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PHYS 153 2007 December Final Solutions

PHYS 153 2007 December Final Solutions - First Letter of...

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Unformatted text preview: First Letter of Last Name: l:l The University of British Columbia December Examination — December T, EDD? Physics 153 Elements of Physics TII'VIE: 2.5” hours CANDEATE‘S NAME: Last Name First Name or Initials SIGNATURE: STUDENT NUMBER: THIS EXAMINATION lETJOI‘NISISTS O 8 QUESTIONS AND TOTAL OF 9 PAGES. CHECK TO ENSURE THAT THIS PAPER. IS COMPLETE. WSTRUCTDR‘S NAME: ——________ _ _ _ _ .. SECTION NUNEBER: {a} Each candidate should he prepared to produce. upon request, his;r her library ,3 AMS card. {1;} Read and observe the following additional rules: No candidate shall be permitted to enter the examination room after the expiration of one—half hour, or leave during the first half-hour of the examination. Candidates are not permitted to ask questions of the invigilators, except in cases of supposed errors or ambiguities in mination questions. CAUTION: Candidates guilty of any of the following, or similar, dishonest practices shall he immediately dismissed from the examination and shall be liable to disciplinary action. {a} Making use of any books, memoranda, programmable calculators, audio or video cassette players or other memory aid devices { ie PDAs or cell phones 3 not specifically authorised by the examiners. Eb) Speaking or: communicating with other candidates. Each lQuestion is worth a total of ID points: — 3 4 ___ 5 E5 _ 'F ti lfl 1!] ID 1D 1D 1D 1D 1D Total Mark out of ED. Total: ED 1. Cine string of a cartain musical instrument is ELK} cm long and has a mass nf 8.75g. It is being plagued-in a mum where the: Speed of suund i5 344- mg’s. To what tension must you adjust. the string an that, when vibrating in its thini haxmnnic (amend overtone}. it. Name: —_—. Student Number: —__ I l produces sound of “valenghh 3.35 cm? ‘ | | F3 2' F 5 fit If} 9 8_j§'3€./a )Lfil’fi'd E) = 507:.3'11‘3 8 (Ur ( a 02;" = Myrna" Name: _— Student Number: 2. A not with a. steel bottom 8.50 mm thick nets on a hot stove. The area. of the bottom of the pot is [1150 mg. The water inside the pot is at 1UD°C . and [1.3913 kg of water is evaporated every 3.12M} min. Find the temperature of the lower surface of the pot1 which is in content with the stove. {Assume all the heat transferred goes. into the water.) Useful constants (you may not need all of the constants} icmfi; = 53.2meff, fist-m = LeWImK, Lflmm) = 334x m3 Jflty, Lutwm} = 2255)»: 103 Jog, sum = dlgflJfkgK", “ice 2 EIUUJXREK Name: __—_ Student Number: 3. Two loudepeakEre ere planed 4.12} m epext on a stage. The epeekere generate identical sound waves when tested in e renge of 2i] He he Efliflflfl Hz. A person eite directlyr in front of one speaker at e dietenee of 1E}.flfl no. The velocity of sound ie 344 mfe. e} What in the loweet frequeney for whieh the pereon will hear a mem'mum intensity? in} What are the two highest {requeneies for which the person will hear a minimum inten- eity'? _I'.‘ T‘ “a 4: aha-v ‘ - E {J ___ _ _ __ _ _ +35%. K (Lia L-‘h— —-—-_ Ah“... d't r37 'F"; r ‘ - :1 :2 i. if ' “ Lip} with»?! do! _ n! ?'r’ Free“! of: r€flf Win. it; ”if; 3131‘ - w - ungai-EESHe K 3M Limo? F— qux5+ 5E é— Zfi fir; H33: 3’33"“ 5 59 5’3 223-35 [4 ‘ .F s 59.5.” .5: ”(F : 8w 223:3er jig—fl Name: — Student Number: 4. An electric: power plant is designed to produce 4 x 1012 W operatingent 3fl% amnion-o}: n] At what rate does the power plant dunrp heat into the nearby cooling lake? 1)} In “enter, the surface area. of the frozen lake is 4 .trm2 and the air is at -5”[3‘. Estimate how long would it take. to Ineit $1.1 orn layer of ioe. if the ice layer is 11 cm thick to start with? {Assume all of the heat dumped from the power plant goes into melting the ice.) o) Is this power plant good for the environment? [Just write a short answer.) Useful constants {you may not need {-1.11 of the constants] $51.33; = 50.2meK, kids = lfiWfimIfl Lflwaterj "—" 334K 103 Ling, Ly'fiwum} = flflofix‘ 133 Jfkg, amm- = elflfljflcgfi. em = 2 l'EJUJfl-igff E n W FW' {GM = @«I'W * 2a: 5 41cm”, 4H5: 157.; _.. L on? F ’2’ chem“ < 3.3) _ $133M: W. :MJW L“ do.) #3?“be 3 Tc? ~_ E awgfl‘ C J o 5' 5' 75;; fl 7.’ ii :Kf‘r TIL) .1 59‘4“” “"_____ M “5 Lil-f (Life— : aawscmfiw. a H ML? 'fl ‘t‘fin ”fix 3344: £0; = fr3§xf§%_r. _,_.— ‘2 at“, Mex- (rim/,9”? — (4’; id: FL" rue/£45; $56 “EJHm'LT ft: 72%:— L42 . T I. has; 5 4;: $414 alflx r01 : 3:65554’3} L? x QIQH: H"— Li?- 7" 547361 {fluff}; agfi-fflrfl {3' , é?!» flier Gm): :Efifij; = Erica-1.3 LC) win} M/WM 49.77%“ ngfifia Wig: meoyflw “flaw jig}; Moi W @311de hw‘ ® maczoio Dc 53:5 5353 4cm H Eume 9” 5m 683023 5 H $93 .90 Em :2 A3 macaw 333? $839 333m h v u WPESWAEV u 3mg. :Ifi... mfilmb rile. a. “a :3 58333 3833 m. 30.35 macaw. : T 3.21va HWQFEN. T5 Wilma a: n: E .25 cm.“ 3833 m 3035 8839. x... u 3:; v "SEAEV n gag. fl w: @ maize: almsmaimfifimm 8ng m E Rub N w a w m m a io n a «a A V m r, T R? w 33%? mm. Kfin: Ni E m “Ea a 1. u |, u|||lurm3|m v Ah v3 Ahwvnn ANHHO1JQHHHoEV H . L. L L a _-II I. I.J, Axum-“VH1...“ “Fianwxxkfl ..1 na.nu.Wamanwfx.%be h EW.%\._A..___._D \(NJHL IIU. #Nlfithx \nfivfl an “Tax an. n fifimém w u N .-- v IIIIIIII r x m. in F .- mm C bum, wax - Kw]; n. \F flLV ATE an km?“ .r‘l. 5 b, 0.33% 3 DH Ea am 95 .B EEBB man .535: mamas. b m3» mica Amy 3 EV Um: acm .5 63.0: .m :w x 3a. 3 Era. um: Em Ema E . Aflv 95. 8 5353 83330: .8 cm a. .m x Ea. legs; u squib AN +1 I H magi” +| Iw if A EL HE ”IMFWQO. a Hgiolm Name; __—__ Student Number: T. Assume that yonere examining the characteristics of the suspension system (info lfiflfllcg automobile. The-suspension “sags” 15cm Twhen the 1weight of the entire automobile is placed on it. In addition? the amplitude of oscillation decreases by 53% during one complete oscillation. Find the values of l: and E} for the springend shock absorber system. "x W @W‘ seas. jaw“? {5'31} 9: “3:3 1‘ £3 ii. on” k: fflurmé I firfifffii-4I%fl_ _________.._.- Grfir ‘_,__._-————'-'_—-'-'. _iéfm the a ’4’ _M i __ 3-h- fbr E 5 ng) NH re T: 34F:,§_T = ZTflrig CW“ h? ‘ ‘72:? W T7,: 111' LEE—4 F GEL casein eiéw' .._, 2,, to $53215»: bi”; £2.“ 2— :“?i’; airing? ‘9”? E: 4"“ bafZééélo 2; Hr— F?” m. AW mmU main: a mi; 3853 Amv wnmnfinaobaxfi aw uwnguw b a- T?” fi$¥La€ v :3 r | Anti ”no 35:552.. who find». wdflu . . {n1} ”Him. wdm. mo 31nd; + Pam m. man ._.n._ be m. dm n H .n 3.3. OEmfl :mm 333 Eémfl woman: man. mm o: :m m Emw 3% cu. in. r‘wx mwmzmsm m: {HP FAR. w. H In E: Ba 3 u u b 35 mi ¢ Jrg g @ Foam m :03 353.2 4.; 9..-? l. #3 ma 3 H3 ”can. 41 H. a H In mimflavaomm n $.9ng An: Em Em;~ :3. any 3.18 D: Em mange .anamm mwmczmzoza 33m EHEWIWQ msnxnlabmoa W wufibmfie. QSMISEEEESE”3%. L. Q? hi £21. $2... $3 . \M ...
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