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Uniform Beam - General Physics - Solved Past Paper, Exams of Physics

This is the Solved Past Paper of General Physics which includes Work Energy Theorem, Specific Object, Specific Interval of Time, Forces Acting on System, Newton’s Second Law Analysis, Nonconservative Forces, Total Mechanical Energy etc. Key important points are: Uniform Beam, Masses of Cable, Hot-Water Heating System, Density of Water, Water Speed and Pressure, Di Erential Equation Governing, Maximum Angle, Function of Time, String of Length

Typology: Exams

2012/2013

Uploaded on 02/25/2013

shoki_sho
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Download Uniform Beam - General Physics - Solved Past Paper and more Exams Physics in PDF only on Docsity! First Name: Last Name: Section: May 2, 2007 Physics 207 EXAM 3 Please print your name and section number (or TA’s name) clearly on all pages. Show alli your work in the space immediately below each problem. Your final answer must be placed in the boxes provided. Problems will be graded on reasoning and intermediate steps as well as on the final answer. Be sure to include units wherever necessary, and the direction of vectors. Each problem is worth 20 points. Try to be neat! Check your answers to see that they have the correct dimensions (units) and are the right order of magnitude. You are allowed one sheet of notes (8.5” x 11”, 2 sides), a calculator, and the constants in this exam booklet. The exam lasts exactly 90 minutes. Constants: Avogadro’s Number: N, = 6.02 x 10°’ molecules/mole Material Density, p ‘Young’s Modulus, Y Ultimate (Breaking) (kg/m’) (10° Nim’) Strength (compression) (10° N/m?) Steel 7860 200 400 Bone 1900 9 170 Wood (Douglas fir) 525 13 50 Mathematics: trigonometric identity: sinA + sinB = 2cos[(A-B)/2]sin|(A+B)/2] small angle approximations: sind = 0 - 67/3 + O(8*) cos@ = 1- 67/2 + O68) Sol vT) OWS 2 OL VN SCORE: Problem]: Problem 2: __ Problem 3: Problem 4: Problem 5: TOTAL: Don't open the exam until you are instructed to start. “Success is not final, failure is not fatal: it is the courage to continue that counts.” Winston Churchill First Name: _. Last Name: Section: PROBLEM 1 a.) The figure shows a safe of mass M = 430 kg, hanging by a rope from a boom with dimensions a= 1.9 mand b=2.5 m. The boom consists of a hinged beam and a horizontal cable. The uniform beam has a mass of m= 85 kg; the masses of the cable and rope are negligible. What is the tension, T. , in the cable? (8 pts) 4F 20 ot SB-0 abeX on pein LNs Commpale 3U oe the Wenge rope 6 loo WV Sanuve? rho zceh the Th al ba: —_ = 6100N oS b. The deep-sea research/diving vessel Alvin can operate at depths of 4500 m. With its ballast tanks empty, the vessel di; s a total volume of 16.8 m’ and has a total mass of 17,000 kg. What minimum volume must the ballast tanks have to permit the vessel to dive? (Ballast tanks are containers within the vessel to which water is added or removed. They are used to increase or zR decrease the mass of the vessel and hence make it sink or float.) (6 pts) Teco) n= 1025 m Booger Fore fe) Vs culo [ef - 13222 Va weight Chee Ww 11,2000 & *E ous ms co, tok rot ly odd 220 My a \V - 28 . Tork pre £6,211 we “ y ot wi 4 b) If θ(t) (the solution to part a) is written as θ(t) = A cos(wt + φ), what is A and w? Express your answers in terms of {vi,m, M, g, L}? [Hint: This problem requires you to know the answer to part a). If you do not know the answer to part a), say how you would obtain w had you known the answer to part b).] (5 pts) ANS: By denition, we have A = θmax = vi m M + m √ 1 Lg Plugging in θ = A cos(wt + φ), into Eq. (1) gives w = √ g L . c) If wind drag slows the mass down such that ~Fdrag = −b~v what is the new dierential equation for θ(t)? (5 pts) ANS: Writing Newton's equation of motion in the horizontal direction just as in part a), we have mLθ̈ = −mgθ − bLθ̇ Hence, θ̈ = −g L θ − b m θ̇. 2 First Name: _ Last Name: Section: _ Problem 3 a. Write down an equation, y(x,t), for a transverse standing wave on a string of length £ = 0.20 m that is fixed at each end. The standing wave has wavelength A = 0.20 m, frequency f = 400 Hz, and amplitude 4 = 4 cm. (5 pts) gid 0.64 bop UD b SMX we at ike “= b. What is the period, 7, of this wave? (2 pts) -,,72 +. te A = psu Te 2-5 Vo s “ £ Hoo c. On the axes below, draw a sketch of the wave at the following times: t = 0,t = 1/4, t= T/2. Clearly label each curve. (2 pts) oy I —o.o% . < [ith d. At which value(s) of x will the transverse acceleration of the wave be largest? Compute the maximum acceleration. (3 pts) L st oniioole J A= A) <=> ok XE 105m) O-)5 m t abi at Ome oh awh sinXh 2 2.5 ™/sr Omay c Ben 0.04 ETS) - First Name: Last Name:_____ Section: _. e. Standing waves can be expressed as the superposition of two traveling waves. Write down the equations for the two traveling waves that make up the standing wave in part “a”. (5 pts) hut) < Ot Sw bx-w+t) + D-0% Sm (yxtwt) we ua er eet % 4 f. Compute the propagation speed of these traveling waves. (3 pts) \ ; \ ft. 0.20 Yea 7 ~ = Fo m/e, ws First Name: Last Name: Section: Problem 5 An insulated cylinder whose volume is controlled by a piston of mass Mp and cross sectional area A contains gas made of atoms of mass m obeying the ideal gas law. There is no gravity nor atmospheric pressure acting on the piston, but there is a spring with spring constant k pushing against the piston in quasistatic equilibrium with the gas pressure inside the cylinder. The spring length is in its unstretched length when the volume of the cylinder is 0. a) Suppose one measures the temperature of the gas to be T0. How fast are typical gas atoms moving? Express your answer in terms of {kB , T0,m}. (5 pts) One estimate of the typical speed of the gas molecule is the rms speed. 1 2 mv2rms = 3 2 kBT0 vrms = √ 3kB m T0 b) Suppose one heats up the gas quasi-statically from Ti to Tf . What is the ratio of the nal volume to the initial volume (express your answer in terms of Ti and Tf )? [Hint: Recall Hooke's law force must balance the force due to the pressure of the gas. Note that there is no gravity nor atmospheric pressure in this problem. Also, remember that Volume = A×height] (5 pts) ANS: Tf Ti = PfVf PiVi where Vi,f = hi,fA. Because of Hooke's law, PfA = khf and PiA = khi. Hence, Tf Ti = hfhf hihi Since h and V are proportional, we nd Vf Vi = √ Tf Ti . 1 c) If the initial volume is given as Vi, how much external work W was done to the gas for the process in part b)? Assume that the answer to part b) is Vf Vi = c1( Tf Ti )α (where Vf is the nal volume) for particular constants α and c1, and express your answer in terms of {k, Tf , Ti, Vi, A, c1, α}. [Hint: Volume = A×height] (5 pts) ANS: The work done to the gas in raising the piston by height h is W = − ∫ dV P = − ∫ hf hi dh(kh) = −k 2 (h2f − h2i ) = − k 2A2 (V 2f − V 2i ) = −kV 2 i 2A2 ([ Vf Vi ]2 − 1) = −kV 2 i 2A2 (c21( Tf Ti )2α − 1). d) If the answer to part c) is W0, how much heat is required for the process described in part b). (Neglect any changes in bulk mechanical energy.) Assume that the answer to part b) is Vf Vi = c1( Tf Ti )α (where Vf is the nal volume) for particular constants α and c1, and express your answer in terms of {W0, k, T f , Ti, Vi, A, c1, α}. [Hint: The number of gas atoms can be expressed in terms of pressure, volume, and temperature.] (5 pts) ANS: Energy conservation gives ∆Eth = W0 + Q Since the thermal energy is characterized by ∆Eth = 3 2 (NkBTf −NkBTi) = 3 2 (PfVf − PiVi) = 3 2 ( khf A Vf − khi A Vi) = 3 2 k A2 (V 2f − V 2i ), heat is then Q = ∆Eth −W0 = 3 2 k A2 V 2i ( V 2f V 2i − 1)−W0 = 3 2 k A2 V 2i (c 2 1( Tf Ti )2α − 1)−W0. 2
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