Exam3Solns - th We/5‘:‘”‘“"“" hEach/2...

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Unformatted text preview: th We /5‘:‘”‘“"“" hEach/2. Physics 132 Exam 3 Va («3,5 W15 Name v12 {Sfc m 1. (10 pts) A prism sits on a lab bench. The prism is made of a new, translucent patented material (index of refraction 2.15). A He—Ne laser (633 nm) is directed into the prism, as shown. (The laser beam is directed parallel to the lab bench surface.) 30" - 60" - 90° Triangle a) Draw the resulting path(s) of the laser beam. Label the angles 9i and Or on the diagram above (relative to the second interface, i.e. the hypotenuse of the triangle). b) Calculate 61-, the refraction angle of the laser beam (relative to the second interface, i.e. the hypotenuse of the triangle). Give the value of the resulting angle of the laser beam and show it in the diagram above. (You can re—draw the figure if desired.) Clearly label only o_ne resulting path of the beam. No conflicting answers. SAM/5 Law. ”15"“91 :— i’i7’.‘3ir\€'}a (Friean [m> (32.4%) Sig/0° '5— \.oo sin 6r ‘/ Z. (Q.i§)(yg) ‘= 6}” ear {Er Ewes/Mo+ Ekisr (Ix/v.5) The. beam ulces n01” NfYM/t lh‘l’u Hm. my] feel-mob it W‘U‘jm 72W Inland/{zflectv'on UFIR) riffith one of HNL 51¢ 5 ._ VHS”! 6 RS 1. (10 pts) A prism sits on a lab bench. The prism is made of a new, translucent patented material (index of refraction 1.96). A He-Ne laser (633 nm) is directed into the prism, as shown. (The laser beam is directed parallel to the lab bench surface.) Physics 132 Exam 3 Name 305 - 60" ~ 90° Triangle "a. “1,31 n 2 I LOO (awe/LB /ncrrvm€/ ‘61; ‘3 ? N: JZLZIf fi‘ :mzlzi 7L!) bead away 90m "Hm hernia/Z Q» s 30 a) Draw the resulting path(s) bf the laser beam. Label the angles 6i and 9r on the diagram above (relative to the second interface, i.e. the hypotenuse of the triangle). b) Calculate Or, the refraction angle of the laser beam (relative to the second interface, i.e. the hypotenuse of the triangle). Give the value of the resulting angle of the laser beam and show it in the diagram above. (You can re-draw the figure if desired.) Clearly label only m resulting path of the beam. No conflicting answers. Small/5 Law... Vl' st-i : ml arm/92 Griffith SO Lu T! OA/ 5Q, l 41 2, 00551634 Physics 132 Exam 3 Name 2. (10 pts) The following ingredients are mixed in an insulating styrofoam container: 0 10.0 g of aluminum (CA1 = 900 J/(kgK)) at 200 °C 0 20.0 g of copper (CCu = 385 J/(kgK)) at an unknown temperature 5‘ T / Ce 1 a) o 5.00 g of ice with an initial temperature of -10.0 °C (Cice = 2050 “(1(ng Cwater = 4180 J/(kgK)) (Lflwater = 3.34x105 J/kg, Lvmter =2.27x106 J/kg) . 19.5 g of ethyl alcohol (CEA = 2400 J/(kgK)) at 20.0 °c The temperature of all components quickly comes to 15.0 °C. A) Write a single calorimetry equation that relates the amount of heat energy that was added (Q) to each of the components in the container. _ fl __, ~\ 3 2— C’lc; ' C (a B) Solve for the initial temperature of the copper. /\‘ ‘ \ h I ,, ' . H" ,_ i‘ l I ‘\ \c a l c e WéfW A‘l’ LUG5 {N7 ‘ICQ/ «bw [\SCC ’2006 l/‘AMAC (bk ('9 C — {6/ + filmy/CM; (w\ MW / M- _/ Jr Mme. Lew-M ‘1” MM CMlmfi3 ~08 1 MBA CE: j}; :1,“ / ma? WM, Henlw’clnder’jlmcl km; (M5 Unfigamrjyj 4- r“ . C’ _ - >26 “7 Set/c177 OA/ J, Physics 132 Exam 3 v 6151 u h Name / 2. (10 pts) The following ingredients are mixed in an insulating styrofoam container: 0 10.0 g of aluminum (C A1 = 900 J/(kgK)) at 200 °C 0 20.0 g of copper (CCu = 385 J/(kgK)) at an unknown temperature _. l o (m C919 “*9 o 5.00 g of ice with an initial temperature of -10.0 °C (Cice = 2050 J/(kgK), Cwater = 4180 J/(kgKD (Lfiwater = 334x105 J/kg, Lv,water =2.27x106 J/kg) o 19.5 g of ethyl alcohol (CE A = 2400 J/(kgK)) at 20.0 °C 11 The temperature of all components quickly comes to 16.0 °C. A) Write a single calorimetry equation that relates the amount of heat energy that was added (Q) to each of the components in the contalner. Z CPL :_ O @ B) Solve for the initial temperature of the copper. 62A]. '1’ QCM + (3.;an + QM + Qwam + QEA : O ‘7 var , 3 ca zca mt wt“ was ANX \. 102/ >1 ‘0 W 0 + M CC¢(H,QC" T) +M1C€.C Icz (00“;001 flk\7 MAKCA IUD'C 200 (M + MlC€ L'uc New + MICQ Cwaw (“DOM 0°C + MEA CEA {H0 20 :0 ___________ /-—-—’—""‘ inf M be, +le Sum m0 afl a? flw. 01W undulinwl w+erm5 (has un'flrs flaw/’73 l M‘ 5 (011W900;@( (—lfiLlfié Jr [MODS/kghwo (ID/g) + (ooefléwxw 31> 4» (009%)(HlaoipwigJ, WK0195&3)(2I400§\<_ 9% 4090 71—107. 53 + twoo’ + 33HHS+ ( 18? ’13135: 20,qu (“fa fl: 0 m +620 @(385 JAM-T» —jVLI : (oLkB(%9/« KBOU TB: ’2037 3 Mr- NL' ’1' /V\cMC Griffith C oppgr ...
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