dynamics1

dynamics1 - \\ Test #1 DYNAMICS Name: - " 1) A...

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Unformatted text preview: \\ Test #1 DYNAMICS Name: - " 1) A freighter of 42,000tousdisplaeemeet4.1103312990111315tests from restmlf the resistance, in pounds, to motion is 7000 times the velocity where the veiocity is in feet per second, and that the force exerted on the ship by the propellers is £20,0001bs. Find: a) an expression for the aooeleration, b) terminal speed, (2) speed at any time, d) the time taken to speed up to nine—tents of the {ermine} speed. {meets mm e 200011976, 3'53 2650”! [be BOX IN YOUR ANSWERS I _,mfl\ . m I ‘ 9.“? M61 Hos. SCORE f 3% ’ [hf-s z . m st.seest6'i“3~iofi‘3w“§* M "(RN21 70be Hi)"; 1 ,Cfl-io *.(><eé‘fi-<-o'i5 “" (j \ mg...“ m . —:. «‘bbfierkg‘g‘fi d” “90qu :N'WFWM W NW“ “NW WM ~-~— =/ ' M, A. " #0 C L (6 ini‘odb} ,0. _, boa-19%: m i I Mb <2. (1% f‘mfi' v.093b‘6t = )0! , mag-k. ‘6 436% ~~. mafia \J r .. Mum—w 5 J2, , 04 (a w. mam! ioqb w, macaw ,ofilw ‘i‘i .f‘a- ‘3 w ‘rwi ’ “ ,..r 0%(53 p903“: J ' . m .l. ._ m. 1j..- ._ . h -1. m ' .____._‘_.I...u.". ..u..;-.;." -' g... “wkw 'Wlmn. . " ' ¢r,m>_p;.amg.%mggf mm W - ¥ 0‘0vi :- 2 . to ‘1 k )0 ‘. , :Whmm- _WJW f. __ V .. ._ “Em-:4“;me a niifléfifilflixg; év 8’3} 3.1V? V w - , . m m“ Mm“ Mm _. “WWWW_M_M~WW@§LV_ WWW__._.__-_§__l_c_l.:W-:m __._a_5_9_1ggs¢a:§_:_gngg.gma_1_.-Q:E-_mywm._-m;~__,_.;.Mg.» - I W “WMWFM “w.........._..._......WW~«uMM...M~.._....._..WMMW.MQMR_~_..A.. , . -n,‘.....wwm....h.... ‘ ———~_—-.ww‘wvm>r—.HMIA—‘H———.- .‘m‘wmw~m.w .wmmn ... -._..~____.__‘m-..,m aggm: fire [H 6 — W x I0 “ igmzré <62» m ‘:~?’?J:§{_C WWW‘“...‘“WI-”__mmwmuwflmmmh...m.m.MW..mmwmwmmnm.-. ..__,_WW ‘ a mar-mummwummMWM... “www.mm Maw .. £3 "3 w WWW”W.M_m.m:.‘...m,.._... NWwfzzmTFgrrfifg_}fiwM.n.....r._....-Wm...wp,uy,...w..w,...,___..-,wm,-...M.Wm.n_~.;.~.d.~..w_.;.,.,;;fl.;n,WJR...-.-“...__mm-...m....,...u.v.W... . H, mm. . a. . z: H5uq.9,w’i,lr.0f.__‘7'.Z,<,{25f x19“? a»: ’W":1::fjffW:IifiifiQ:"71:3:I; u m a w wfiwv,_h_.._‘._}“’3z_§fgx w m r 5, be , i _ (flu .2)“; a: O %=O {iv/Lg: \[LVB 2) A batter hits a baseball A with an initial velocity of VC) 2 100 file directly toward fielder B at an angie of 30° to the horizontal; the initial. position of the bail is 3 feet above ground , leveh Fieider B requires 0.25 Seconds to judge where the ball should be caught and begins to move to that position with constant speed. Fielder B cheeses his running speed so that he arrives at the “catch position” simultaneously with the baseball. The catch position is , ‘ - ‘ ; _ the field location at which the ba11_a1titudeis7 feet. Find: a) time of flight ,of the ball, b)_ speed of the fieider B, c) veiocity of the ball relative to the fielder, 'eiJraditisef curvature ' atZSeeends. ‘ . - - ‘ ' BOXIN YOUR ANSWERS I i V ‘ . ’ e - 4- - ‘ ,t ' £3 “‘EOQQEIN - ' .; - - I X :0 \{0 ‘t “ O.» ‘ {GIN _ is “3193*” t , W i ' {.9 M ___:-.-;:;ii.... ' ‘ at; a": Knows ‘ I WV". 39’1") ' “' ‘ " ' ‘ 23)IS‘9{.‘CWS " 423% - 7 Mess em 9 35'” 5.- we?“ in ass him ‘i'if\if\5;‘§3if. \t 2 o it": 6i": \7 ii?- iii $3.533 «it «2 A Le it?" wit-"fit? ic- _ Q :~\\s\\4{?’%%’_0*«” I Afiwmt ' . ‘ \Q ' 990 W Yofifl€{ mm) Vfigpfix‘fifi Ugwwxfi) \JMB .4" ' w. _l W E_"\‘ . 7 W. a web”? teemfim“ 3.0336 ‘Sfitcmie in??? EVE ‘ “’5 5’ 2 WI .3“ meg-m \iflv‘tfl «SSW-39Cst _ ' ' 47l5 ‘ “fig; ‘Nt MS am if 8 to mom i ‘ .57 54 CW? ' arts {05% "I 63* Mme-Ii ‘50 \JWN’}€? g“... agflqv tbjmpgx L ' ' to, \HWP i - MAM» 1935 Ste '36 ” D h ' “f 993W? 3 “:7 -~....—-\ I 1 * E E f“ g j X—dirediowl . ® 3.. cureaw‘on "i 61Xer ‘ 0% :“39t52 WW 3:22:52“ 1. 5- w . , X: %6.60t ‘ ' 7 ‘ ‘ I: I‘ i __.. . ‘ ,, l5: mxgdl‘t} 4»- 50C f3) (D) Xw-(ngMmqa): .jélfih‘a‘j ‘ 9:21» - , ‘ ' , ' <1 w: 361.5%an = £41.55 ©f; : £16.: 64—1501: +3 b: !’3)o&>~—01 as = Q'Ham .._ _: 4646 1‘ Jo't- ~ _ v wafg; 2—- 41.53 s . _ 03-”16_.It1.+_5¢fi;”‘/. , A. ‘ _ ‘1 a???" . . ‘ . _ . or t6..st?ee_.5qt M , . . - ' - . . - ‘ -' ‘ ‘ V __..._:+b:‘“\}‘ blaqqc, ‘ ' JSOKHCWJWMDX Wm ? vaii€\¢af_.“‘{" .VWF‘ .‘ . . ‘ “539, _' 3,4! 73,35 ‘ @6954 Mimi :a my; +V.r3/;:._._n._ 6":- 0? 94.35 ;:> 3x29? ‘oiho'icrégsflm V533. 53-,;\(3.o;‘wm ; +wr.qq - -- ~ . ,,‘__,./10’“_Mz'sh‘o Vssx :g-‘fi ;»-€0 ._ '1'" 090. 5122.49} ' ' V5.1: Mia "M “*7” ' I ‘“9H5G7Lffl'figéiz60~éfimlqflfigi . ,8 «91:23qu (fa-£04) al9£.,_.‘j'!§~c3 :WMmr: . ‘ fix?» w . E#917ny'.+.‘.?b§.l.p§_ -_p.“-*=;,Ma.;;p JV. \4 4—H,qu 7L was]? a“ V3”: 2 ELBMflHJ—I oi ' 53,5fl'qgtgm WE A F?" 0. :Ma.-. 0 _ $8.6)QBQJ35 1:: " Q48395Q- _ A3 36-6.}.1‘ w: .11: ‘5 "*9er {M3 7 £3 ,1 maan =7 - $162. $4 «Him; 3) At a Certain instant after jumping from the airpiane A, a skydiver B is in the position ‘ shown with a speed of 30 m/sand. stitlaccelerating at 9.81 Ell/82 downward. The skydivers ‘ I altitude is 10,000 feet for the conditions stated above. The airplane(A) is in a circular path- of radius 2000 mat a speed of 50 this and is increasing its speed at 2 mfs. A radar on the ground is tracking the skyd-i‘ve'r.‘ The angle of the radar is 30° when the skydiver is at '_"\_ ‘ ., 10,000 feet, Find: a) how-quickly the radius is changing for the radar to follow the sky mum—«3.: . diver, b) angular speed of the radafic) the radial‘aoceleration of the radar, d)'the velocity I ' ‘ " of the plane as observed by th 'skydiver; e) the ac eleration of the plane as observed by . . the skydiver. ' ' .- I BOX INYOUR ’NSWE . . H . “a: a ‘N‘ _ , N _ _ _ a ‘ ff \, o 7 _ I _ H theme) : 3&0 +x..\:em\ . 7 P g t a - .h j - ~' -’- * ' ~ ‘ «l mm me u 9 WW {fq‘céi ., ' a {5356’ I - :{Q wag)»: in} t “3-K ‘ _ a >. x V K , I ‘ . (WWW-(em? _ . M \ am 4” («Wk ‘30 ((0% ‘50) : O'Vfl‘Jfl‘Fbx ‘ t "W ‘ ~ 7 \ l\?}e§@w~\=2%{j) +2C6i0‘13) e " t v I’ -' -‘ ems» (WW r W M ‘ [9‘ 5 «Costa»; «ms: WWWWWWWW a - q: a» c1 “(:23 \v\ = (p%~“~Pa.=‘“'=°-s '-i‘-'-_.s.sc‘\*2“ “if: :‘3 1 3 Z -' __ WWW;__._____.__;_w___m_m_w. A ‘ - '0 20.0. i I L a”? qgwmm . . . 1 , . ..‘ '_ -_. .-._;_4W7:::i __..____._....‘ww+mmm7mm__._._.;n 1' -'in.“Wm~;_..;4m.gmf“MWwmwwmfwrm MMWWWWW‘M.mum"-..I 3' r . . _ ‘ '- ' ‘W‘»».'.m____.A__.z‘.—..Z;l_'_...,......... 'H,‘ A “mm ’, r50 2 . I , “W. - ‘ H “2.1”. . mam?) my: w a 2 - 62:1 0 ' “1% m - gee. ' e {3’ am, ’3 _ 4) ' Fer the arrangement Shown, block A weighs 64.4 lbssmd‘? ’18 weighs 96.611335 If the: ‘ coefficient of frictiOn between block A andrthe $e=i®géty Biimi'sa‘i’afia‘ aeCeleration of each block,.b) tension in the cable. - L '-. - '. BOXINYOURANSWERS We {é'qrq/bs as. ,q& Tee ‘ ‘ bug) z-Qeerbs, ' M M #:xmA/zm/xr ' " ...
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dynamics1 - \\ Test #1 DYNAMICS Name: - &amp;quot; 1) A...

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