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Unformatted text preview: PHYS litil — Final Exam. "£002 NAME: i493 .
cat«TEE: STUDENT NUMBER: OKANAGAN UNIVERSITY COLLEGE
PHYSICS DEPARTMENT
PHYSICS 122
F INAL EXAB‘IINATION
Wednesday, 11' April, 2002
TIME: 9:00 am.  12:00 p.m. Instrucmrs: Chris Melin. D. O'Brien, T. Wrzesniewslti mm: 
Calculators and a two+sideri formula sheet can be used. Computers and QWERTY board calculators are not permitted. Answer the examination questions in the space provided, or clearly indicate the location of your solution.
You can use the backs of pages for calculations if necessary. You must clearly show your method to receive
ﬁill credit. if you cannot solve a port of a question and the resiilt of that question is needed to do the
remainder of that question, you mag.r invent an answer for that question part, and clearly State that the answer
is invented, and proceed, with the remainder of that question. The pages in this examination are numbered from 1 to 15, including this title page.
This examination is our of [$0 points. ’ Note to jgvigilator: This examination does not require examination booklets. QM: k e seems” smite: e, = assets“ citith qimn = 1.602(10'19 C nu=4rtxllTTTran g=9.31m’sec= mesmﬁ 9.11::10‘“ ks
Po= LIIr'litllll1 Pa “Imam: 1_6?x10v21kg I" J’ c. e. b D T T C c 'T' i"
thiﬂnl‘ Elia” #9 #1" all ﬁles possible a . l
n 1
Grade obtained PHYS [22  Final Exam, 2002 Marks
Part I: Multiple choice questions Circle the correct answer for each queSIion. There is only one correct answer in each case, A blank
answer is a wrong answer. 1 1. If the distance between two point charges is tripled. the mutual Electric fDrCB bBtWCEﬂ the two
point charges will be changed by what factor".I a. 9.0
b. 3.0
c. 0.33 @to I E. An electron enters an electric ﬁeld with a velocity 1. . With respect to the ﬁeld, the electron
experiences a force acting: a. parallel to the electric ﬁeld and pointing in the same direction as the electric ﬁeld
@lparallel to the electric ﬁeld and pointing in the direction opposite to the electric ﬁeld c. perpendicular to the electric ﬁeld. cl. always in the direction of electron's velocity. 1 3. Consider the electric field between the two parallel plates of a charged capacitor. Which of the
following statements is correct?
a. the electric field is strongest near the posnwe plate b. the electric ﬁeld is strongest near the negative plate
c. the electric ﬁeld is strongest miditray.r between the two plate:
@the electric ﬁeld is constant throughout all of the space between the two plates 4. The direction of the magnetic force on a current: carrying wire located in an external magnetic
field is: h—t a. perpendicular to the current b2 perpendicular to the ﬁeld EDDﬁlth answers [a] and (b) are correct
. neither (a) nor (b) are correct 1 5. The path of a charged particle moving parallel to a uniform magnetic field will be: g} Straight line . a circle
c. an ellipse
d. a parabola It.) PHYS 122  Final Ham. 2001 Part1]: Solve the following Emblems #1. The point charges Q1 = 2.00 ttC, Q; = 6.00 tit:
and Q; = 4.00 ac are located as shown in the accom
panying diagram. 4 a} Calculate both the magnitude and direction of the
electric ﬁeld at the origin due to the three point charges. El: (F1 :0) = {H.5'al03l D J “III: ‘2 k: i
5 [0: Fa‘) : ( D I 13.53;“.95) U/C PE? 3 h] A point charge Q4 = S.{}D tilt: is placed at the origin. Calculate both the Enamel; and direction
of the electric force on the point charge (2.; due to the original three point charges. _. .— 5 _,
Fur g I? : {— 5.00:: I0 6,43,] sigh/1'0 y; a} ’22:.) o
.. / Wrath
j lJ e} The point charge (1.. = 6.00 w: is removed to an inﬁnite distance from the origin. Calculate the
absolute electric potential at the origin due to the three original point charges Q1, Q: and Q3. kﬂl Rat; ICQ3_ It“.
Vo=u,+uy”3" 3r“ 1' '1'" new a) PHYS 121  Final Exam. 200?. 2 d] A point change Q5 = ]ﬂ.ﬂ LLC of mass. 2.0mm“ kg is releascd at rest at the Origin. Calculatc
the speed of the: paint charge Q; at an inﬁnite distance from the origin. ¢?Eu +¢KE :0. t
ﬂgfvr—Uo) + "ifﬂ—O) = 5'
.ﬁ'w‘f'o—IH..QJE__ (10.ovmorﬁ){3Y‘?6“03) ; {U}: 6:0ng “143'. 2 e} The point charge Q; is remand to an inﬁnite distance from the origin. Where should a point
charge Q5 = EDI} M; h: placed on the positive xaxis such that the absolutc electric potential at the origin is zero. PHYS 112  Final Exam,2002 In.) #2. There is a unifom electric field between the plates of a charged parallel plate capacitor as shown in the autumn
partying diagram. The distance hetwe en the plates of the
capacitor is 5.00 mm. a point charge Q = 2.03 cc of mass
moxie” kg has initial upward velocity vi, = acetic mfsec
very' close to the lower capacitor plate. The acceleration of
the point charge Q pcintc dcwnward and has magnitude "
1.0ﬂx1ﬂm mate. a) Does the electric ﬁeld point upwards or
do we b) Calculate the magnitude of the electric ﬁeld. E E 159' a " m.
E = '3’ J I, u“
0
a2.»— _ Lona (LI9W .5.
M f‘mplo'q c] Calculate the electric potential difference between the capacttcr p a es. E: cg
I
ﬂu? Fﬂ'ﬂ? [01:03:11 ¥(19905)““2$in0 V d) Which capacitor plate, the upper plate or the [exact plate, is at the higher electric potential? " UPPEV’ :2) Calculate the maximum vertical height above the lcwet plate that the point charge Q needs before
itccmcs m reSL w t: giwrcaﬁ : 2 y LEO H9"? I0: ‘3
“3"” m, memo m
Jeane ‘3 PHYS 1‘12 — Final Exam, 2002 1 r0 #3. Four resistors, R; = 5.00 (1. R: = 300 511
R3 = 20.!) ﬂ and R4 (the valor: of which is unknown} are connected to a orator}.r of voltage
155.0 V as shown in tho accompanying diagram. 3 a} The equivalent resistance of the four resistors
is 13.0 1.1 Find the value of rho unknown rcsistor R4. Hora—ER=E_‘3:.
J. _ .L J sti g
3 8’0 29 *9
ﬂL~%=ir
so *9 '9 5v Rq’ "£2" ’64:: : ﬂ
2 b) The voltage across the resistor R. is 15.0 V. What is the current passing through tho resistor R1?
V ZI. DU
I" 'E ' ~ [534 5‘3
2 c} Find the current passing through the resistor R1.
1‘52: 55V25U:WV =' “’5 '5 Vi! ' U _, W” s 60.5%
Its—£23" 3’93 2 d} What is the power lost {dissipatodj in the resistor R3? U3 you
: —— " I—— n ?3... 1392;: (2.0)?, 29.12 _. Wm; } PHYS 121  Final Exam. ENE 2 e) 1What is the total power provided by the battery? .4 .— SV>LSA= Far"F— J_F_— I I 2 D What is the tetai power lust in ail of the resistors? 5 #4. Consider the e1e¢lric circuit presented item. Find
the currents I! and I; and the unknown electromntive force 2. Use the Currents with their assigned directions
as shown in the diagram. in) il+i2 =2.
(a —15'+(2r§)+}'fl=0
C3) g2_ra—io=o —$+¥ff0
52(2—1“) —t0=' 0
~5+¥Ir=0 E+2f}19=0 f2 = 2 — eatmi = 1.29614. 8 = IO + .251 = i0 + attzé’c): 12.51v i
= . A i
1: 1196 f“
i  haﬂv j
_,_ ,;._rJ¥= PHYS 12?. « Final Exam. 211}? #5. A resistor R = 100 o and acapacitor C = Lt'ioslttr1 F are connected to a 100 V hatter}:r as shown in the diagram. The switch 5 is initially open and the capacitor is initiain unchargedr R gm
At time t = 0.00 see. switch 5 is closed. Imv 2 a} What is the current passing through the resistor R after switch C
5 has been closed for a very long time? 0/51 2 b] What is the charge on the capacitor after switch 5 has been closed for a very long time? ' Vﬁeioou
G=urg =aovvaotr . [.000
L. 2 c] In the coordinate system provided 1:10! the charge on the capacitor versus time after the switch 5
is closed. 2 d.) .Calcuiate the voltage across the capacitor after switch 5 has been cioscd for LDD sec.
" it. ""
v“). ummw ) t.ac=t.oo_<, Vfl) = i00u( 194)
WI] .~ 'ODufiEP»368)= 631v PHYS 112 — Final Exam, 101i}? After a very long time. switch 5 is opened. And next a dielectric of dielectric constant 1: = 4 is,
inserted into and czttnpletclj,r ﬁlls the capacitor. 2 fl What is. the voltage across the capacitor after the dielectric has been inserted?
cl: "It. to .— can no"; 1“
GI I: a! D a . Lam? F: 251/
VI = “a.” mounts“? #6. Two infinitely long wires. carry electric
currents I. and [3 both of magnitude 4.00 A into the paper as shown in the accompanying
diagram. A third wire of length 5.00 m carries a current [3 = 6.00 A out of the paper. 2 a) Calculate the magnitude of ihe magnetic ﬁeld
due to 11 at the location of wire [3. e 2 4 _.
i3: $3" “"0 ' unto?
tr. My 2 b) What is. the direction (Le. angle} of the magnetic
ﬁeld due to the current IL at the location of wire 13.“?
l
309 K  is: IT
3 c} Calculate the magnitude of the net magnetic ﬁeld at the location of wire [3 due to the currents
I; and I2.
45 “'3
z B '3 Hr: It ;
131] I1] BM: ‘5‘: £321"? Fit$339,
ET = 0 ii!
is 1' 15
i. 3L PHYS [22 — Final Exam, 201)”: 3 d) Calculate both the magnitggig and directiﬁn of the net magnetic force due to It and I; on the currentig. _,: ___
tar Law)
lg“; hr
_ J  t M.
1 any}: (3.25m r 6.90.4) g (SQOM), 1.09::me
id, c (of Laura‘ru) #7. A singly charged positive ion has mass rn = 3.201(10‘26 kg and velocity 'u" = 4.ti[inl{l5 W5. It enters a uniform magnetic ﬁeld
B = 3&0 T which is perpendicular to its velocity and directed into
the page as indicated in the accompanying diagram. 2 :1} Indicate in the diagram the direction of the magnetic force acting
on the ion when it enters the magnetic ﬁeld and sketch its trajectory. Ix.) h) Calculate the radius of the path of the ion in the magnetic ﬁeld. ll] PHYS 122— Final Exam, 1002 #3. Consider the accompanying diagram. A
conducting he: of a. length i = [.5 m slides at
constant speed v without friction on the corn
dueting rails connected to the resismnce R = it) It. The resistance of the sliding rod
and the reiis is negligible. A uniform magnetic
field B = 2.5 T is directed into the page; t a} What is the direction of the induced current I in I
the resistor 3'? Circle the correct answer of the two following statements: '1' he induced current I is directed upward in the Esistor R. The induced current [is directed downward in the resi tor R. 7 h} At what speed it should the conducting her move to produce an induced current of I = 0.50 A? grad“Dr 31.2 = IR_ (new) 50m _ gm
1; iii—t." 06 ’5 2 c) What external force is required to move, the her with the constant speed It culminated
in part b above? ?e : q’eti: W Feet: BIL IN? = Par : (rsrﬂosanﬂisc) = ‘f = 1.9155.) PHYS In — Final Eaa mi 2002 #9. A cylinder of Sﬂlid uranium weighs 9 [.5 N in air. When
attached to a spring scale and completely immersed in water
(see the accompanying diagram], the cylinder weighs 36.6 N.
The density of water = LEI]: [El] kgfmj. 2 a) Draw ti free body diagram of the cylinder of uranium.
Show all the forces acting on the cylinder. '5
2 1:) Calculate the volume cf the. cylinder?
T+FB.—u F=”'T' H”
FB= gea‘unqt ? _t. 5
F3 _ emu ‘ 501103"
"e!  may Walter M W
1 c) What is the density.r of the cylinder? '4 k
{.5 A! _ Ho
31% .. —.l'd—_—.:' 9 'ir 3 I a 3' 2 d} The cylinder of uranium is next completely immersed in ancthet liquid and the scale reading
is now 24.9 N. What is the density cfthis liquid? = etc2H . (6.610 = gitlg; «3. w #1 «£01 .. MW: iﬁéxlﬂl' l! 5
S" H (5,0;ro"ﬁu.‘){wh/s') 3/” l2. PHYS 12?.  Final Exam RUDE #10. Water is flnwing Hu'nﬁg
through a cylindric 31 pipe 1’1. of varying cross section and elevation as shown in
the accompanying diagram.
The lower section ef the
pipe at {he left of the die;
mam has :1 cross section e * “A; = 1.50 em2 anda speed Sat"nil 2'50”)
of water through this seclien V] = 6.00 1111's. The narrew P1,
upper section of the pipe h a crass section A: = 15¢} em2 1,. 1: O
and the speed of water through the narrow section is #2. The difference in elevation between the two sections of the pipe is y = 153 m. :1} Calculate the speed Va.
A. 11' i = ' :4 a ‘1; 1 0a a MI: .2’SO)( an : M/S 7 h) Calculate Lhe difference of the pressures 3p : p; ~ [:1 between points I. and 2. [3 P, W7": w‘ﬂzr e «535.: eye? PR‘PJ ‘_ ijtJ'z) 4if(¢52‘%£2/)
0 192‘?! ‘= WWW“) (4.5) 4 ﬁ'f_0ria3(§a ml)
3 p, .7(;.om")[ac.;Je,=.qc.1 km W PHYS m  Final Exam. 1001 3 #11. A small lamp is 1G m below the sulfate of a lake. It emits light in all directions. A heat goes from directly above the lamp to the point when: the lamp can no longer be
seen. How far does the heat travel? The index of refraction of water. :1 = 1.33. Sketch a diagram illushatlng your solution. , ml Sign.
’F’ e.
f ,‘ifcs‘iﬂg‘m'lw x=ii+au9 4r “9" imug'q
'1 I}; =' :‘1 he. i
#12. Considerﬂue diagrambelhw. Aconcave milmt with radius ofeurvnrurc = . cmis placed 5.00 em to the right of a converging lens of a focal length f = 2.00 cm. A 0.90 cm high object is
place:1 5.00 cm to the left of the lens. The points labeled f in Ihe diagram are the fecal points of
the thin lens. The point labeled 0 in the diagram is the center of curvature of the concave mirror. 2 :1] Draw a ray trace in the diagram above to ﬁnd the position of the image formed by the lens.
Label it 11. PHYS I23 — Final Exam. 2cm 2 b) Using the thin lens equation calculate the position of the image formed by the thin lens.
Note: your answers to parts a] and b] should be consistent. '
.L + J” 5 i:
a “h '
J, a _ J  J7.
3— : ' ' 1 5
t a i”:
(ll ; '55. 153 can 2 c} Calculate the magniﬁcation of the image produoccl by the thin lens. _ art _ H333 HI: Tap 2 :1) Draw a ray trace in the diagram above to find the position of the image formed by the mirror. it I3. ¥l=g~= Law 2 e) Calculate the position of I; with respect to the mirror. 2 Pi: 6.00  “5.32_,. 2.63m. L+Jc_c.t. $12!.GCM}
‘1 P1 'E z
I_  .1. — i
1:  .2 E?
2 E) Calculate the height of the image 1;.
Hint: check that your calculations and drawings are consistent. .. $2, LG
H : __f_ = __,_,_ i was
a 2 Pa HIN : Ht “1 H1 _ {C— 0.6?)(10. 6) : LI 0 h'= h“taﬁ 0.!30Lh * UHG: !O."56CM4I IS ...
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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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