween final and ini rial angles A66f 6i SI uni r radian r39ad PHY231 Angular velocity Average angular veloci ry A A6 6f 6i a Ar Ar SI uni r rads The angular39 veloci ry Tells you how fas r an objec r is r39o ra ring IT is some rimes expressed in r39pm r39o ra rion per39 minu re 1er 21160 O1047r39ads PHY231 Tangen139ial angular acceleration Aver39age Tangen rial angular39 acceler39a rion At gt 06 Au Hf 001 At SI uni r r39ads2 The angular39 acceler39a rion Tells you how much The angular veloci ry is changing Through Time on is r39ela red To a1 Tangen rial acceler39a rion a1 ocr Do no r confuse Them wi rh aC cen rr39ipe ral acceler39a rion PHY231 Displacemenf and angular displacement For a poin r on a circle of radius r The displacemen r As along The circular pa rh is As rAH SI uni rs As me rer m r me rer m A6 radian rad PH231 3 K r AS rAH A0 e Cayuga Leavan Tangen139ial velocity and angular velocity Tangen rial veloci ry and angular veloci ry ar39e no r The same Thing y AS rAQ ax v ra v Ar Ar P r A5 IV real 0 M j x aaaaaaaaaaaaaa ng An objec r moves on a circle of radius r3920 m a r an angular39 veloci ry 0010 r39ads I rs veloci ry is vrw201020mS PHv231 10 RoTaTion of an objecT For39 a solid objecT r39oTaTing All The poinTs of The objecT have equal angular velociTy a same for all points The poinTs far39Ther39 away from The axis of r39oTaTion have gr39eaTer39 TangenTial velociTy v wr depends on r PHY231 TangenTial acceleraTion is r39elaTed To angular acceleraTion and Tells us in ms2 if The r39oTaTion of an objecT speedsup 090 or39 slows down aTltO or39 has consTanT r39oTaTional speed aTO lat r05 CenTr39ipeTal acceleraTion is The one Telling us The objecT is r39oTaTing on a circular paTh 2 V Constant acceleration There is a direct parallel between linear motion and rotational motion a constant a constant Av aAt An ocAt Ax viAt gum2 A6 wiAt aAtz v vi2 Zan a 0012 ZaAH PHY231 Gravitation general formulation mlmZ 2 Iquot FgG G 667x1039 Nm2 kg2 Near Earth39s surface Fg can be approxima red by Fgmg g 981 ms2 Gravitational Potential Energy GPE Near Earth39s surface we had define GPE mgh This formula is only a good approximation when h is small compared to Earth39s radius In general GPE of an object at distance r from Earth39s center is MEm I GPE G Remember that important quantity is not GPE but AGPE the change in GPE PH231 15 Escape speed We apply conserva rion of energy IKEZ GPEZ KEf GPEfI Af rer39 some algebra we find 2GM vesc R Calcula ring for39 Ear39Th we would find vesc112 kms PH231 16 Kepler39s laws Third law For a planeT around The sun of mass M5 The period of revoluTion squared is proporTionaI To The cube of The radius of revoluTion 432 3 r GMS 112 T is The period of revoluTion which means The Time The planeT Takes To do one revoluTion around The sun PH231 18 ExTended objecTs EXTended objecTs have Two moTions TranslaTionaI moTion Fma wiTh F The neT force acTing on The objecT and a The acceleraTion of The objecT s cenTer of mass 0 RoTaTionaI moTion x WKWN 1101 f T 5 Themmmm mmhawmmd Demumm cmm K J mam3 39 CenTer39 of mass or graviTy The 2D posiTion of The objecT39s cenTer39 of mass is defined as WiTh IT is differenT Than The objecT39s gzghieTric cenTer If you hang The objecT aT r39CM iT doesn39T move Torque Define axis of roTaTion To creaTe Torque The force needs To acT aT a cerTain disTance from The axis of roTaTion force acTing on The axis have zero Torque Be perpendicular To The posiTion vecTor r linking The axis of roTaTion and The locaTion where The force is applied Be perpendicular To The axis of roTaTion Mechanical equilibrium for an object An objec r is in mechanical equilibrium if The ne r ex rer39nal force is zero Fner 2F 0 The ne r ex rer39nal Tor39que is zero I I Br 0 The fir39s r condi rion means The Tr39ansla rional acceler39a rion is zero The second condi rion means The r39o ra rional acceler39a rion is zero PH231 22 MomenT of inerTia I Valid for rigid objecT RelevanT To esTimaTe angular acceleraTion using The neT Torque EquivalenT of The mass for TransIaTionaI moTion Depends on The choice of roTaTionaI axis Depends on The mass disTribuTion The larger I The more difficulT iT is To creaTe roTaTionaI acceleraTion PHY231 23 Parallel noTions There is a direcT parallel beTween governing equaTion for TranslaTional moTion and for roTaTional moTion F ma a F m AcceleraTion proporTional To force The larger The mass The smaller The acceleraTion arI Angular acceleraTion proporTional To Torque The larger The momenT of inerTia I The smaller The angular acceleraTion 0c Rotational kinetic energy A particle of mass m moving through Space with speed v has kinetic energy KE lmv2 2 A rigid object in rotation is made of many particles moving with same angular velocity u and the rotational energy is KEr 11002 2 Notice again that it is easily obtained by replacing velocity v by angular velocity 00 Mass m by moment of inertia I 25 Angular momentum As for39 Transla rional mo rion There is a no rion of momenTum for39 r39o ra rional mo rion IT is called angular momen rum and no red L L210 AL Tnet At PHY231 Conservation of L If There is no ne r ex rer39nal ror39que ac ring on a sys rem The To ral angular momen rum is conserved L L total i total f Two objec rs in r39o ra rion pu r in ro con rac r L Ltotalf LU L21 LLf LU total i Ilwu 120021 11001 200210 27 TranslaTiona mo rion RoTaTiona mo rion Displacemen r Ax Ang displacement A6 Vdocny V AxAt Ang veloci ry a AQAl Acceler39a rion a AVAt Ang accele a AgoAt Nass m Nomen r of Iner Tia I Emjrf Force F ma Tor39que L39 VFi 06 KineTic energy KE mv22 Rotational KE KEr la2 2 MomenTum 9 W Ang momentum L 00 28 Deformation of Solids ElasTic properTies of solids STress Force per uniT area causing deformaTion STrain Measure of The amounT of deformaTion STress is proporTional To sTrain stress elastic modulus x strain Three primary Types of deformaTion and modulus Young s modulus Y ElasTiciTy in lengTh Shear modulus S ElasTiciTy of shape Bulk modulus B ElasTiciTy of Volume Large sTress can lead To The breaking poinT of a solid elasTic behavior noT valid for ThaT poinT Young39s modulus Y Elas rici ry in leng rh Solid deforms because of The ex rernal perpendicular force F Tensile force applied on its cross sec rion A In rernal force increases un ril i r balances F AT equilibrium one iiiiiiiii F quan fies Stress MP TenSIle sTress FA a Tensile s rrain ALLO 400 E t 300 1aSIC Breaking During EiClS I39IC regime 1m point 200 F AL 100 Elastic Y yNm2Pa behavior A L I I I I Strain 0 pHy231 0 0002 0004 0006 0008 001 eeeeeeeeeee mmg Shear modulus S 39 Elas rici ry of shape Shear39 s rr39ess One face held fixed Parallel force applied To opposiTe face 7 No change in volume 5 L7 Fixed fare Shear39 s rr39ess F A I Shear39 s rr39ain AX h E 5 SNm2Pa A h Large 5 means objec r is hard To bend PM b Dug yyyyyyy rug Bulk modulus B Volume elas rici ry Solid subjects To uniform force V per mil area i Example Objec r immersed in T al a flund Volume s rress AP AFA Change in Pressure Volume s rrain AVV AV AP B B Nm2Pa V Large B means objec r is hard To compress PHY231 32 Density For39 objec r wi rh uniform composi rion and Mass M and Volume V M SI 39T k3 Uhl pgm p V We can also wr39i re M pV SI uni r M kg Impor ran r for39 buoyan r force I PH231 33 Pressure F is The magniTude of a force exerTed perpendicular To a given surface of area A The pressure P is The raTio force over surface 135 A SI uniT P Nm2 Pa Pascal US 1 psi 6894757 Pa 689 kPa DeformaTion of solids depends on FA so The larger The area where The force is applied The smaller The risk To exceed The uTimaTe sTrengTh of an objecT PH231 34 Pressure with depth IP P0 pghl Shape of The con rainer39 doesn39T ma r rer39 only dep rh Type Of flUId ma r rer39s Pressure measurements Open Tube manome rer Barome rer Measures P of a fluid Measures P0 P0 3 on T PP0pgh Absolu re pressure P Gauge pressure P PO 36 Pascal39s principle A change in pressure applied To an enclosed fluid is TransmiTTed undiminished To every poinT of The fluid and To The walls of The conTainer NeglecTing The heighT difference beTween lefT and righT pisTons we have P1P2 55 1 A1 A2 eeeeeeeeee mm PHY231 Review Final Skip problem 84 Problem 62 was done but don t worry about it Problem 61 6367 7 07 173 see Ioncapa 7475 see 4950 A few questions concerning midterm 2 have been added A hinmtullir strip igt huld xed at tho lmmun 0111 21gt lum 11 in the liE39l 111 Till39 lllt lul 11 HIV lx39ft 11 1 L39nvl l li39ivlll ul illll39nl39 ilt39l L39Xpulbinli nt nHT 12 x 10 lquotI39 thr 111rt11 011 thiv right lm HIM Shit lil lflx39 When the gtt139ip igt limmil it Will complem till MJIH39CLIL L39 ALOLLO 47 AC bum loft Bf Ivinnin straight le 110ml right The right strip has a higher coef cient of expansion so will increase its length more than the left strip when heating Hence the strip must bend left 8131 111013 of Nitrogen at 230 3 tonmormuro and 330 mm prosmre cont 3111 how 111211137 1110105 48344 0995 BC 11133 311 1367 DC 1599 561 Hfquot 2997 E11 3 1871 FC39I 2189 G39 1 2 Use pVnRT ideal gas law npVRT p330 atm 330 X 1015X105 Pa 334950 Pa V86 L 86 X 1X10393 m386X10393 m3 T230 0C 2302731529615 K R 831 JmolK So 11 334950 X 86X10393 831 X 29615 117 mol A hot 300 K and 1 gold 200 K hum reservoirs are uullllmrtoll 10 until ulh r hf two llll llllL Lll Llllll lllmlnl bars in two different ways 11 shnwn in the figure Use P kAThTCAxkAATAx For39 left side k kaluminum Cramp42ml Tn The left foll glll39ill39lull The 12m of hour transfer 20 all area of Single bar in The T mn gm39nrinn ix 25 high V mif r h the I ThTc600 K Ax Lengrh L of single bar39 em 1100 I 49 Ali nur liz39xlf om t39hird For side G 1110 tourth DC Viv k kaluminum film titevi s A 0 WM 0quot area of single bar39 r I OL39 111105 ThTc600 K Ax 2 x Lengrh L of single bar39 PM k x 2a x 600 L 1200 kaL Fright k x a x 600 2L 300 kaL So Pright is 3001200 l4 times that of PM Make sure to read carefully don t turn question around Problem 50 Similar A cylinder contaim CO2 gas at 4609C tC39lllpCl iitlll39C39 Vtht is the RMS hpC39C39Ll 0f the gm 111x110 ul39 The 1112 of a C02 molecule 15 730 IU ZH kg m m s 1 101 Bit 340 109 cf 425 102 L a 102 Bi 1555 x 102 FQ 531 102 1 104103 HQ 130 x 103 Vrms 3kBTm V 3RTM kB Boltzmann constant 138x103923 J K T Temperature Kelvin m mass in kg of 1 molecule M molar39 mass R 831 JmolK Her39e T4627315 31915 m73x1o26 kg 50 v425 ms uvc much heat is required to convert 220 kg of solid ice from a temperature of711U C to liquid water at a 101117 pemmre of 715 C39 The speci c heat of ice is cm 050 kcalkg and the 11th of msion 10139 water 15 L 707 healkg 271 m1 cwmr10 kcaIkg C 52AQ 538 x 101 B0 730 x 10 CO 113 x 102 DQ164gtlt1112 1302 me F0 345x102 GO 53900 X 10 HO 73925 X J mammga Q22 x 05 x 11 121 kcal Example Healing 0le9 Q22 X 797 1749 kcal c Q22x1x715 1573 39 mam Total 3443 kcal Iaiscs T of 105 melt ICE mmei T of Wnler vaponzcs m cr Q m MAT mix T 01 steam Moanui cnmnts 011 tm st 21 s indicate that Star X 11225 a urfacc I39Olllpti39l ai Lu C39 of 6300 K and Star Y hELb a nurfaco t0111porature of 1230 Ix39 If both stars have the same radius what is the ratio of the lumincmity toml power output of Star Y to tho luminobity of Star X Both stars can he ilt11idr31 c1 to have an emissivity of 10 53A Ex 1 695 134 759 0 89L DQ 1007 j LS8 FQ 1230 931453 HQ 342 PcsAeT4 Stefan39s law Js 556696X10398 Wm2K4 A surface area e object dependent constant emissivity 01 T Temperature K P energy radiated per second PX er63004 Py csAe123004 Note that aAe are the same for both So Pysz 123004 63004 1453 A train has 11 11111 of 7137 10 kg and is mm39ing 3911 21 speed 139 The engineer applies the brakes which results not l39liui l V39Ill Cl force of 147 1U X on the train The are held on for n time period of 1M 5 Ylmt is the new spl wil of the train in ms 541411 Imam ml Bx Insx In1 Cy LDOX In1 D 214 x 101 E i 212 gt 101 Ft 21 to1 GQ 309 x ml s u x 101 w How 111139 1005 the train travel luring this period 71 m 55A 380 x m2 B 51 x I09 CM 053 x in Dr 5103 x 102 El L21 x 103 Paul ml gtlt I03 1 214 x 103 HQ 25 x I03 54 Change in momentum ApmAvFAt mass constant AvFAt m147X106 X 186776 X 106 354 ms slowing down Initial velocity 811 kmhr 811 X 10003600227 ms Final velocity 227354 191 ms 55 Acceleration a AvAt 354186 O19 ms2 Distance traveled Axv0t05at2227X186O5XO19X 18623 89 m An mummy shell is launched on a at horizontal eld M 2m augle or 11 2139 with mp1 Lu Lhe huuwuml 111111 ParquIIC mo on nth an initial speed of VD 338 1115 tht is the honzontal ho mzon39ral distance covered by the shell after r rquot ht 9 s 0 1g 771 m X39rxoV0x39r 56VAQ 148 x 103 BO 1 73 x 103 co 2 113 x 103 X rvolt 37 x 103 E0 273 x103 PO 3 25 x 103 GO 380 X 103 HQ 445 x 103 Verhcal 2 y r yoVOy r 059 r 7 pr 39hat is the height of the shell at this moment 2 m m VYT v0y 9T 9 981 ms 57AO 5 71 x 102 BO 759 x 102 Co 101 X 103 VOXaocos nitial angle D0 1 34 x 103 E0 1 79 x 103 FQ 237 x 103 V W Sin ni al an 16 GO 310 X 103 HQ 4 20 x 103 0y 0 g 56 x rx0v0X r O 338cos381 x 557 14815 m 57 y139y0v0y13905g20338sin381x557 05x98x5572 62 152 1010 m Tofal Disfance Traveled in horizon ral direcfion Y1390 so 338sin381139 05gtlt981392 0 209 4920 209491390 So 1390 sfar f or39 14265 so x4265338cos381x426511344 m Height 11139 highest point Vy1390 11139 highest point so 0338sin381 981390 139213 s Y213338sin381x213 05x98x2132444222232219 m A 328 kg sphere undies it perfectly inelastic Collision with u set0111 sphere that is initially at rest The composite systum 111015 with a hptftltl gqual to um third the original SpCCt39l of the 323 kg SpllCl39tJ Vlia it is the mass of the NZ C39LHUl split1 0 gm kg sassy m Bf 31 Cu 356 DC 402 455 FrjS 514 3Q 551 HQ 05 Perfect inelastic collision objects stick together after collision and only momentum is conserved m1V1m2V2m1m2Vf Here m1328 kg m2 V1 V2O VfV13 So 328V1 328m2V13 V1 drops out m2 33283283 656 kg An 21111111113 m iuuning 11 11 must11m spur1L 13911 L391 2 1 1ii111111 1 of 137 111 in 11 ti111139 period of 215 111i11111cs The drug f13911391391 19111111 by 1111 1111111 1111 the 511111111111 is 13811 N 12711 1111rm 1111 13911 11 1 1 1 1111 5113911111111 139 must provide in 1 39quot0139C1111111g that 1111113 1111 111 59111 1111 1 11139 B1 2112 1111 D1 3115 I 111 E1 3111 1111 G1 11111 1111 HQ 7711 1111 r 11L 111l WFsz work performed against the drag force when swimming AX PWAt Power needed work per second carried out W58X1377964 P7964215 x 60616 Js W A proposed space statian includes living quarters in a circular ring 51 111 in diameter At what angular speed should the ring rotate so the occupants that they have the same weight they do on Earth in radSJ To have the same weight as on earth the centripetal acceleration must match the earth s gravitational acceleration 981 ms2 For objects rotating at constant angular velocity lineartangential speed Fmac F force can be tension friction normal force gravity accentripetal accelerationv2r 032r used 03vr Here F is provided by the normal force of the hull of the space station but must match the gravitational force on earth Fnmg mg moazr 0gr 98255062 rads don t forget conversion from diameter to radius A 0400kg object is swung in a circular path and in a vertical plane on a 0500mlength string If the angular speed at the bottom is 800 rads What is the tension in the string when the object is at the bottom of the circle For objects rotating at constant angular velocity lineartangential speed Fmac F force can be tension friction I I normal force gravity I accentripetal accelerationv2r 032r T used 03vr At the bottom FTFgmvmac F So Tmgmac and Tmgm032r And Tmg 032139049882X05 Tl67 N Note if mass at the top Tmgm032r If the mass of Mars is 0107 times that of Earth and its radius is 05 30 that of Earth estimate the gravitational acceleration g at the surface of Mars gearth 980 ms2 2 planetrplanet Fmg and FGmm SO gGrn tr 2 plane planet gearthC nearthrearth2 Grn r 2G X 0107 rnearth053Orearth2 gmars mars mars gearm010705302 038 gmars gmars038gemh038x983 733 ms2 PNun l in this iigiirv 11m u riiumvivr of 192 m 68 Force on piston 2 mg81 Piston 2 her n llii lllllftljl of SLLil 111 In the absence of frivtinn illitvl39luilu the form l39r tilllil ixl uu Ii TUIL l 1139139lt211quot39 TL support an object With 1 1111 of 818 kg DL39H L39LI on piston 2 Neglect the height lij fi139r11nv between TILE IIUTTUIII of the tu39o pisfulb nnd 1gt 111nr3 that the lii ttjllgt are 11111551055 That is the 11mgnitude of the force F required at the end of the handle Pascal39s principle pressure in an enclosed volume is The same everywhere PF1A1F2A2 8x981 8024 N A11Cr121t0019222 29xlO394 m2 A2nr12n009422694x10393 1112 So F1F2AlA2335 N 69 To make a torque tF1d at a distance d2 from the rotation point a torque CZFhanddhand is required by the hand d12 A uniform horizontal beam of length 60 m and weight 120 N is attached at one end to a wall by a pin connection so that it may rotate A cable attached to the wall above the pin supports the opposite end The cable makes an angle of 60 with the horizontal What is the tension in the cable needed to maintain the beam in equilibrium I F Equilibrium sum of torques must be zero vert ILbeam JccableO ILbeamZFgravdcenterofgravity120 X 3 3 N W gable FvertdTsin600 X 6 Solve for T 360 Tsin60 X 6 T60sin60 693 N A solid cylinder 1 MR22 has a string wrapped around it many times When I release the cylinder holding on to the string the cylinder falls and spins as the string unwinds What is the downward acceleration of the cylinder as it falls 339 IVERYHARD T c 65 ms2 1 98 ms2 Newton s law 2Fma so Tmg ma a will be a positive number the sign indicates It goes down Torque rlocIarmr2a2r Also gt combine Tma2 Torque TZF X d T X r So Tmg ma And ma2mg ma So mgl5ma and agl598l565 ms2 A large tree trunk is oating in the sea The density of the sea water is 1040 kg m the density of the trunk is 770 kgmz What fraction of the trunks Volume 15 under the surface of the water 70140 0315 130 05350 CC 0102 PO 0580 GO 05055 Du a 431 051 H0 0740 For oating gravitation force buoyancy force So weight mg weight of displaced water mwamg pwoodeoodg pmNdisPlacedg used PMV SD M W So vdisplacedvwood pwoodpwater770104003974 This is the fraction of the tree under water An 141011 hour Lug ints nibol39l 035 id of hunt and ex humts 7551 kl of him in with 1310 tht igt thv vt rir ncy Hf rho rn39l11lt 77Aquot 020 4 II 2 B um 0 3 cg 132 1n l DQ 1m HF Eg 277 10l Pg 4m r 111 G 352 y 1u1 H 544 r 0 1 How much Work is done in u Eyck in M 781 5110 BU 11m ml 10 145 r ml Du 17114 13quot 22 101 Fig 2w 10 G 350 v m HQ L37 101 77 Ef ciency of an engine any lQcoldQh0t1759938O 191 Note that for a Carnot engine only ef ciency lTcoldThot 78 Nothing is known about the volume pressure or temperature internal energy so here one must use Ef ciency QhotQcoldyQhot WQhot so 0191 W938 W179 kJ Easier is W QhotQc01d938759179 Tlu grupll gt11HWgt hr 3941ilm39lt11w1t us 1 t39llusstinn of tiuu for u 1111 llt39lllul39Ulllil39lil mullmum simpll harmonic 111x yrinu 1 xun r3 r1 o m a OI2345B78EYIGIl39IZl3I4IEIG HS Tliigt l39llur llnu mu In lmrrihvtl l1 Illt following l39unnlllu Xlll 99111th Whrn X and A 111 mavmu ml 111 moter 1 i lumxuml in wmuxls 2 igt llll39zllll39quotl in huhN 79 What is the amplitude 80 What is the period 81 What is the angular frequency 0 82 what is the frequency f 79 Amplitude distance between equilibrium position x0 and maXimum displacement Read from graph 110 m 80 Period time that it takes to make one full oscillation Read from graph T46 81 oa2nT 27546 137 rads 82 flT l46 0217 Hz 1 n A 304 11m is Mb 7011th from a 9 11 1ng with u s ring I V V Constant of 179 39rquot111 At What hum11 39cqm1u 39 will Thf sysfmll osti39iurlt397 in Hz 83Anr 5125 10 Bi 134 1nL CQ 105 DJ 119 EQ 135 Pg 152 GQ 172 HQ 1J l oakm 1790664 519 rads oa2nf so foa27 and f0826 Hz Vllut 3915 rho bound level of u mund with an intensity of I 100 x 1U i 7591112 Give fmm HHMTC39I 111 113 units 8613ij 1157 BC 1965 IQ 2554 Dl jtn 4135 Eli 5000 Fquot 3Q 12615 HQ 15202 57 UU Sound level B10ogIIO 10103912 Wm2 10log1x103961x103912 60 dB qu the intensity of tlligt sound is i11tirc39uscwl to a value of 1260 timw of its u39iginal inttgns it Vlmt igt thr 110W increased sound love Give your mmvcr in dB IlllltF 87Aquot 4192 Bag 5575 11le 7415 DC 11862 EL 13115 Fug 17145 ij 25202 HQ 30555 New Sound level 10log26x103961x103912 7415 dB Or39 use logab logalogb New sound level 10log26 60 1414607415

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