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Phys1-T2 - PHY 2425 Engineering Physics 1 fiffi Exam 2-3...

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Unformatted text preview: PHY 2425: Engineering Physics 1 fiffi Exam 2-3.: 22 October, Zflfll Name: LDC- DU SSN: 52 '4' Y [Last 4 Digits) 1. } Three crates of masses, ml— — 25 Kg, n11 = 35 Kg, m3 = til} Kg respectively, are tied together and are sitting on a frozen lake. We can assume that the coefficients of fi'iction,|.1., and 11,, are ze_,ro and that the" we is level. La} If a horizontal force, T3 = 24:21 N, is applied to the system, what is the acceleration, am, of the system? T3=24DN QM m1=25Kg n12-35K.g m3=fiflKg 45;, F: mo /‘7 Ji—r:II = mtws.‘ flg‘lrfi / 1’6 T T tight-D N _ ”Em/:1 q ___ __W’ : __——————--—— = F at, mil-It}; (no. 4 m1. -I ma) {saw} -I “FEE! + Hot?) Lb} What are the accelerations, a1, a1, and a3, ofthe masses, 1n. = 25 Kg, m1 = 35 Kg, and r11: = fill Kg respectively? fll=q1=QEIGEgfl ZE— Lc) What is the horizontal tension, T, , in SI units? T.= c m, : Qmfiti‘fi as“: 3 I“ N 1.11) Wat is the horizontal tension, T1 , in SI units? +ma mg 4m“? . 1 T1 :sm/stf-E'Eti-Icssg) = “tori-i 2.) An automobile of mass, m -= 3500 kg, goes “out of control“ on an Interstate Highway»r while managing to go in a straight line! The car leaves a skid mark on a level road of length, x = 15!] 131, before coming to rest {v={l}. The coefficient of kinetic friction between the road and tires is determined to be ll»: [1.35. We can assume that the tires are “locked" during the skid. T“: 3561': t§_ a=v _ "A” ‘11: team Q #J U: o g '—— Mir-o 59.5 3‘) .- 11 x=15flm 2.a} What was the deceleration of the automobiie during the skid? {Eli—J! EL: ma“ / [ii-.3 a : mall = C] .. 4.; -— mas a All: N ': r001 / 1N "“1 the = M I; m 13} jut-r / —|. /" New: 0 ~=N=Ncmfl e : 0-215 at 61-2 ”is? 2b] What was the initial speed vu of the automobile at the beginning of the skid? 1 l ”a" :' ”a1+ Ra (T'xfi) 1—} “a I -' 2a ‘1 4—} #5 ___ r-Efitk elfi-e Ll-H‘l'fgl)( lflm) “-3 U] 10 mrs. / \Ja '2 '52 mfsi/ do] How much longer would the skid mark be if the automobile had been twice as massive, m = 7flflfl kg, assuming the same initial speed? ’It will be tfi; 3mm: ; dollar-iii ribofla- haw 3.) A ear weighing EELD RN (Frag) and traveling at 10.0 mie attempts to make a left turn with radius 50.0 m on a level read at an interaeetiun. 3.3} 'Whet Lee of frietiou-r‘5 is required to keep the ear in its circular path? Him, we eentrlpetaI force F: tewards the center ef curvature is previded by the static frietienfi between the read and the tires ef'the ear and acts en the “aide walla“ uf the tires. W .__ 1e. 0 It N U : 40.0W‘U'S I_.=.| FL: Gil-Om W W :: Wig -? f1"- = 13—. ' ./ “Fr- =__ "l' i1 (3! My?) (Jew/9L #IOEEKC hega'w v=lflmfs l1 tl 3J3) What should be the (minimum) value of the euefiieieut of static friction between the tires and read, its, in ertler for the left turn he be aueeessfifl ‘9' 4a) Calculate the force, F, on an artificial earth communication satellite in a circular orbit which is exactly three earth radii ([1 = 3R3} above the surface of the earth. Hint use Newton’s Law of Gravitation. Use SI units! F : QflfL= G m'm’mst’m' Nm/rg‘ 1000': “3mm” l I D H LI“ "I‘IFI')L [(15.13 filfié W\+5(fiv affil'il] H3! F I"; fiflllfi {”5 : :ic-Is'amo‘fia) / 4.1:} Calculate the acceleration of gravity, a. for this very same satellite. 1‘5 CT ME :56 éTvm mm #59qu ‘3?“‘9 ml) EMA? a '10 29 9‘0 Q'JELLML‘ {fat 5.} Using Kepler’s Third Law, calculate the perk-d If the very same satellite in Prob, 4 31min above. 4 17" . . T 1: f Y? '| / CT M E . / 3 1 14 :4 (ML-01 " (EFQF’rhm‘) “Ljr . I. GulE'bTiID-"Nmi/kql) '3’ (5.:er ma ”$5 “ (L) QEQED’MO'!) / # $51.61 5101'; I, T 2 W : #07675 / I'Hr #07675 g : 111.?) hr 5am); ...
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