prac exam 4 - General Physics 11 - Spring 2001 Practice...

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Unformatted text preview: General Physics 11 - Spring 2001 Practice Exam 11 Solutions file:///l-l|/My Web Files/teach/teach/ng-pceZ-sol-sOl.htm GENERAL PHYSICS 11 - PRACTICE EXAM 11 SPRING QUARTER 2001 - SOLUTIONS 6: 3. A, A wire of length 2.35 m and diameter of 1.63 mm carries a current of 1.24 At The wire dissipates thermal cnergy at a rate of 48.5 mW. What is the resistance of the wire? What is the resistivity of the wire? F‘L'LR —3 P, = yam/0 w : /57,03/5‘JI___ @<i rt; /»’ - 1" a p v f A ”[/./:}xlvi)z[p,0315).n_ M fiat-SLR : @D W ? = -B:g’= L ' 2.35;, h”_t_..____ A lof5 5/17/20014111PM General Physics 11 - Spring 2001 Practice Exam 11 Solutions file:///H|/My Web Files/teach/teach/ng-pce2-sol-s0l.htm 2A. cylindrical mpucilor consistta of two thin. COZIXlLU metal shells, each 3 meters long. The inner shell is at. radius 4 cm and the outer shell is at ratlim 5 cm. The region between the shells is it VQCUUIHV a) What is the capacitance ofthis system. b) If the two shells are connected acres}; the terminals of at 20 volt battery. how much energy is stored in the capacitor? ilk « xi, LL I A —.->~ 7:; Q . o ‘ ‘0) fDYQ/éo ', ECAZTTFJ‘ /69, E03): mil 7. ' AV= Lad/Q s Q EL... r: 0“! arrival F ~ _ 3. - CV" Q £3 C" 15v 19 1 0R “ “From ¥eovn I c g ‘3» :- 211’ GO ' 2M - /n ("/00 = t \ 2 “'1 fix Iow'm/c‘ 85": 1 mu+ gr cleriuocknvx J {'F time more. «For carrw+ numilfplax re..w~l+, t cvxt‘b’ 1? “Murllkd "L. _. Q“ U = ‘3: CV ‘ '1; (‘1.‘16X10 "’F W. ‘-l P413 3/9 43" V1 CV ‘lH fir WUMLSET‘A /‘m9"$ \/ 20f5 5/17/20014IIIPM l. 3of5 iPhysics II - Spring 2001 Practice Exam II Solutions 23. C alculute the equivalent capacitance of the circuit shown below. What is the charge on the 12.0 M7 capacitor? What is the voltage across the E60 uF capacitor? C‘i: C‘ *‘ C1. ‘3 (8 +133/MF ‘7 (in Pu!‘\\\{\3 CA.“ :. L + .\_ ,. l ‘ 3 ‘13 Cu. C - (—- V }: -\- 3: :- : w '3 22a? 1);: u RAB/c: 2;“;— => cm = 8/5 Cum Viz; ~ 0111 - 03 (£9?an =1 ‘if/aC. 1 C3 V = Q: \/3 : mc: 81/ IZ/HF file:///H]/My Web Files/teach/teach/ng-pce2-sol-sOl .htm 5/17/2001 4:11 PM J General Physics 11 - Spring 2001 Practice Exam 11 Solutions 4of5 3. A. In the circuit diagram shown below, R, = 39. R: = 651R1 = 40, RJ 2 25‘), R5 = 20, §, = 16 V and £2, cg: = 16V, Both batteries are ideal. . v . t h / (3) Calculate the currents 1,, 11. and 1,; as shown m the Circuit diagram. ((1‘) (‘me‘jfla 2,5393. ; «3/ @ ‘7. {K 3‘31. 3.1. K)? 3/. (b) Calculate the power dissipated in resistor R“ E‘: f» : fit: 1 2‘ “a” / I r :k ‘ @ Afflyfllnf PINE Jrh‘f lei‘T [08}? f, ‘2‘3"; ’L,€tz.:0 M" ‘ ~z; =0 . . y” fable, 1"” “57” "”"F‘ it); for? _ f (3; 51 “‘3R3”A¢i@fl”fis)=0 /e—qzbiy;;=op an) \ 5L _ - _ t) 11 . . ital-Trail" fife-r Rytfi' 1m?» r9!”- 0? F, b Lgfwz 5:5”; ’22. 4» tuba? J V 22’) = ill”; \ 0r "b: yfis‘ in -Z‘ /"‘: 21C)... 9 ‘ ‘ -Z‘+‘:3L \i Y.” (2g firm emf rule, 1—3 = MN?“ J \3 t“ 1H: z 3 43H,” {eight-loaf 67572 /(o-443-9~2.'0 r)“ .7‘ at {lg/31% {3.0. “’“ilb’fi‘ ‘3 :0 . ,rs 3LT ‘9 3.. "7 /g _ 17’ ( w} F) 7" " k} - “#3:”, x i“, . Z ‘ (a fi/ it. = IA - .1132»; My»:- 3"*2' 3A.- t WM .. J l : M ~- ~ ‘ 31 B. The circuit shown in theg'diagram is made using identical light bulbs and an ideal battery. The greater the current through a light bulb the greater the brightness of the bulb. For each of the following statements circle the correct answer. “M A“ curtail" in kid'- 2:) Bulb A compared to bulb B is. brier dimmer, the same brightness. Egan-k I’miA—m b) Bulb A compared to bulb D is n ‘1' m dim r the same brightngs‘gffw- a; ’5 c) Bulb B compared to Bulb D is: brighter, the same brighmess_ ‘ cs. @ in 1/, So here . ‘ coy»;th Tiara Bulb B Is unscrewed leaving the branch aiming B open (disconnected). ‘ A (1) Does bulb A become: brighter,@ stay the same brightness V, é‘c'zg—Swi—g-v; 32mm» 2... ’ 6v? Vacrass B 35 225:0) <7}, Since. g6, .26. aw; at w WM 59 finale-Y £35; 15 Vt‘rV’se 1 v m kg? Burma; 3:) )is 5, EMMW ._t .. {tgrri’V‘T' W-.V‘ 5/17/20014:11 PM file:///H|/My Web Files/teach/teach/ng-pceZ-so1-sO 1 .htm /_ General Pfiysics II - Spring 2001 Practice Exam II Solutions 50f5 (b) (0) (cl) KEle The figure below shows the essentials of a mass spectrometer which can be used to measure the-miss of an ion. The ion whose mass is to be measured is produced in the source S, The initially stationary ion is accelerated by the electric field due to a potential difference V. The ion leaves S and enters a separator chamber in which a uniform magnetic field B is perpendicular to the path of the ion. The ion strikes a photographic plate at a distance x from the entry slit. Let B = 75 mT , V = 1200 V , the charge of the ions q = 1.6022 x 10‘” C . and let x = 12654 in. Derive an expression in terms of the potential difference, the charge of the ion, the magnetic field, and the mass of the ion that would allow you to find the velocity of the ion as it entered the separator chamber. Derive an expression that would allow you to find the radius of curvature of the semicircle the ion moves on in the separator chamber in terms of the potential difference, the charge of the ion, the magnetic field , and the mass of the ion. From parts a) and b). derive an expression that would allow you to find the mass of the ion. What are the velocity, the radius of curvature. and the mass of the ion based on the data given? = (\tS‘ytolcl (SOG‘FWKW 20‘ (:5 MA. M («t Jews to“ KC? 54 to" 33 file:///l-I|/My Web Files/teach/teach/gp2-pce2-sol-SO1.htm 5/17/2001 4:11 PM / ...
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This note was uploaded on 10/12/2009 for the course PHYS physics 2 taught by Professor Kogan during the Spring '08 term at University of Cincinnati.

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prac exam 4 - General Physics 11 - Spring 2001 Practice...

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