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Unformatted text preview: Friday. Apn‘l tt, 1003 PHYS 121 — Final Exam. 2003 Marks #1. Three point charges Q1 = 32.0 ttC ,
Qz = 103 [AC and Q; = 35.0 LLC at (x, y} = (3.00, 4.00} m are lccated as
shown in the ececmpanying diagram. 4 3) Calculate both the mgﬂ't‘ude land gm mint; ofthe electric ﬁeld at thaoﬁgin due to the three point charges.
J EE'J‘WO (5293“ch _. (£003?
nﬁ' 5 “g t.. 0 4103 In 3
:(,+_Gr1.l\) iEwizlgxneqz 7!‘15’ ,5 E
J.._—————p—""
admit VMJW e 4mg. 4.62 9,“:st (gar6") —5.001.LC cfmass 4.503(10'a tag is placed at the origin. Calculate both a
the original three 3 b) A point charge Q4 =
the Law'er and directign or" the electric fence on the point charge Q4 due to J POinghmges. I 3 '3 N
{Flaws}: (5m; )(1q_3rro_}__ qqm a '1:
ca '1 )3:
9:: Q +igﬂﬂ: f n._.II"""'_II
#—I'"'——
1'. c] Calculate the absolute electric potential at the origin due to the three point charges Q1, Q; L...
andQ3 Eat k 52: It .535} q ﬂ £~I£€J+E “.6
U9: “IiUlriUb 1‘ ?! f '3'? #3 :( 1 (p L V (J ; Qt03[?4(IE)+1I) .~ ﬂanme 1/ PHYS. 122 —Fi.nal Exam, nan Friday; April 11, tour 2 d) The point charge {2; is released at rest at the origin. Calculate the speed nt‘the point charge 1%
I an 'aﬁnit '31 fr the ' ' . v
Q” 1 m m "m mm 4, r (:2 at rxm ‘ dawn” APE 5 a
maﬁa aﬂ HOMO? has been removed to an inﬁnite distance from the origin. Then a point
charge Q; = 20.0 L“: is placed such that the net electric ﬁeld at the origin is zero. Calculate the distance of Q: from the origin. (q x i 0 r3} wx [ 04 3
‘Elatgag; mama  “a
5 (€xlﬂqlllf20el0h‘) I 4'" : l‘i.€x!03 ‘ 2 e) The point charge Qt #2. There is a uniform electric ﬁeld E = lS,llDD MC.
between the plates of a charged parallel plate capacitor.
The distance between the plates of the capacitor is
3.06 m. A proton is launched at time t=CLDU sec at 1.0:) m alcove the lower plate with an initial speed of
1.2. = inﬁlirtliil6 misec at an angle of53. l°ahone the
horizontal as 511.0th in the aecannpanpingr diagram. Neglectgravity.
2 a) Calculate the acceleration ofthc proton.
F  LE 'Lﬂ’L—"QWWQ {Swag}  Hair to"? W:
a": m' M ' LeaHG'HHS “' S ta
E7: Q 0’ .. .Lllt.£l0 I"4/5? E b) Calculate the maximum vertical height reached by the proton above its launch point. PHYS 122 — Final Exam. ZCIUZ Friday, April II, 2003 4 c) Calculate the x and y coordinates of the location ofrhe proton at time t=2.ﬂﬂxlﬂ“ sec. 2 (model’er €3.I°)rraoosre") 4 LEO»:
a]? Hot? 91.. 55.1")(2rto'g) ..(.I:Iu!c'9(¢.anrm
2. _. [fagt u j : uajt— age" =(I.SO B: "0H9M
(201) = LL90, —0.lr8)m #3. Five resistors are connected to an 18.0 'v‘ battery as 4 shown in the accompr diagram; Calculate the
equivalent resistance of the ﬁve resistors. .J_ .L
4' =g‘gJIszl [216?5 #4. A copper wire of resistance 3.5!] ﬂ and cross sectional
area A=2.00x 1 CI" tn2 is connected to a. 11.0: ‘it‘ battery. The resisﬁvity of copper is p = 1.1'lelﬂ'it ﬂm at 20111 “C. 2 :1) Calculate the power lost in the copper wire at 201:: “C. ?_ “2.
1?: 11R 5w 2 b) Calculate the length ofthe copper wire at 20.13 “C. _£
2=§pf 3 2 tree. 15W:th It E T ‘ LWKHD'E 13"“ PHYS 121 — F‘mel Exam, 2003 I5 I' #5. Cansider the accompanying electric circuit. a] (Stalemate the unknown currents II, [1 and 1;.
Use the currents with their assigned directions as shown in the diagram.
{53.1 b:
lCL 157 50 L @J 60 +rs—Isl'z 1 L2 (3;) *IS—IO ~55 “$1,150 LIJI 35952;; 15:50
".5174 : “:5 ﬂ I51; 353“ 15;;4'53—3 =6
5135 15+ v51; 3:4 = '7 5351:: 35 I
Ii: 5.— :‘1 .15: —5'=:la_
Iaiétriﬁéej
:3: 1355 " 153*
:(5—:&)_(.5+333 Egg13
I; imp1:9; 51.1 b} Catculate the voltage ﬁ'em point a to point I: in the
potential. (gwuﬁ b  C...  ﬁt. PM‘H') Vqﬂub : IC' 5[I') = —l.‘:: U “157% ’3 ,I 3,
'5':L I»? l 5J1
C. cireuit. Which point, a or b, is at the higher PHYS £12  Final Exam, 2003 Friday. April 11. 2003 #6. Two resistors R1 = 1000 H and Ba, = 3000 [l 5 5s
and a capacitor C = 100 LLF arc connecth to a ] 100 V hattcr’g.r as shovm in the diagram. The I m h, 
switches 53 and S: are both initially 0pm and R _.. fawn
the capacitor is initiain uncharged. At time i t= 0.00 soc: switch 51 is closed. 2 a) What is the charge on the capacitor immediately after time t = 0.00 scc. O (25:20) 1 b) What is the ciectric current passing through resistor R] immediately aﬁcr time t = 0.00 Sec. "' " {001.} [‘4
.1: : hug———'\ : 0!
[00052 2 c) Calculate the charge on [he capacitor at time t = {LOSE} sec. 1 é“? 1‘.. Ettat5
(its): G‘me ' t ._ 0. “‘1’
62(19): am “'3 %?_)§_m" W W  ~2 2' : _Qtu:oc: 0'0];
iii“0.05) . 510? (f i I 53 d05cll Li).
L d} What is thc clcctric current passing through rcsistor R. after switch 5. has bccn closed For a
very long time. i 0 ZERO 1/" Aﬁcr the capacitor can be considered to be fully charged, switch 5; is opened and then switch 52 is closed.
3 e] Calculate ths voltagc across resistor R, immediame aster switch 5: has been (:10sz
1 ‘ f
‘ tH csruucsmcttotz It
Pr 5 1'r M I00” : [El—#:00er
Limos _,_A lw : “1421 J
U:Iw~ﬂ~' ifsv i, z 2 PHYS 122 —Final Exam, 2003 Friday, April 1 l, 2003 2 t) Calculate the electric current passing through resistor R: when 0.100 sec have elapsed inee
th‘hSzhasbeeﬂdostd. .t. (E = Giulinz) :(lmimﬂ)(cm) r v 1a, 2p
Itt‘);lw€’ 0,90 L I(o,ee): Janna, @— “"’° = aim “,4 L: L15 I 1. g) Calculate the voltage across the capacitor alter a very long time.
or 0 ( ZEQO>
IL ‘
#7. Three inﬁnitely long straight paralch _ _
wires calr3.r electric contents Ii=4ﬂﬂ A, TI Fqﬁ Tit‘25 TIB 3A
t=2.oo Aand 13=a.oo a T5: 53
. . SJ 9
2 a} Calculate the magmtude of the net magnetic
a ﬁeld at the location of I; dueqto 11 and I} .1‘
a II _ “i if “9" Y _ _ i !
3 = ﬂ ‘ ~ LII”? r @ 1I+€—~—gm—+ rIn—.E
l  i JITJL mi ,2 {I}
"" la a; :I «.1
52 = if?! t in: ft? we? ' [E'er T. 6) a 2.11" (I) '6 Lilr10 T 5' J"
liql"
K
0I
&I
i1
I“
Fa
TL
0 r
‘11.
.I
H t h) What is the direction of the net magnetic ﬁeld? Circle one ofthe following: into the paper towards the right towards the top of the page out of the paper towards the left towards the bottom of the page 2 c) Calculate the magnitude of the net magnetic force per unit length on 1: due to I i and 13..
" 3
F : E)! L F u; u 12!? F
E= E) if : 62m? T)(E4): .2th0 a“: 0—.)
L M l .2 hard6 H E’
‘ ‘55 l d) What is the direction of the net magnetic force? Circle one of the following: into the paper towards the top of the page out ot'the paper towards the left towards the bottom of the page
7 4r PHYS 122. — Final Emilrt= 1003 Friday, April 11. 2003 #3. A single loop of r.irire of side lengths 0.100 m and 0.200 m moves at constant speed it into a
region of uniform magnetic ﬁeld B = 0.200 T that points out of the page. The resistance of the loop is 0.00 51. At time I = 0.00 see, the loop is just about to enter the region of uniform magnetic ﬁeld.
and at time t = 0.250 see, the loop is halfway into the region of unifom‘: magnetic ﬁeld. ‘ 120.00 sec 1"" Digger: l a] What is the direction of the induced current in the loop of wire. llilircle one ofthe following
possible answers. counterclockwise 2 b] Calculate the “ha“Se in Ina30303 ﬂux Ihfﬂugh the loop from 0.00 sec to 0.250 sec.
Mllzeiez— of, _ on?» f7.({9.i/)(0J) c. 0
0 : QKl0‘3 3007* 00“: 2 0) Calculate the magnitude of the induced voltage in the loop From 0.00 sec to 0.250 sec. ~3 lab _
5; £53, 2"”? . nglOsl/
of 0.2505 ' 2 0) Calculate the induced current in the loop from 0.00 sec to 0.250 sec.
.. '5 L; __
I: :5 .u 3% . Lot ro x4
‘3 2? r2
2 e} Calculate the speed it ot'the loop.
63—4": : {j— 05' g '5 E) L U :5 U
M e 9’ 0 M
v __ _._...— .‘P ____________.i all“ 0.255 : 0'9 W5 CIGQM/f 8/? PHYS 122 — Final Exam, 2003 thy', April 11. 2003 #9. A cube of copper of Side length 10.0 cm sits at . I: the bottom of a tank of water as in the accompanying ‘ﬂh
diagram. The density ofoopper is same L:ng and the 7
density of water is 1000 kgfma. 1 :3) Draw a ﬁne body diagram of the copper cube. I Cu l L}
F” mg
2 b) Calculate the net force exerted on the cube of copper.
IU
253 0.
F ra “‘5
1 c) A cube of wood of side length 10.0' cm ﬂoats at the surface of' the tank of water, The density of
the wood is 40:] 143/1113. Draw a free body diagram of the cube of wood. U _ 5
FE MT Fb= “43: FM if 2 d) Calculate the depth D of the bottom of the wooden euhe below the surface oftht: water. PHYS ll? — Final Exam, ill['13 Friday. April I], 10133 #113. A Ushaped tube ufunifonn cross
sectinnal area. initially contains only water
as sham in the accnmpanying diagram.
Oil is then poured into thc right—hand side of the lube and the 011 ﬂoats as a 2.00 m
tall column above the water. The height ofthe water in the: left hand side nfthe
tube is h = 4.00 m. The density of water
is p = 100:) 1:3me and the density :1me
oil is p = 300 @121? 2 a] Calculate the pressure in the water at paint A.
.. := 5 . a: fmojbn‘ lﬂméf/Nxﬁfﬁ/f“ q'ﬂ‘ $2 2 b) What is the pressme in the water at point B. ‘llu: SM M ‘1‘“ AF 2 (1) Calculate 11:13 height ha 0f the water in the right hand side of ﬂu?» tube. ' 3 EH30 (9K) = 539,; +2
995:30 “' X31430 ~= 230ml. 1E! PHYS 121 — Final Exam, 2003 Friday. Ami] ll, 2003 #11. Water is ﬂowing through a pipe efvarying Va Qua. ._
Cross section and elevation as shown in the
aeeempanying diagram. Water is flowing at
paint A in the pipe at a speed :3va =15.0 mi’s.
The diameter ofthe pipe at paint A is 0.600 em. WI
The water emerges inte the air at paint B at a
speed v“. The pipe at point B has a diameter {if 0.500 cm and is located 15.0 tn below point A. _
The pressure at point B is 1.00 athISpl'leTe. B 2 3) Calculate the speed v.3 ef the water at peinl B. 0,5 = U A s n. a e
a Vail» '30 ‘ﬁ/m'em’) — ﬂight/5 u: . r
Path 2 b) Calculate the pressure in the water at point A. a
t '5 5’ VB: 0; +€3ht+£€hi = Pytﬂﬂ
' L:
Pa = 'Pe " ‘3'(“e‘hn) tiffuez'uﬂvr' 205110 Pm” /L/ LDI nor 4 t Lamar) + i. an a no 5'
4 #12. A marble is located 2.00 m heiew the surface of a pool I i...
ofwater. It is viewed by an observer at an angle 0f35.0° te q
the vertin as shown in the diagram. Calculate the apparent
depth :1 measured vertically of the marble below the Surface of the water. The index efreﬁaetien efwater is n = 1.33
and. the index {if reﬂection of air is r1 r 1.00. .0
lat = I ﬁguresﬂ
1.1—: ' ‘5'
5.: laws? 2:: =4m25.5s‘ PHYS [22 —Final Exam. 2003 Friday, April I l. 2003 #IS. A small toy bee: labeled 0 of height 2.00 em is placed 3.00 cm to the leﬁ of a thin converging
15. The focal length of the converging lens is E = 4.00 cm. A thin diverging lens is located 12.0 cm to the right of the converging lens. The focal length of the diverging lens is f: = 4.00 cm. The points
labeled F1 and F; in the diagtam are the focal points of the thin lenses. 2 3) Draw a my traoe in the diagram abete to ﬁnd the position of the image fanned by the converging
lens. Label1th. 2 1]) Calculate the pesitien of the image It formed by the converging lens. +r We
_ J. _ J. = J— . J.—
‘é’r ' .f' ft (1.00 3.00 I sir 3.005» 2 0) Calculate the magniﬁcation of the image I. produced by the converging lens. Ml.. _i—I=' Fig—= "'"i P. 5’ == PHYS 111  Final Exam, 2903 Friday, April 11,2131} 2 d) Draw a ray trace in the diagram above to ﬁnd the positien of the image formed by the diverging
lens. Label it 1;. 2 e) Calculate the pusltien of the image I; with respect to the diverging lens.
P2... lz'go— 9.09; ‘1.a0 cw .L..J——J—. J—— —_E.
q; ' {a P2 fame) 4m: 2 1‘) Calculate the height efthe image [2. M3: 115% “(433) inﬁlas Pa mane?
HM = MI “43: (—l):(+0_33)~— — 035
cﬂééaq he : MML :4 h: (2.00)“ «(—51. ave7
mg, 3 g] Circle all of the correct ehareeteristiee ofthe ﬁnal image I; in the following list:
E hug—Elna in front ofthe diverging lens 3 image in back of the diverging lens 4: real image virtual image ...
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This note was uploaded on 04/18/2008 for the course PHYSICS 123 taught by Professor Wrzesneski during the Winter '08 term at The University of British Columbia.
 Winter '08
 Wrzesneski
 Physics

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