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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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