homework 7 - PH 132 SPRING 2005 HOMEWORK # 7 Assigned:...

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Unformatted text preview: PH 132 SPRING 2005 HOMEWORK # 7 Assigned: 02/21/05 C’k2?I7.;//2/32152/572 50) 3”??qu 2%}, r ________‘_,_..»— :.-—-—‘-—‘ CL-[I/‘b’ZZ? %/ 3" Cw? Law 20/9; twp; WP 50 SHEETS 22-! 44 200 SHEETS 22-141 _ 22-142 1005145275 5) AMPAEI { A car battery with a 12 V" emf and an internal resistance of L 0.040 (1 is being charged with a current of 50 A. (a) What is the potential difference across its terminals? (to) At what rate is energy being dissipated as thermal energy in the battery? (c) At what rate is electric energy being converted tc'chemical energy? (d) What are the answers to (a) and (b) when the battery is used to supply 50 A to the starter motor? 410' I Cjt‘vm: 5 -= {2v internal mustang r: 0.0452, t'= 50H End? a.) V ' b) m4? at thermal energy , dr'SfiPm‘xan ,% c) wife of ekdinkza/ energy twister, 93 a) mm) mm new I SWl'es 5N! Rtenfihi dtiqérence in caer hops 3 Power. 67/) Write a/oop egg/Mm: V: “E - (p Use the car/mt ai’z'e’cfi'm 7i) ad V: *5 .1}. :”/2v_ : 51””. M W wk! 6 Mimi 7%! CI/Wc’i” WW4 #16 cur/m6 .9 we k} _r. we WMEZ/ we wf/f {Mir #Ie £22497 act?! - céééreme} ya we W-mfifnam. “6,74%: Mk (iLdW/WK - ANS; ‘ _ ' raga}: Jég rift—WW 602% #276 car/mt ‘ . - ' E't‘r . QR = liar: (wit) 1/0. am, -4 Ida- av: (50am v)=ANs. W: : t: 4 32 this may be the case, or +14: 0227 Macaw mare.” C artifir i New 14% cameer WW fan fit 1%: Meraé?€cfi&t : 777:3 0'ka a flaw bop é’fmfiax: V= 6—5.» = /2v' ~ (0.o#@)(so,4) rims. Ph-‘i'zr = (b‘on)‘(a.o%r2;) rpm. 22-!41 50 SHEETS i} MMIJIMQ 21—142 100 SHEETS 22-144 200 SHEETS i poté rial at point Q? r The. @4th dfééreme flan Q 'I \ ‘,. _In Fig. 28—25. if the potential at poinI_F is 100 V, what is the 400p. Eula! I ' "£1 +L‘E, + +L‘R2 :0 \ [03: +22) = €524} '--=7 fnpxis aft/an: Up: / r2,= 3.on , 21: 2.042, EM: vq 5mm Kiroh‘hqfrg ms ,Pomra/ wreer \ . We WEN Had Me Chg/W! c‘. Cho cue H5 a??ea'>'2m as skew/2. (8; >52) (I: f; = [elf-R2, 24?? Sum mp 79% pcxbffi’ca/ a’r‘Fé’rmccs My '6?de am e 76w : VQ “VP an. ER; '5, Va = (Pang, + up = (20A)(2.o ) —- xsov How: ANS. fl Nb;sz ': (EH?!) 2 '1 “1 .__.‘ q / 3: I n. Eh. ._ '0 : 4 .07 A /.2x M '1”) A <—*. 3H“ :20 f2_, J 00V é=f50v,fz=50v 1 533. gram!— “3 +734- (F; +6? 6” 62‘ ((-1. U 2 -— gg+£2 Mm? £3 ‘— ( J a. J i ;+ £2 f; (“,4 {9+9 g "7’1 ‘ ( £19“? {XE ': 6px} 7‘ 5/13 ‘“ .— "'" EJYQ— *— H (Miami- % «ha i a r: df’y-ééi: 2’4) '1 22: g 72: a 9 (fig, : [20*06/4 \j, V: Um 63+ (13L) ‘ {2’ 2‘ 63+ {2 .3 P . i F me: :2, #005135 a: 50.2,, th75LQr , 5%.0v EM 61) REG, b)(’l ? L2) £5) {‘4 :3) in Fig, 28-31.what is the in each “5mm? PL“ h! f gfifiivéient esistance-of the net— 9’ R2 = R3 = 50 0‘ R4 — E I I'LL-'0, and% = 6.0 V; assume the fitter); is ideai. 50 SHEETS 22 142 we SHEETS 22 144 200 SHEETS V .30 In} 6:“ E 3* Sb x N In... 3 5% N 6‘ “a #i_.:J...J l - ' R; R5 kRuI —-)‘> ngq 22-141 6) . ammo. Now #78 Cr'rcou’z’ Ago/<1 We 1W5: + K and €257 arr r'n semis: Race : 21+ 2234- 'A'us' 63- I b .Juflcfi'on Rule. (pow c) l L', 3 £2 + £5 + L}! (0 Loop Rule: ace€= “d,P"L'zR2 +5:0 {2) : +62 R2 :0 (3) 5”“ ‘ “3 Rs +63! PL! :0 (q) ._ E : (moi _ ,2 LI " A gas. Uéa'na 6f. (2»): (2 F31: 5'51Q: i 1 I 50 S H EETS 22-142 100 SHEETS 22 144 200 SHEETS 2244] Combmmj (5) and (q):- “63R3 ’H‘qRq: . + “LIIRH :O . 5C3 RS 1" (IZRZ =0 s a: = cezma — . _ Ams. USE/fl Ct): . (“LIT Cl - 52-133 = firms.- 36* (M) 152, Off 6117 a (5E0?) r‘: F M F153 ’19 z + « 77:7 1% {(5142: ) a U0 “‘3 38E. A circuit containing five resistors connected to a balmy with a 12.0 V emf is shown in Fig. 28-38. What is the potential difference across the 5.0 fl resistor? . 50 SHEETS 22-1412 100 SHEETS 22-144 200 SHEETS FIGURE 28-38 Exercisa 38. E, 21441 {#13 C/wa‘ly There are man' efmm we tau/d punk 6r kw}? 731*: K from/10W '5 I? 16.5 , however keep In mfiw/ W! w! m: we We @0er “fir 1/; ,50 mate a bop efmfiim Had xbmgomyq 122.1009 1‘ mg; 01,0" 2:) ma cmmw More: mam mm M one MWMW 4'6” #34:)" A’O‘P' CL} :3 #26 Cat/rad ()1 Amp T) ’0') “MM” a—ga—gg=o Jr (2, may-=5 >22: 5 q...._-___- => 61: : 9323 6 __.—-—I '4' R2 Q2.on 52067;) 2 pm 3053?..- +5.06}, '..I \4; 4122 :( } _ i PH 132 Suggested Problems a. E 4.2, I I . #1 . A certain car battery with a 12 V emf has an ininal charge of I 120 A - h. Assuming that the potential across the terminals stays constant until the battery is completely discharged, for how long can it deliver energy at the rate of 100 W? 55m 7772':— BATD—é/E‘? bees NJ AMT. a}: “mag £600;ch To 77:15“ 5%? :7“ Dex—male; . 7w; Ram-AT WHJC/yl 77%, 3:5 ACLDMPLJ 9/453 :9 was mg Pbmee: F02. 77415 Erase} THE 338 l R97“. mFF. I; THE g "M E 9*)": (a) 0?: Tag cunt ; a... ' l :23 E Miran-f, ; “NH : Z i . E. T METE- T—HE 90532}; i M“ M {a scan 72am, 7‘63 z MOUEcHAP—GE ;; .t 2,... Am A 3 AT ’— ? 'PEDT. bth/T‘: 5:5 Aw : 343V C a gas“- pmg: wen we «~17? “it i. v’ - ’ g : , ' a 22. d “3 £1 a U6??? [2} T EL H. L Wf’ Cafe-79 {Lt-- 791/ CO: D’ng = ‘22 E M _ _ z .9 f , a All ' 6‘ L 3 E» /£C 3 . I 6‘7 2755‘? am ca [W ' Va, T); M; .: weed/kc “0044 H ’4) (‘3; js'mxv-r“ F6 - 6~/¢v W0 50 5 HE E'I'S 22-!42 100 SHEETS 21-144 200 SHEETS 22-]41 '69 1|an i .l_......._._. 20?. Wires A and B. having equal lengths of 40.0 m and equal . , diameters of 2.60 mm, are. connected in series. A potential .iifi‘er- 67‘ M ‘ j :1 5 (/0: 0M ence of 60.0 V is apgfied between the ends of the composite fire. A 3 The resistances of in: wires are 0.12? and 0.729 0 respeCEi-Tely -3 ‘ ' .1 = X Determine (a) the camera density in each wire and (b) the guerr- 0:0. (a /o M tial difference across each wire. (c) Identify the wire materials. See Table 27—1. 5 3 60. 0 l/ 1 R9 0.32m, ) 23:0.729‘19, a) 5!; 3 '33 b) V» 2 Vs 5))??44311’45 of A *8 arrw may; , om Law,_£e.s.:smg +he Curran! cabana) we Md Me carmf‘ ‘* E 3:: <27“ 5 = The wires an: I?) Sena) So Peg. = R5 fRs it; . L J: E); and dfi=cfg 35¢ $24-23} ‘_ ._5_ _ L' area" P = , t 9.5.4570») -. 3* ‘IB IVE-’5‘ w. Mam-w" 3“? ’" ’9’” 1 Eb) Ohm‘s Law =- VPr = (RR = {702009.12 7152,) = m. } U3: Ease (70A)(0.72¢252,)= ms. N075: Z = VA +V5 as H aha/d! r ' 'fliel "’},'}RJ/o.—€"’?=E_& (3.10%!)6 1%! Wkndé as)? (f {ensfiw er A /0 L {03* BEE)“: “(0.97%){2-6N0'3MZ’: KM X/b'ig-m LA ‘t’(‘{0m) Usfnj Rick 27%, p. (058’, we (an sec fia—l buff: )4 :2: ANS- . {93 3 Rs rT(0- 724 23.3(2. (o f/O"3nf\}2, ‘Z 628* /0““i$2:- M 3 L3 4/5/01“) - Wire. 5 :5 Am. ,5 _ In an RC series circuit, % = 12.0 V, R = 1.40 MB, and C = 1 1.80 p.13. (a) Calculate the time constant. (b) Find the maximum w charge that will appear on the capacitor during charging. (c) How long does it take for the charge to build up to 16.0 “C? E i ‘2 O V' E 5 LL“) HOE—[Z c, s 130 xto‘é F l l l 5 E I l 't F a I 50 SHIEETS 22 ‘42 100 SHEETS 224 44 200 SHEETS 12-14] ~33) '. hMF‘flLI {I \ l ' be five times that of the series combination. HR; = 100 Q. what J i i E E E l | i ! L..._.___,i___.__.__ . .. .._..._.. art/en: R! #00152, .J, a 5g=g 4BP. Two resistors R1 and K; may be connected either in series or in parallel mugs an ideal battery with emf ‘8. We desire the rate of electrical energy dissipation of the parallel combinaxicn to is R: '9‘ (Him: There are two answers.) ifiolwfibm fResfsiom in Sen'es and hi Para/M Time are iwa poasfble' Cmnccei‘ans : sane; and para/{cf I 525: PP ii .1: flRfl'RL) Rl‘rRz 21R: (RFRz): 3' 52‘121 785+ 22,21 *Rf 61211250 R12 “gelR14-R'L1ro 3 R21 —3R1RZ+RIZ _ Use €uadmfi'c Em:qu R1 = 512 i (320‘ “L! 212' -—J—._..____._._ 2 {22:58:51. 0} 290a, ANS. ...
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This note was uploaded on 11/04/2009 for the course PHYSICS PH13100 taught by Professor Dr.wick during the Spring '09 term at Clarkson University .

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homework 7 - PH 132 SPRING 2005 HOMEWORK # 7 Assigned:...

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