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Physics 101 Midterm 04W

# Physics 101 Midterm 04W - 17:18 IFAX Formulae mm Av‘ =...

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Unformatted text preview: 10/05/2007 17:18 IFAX ' Formulae mm:, Av‘ = coast, a Javed Iqbal 001/002 1 . Student No. _ . ._ - Pns Clicker No. Last Name First Name The University of British Columbia Midterm 1 Physics 101, Section 102 - Oct 15, 2004 Javed Iqbal No books or notes permitted. Graphical calculators are allowed,'but formulae may not be stored . Do all four questions. P+1l2p +pgh= coast. x(t) = Acos(cot + 41). v(t) = -ccA sin(cot + it), a(t) = - co" x(t) I mass-spring system: a; = gem)“, pendulum; co = (g/ij', T = 2m, K = lama. U = 1/2 1:22, o = (Pl-szcﬂ/‘snL. 1 atm = 1.01me N/mz, water density '= 1000 lag/mi, g = 9.81m/s7' Problem I. F=-kx, A pipe of CIUSS7SBCﬁODa] area A is joined to a loner pipe of cross-sectional area A12. The entire pipe is full of liquid with density p, and the leﬂ: end of the pipe is ht an atmospheric pressure PD. A small open tube extends upwards ﬁ'om the lower pipe. Find the height h; of the liquid'in the small tube when: (a) the right end of the lower pipe is closed, so that the liquid ' l4 is in a. hydrostatic equilibrium. (2 points) _ _' ﬂppia BQ—Iwculil.,5 61M ail—Q q—é) '. . *9 M9 in; =P2+°*O ° + a = ﬁtﬁ’ahz (b) when the right end is open and the liquid ﬂows with speed ir in the upper pipe. (4 points) . No on He ghai xiv—\o ‘iiﬁouéin J‘ke- ~05": is “cm - ism ﬁend-61% Cami-inuch Eﬁue-En‘ a]: @4— Q Asv : Lia—13"“; =9 \Iz:’2\I We aim-“- at (ESQ) mo _,PD* Lieu-«Lieak‘ 2 'Pl‘ar—iehl '. ?9_ = - Bu): “Pl -. "P0 P0 '\' “- ?D ' [Problem 2 Consider an object ﬂoating in a container of water. If the container is placed in an elevator thaIaCCeleratesupward C‘hawﬁe {in egeé-w'e ‘6 .' act‘s \oo’ﬂx “Ae— Dhigeet =1» Jr‘et that. C‘i ' l. more of the object is below water. 2. less of the object is below'water. @there is no difference. 4. There is not sufﬁcient information to answer this question. (Circle only 0_llq answer, multiply selected answers will not be given any marks) 10/05/2007 17:18 IFAX ‘- .I (b) the phase constant 4: of the motion. (2 points) » Javed Iqbal 002/002 Problem 3 The position vs. time graph of a particle is shown in’the ﬁgure. Determine (a) the angular frequency of the motion. (1 point) )1 T 1 115 T = ‘3 s 21‘: 1: T ‘3 x (cm) Lo: XLt'): o.\0mCosL1—E-i:'* :__'10 _ ‘ . . 0. tom c0505?) ————— xu'): 0'05 m :— \ (opt—ted:- — l _ -il 41 comes-— :‘f’i = is ‘i was (33 'Clﬁ-“HSW (o) the velocity of the particle at: = 1.0 s (3 points) T KHZ): D‘\l . “if T . '0- 1 m \I U): "‘k0 ‘01:) 5' LLl 3 . r . T.1T m um = - EJS‘“(‘K“3‘) ,5 ~— +O-02~ “I5 EMA A 0.10 kg ball, attached to a spring with spring constant 2.5 me, oscillates horizontally on a ﬁ'ictionlcss surface. When the ball is 0.05m from the equilibrium position its velocity is 0.2 m/s (:1) What is the total'energy of the ball-spring syStem? (2 points) E 3 li- k X L ‘t -L "N‘ \J L 2. r 1 r .LXQ-Sx(O-OS)Z+ liJLo-‘LXLO-l) = 0.00513“ 3. S (b) What is the amplitude of oscillations? (2 points) E:— ,__.——--\ - 2—E = t l K , (c) What is the speed of the ball when it is at x = 0.06 m from the equﬂibrium point? (2 points) _ 1 \ L t; — 55“" *EWU we” x3. \‘ = is: Te ﬂ 2: i _ END-“=7 ’\o‘ ’5“ 11 yo )L D .0 o L“ 0N d-F’ ...
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