ExS2F10 - Physics 123 - Analytical Physics a proctor will...

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Unformatted text preview: Physics 123 - Analytical Physics a proctor will help you. Also raise your hand during the exam if you SECOND COMMON HOUR EXAM g E“ a (81:33”? 1 1 1 1 1 r J r r y ’_ . . case t 1e cover s leet unr er your name sticmr ant 1ave your ” hlondd‘y’ Novenlbel 8‘ 2010 student ID ready to show to the proctor during the exam. Pl‘OlGSSOI Bartynsk1 10. If needed, the acceleration due to gravity 011 earth may be take as g = 7‘ 9.81 Ill/S2. V Your name sticker with exam code. SIGN HERE: 1. The exam will last from 9:40 — 11:00 p.111. Use a #2 pencil to make entries 011 the answer sheet. Enterthe following ID information now, before the exam starts. . 2. In the section labeled NAME (Last, First, MI.) enter your last name, then fill in the empty circle for a blank, then enter your first name, another blank, and finally your middle initial. 3. Under STUDENT # enter your 9—digit RUID Number. 4. Enter 123 under COURSE, and your section number (see label above) under SEC. 7 5. Under CODE enter the exam code given above. 6. During the exam, you may use pencils, a calculator, and one 8.5 x 11 inch sheet (both sides) with formulas and notes. 7. There are 15 multiple-choice questions on the exam. For each question, mark only one answer on the answer sheet. There is no deduction of points for an incorrect answer, so even if you cannot work out the answer to a question, you should make an educated guess. At the end of the exam, hand in the answer sheet and the cover page. Retain this question paper fOr future reference and study. 8. When you are asked to open the exam, make sure that your copy contains all 15 questions. Raise your hand if this is not the case, and 1. A grinding wheel 0.500 111 in diameter rotates at a rate of 8.00X102 5- A 111353 M = 55 k” 011 a horimllml table is Duued by a llOl'imllttil revolutions per minute. Find the magnitude of the acceleration of a '. m string that passes over a frictionless pulley to a free~hanging mass 111 speck of metal caught in the outer edge of the wheel. 3.4 kg. The coefficient of friction between M and the table is 0.28, 2 The acceleration of M is ‘ 3r10 «2 , u" at} K) 2 r 5~ fig 209 a- T :L R :(L-flg) K a) 3.7 111/s2 '9 «T l T ‘T...-§- : M CL . ‘ ‘2 — 1 c) 8.00x102 111/s2 g; : gigyx \07' a '5 3 a-\ ® 2.0 111/s2 "£7 7 T Mgr MA d) 1580 m/52 e O S _ ' c) 2.2 n1/s 2 l f mj_ T : ,ma’ @ 1750 111/52 R = O. 230 "" r d) 0.20 111/s2 mg ‘1 m — MIA) jzmfl’mDQ av: I‘jsx‘oifi rim/5:. . e) 0.49111/5 CL «hep/1M) 3Mvhsq) 8, Zgom : ‘ ‘ _ _.._—-—__—> A ‘ z. e r. 2. A highway curve has a radius of 140 111 and is unbanked. A car weighing " ‘ no ’I' M g —( ' 9.0 q 3 7‘ (3. Two forces are applied to a 5.0—kg object. one is 0.0 N to the north and the other is 8.0 N to the west. The magnitude of the acceleration ’\ 12,000 N goes around the curve at a speed of 24 111/s without slipping. \Vhat is the magnitude of the horizontal force of the road 011 the car? ‘ of the object is: 1 gm ‘9 4A a 12,000N ,w u . =~35 + ' 1i 17000N l? F: "M" 43—)? ‘ a) 0-50111/82 i: I?1 “J ‘ """ ) ’ 1 ' @ mini/52 3” a: e = we a + \«11 (31> c 13.000N : \ZOOD’KLZZD ) 2'8 /2 m mgN ' 12! (I . 1118 7 :S'OX {0 N e) 50111/s2 3. A 5.0 kg object is suspended by a string from the ceiling of an elevator that is accelerating upward at 2.0 111/s2. What is the tension in the ' 7- A boy PUHS 3- WOOden box 310112; a Tough hOTiZOHtal floor at constant .9 string? TA; W W , speed by means of a'force P as shown. Which of the following must be T true (f is the magnitude of the force of friction, N1 is the magnitude of m— a) 49 N b) 36 N @ 62 N d) 13 N 9) 52 N the normal force7 and W is the 111agnitude of the weight): «n57 Twm9= ’mCL =9 T=m(j+av3=5‘°(q‘3l+u):62“ 4. A boy 011 the edge of a vertical cliff 20 111 high throws a stone horizon- tally outwards with a speed of 20 111/s. It strikes the ground at what horizontal distance from the foot of the cliff. a% P : f and N Z W b P=fandN>lV 12. a) 10m Sz-é-g‘tz f: .5 ; w ©P>fandN<W szeight “£31 40 n1 r g d) P > f and N = W f: friction C) 50 111 -_—, 2, 0 A _ e) none of these N ; normal d) 1.1x102 m ' 2 e) none of these q~ Chasm 9W :7 +a¥ l’ : O a: fimm/s ’X:’U.:‘t -: w$2.0= Dim ,‘ e: :———- :7 T // . l’ (:06 *3 n P We P>§- L 4 l 8. A boy 011 a skate board skates off a horizontal bench at a velocity of 10 12. How much work is done by a person lifting a 2.0—kg object from the 111 One tenth of a second after he leaves the bench to two significant bottom of a well at a constant speed of 2.0 111 /s for 5.0 s? figures the 111agnitudes of his velocity and acfglexr/itéion t 4 - 0.15 k] ® 020 kJ‘ C) “.25 k} d) 030 k'] 10111/s;9.8 111/s2 - ‘—"’ ‘ r e) 0.351;] W : F. d z 2 r 0 P‘ Cl-gl " m- V b) 9.0 111/s; 9.8 111/s2 l m/ . : 9.00 3' = 08 247 k: c) 9.0 111/3; 9.0 111/s2 'U‘" z [0 ’2‘ 8H 0" = 0.032% 13. Peter is driving cast in the right lane of a highway at a speed of 31 d) 1.0 111/s; 9.0 11‘1/s2 v5 : %t - ‘ M 111/s. George is in the left lane and driving west at 18 111/s. What is e) 1.0 111/s; 9.8 111/52 ("OW = W 5 lo 3' +0 2 Peter’s speed, in 111/s7 relative to George? a; W (“6‘ W ' a) 49 \V‘st PM 3\ W5 9. In order for you to jump off the floor7 the floor must exert a force 011 b) 13 West, Q 4 \8 m5 you I“? @ 49 East V = me + Vn'ka al a) in the direction of and equal to your weight. K d) 31 East Pa .. 5 l + 13 = 4 qg’M b) opposite to and equal to your weight. ‘3) 13 EaSt C) in the direction 0f and less than your Wejght‘ 14. When a car goes around a circular curve on a level road, d) opposite to and less than your weight. _, ‘ _ _ . . @ Opposite to and greater than your weight. l M >ml a) no fr1ct1onal force 1s needed because the (tar snnply follows the . . o . _ b) the frictional force of the road on the car increases when the 10. 0.20—kg omect attached to the end of a string swmgs 111'a vertlcal Gains Speed decreases. D as; :r e Cll‘file (“lde 2 80 .Cm)‘ At the mp 0f the Glide the Fensmn m the the frictional force of the road on the car increases when the strlng drops to zero mstantaneously7 but the object continues to travel caras Speed increases. F _; ma : “(1%; ’ W M A; 7‘ m 8‘ Clrcle' \Vha‘t IS the Speed Of the Owed at thls Dogma“? 5 d) the frictional force of the road on he car increases when the car a.) 0.95 111/8 4 .9 T + W = “m a/ L moves to the outside of the/\curve. b) 3.6 111/s “I'll W T z 0 p 9 Q r. {Er-Z e) there is no net frictional force because the road and the car exert @ 2.8 111/s _W s ,W‘ 4/ z m 1;; equal and opposite forces on each other. (:3 ,U— = R‘ :W 15. A 2.0-kg block slides 011 a rough horizontal surface. A force (P = 6.0 ' m N) is applied to the block as shown. The magnitude of the block’s : 2‘3 ’3 acceleration is 1.2 111/ 32 to the right. What is the magnitude of the 11. A 2.5—kg object falls vertically downward in a Viscous medium at a force Of friction acting on the block? constant speed of 2.5 111/3. How much work is done by the Viscous force 011 the Object as it falls 80 cm? P a) +2.0 J 1)) +20 J c) —2.0 .1 Q1) —20 .1 e) +40 J g C9“S‘l”a«“.’ a; z) err = «3 = 2.5% ‘13! : .0 .7 4 .Tfr w: l'x: *ZJX‘iJ/XOfi D 0 mm ‘7? :.203_ (m_w+?@éo. ma : a) 94/“ £0 —" 5: 57 -.: P511060." ma 3 : évDXO-87 'Z.OX"Z=2\9N ...
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ExS2F10 - Physics 123 - Analytical Physics a proctor will...

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