hw2 - so swans nm 1.00 sums 33.144 we saw 214" Q...

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Unformatted text preview: so swans nm 1.00 sums 33.144 we saw: 214-" Q 1‘“ a) With what speed must a ball be thrown vertically from B O chaur‘r’ BE am, 9:, UéE ; _ ___.. ... ‘—‘—r gram level :0 rise to a maximum height of 50 m? (b) How long y =50 ”4 willitbeinlhe_air?(c)8kc3chgraphsofy,v,andaversusrfm" V. _ M the ball. 0n the first two graphs, indicate the time at which so In - O “‘5 is reached. am .. ., a .. .. . 3 “£5. y' l fll wig burgh! 7-0 £35.31; 2215 JwIT/AL ”EMT: V; ganja 71%: 49:93:77th THAT THE FIAafiL— 7- sz» THE TIME 774.9: 31414.. SPENbS 1A5 1.5 ML (-4.95 cm 05;: _W° “9 Lb. n"’5= t; zuu : 9.2.9 at. A3035 MULTIPLJ)’ BY 2-: 2—“ ____i_.--..q....t__' . __ HEW 6 52A stone is dropped into a river from a bridge 43.9 m above _ '. the water. Another stone is tluown Venically down LOO s after the i first is dropped. Both stones strike the wane: at the same time. (a) What is the initial speed of the second stone? (1)) Pint velocity vetsus time on a graph for each stone, taking zen: time as the instant the first stone is released. PM?” 2224“: A PIWRE #- Loam—E A' 25?- szl I must?) Peon mimosa min-t 33.3.1: , EEE 0' Q 17.5w AT 1.10. vat... AT' 1—5.5, 3.3} .' EEE i .x. (a A car moving with cmslant nucleation covered the distance GIUEM o between two points 60.0 m apattiin 6m 5. Its speed as it passes' the second point was 15.0 mls. {ail W'hal was the speed at the first point? ('3) What was the acceleration? (c) At what prior distance from the firm pair“ was ihe car at test? ((1) Graph 1 versus 1' and v Ivmusrforfllecax. from rest (r = 0). um um: 1924;,— MwA PIJURE + Loam-E A “Ry-T Pkwy, sweat «a; A25: gym oz. AQKEZS IHFDRM'DDN. AT” TM 5) FF? P75 macaw-$955 A BC; T'a DENaTE 7W5M. 50 SHEETS 21-142 lofl SHEETH' 21-1‘4 300 5'4““ A -- . ‘.__..L_,__._........ _.-..“. "firm" . ® 50 SHEETS 22-142 :00 SHEETS 22-144 200 SHIHS 22—!“ C-HZ‘x',' ' HKWS [OS-A pamchutisl bails out and freely falls 50 111. Then the pan» chute opens. and thereafter she deceleratcs at 2.0 mfsz. She reaches the ground with a speed 013.0 mix. (a) How long is the parachukist in the air? (b) {M what height (10:5 the fall begin? a B “'C- = + 2:53:31 ' 8“ VA - O "5‘ - 31‘... V‘- "'-3! D 5 . m2; ‘EDTA-LJ'ME PARAanm- 1:; AIR. man 2 'b E»; $EM‘25T5LV: 5-D 7H2: WALT/ME: IS: T=n43+1§+c : ”' :-AWL [B a Fm» 7795: 779714;.- bleANCE we '71:: FIMD XE min—57: SD_DSEi §c> .7345 TOTAL 13:97. Is: 260.“ + Sou. = W 50 ‘SHEE‘I’S 22441 100 sums 31-1“ 2003"!” . 22-141 @_ Cal-{3' a‘=(3.0m)i+(4.0m)j aml b=(5.0m)i+( 2.0m)j give 5’ + b m (a) unit- vegtot mttation an_d as (b) a magnitude ant (c) an angle (relative to E). New give {3 — a in (d) unit-vector notation and as (e) a magnihtde and (f) an angle '41; Forthcvectors - _ I | HKWS I h I 'E -—- .0] +Z|o aim : (s.o\‘+(a.c§§)?' :2: M 77;;- omeéamflom (mum? 715m: Pas x- Axis) rs Galvan my; ._...—-' fl, (2' :23— -< Dem— 13-2 (5‘0 Fwy-(30' 03 ::.(Zo.—é=oi)w W; M?" ##3#va 5:5. éJVEuB‘J’: )Tj‘fi - . ' ' Ira-'61 =-‘!Cz 031+ (-6.) o -— _-5M& ofi- Emu-rm” (BEA—AT 3% 7° 71:15- 7’05 x—Aws) rs. GIVEM an 1' E 9 i? I’m 2 O =®fl 50 '5 NE E15 22 I42 I00 SHEETS 22-1‘4 200 SHEETS 22-141 ® ‘3: “we vectors are given by ' a" = {4.0 m)? — (3.0 mfi + (1.0 mi: and B“ = (—1.0 m)? + (LOmfi + (4.0.1011. ln unit-vector noiation, 11nd (a) a‘ + 5'. (b) a’ — if, and (c) a third vectori’suchlhatfif- b + E’= 0. m LNEN" 3t: ’1 ?-‘53 1- L =—T+I+HL Tm; 1's. :rug-r 779g NEGATIVE 0F 7»: mswék- r“ “Pm-r ._ “1‘7. -_._AI__._.____.__.~..' __ _ I HRLO Q I a T H.111: tWDIVectors E” and b in Fig. 3-29 have equal magnitudes . i' of 10.0 In. Find (a) the 1 component and (b) the ) component of I ' N OTE -' [as - £90 3 S E I their vector sum F'. (c) the mag- J, .hitude of F. and (d) the angle F‘ makes with the positive direction of the: axis. sun ttw m Fig. 5-29 Problem 2]. 50 SHEETS at =§ _ - _ -_J_ _. §§ FIR???" F’mb 7mg COMPDMEMt—S fall a AND bf. 23 I at: _ .1 22-141 &y =' A WCw") :: 5.00“ I I i I I I i I Ax=AMC3O§=KIG€Iu§ g K'ax?+6{7’\ ' ' I I I I E] Vx=¢x+I>I<=W=Q§ “" ' r A' I ' l"=-Fx?+ A f‘ =A ‘1‘ :...:Q 7 I I y 7 L7 =Q.§??+ 207’: v-It /—~._ 50 SHEETS 22142 100' SHEET: 22-144 200 SHEITS 22-!“ ® _ 3‘ J Use the definition of scalar product, E- 5. = ab :05 8. and the fact that E- F = all), + ayb’; + art): [see Exercise 30).“: Calculale thc_angle Eetwetq the My \rcctors given by a‘ = 3.0i ~I_- 3.0j + 3.0:: and b = 2.0: + 1.0,- + mi. ssrn nu ma: fl . d “I! E- :5 Mb)! *arby+qal’-z 2' (33 (2—3 + 0&6 + (93 (a3 .HRkJé_ Fm new . Q 21-! :I 'ggggus . . Q 09 _ i 5 was 209:er CH3 — ' HRLAJS Sé'Awtl'tee-lw‘itharaaiusnféifl I ~ I I- ._ -' . $5,723 EIQMCEANEMT IS A cm rolls without slipping along a P vac-3'0:- J 50 ”5' M horizontal floor (Fig. 3-27). At . - - sPszJ F 1' ET (WWBIY time n, thedoleainwdonthe usiflfiv CDHMNSEHT" rim of the wheel is at the point of .. ppm .912. MONITUhE contact hettimcn the wheel and -; AN}: mIEWATqufim. the floor. At. a later time :2. the 3.5.1:. ;--...... ._. __ ___ . __ wheel has rolled throughbne-half At time ,1 A. time ,2 _ EH. K "' q 5‘ O "' M of a revolution. What are (a) the _ ' ' _.._. 0, ”50 M magnitude and (b) the angle (set R" 5 H Problem 7' alive to the floor) orthetlisplaccment odeuI-ing this interval? sum t—mr-E' bLoaATEAfiEJ-TFW‘ X l - . fifimfiz @ I” mennvbf Rot; OXIENSTAT - 131 ' = — W 5:) =63 ' M 6‘93, :2) e:'7;q-I( r—H- $33 1§§ man 0 m§§ 333 ' '77? 3332 ® F... 45: t vector If of magnitude l0 units and another vector (-1. of magnitude 6.0 units differ in directions by 60'. Find (a) the scalar - product of Ihgtwu Vectoe's' and (b) the magnitude of the vector product i” x b. 55th E’T: = flame! Com m '60") -30 EEEEE = 003(Q ALW‘CGD") I 2'- SZ Hmoé 50 sums 22-1l2 100 SHIE‘IS 22-!“ 200 SHIITS 21-1 41 [1:] m1. Ashipselsuultosailtoapoiml20kmduenorth.An GWEM- 6L. = [2.0km unexpected storm blows the ship to a point 100 km due east of its _ l: I starting point. How fax. and in what din-action. must it now sail to b "' I ‘90 reach its original destination? k L) ” 9 \ n.1,...z THE 2:; TAMt-é.’ 77%;: gm? Husrs An. 15: \ I l \ . S I \ . . a} \ c MW 1 \\ 1'3 can/En. 3, : ‘ .' \ ' e :m'C—Q: -1 I g ' ....... \ 9 ' - 50 TM SHWMUE.) I V”! salt. It!" 3:13" weer ' of: NOR—TH Fall A m7.— ”_—I:___ or: 154 P—M. 50 5mm 22-“: too sums 11-!“ no stuns 12-14! c443; -— 2‘5? . Maw; 28?. AgolfcrummmepunstoLeuheballinwchehahThe firstputtdispiaocsdleballl2finoflhdnsecond6£flsoum E: h Ahab, andthelhird 3.0t'ISouthwest. What displacement was neuiedta ' J - A gameballinlolheboleonunfimputt? Y; _ +36%)” (10,); “Araf- 1 : raw 81); mm )’ WEWLAJZNE LEA-64'! yam—or. 1132M; - c5}: 27:75 coMPower-s S. A's: “EL": Q3106” “”5 W“ I; [email protected],. 0&4», 95")?“ -— (6 cm 5.405"): §__uT' “'15“— m‘iS": E. =3.oJif‘T (A -33 w c.—-‘(3:0‘r"f MHS°>?-C3'°fi may): 2-3 0 2571+“,- C, +33 7—745 MET):$?Mc£MEMT F'- ze‘. mus. swam: In: “mamas 9F memmba my; ORIENTATIOH; mm M5 W(:.6:>eb 03$: . 50 77!; bis-PLWHIEW" Means—2b 7'0 ea?” I5 6.0Pr AT 2l°£A$T oF Mon-TH. 50 'SHIE‘IS 1244! Ion IHIIYS 21444 200 SHEEI’S 22-1 41 ® LU a 5-— 145‘ “E. fFor'ihe mots in Fig. 3.35. calculate-(a) n- b. (b) 3-1:, and (c) hoe. - ' .F' M” FIND “5:. I3: I . —. _ Z _- d o‘- 1: l q 30 1E NEXT we: FM) 0“; - 130 6 . E: =CHXS) @093“): ._A__________ Nana-1 .Ap A .4. c. H3 ...
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