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1051test1answersW09 - Circle instructor Poduska or Morrow 1...

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Unformatted text preview: Circle instructor: Poduska or Morrow 1 Name: Lab period: Student Number: MEMORIAL UNIVERSITY OF NEWFOUNDLAND DEPARTMENT OF PHYSICS AND PHYSICAL OCEANOGRAPHY Physics 1051 Winter 2009 Term Test 1 February 13, 2009 INSTRUCTIONS: 1. Do all questions. Marks are indicated in the left margin. Budget time accordingly. 2. Write your name and student number on each page. 3. You may use a calculator. All other aids are prohibited. 4. Write answers neatly in space provided. If necessary, continue onto the back of the page. 5. Do not erase or use “whiteout” to correct answers. Draw a line neatly through material to be replaced and continue with correction. 6. Assume all information given is accurate to 3 significant figures. l 7. Don’t panic. If something isn’t clear, ASK! l ANSWERS SEE LAST PAGE FOR SOME POTENTIALLY USEFUL FORMULAE AND CONST ANTS Circle instructor: Poduska or Morrow 2 Name: Lab period: Student Number: ‘ [10] 1. (a) Pendulum a is a simple pendulum consisting of a 0.5-kg mass suspended from a massless string of length 0.80 m. Pendulum b is a uniform 1.0-kg rod of length L hanging from a pivot at one end. Its moment of inertia about that pivot is I = §mL2. Its period is lengfh=0.8 m ‘ ‘ mass=1.0 kg moss=0.5 kg . L: ?;E the same as that of pendulum a. What is L? - /F 74.: 7; ) H501 Wag 4/6 {:3 WM? MD 4,: new" , 33 La ‘ I - 315 r 3 (Iva/C 50: 1%) £0, 2%: 4/ 273’ Jami/WP”; 6 +0 fid/INDSS) "‘ [5 3 El? " 5303‘”) l .45 :AaMM} (b) A 0.2-kg mass attached to a spring is oscillating on a horizontal frictionless surface with a period of 0.7 s and an amplitude of 2.3 cm. (i) What is the force constant of the spring? (ii) What is the total mechanical energy of the system? (iii) If the position of the mass is x= 2. 3 cm at t- — O s, what is the earliest positive time for which the velocity of the mass is v — -+20. 6cm/s? [a] (DA/LIZ mm; burg—“833's" ”/4 ML.) 02th M7“? an; m .. —3 L27 W é’m = 5 éfl’: g (/54 %){0.02 3m)’: 4,25 “9 I (Hi) fissamt 7r: 14 C05 (”#0) -1 7((fw) r0.0,23 M .: 0 023m: 0.023": w;(0:4>) 3, 45%“ [0: 0 6‘. 7/69 : w/flsl‘aw" «2:32): 9 W999 040w; r _ sagsfiaozzm x grew-75’s l" t) 04.06”; = — 0.205 '3'; gm (their (‘34!) 5/‘~"('/)-' ”El a} . - , 37F clxwre. S‘M 7’0 ’ + 'E. a r / [4" ’- I I) 6 mg?! 5 ( I) J30” Cam/leg,‘ papil/‘L é) ., J, 3” ll 395/9” X i “,_..iflt-~.w.m_...~...m..WWW. ry..WW~.W.._«.—~ «a» Circle instructor: Poduska or Morrow 3 Name: Lab period: Student Number: [10] 2. (a) The diagram represents displacement amplitude versus position along the tube for a mode in a tube that is open at one end and closed at the other. The L ‘ ‘ X: frequency of this mode is found to be 440 Hz. x=0 L (i) If the speed of sound in air is 343.2 m/s, what is the length of the tube? (ii) Assuming the same speed of sound, what is the lowest frequency at which this tube will resonate? , ' A TUBE bum»: ONE 5WD 0‘99": [3] (L) gm .4:- 5 U 5(34312M/9) H V W _, L: L : : TL, WW "t Z; :3431m/s *5 174): 9/4/40 Hz) 1 0 H 2 -- k.” [2 . I - 7r 3 4/3. 2 "4/5 3(“3f1:_ : = 58.0 HZ ’7‘ L 4/ { a. 973%) (b) Two speakers, located as shown, emit sound waves in phase. Speaker 1 is at the origin and speaker 2 is 8.0 m along the y-axis as shown. An observer starts from speaker I and walks away from it along the x axis. The observer first notices a minimum in the sound intensity when 34.0 m from speaker 1. Take the speed of sound in air to be 343.2 m/s. Y (i) What is the wavelength of the sound? speaks, 2 (ii) The observer now stays in place as the 1‘ frequency emitted from the speakers is increased. 8.0m \ observer What is the frequency when the sound at the l _ X location of the observer is a maximum? speakerl g 34.0m —---—> 1 ‘ , v, I TH mrFemch $0 [a] (l) MIN/MOM [NR—”5,77 Mall/S 2 PET" 096 ) AY :‘/(3L/M)Zfl (8002 ' 34m = 0.928”! 2. / V , l I g A].- 72 M Q [2] (it) Fort Mix/MUM ’NWS/W’ 343.2 m/5 : 370 Hz V7 {‘7' 0923'" Circle instructor: Poduska or Morrow 4 Name: Lab period: Student Number: [10] 3. A sinusoidal wave propagates along a very long string in the positive x-direction. The wavelength is observed to be 2.7 m and the period is 0.08 s. The linear density (mass per unit length) of the string is 0.0065 kg/m. (a) What is the speed of propagation for this wave? (b) If the maximum transverse speed of a point on the string is 4.0 m/s, what is the amplitude of the wave? (c) What is the tension in the string? (d) A second string is made of the same material but with twice the cross- sectional area of the original string. The second string is stretched with half of the tension applied to the original string. What is the wavelength for the second string if the period is kept at 0.08 s? (Show all steps in your calculation.) [fl 5 deWA/so rig/7,7,??? ’ 7, [SE M [is © 7;: if. 90 0}“ I 7— ,(g; 3,433 (0 0055 kg/m) : 7.4/0 52 717m, “Eu—~71 ‘ 71+ ’ I 2% if 4 1m )NbV5i/Z‘?MS Circle instructor: Poduska or Morrow 5 Name: Lab period: Student Number: [10] 4. One charge, q A = +2 uC , is located at the origin. A second charge, qB = +3 uC, is located at x = 1.5 m, y = O m. Point P is located on the y-axis aty = 1.5 m. (a) Find the location on the x-axis where the electric field is zero. (b) On the diagram, draw vectors representing EA and EB at point P. (c) What would be the magnitude of the force on a charge qc = —0. 15 uC , placed at point P, due to charge qB ? (d) Imagine a spherical surface, centred on the origin, with a radius larger than 1.5 m. What is the total electric flux through that surface due to charges 61A and (IE? “‘ EA (“MEI-5‘2 5., a}: Rae—a r?— VIZ V12 9:4 V‘z : (2%": a NM >(0.zsno'° C) (3“ ’D’Ld (<3 [FF- kigfic ’ (ammo j: V; ( (15'th (I‘M: Circle instructor: Poduska or Morrow 6 Name: Lab period: Student Number: Some Potentially Useful Formulae and Constants: dzx " q q A dt2 = —a)2x E2 =ke :12 r12 F; = —kspmgx (Hooke's Law) E ___ k l r” e 2 r a) - 2—” (angular frequenc ) T y E = keZq—g: i r: wZ = kspring m v d=331m/s+0.6 m xTD soun S'OC C 2 _§ _. .. w ‘ L $15 = jE-dA wz : M qinside I (DE 2 80 27: k = 7 (angular wave number) V = ke ff v = fl (wave speed) V = keZ—qi i ri v = Z V _ 4 3 [U sphere _ Efl-r 62 1 a2 2 ___)i = ___y ASphere = 4727 2 2 2 8x v 61‘ Physical constants: _ 1 _ 9 2 2 ke—47r80—8.99X10 N m /C 6:1.60X10_19C 80 =8.85><10“2 CZ/N-mz g=9.81m/s2 ...
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