2009 Physics Bowl Solutions

Transcription

1 9 Phyc Bowl Soluton # An # An # An # An # An D B C A E B D D E D A E C A B C B B E C 5 D 5 C 5 E 5 A 5 A 6 D 6 A 6 D 6 D 6 D 7 B 7 D 7 C 7 A 7 E C E E B B 9 A 9 B 9 B 9 D 9 C E C A C 5 D yr 65dy hr 6 n D Convertng unt 5 ~ yr dy hr n B Convertng unt 5 c c 5c A Fro Newton, we have F net a The MKS unt of a are kg wth a the MKS unt kg of acceleraton Puttng the together gve u that the unt of force can be wrtten a C n a eaureent, all non-zero value are gnfcant, a well a any zeroe at the end of a eaureent after the decal Therefore, the gnfcant dgt are bolded: 5 D Knetc energy coputed a v ( 5kg)( ) J 6 D Antnode are eparated by a dtance of one-half of a wavelength Hence, the wavelength c λ c here For wave on a trng, we copute v fλ f v Hz λ c 7 B The retor n the crcut are connected n parallel Hence, Req Ω 9Ω R R R R Ω Ω Ω Ω eq C The peed of the object greatet when the lope of the lne tangent to the poton v te graph ha the greatet lope Th occur at te t The graph how tangent lne at the pont of nteret n the anwer 9 A For the trajectory hown, the velocty tangent to the path whle the acceleraton n the proble that due to gravty (traght downward) E Ung the conervaton of lnear oentu for the r r r r r r collon, we wrte p v v v v where we dentfy M pnt fnal f f,

2 truck X and call to the Wet a the potve drecton Th gve M( V) X( V ) M( V) X( V) 6MV XV X 6M B Durng a lunar eclpe, unlght blocked fro the Moon by the Earth Hence, the Earth between the Sun and the Moon D The gravtatonal force on the phere by the Earth g ( 5 kg)( ) 5 N By Newton Thrd Law, th the agntude of the gravtatonal force exerted on the Earth by the phere E Ung contant acceleraton kneatc, we wrte ( 5 ) ( 5 ) v v v v a a 975 The queton aked ( ) for the agntude (potve value) of the acceleraton B Work done on the box for any dplaceent parallel to the drecton of the appled force Hence, nce the force appled to the left a t ove eter to the left, then the work potve (leftward force and leftward dplaceent) and ha value W Fd ( N)( ) 9 J L at 5 C Ung the deal ga equaton, we wrte PV nrt recognzng that R (ee ol K contant heet) to utlze the preure gven n atophere and volue n lter Alo, the PV ( 5at)( 6 L) teperature ut be n Kelvn Th lead to n 9 ol RT L at ( 7 K) ol K 6 A The expreon x v t at only applcable when the lnear acceleraton contant θ ω t t α only applcable when the angular acceleraton contant 7 D Average peed coputed a dtance dvded by te The total dtance traveled eter for the frt part and then an addtonal x v ( t 9 )( ) for a total of Hence, the average peed 7 E The agnetc feld aocated wth the current n the wre drected out of the plane of the page at the locaton of the electron (Rght-hand rule rght thub along current, fnger wrap n the ene of the feld) Ung the rght hand rule for the force on a charged partcle n a feld rght fnger pont to the left (velocty), the fnger are curled toward the feld (out of paper) and the rght thub pont toward the top of the paper except that nce the charge negatve (electron), the hand flpped degree gvng the fnal drecton of the force to be toward the botto of the page 9 B Energy tranfer fro the bulk oton of a flud bet related to the ter convecton C Th a -D contant acceleraton kneatc proble Horzontally, there no acceleraton, o one ha x v t a t ( ) v ( ) v x x x x By yetry, the vertcal coponent of the velocty ntantaneouly zero halfway through the trp Hence, v ( )( ) v y v y a yt v y y Ung the Pythagorean Theore to v v v x 6 fnd the total ntal peed yeld y ( ) ( )

3 C Snce the object ddn t change t oton, t ha an acceleraton of zero Wth only the appled force and frcton actng on the a horzontally, thee force ut have equal ze by F net a The frcton force therefore 6 N D The perod of a ple pendulu coputed a T L π By hortenng the length of the g trng to L, the perod now ½ a large becaue of the quare root The new perod T C Proton are confned to the nucle of the ato and are not oble, o the chargng done va electron tranfer That the cob becoe negatve ean that electron were accepted by the cob fro the har B The knetc energy of the a KE v v 56 The net force actng on the object found fro Newton Second Law a F v net 56 N nce the r acceleraton toward the center of crcle and ha agntude v r 5 E The oentu of a photon drectly related to the energy of the photon the ore energetc the photon, the ore oentu t ha Of thoe lted, the gaa ray the ot energetc 6 D By addng the hortng wre, bulb now bypaed Th reduce the overall retance of the crcut (check t: before hort: 5 R, after hort: R ) By decreang the retance, the current n the crcut ncreae Ung Oh Law, bulb now ha ore current and therefore there a larger power aocated wth t ( P R) and t burn ore brghtly Bulb #, however, d ung Krchhoff Loop Rule the battery voltage unchanged whle bulb # ha a larger voltage eanng that the voltage acro bulb ut decreae and t therefore der ( P V ) Fnally, R nce there a bt ore current n the crcut and the branch wth bulb ha a greatly reduced retance ore of the current wll be drected th way, thereby brghtenng bulb 7 C By the rght-hand rule, there a agnetc force on the proton drected nto the plane (Fnger pont up the page, curl the to the rght thub pont nward) Snce the proton undeflected, r r there ut be an electrc force of equal ze drected out of the plane of the page Ung F qe, the drecton of the force on the proton and the feld that t n ut be n the ae drecton nce the charge potve Hence, there a coponent of electrc feld drected out of the page E The deal yellow pgent reflectng the red and green lght Cyan produced wth green and blue flter Conequently, when th lght hne on the yellow pgent, the green lght reflected and the blue lght aborbed The pgent appear green 9 B To tranton to the ev tate wth only two photon eon, the only opton are for the electron to ake the followng tranton: ev ev ev gvng u photon of energy ev and 9eV or ev 7 ev ev gvng photon of energy 6eV and 5 ev Th ean that the ev photon not poble wth only two tranton A The ot traght-forward approach to ue that the expreon for power fro a retor n the kg J P W for R whch yeld unt of N kg Note that the bae A A A A A MKS unt the Apere and not the Coulob

4 A There are a couple of opton for olvng th proble Method : Wrtng the contant acceleraton kneatc expreon for each leg of the trp yeld x v t a t x V( T) For the econd leg, we have V T x v t a t x V( T) VT where the acceleraton wa T v V coputed durng the nterval a a Puttng th nforaton together gve the rato of t T VT the dtance traveled a VT 6 Method : Graph the oton a velocty v te The area under the curve wll gve the change n poton and nce th purely a D proble, the ze of the dplaceent the dtance traveled So, the proble reduce to the rato of the area of a rectangle to that of a trangle That, LW bh ( T)( V) 6 ( T)( V) E By placng the cube n water and drawng the free body dagra, we have only a gravtatonal Vdplace ρcube force and buoyancy Equatng thee gve ρ water gvdplace ρcubevcube g 6 Vcube ρ water So, the denty of the cube found to be 6 kg Ung th ae logc for the cube n the ol, Vdplace ρ 6 we fnd cube 75 75% of the cube uberged n the ol Vcube ρ ol A The PV dagra how qualtatvely what the procee of nteret look lke Snce the frt proce contant preure, the contant volue coolng reduce the preure fro the ntal value, akng (A) true The volue clearly ncreae, akng (B) wrong The teperature could end up greater, le than, or the ae a t tarted dependng on the exact nature of the procee The ae true of the total heat aocated wth the procee Fnally, the nternal energy not changed ONLY f the ntal and fnal teperature are exactly the ae E Lookng at the lght Medu, we ue Snell Law to ee that the ndex of refracton greater n Medu than n ether Medu or Snce the lght bend o everely away fro the noral n Medu, the ndex uch lower than t n Medu So, n < n < n The peed of lght n thee eda therefore expreed a v < v < v nce v c Th alo ean that the n wavelength are ordered a λ < λ < λ a the frequency of the lght the ae n all of the eda

5 5 A The contnuty equaton needed a the total flud ovng through Regon ut equal the flud through Regon Hence, we have A v A v Snce the area n regon te the area n regon (double the radu area π r ), then the flud peed greater n regon by a factor of 6 D Ung Krchhoff Loop Rule for the crcut, we have V V V v V R V v 5A Ω ( ) ( ) ( ) ( )( ) v B C R C C 7 The potental dfference acro the capactor -7 volt gvng the ze of the voltage a 7 volt 7 A Conventonal current n the left-hand crcut drected clockwe By a rght-hand rule, the agnetc feld nteror to the loop wth current therefore NTO the page Snce agnetc feld lne for cloed loop, outde the current-carryng loop, the feld ha a coponent drected OUT of the plane of the page Snce the retance n the left-hand crcut ncreang, there a decreae n the current and hence a decreae n the agnetc feld trength By Lenz Law n the rght-hand crcut, there an nduced current n the wre to generate a feld drected out of the plane of the page Th ean that there a counterclockwe current n the wre and the current fro B to A through the retor n the dagra B There are force actng on the a a gravtatonal force, the force fro the trng, and an electrc force (ee the FBD to the rght) Ung Newton Second Law n equlbru, we have r F gr r r r r r F F F F F That, the Str Elec u of the gravtatonal and electrc force equal and oppote to the force fro the trng By reovng the trng, there therefore a net force drected oppotely to the trng force Th gve the drecton of the contant acceleraton actng on the object and nce t tart fro ret, the a wll ove n a traght lne 9 D Effcency can be coputed a what you get dvded by what you pad for Here, we got the a to re to the top of the nclne We pad for t by applyng a force over oe dtance and puttng work nto the yte Wrtng th atheatcally, we have ( ) ( )( g d g kg ) Wout e n n Note, we took the heght Wn F h F 75 N d of the nclne to be d and the hypotenue (how far we appled a force) a h leadng ton h C Conervaton of echancal energy can be eployed here Th gve KE PE and we note that we have two for of knetc energy (tranlatonal and rotatonal) ncorporatng both, we wrte ( ) ( ) ( KEtr KErot PE v ) f ω f kx Notng fro the equaton heet that cyl MR, we can wrte, ung v rω, that v ( MR ) kx v v kx v k x f ω c 5 E Th a queton of te contant The crcut wth the allet te contant wll charge to 9% the fatet The te contant are RC, RC, RC, RC, and ½ RC for the crcut gven D When the object coe together, there conervaton of angular oentu (no net external torque) Hence, we wrte L L ( )ω where r leadng to ω ω gr f ω MR MR ( M)( R ) ω Elec Str ω ω

6 B Becaue the object placed outde the focu of the len, the age fored wll be real for th real object The agnfcaton, though, unknown f the object placed between the center of curvature and the focu, then the age larger f the object located at the center of curvature, then the age the ae ze a the object Fnally, placng the object outde of the center of curvature reult n a nfed age C The LC crcut ha charge ocllate a a ple haronc ocllator Conequently, t ¼ of a perod for energy to wtch fro the nductor to the capactor 5 A t wa Paul who conjectured the extence of the neutrno (Fer later worked out the theory of beta decay and naed the partcle) 6 D n order for the 6 electrc feld to copletely cancel, one can draw a unt crcle ( π rad) and dvde t equally nto 6 pece Dong o gve an angle of π for each pece Th the phae dfference for each of the feld o that when added together (n a phaor dagra), the total feld wll be zero Conder that co ( π ) co( π ) co( π ) co( π ) co( π ) co( 5π ) 7 E The te-changng electrc feld nduce a agnetc feld Snce the electrc feld out of the page and ncreang n te, then by the rght-hand rule, the te-changng feld act lke a current (a dplaceent current) and the agnetc feld aocated wth a current out of the plane of the page drected n a counterclockwe anner around the crcular regon At the locaton of the proton, there a agnetc force therefore drected nto the page (fnger pont to the rght, curl down the page, the rght thub pont nto the page) B A travelng wave ha the for f( kr± ω t) where the drecton of wave travel would be r For the feld gven, th ean that the wave travelng n the x drecton 9 C The drecton of energy flow coputed by the Poyntng Vector whch related to the cro product of the electrc and agnetc feld At the orgn at te t, the electrc feld ha value 6 ẑ We ut therefore olve the proble x ˆ zˆ?? The agnetc feld perpendcular to the electrc feld and fro the rule of cro product, zˆ ( yˆ ) xˆ The agnetc feld drected along y 5 D Ung clacal phyc, we copute KE v ( 9 ) v v c That, the electron ovng at te the peed of lght!!! Concluon we need relatvtc phyc to anwer th queton So, we approach t a ether KE γ o c γ 9 Solvng for Method #: ( ) ( )( )( ) γ yeld γ 97 γ 97 ( β ) β 979 β β v β o we can fnally wrte pγ v ( )( )( c) kg c Method #: Ung Enten equaton for energy, we can wrte ( c ) ( KE c ) p c c KE ( KE)( c ) c p c c E γ Splfyng th expreon gve u p KE c c KE o o p ( )( ) 9 kg