Physics 101 Fall 2002 Notes

# Physics 101 Fall 2002 Notes - E A Edi—kg load is being...

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Unformatted text preview: E:[ A Edi—kg load is being lifted with constant speed using the pulley i : ._- 5 C : arrangement shown in the dent. What is the magiutude of the force F? [3.1 15 N [ E] 39 N C' 14? N [D] 204 3; [E] 533 I hj-i— - [2] A lﬂﬂD—hg car is picking up speed 3.3 it goes around a. horizontal curve whose radius is 19!] in. The coefﬁcient of static {ﬁction between the tires and the road is {3.35. :1! what 5 reed will the ear begin to skid sideways? 613.3 mg: [13133.3 mia [of 23.3 mfe [o] 34.3 en's [E] 33.0 no [3! A person lifts a. 153.55 of 20 kg vertically ugwnrcle at a constant speed of U329 mi'a through a. distance of [1.31] ELL. What is the wort: (Lone hy the person? [0.13.3.1 [B] 13 .1 [o] 50 .1 @133 .1 [E1200 J [at] A Elli-H horizontal force pushes an. object Moog a rough ﬂour Ho that the ohject mt‘nfEﬁ with a ttoriﬁteut velocity of M] mfs. How much. power does this force deliver to the Object? |A_ II} W’ [BE 51.0 W @30 W [i1E 160 'W [E] 785 W" [-5] A. miter coaster of tﬂui 8‘3 RE is waving With A a, speed n[ 39 mfg :1: position A as shown in the ﬁgure. The Vertical height at position 9 200m above ground level is 200 In. Neglect friction Jlsom end use 9- = 10 mien, What is the trite] __E ______ __.__ -__?n|:nﬁn]Er.er331-I:I energy of the roller coaster at point '13? [it 1.3.; x 10't .1 [B]2.1}2 3111“ J [\$1.00 x 10H [n11.33x 1053 .133 x 105.1 ['6] A Dido-leg baseball is dropped from rest. It has a speed of 1.20 tufts just before it hits the ground. It Dehonndﬁ with u xpeed of LEI-l] mg'e. The hall in in contact with the ground for 0.0149 3. Vii-That is the average force. exerted by the ground on the hail during that timid? [A 2.00 N upwards [E] [LEE] N downwards [C] 12.9 N upwards @210 N upwards EE] 22.0 if downwards [TE A EDD-kg cannon fires 3 4.0.1.23 projectile with a velocity of hill] mf's relative to the ground. "What is the recoil Speed of the cannon? E11: 2.0 1113's .49 mfs [F] 5.0 131,33 [D] 8.0 mfs EEE 143.9me E5} A wheel that is rotating at 33.3 rndje is given an angular acceleration of 2.15 mtlfsa. Through what angte has the wheat turned: when it: angular speed reaches T‘Eﬂ Fair's? 3.5.; 35.3 rad [B] 33.2 me {o} 223 red in] 333 rad @003 red E9] A child is riding a IIltrrergi)—Trnl]tft which corngitetcs one revolution every 8.39 s. The child is stamii an; 9.55 no £1:le the center ufthe mori'ngwxound. “that is the tangential speed of the child? [.t‘: 0.353 my; 111‘ 1.30 this @343 nijs [D] 3.34 mfs [E] 33.3 rule's [In] The moment of inertia. of a uniform rod about its center is given by I = M'Lzlfl‘l. What is the kinetic energy of e 121]ow rod with .1 13131.35 of 451] g rotating about its Center at @0330 J [13] 0.700 J [C] 2.10 J [o] 4.20 J [E] 5.13 J [13] A block of ms in = 1.0 kg is held at rest against a spring wiih a. force constant 3.- = iﬂﬂﬂ me. Initially the spring is compresseci a. distance a! = 1r] e111. When the block is released it slides across a. surface that is ﬂitdiioﬂiessl1 except for a. rough-patch section of width .5 2 '20 cm that has a. coefﬁcieni of kinetic friction pi: = 13.50. Take 5| = ID mfeg, and ﬁnd {3.) The initial mechanical energy of the block. [12:1 The work done on the biock by friction when ii. crosses the rough patch, it} the speed 1) of Hue block after crossing the rough patch. E uihbrnum ML=ii<Xitii<bil ciaosition L l l h—ci—I-I E =14. more: i“ We“: ! ...: 1. l. LI 4 i _ i rm Li L _ A -r--».-'»'..--..-ea.alx‘.em:-4n~ﬁ=:.-eaawo EU 4. Law l—34i i u . in) W}_-}lk5, lie-fa-MJ {ﬁazmvmjn} _:J szj’ WP: —‘l,|1krlbijv5 :-D‘33.f’0.jD-U,Aoj:- to} “H ‘u I 0 1; FORMIULAE SHEET Quadratic equation: I131: -|- 52 -i- c: U =35 Z = 44—. 2: 4“ Vectors: J: —‘1:-i:‘ + “H.731 .11: = fin-3853” .11” = Asin 3,41 .11: MIA: —E— Aii' tennis. = Er 5+ 15? ={A1 _ 5,): + [143, + am, .i— E = ..i+ [—E}, as: F, —s. a a Relative velocity: if“ = '5'” + ”an: 'LIEHIL = —'l-."*h Motion-at constant acceleration: en E EU 2 l1}, 1TB E 1TH: [1], Eu 2 22.32:. . — . I II: I 1.2.3: +ﬂ1ﬁ. z 2 1'0 1-35.13 z : zu +1135! -r 13413112, 1? . ... . '1 u: :Iio 5‘ Free fall and projectile motion: :1 = g(—y]. 5r = 9.31 mfs’, R = 43¢! 5‘ Newton’s Laws: F“; 2 :12: = m5, I35“ = —1_'—1AE Weight: i-‘V = mg[-—1}]I friction: kinetic: fa. 2 mi“. static: f. E ,u.N -' J Springs: F = —i:*."., circular motion: 5:9 = ”Elf—T], fun I 7": IL 1*) [13: A bullet with a mass m1 = 5.1] g, and a speed 1-3.- : SUE] Info is ﬁred at a block of “fund with a. mass 771;; = [1.100 kg. The block restn on a. frictionless surface, and is thin enough that the bullet passes completely through it. Immediately Elite: the bullet exits the block, the speed of the block is on = 2t] mfg, nod the},r are moving in the same direction. {a} What is the initial total linear momentum of the bullet-block System? {1:} What is the speed of the bullet when it exits the biotic? (1:) Find by calculating the initial and ﬁnal kinetic energies of the bullet-bloc]: system how much mechanical energy not lost when the bullet passed through the biock. ' .2. . >2 “‘3 0‘) E30; : Maw” If; btrur E? if I 1 (ﬂxw‘get) ( tea FHA) 53 :- Zﬂligr f; 32 .5 :1 J A -_'> .1: ; b) 2P; : 5* F'.‘ '3 mettltrvﬁtmtﬁ ”(that ”a: blank 1 p. J (SEX ”ﬂit-tr) {it}; tour) + (1.0%” erZGH/‘EJ—l “; Eﬂke “13 J “harm/é "- Mic-butler : I F; ‘9‘: 5‘on 15.} Vf’hilt'ﬂ" : ZOO ”/5 L- ﬂood I n- "; LOQHAJ .- 5) l<E."=J'L?-1v1': 363.07le l‘rN . __ UL :: lath] L I bin—J‘- .. i .- m K E41 ‘ 1- ”lib-”J when” f 1 19"”: 5),sz : Ilo‘i‘r‘ioaf 2—mon— "75’DT I'D—ﬁt Impulse and Momentum: 3'3: mil _' I = Fm = \$33 If the net external impulse is zero men P510: = L33 f ‘301151 . t' . .. _ if Two particle collision: ml ‘31,- -;— mgfl'g: = 11111:” + ﬁL2U3f|.c&nlpletE1]r meta-e 1c. “I"?! — 2; Elastic collision: K, = Ki: in one dimension: 1:“ —-u1_,« = 111,; — 1:23“ Center of mites: It“: = Elm-.3143 Emi: film = pm“ rocket: F 7— PF _ a Rotational Eviction: 2m.- rad. = 1 rev = 360“, .s = r31 E: = E; M? 'E" o = n: wmwu-l—od] 51:35 +£L-“é'ﬂ H=Hu +wﬁ+é~o£21 w‘=wﬁ+3a(5—~5ﬂ} j I t '— -' = 4- “ ralli motion: r,- = 11.. ﬂ = (1.: II; .= rm, ac]: : TU I L'.‘.-_ — T'Eﬁ. I]. at? . at, 11;; .1 - I w: ' I _ _ .. ,2 = L1 m L Jun—I Rotational kinetn: energy: Kmt = Z: , I —- E mm” H n 2 :15 g. ...
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