HW-7-sol

HW-7-sol - PHYS 34M Hw ’7 SoLUTtONg 1 Gianco|i6 6.P.060[3531051 0/1 points Show Details A 1100 kg sports car accelerates from rest tokm/h in 7.4

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Unformatted text preview: PHYS 34M Hw ’7- SoLUTtONg.) 1. Gianco|i6 6.P.060. [3531051 0/1 points Show Details A 1100 kg sports car accelerates from rest tokm/h in 7.4 3. What is the average gpowerdeliveredb the engine? VI = Em J M .000", x? W M 36003 ‘1 km 3? 80L: AV_ Powgh : 1 EnCY‘flZ tYQMSFsYWC’ v Mme” time F7 = 14. =. 53 z F’v‘ t_,+*———->. .Wcm t t F —~r-»v~~--~-——-+ cl 1 1 m ' - accdeYoJm’w a = vl—yf” : [29qu /S = 3‘57 m/éz t 2-H s V = VFW; z 0+!D6-3‘1‘m/s : wigs, m/S 3” '2» AW‘TW Powev _ ‘P‘ :Lkg)([@%)(@%) = FEES-1w \{2 = Fag—53'} Av' ____,_... Altev'nahvc wax P CY _, [5,“th tmeowmed ' EFF: 0W » , tt'mfi Ma) rs 019 cm pmn {>01le H» t“ 09,43 eneraj change; ehe'rav , St'nce 't/ Potential mm '8 I 4 { HA K CHQHC 54' #3: W o} 50% F '1 D 1 Y lrhvt'lHfi/f _ 1(l‘j)(26'e’c‘ 955 '1’ I __ AK - LP?!) 1 P —; T .. FI.L‘ S t "5 -; [517“.3l'w : ‘s‘litoolw \‘// t 2. Giancolie 6.P.067. [3541511 0/1 points Show Details l 5 During a workout, the football players at State U. ran up the stadium stairs infils. The istairs are @m long and inclined at an angle of 30°. If a typical player has a mass of i 105 kg, estimate the average power output on the way up. Ignore friction and air a resistance. 1 i g3} (Maw cl = m l hfk G = 30° time. ‘b = E, M k mags M = los‘ l4? Avwra‘gx PoweT R_-___‘__[L,,_ TS _. my brains Prrmtcl ‘_ L = sine q70unol firm a :3 : dSl'he'l. _ l l kl) Energy Hansfiwmecl (t)ng Flawless) Fvo'm Fern? I by F01 Foknh‘cd, SYNC: Hacj Shawl-Cal From All" is Ohk’ - l j PMM (‘1 (ll/loll Al—DFFUJ on Teaching Fedhi' 2 O r '_ m L, It mg cl sine E 1' a.) I mgkll m/ ’ _ :31... it t 1: 1m] w. 3. Giancoli6 6.P.076. [353168] 0/3 points Show Details ‘ An airplane pilot felll355lm after jumping from an aircraft without his parachute opening. 1 He landed in a snowbank, creating a craterl1.1|m deep, but survived with only minor \ injuries. Assuming the pilot's mass waskg and his terminal velocity was 50 m/s, estimate the following. 203'. (a) the work done b the snow in bringing him to the rest r—xrfiafl J (b) the average force exerted on him by the snow to stop him r“_— xm N (c) the work done on him b the air resistance as he fell r—xmfii J ' l J ' J. V : 0 Pilot tau fiom pom: L +0 amourwluxxmtl) . ha: h On a Show bo‘hk . anal leecl at Cretch h . l s. ‘y’ enigma olul; in law.» a Cw;th moss Yh: F kl: o Vstom/S. c. h = rh — -— ._ TL -—,'r—-~ qVOUw \/ = ii ll m ‘gly/ WV : Va: 5° m/s “32'5’ 3 O Wonk ’G‘r‘re‘rgj lwmm‘lrxte is AU — W A K + NC- Kfi Non ’LOhge—(Vaiwvt wwk 5/ Eleanor, in 0“”— i’o Ph’cin’ohoJ W - Folihlo’oui) Enema 3 \thc : " @ oi f e'n'ehqicg La) l/Jkuq Flick Dread:te Oral-w in Show he axis . O k ’FYVCHOYlQJ p070. aim in Show QQHAJO ukvo diet 50 year 5'0 1 s‘houo is Tron- Congewukve. Y) b o J 1 L mg/Zo' N ’ N W, __ - -+ 93 L NC gall”, losnovo I (/3 (b) 1 N3 :- —-_lm\/: -|— mg(..(y) : -m(ivl + '1— : “(whfl‘ha + (q-gaflw) — \/ : ’iw ‘3 =‘- —!l-o;>_xlo‘°’3' / Fara Qxefl—Qd [r33 S'hOyo on 13"(01: masmlhloh offimbhu '3 Now/Hm. . - : —— F A = __ CHI, N5 ’ NM; :0 5‘\35 Para {XETkfls Show F : — NNC S \/ F ([email protected]) : H18?)an N z ,_ S / mm = HQQOO —» servahve b CU'Y Yevcl'y'CmCJL 1'5 adsO hon urn NOWk C‘Ohe 3 ’f(!) +0 (ll Qho| L'E 1's mlermf Pam Fem Ak ’\ AU '/ WNC O L- i v + W}- man) N '5 J3:sz ) ‘ AiY m (Lg?! z va— mats ; a, ': .Z 2, > 2 V 5. m _ Mm las’l > NA ; "%)( —.Ii(5—° L3) C 5>>< VT / -m T I flv \/ / m j H n If you stand on a bathroom scale, the spring inside the scale compresses 0.50 mm, and it tells you your weight is 700 N. Now if you jump on the scale from a height of{1.1lm, what 'L ‘does the scale read at its peak? x -; 0.57; mm : 5xm-H m. i V 3% Ell-P Bathroom Scaie veacis “floormoJ 166ch (N3 / voiu‘ck is~ N = We} Normal flora is Cori tract Forge i/oiVH'x mkrck Scale rs Flushing, You Back ’ W ~ o'moudot. Slhu you are Ismassinj sate Wit/k Somme . i—o Hm: Wins hovmoj Fore. [S tr» Fuvn OTISM’ia’ O‘UQ Sim-Una} CX‘LQWQHQIhfl because of you? Compression fa‘YQSSih Ct SLfi'haUn% on “2.).4 a 6.3731]? CQAS'fiY‘JC- so} I F kfl pokQ'ye I )( ('5 Comb-trustw, 1 N 1:. 5 : x L! i‘v‘x ‘7' : )l ‘5 ii m___‘% ’ - k ’ X s‘xi6'"'m . ‘ ) \AHieh \fou flow {Sean} 1. F0 on SquQ you Compress comp—ru/Siwv x So State Wm TC“; IS ‘ , _ 5 K >< fiew’FS ‘ we Mm iv “Liner 'x,’ JP-vsi—v Since energj l’s congewvw m Hus Process- AE : AKJrAU = 0 AK —r Ava -r AUS : 0 z \ SW amw'wh'svno} fool-can Hob» Q‘OCW%H ’3 o k“ flurry} 3. U5 ‘ .7L'kx ,4 O O O l Jiyflz’ $745+ 93%;.- mgk, + il-kx,» = o -, WWW” "\. “xv .... _.,\ _L kxll = mgk 71w; Fol-embed AHCTfij you/08% L [206% ED Sh:ch i50|€vaUJ X‘ 1. Q ehe'r‘afl In 5Ffl'nj k (kg/va WMflAfl)~_/’ , ><.: QKVOONXE") , 0432,59. m, I ,, .. _________,.._—-— \ H woe N/m ' Soak ’5' so have. “rage/Una on ’ UL’XDDG’N (jo-oaaz)m) N = E = ’0‘} ' 7" MCVO : WEB N _/ \// 5. Giancoli6 6.P.053. [3541521 0/1 points Show Details Suppose the roller coaster in Fig. 6-41 (h1 =m, h2 =m, h3 =‘20b passes point 1 iwith a speed ofm/s. If the average force of friction is equal to onef its weight, with whats eed will it reach point 2? The distance traveled ism. X ‘E‘ m/s 3912 ‘Alork thway Pflhct'w l‘s AK + AU: Wm Nm conservative New); r5 AM ha chh‘onodl 9376:: which OJUOCA‘j OFPOSQ$ all: Placement- WNL: Falclcos¢> 2 gains—H80" 5m \3\ vi, ll Vl— \§s V F: a MS V l El 6. Giancoli6 6.P.055. [3541541 0/1 points Show Details ‘ 0.580 g wood block is firmly attached to a very light horizontal spring (k=l GOIN/m) i j as shown in Fig. 6-40. it is noted that the block-spring system, when compressed 5.0 cm l and released, stretches out 2.3 cm beyond the equilibrium position before stopping and Eturning back. What is the coefficient of kinetic friction between the block and the table? 3cm 309}. Given: mass of— bloclc m: ID'S‘a‘olkg spring consEqu k t [Ell m. » -1 )g: 13-0 cm = EOXlo m _L X1: 3'5 Cm : Q'BXlQ m. N0on 7 supra? Fm'r‘flble ’3 ~ AK +AU = WM Norx— conservka WWIL Lg elm h: F :- Bother» between block anal +0-b’c' 3% Z H A .Y‘ C 'M 3 0‘ L03 ::. —- F Cog rare” 010 kl C, L V0ch Wf E? 3‘3 H S‘mu we awe Co‘hgloh')‘ block—Sb'hnf l-ooelvlne'r _ i F A , QC % x ‘l’ ) '1' $)Sb7u\f i Aka) CK 1 Hook 1 (x +x K o k : fl/u mg l 1 0 O O O + J- k X:’ Ji X‘ K J—W‘ ; '3]: l + m 1 — fl) 3’ Jr __ m (X +><2 . k 04— X. ) = Pk 3r ‘ ’7: 7. Giancoli6 7.P.003. [355785] 0/1 points Show Details §A:0.145‘ kg baseball pitched at|36.0[m/s is hit on a horizontal line drive straight back i -3 gtoward the pitcher at 52.0 m/s. if the contact time between bat and ball isli .00 x 10 is, 3 calculate the avera e force between the ball and bat during contact. ‘Ejfl N 904'. qi'vgh‘. Mass 04‘ base {34.04 m; 0qu kc? VtiocH-aj (Vb-l : m/S ~e...__.._———- quncd VeJou’i—j [V5,] : ‘507‘0 "fig —3 + é—d—‘ afiv‘reck'w’ nerd—ed Hme t = {ilooxm is. Foyce z r3 Chan? in At; F : OFPogll-Q drreclvsw At F = mtvflvd Lo-msk‘aflrn'og) “MEEWS )> /__________________. At (lboo K153 '3 'T\ u E (b ‘0 0 Z 3. child in a boat throws [email protected] package out horizontally with a speed of 10.0 m/s, Fig. 7-31. Calculate the velocity of the boat immediately after, assuming it was initially at rest. éThe mass of the child isg and that of the boat isEflkg. Ignore water resistance. Magnitude m/s Direction ( (o) in the opposite direction to the package t” (_) in the direction of the package x 90415 grew. mass of Fackoflg mF ; lé-oolkt} mass 0; Chick ma: [Bo‘olkg Mass 0; E0031” mg: k3,. Haw“) VP, :1 ‘l’ ‘0 m/s (“q posih've d-H'QCLYW ) S‘nu “were rs no hel— Fara?— aal-mi, momenl-um I'S \ A CowSeYVeoi l- e, . AF = 5’15" = O :- 0‘ m \/ bk :1 m3 VE> 'l' The/Va + F F / I ' m V, + mt. Vc at mPVF P; r5 9: i I Velou’l-«Jr» VQ, 7 Va 0 : Lam‘s-i» me) VB] -’r “‘13 VP l m VP’ (“3” '0 m/s) = , b-Voélm/s V :. "’ P 2 ’ & mam (kg, +kg> / IE is in oFFogilrL olivech'mx to H“), package/L' 9. Giancoli6 7.P.008. [355783] 0/1 points Show Details 9 boxcar traveling atm/s strikes a second boxcar at rest. The two stick ; ge er and move Off With a speed of [Em/s. What is the mass of the second car? immfifiaww 80/0»; five/7): TY“ I ——> I \II : ill} VHS 5] ~ El : 0 ml . .- - - _ ’ ’ v" g aollrstem I'lm, f m1?) \/|=\/1 :V :ll’BIm/g ‘”'“’” V hm ; 7 mi "la—‘5 'Jlu-s I‘F inclqska (Ollislo'h. SMCQ he} Fay-u. acid“? on Mo c. onSeYVfiCJ . box cgwg rs y’m ) Momehl-um rs O I ) mlvl + m% '2 m‘Vl ‘l‘ mLV> m‘v} — m, v + m; V —— m 2 m ( V\‘V) é may : m‘ \l‘ ,V l W: mmw>e(flflemmg—fl%) l 10. Giancoli6 7.P.010. [3557951 0/1 points Show Details ~ , iAiSSBOI kg open railroad car coasts along with a constant speed of 8.60 m/s on a level ‘ gtrack. Sndw begins to fall vertically and fills the car at a rate of 3.50 kg/min. Ignoring gfriction with the tracks, what is the speed of the car afte. 85. min? Kiri—3.41] m/s ' Y R: - , . 46M. V .33,— Smu Show is rm} IMM‘JH ax TAX Va‘rl'l'clej clown J net . ‘ ma i———-> - AF’rsTR'S‘ofmn {rs-7% in )(PoQJ‘YQQ‘fl‘W ‘S unabated-col bi, tritium} pO‘a'CA Email} Vie} [h \i»o\l‘f€.C\fi'W 1 So momenl-um In x—c/UTECHGM is sh‘ll dowa‘YVLo’ LUCY ma : (5360; Kai, Vex: ? Co m/S QHOVO ms LB'foJfl 3 ma ch 7’ me+ m5 ) V§x V} : W‘c ch : Q '43,)( 3'5" ’95) x me+ms) (167860] k? + lam-gt k3) 11 . Giancoli6 7.P.O17. [3529201 0/2 points Show Details of mass m =\0.055 kg and speed v =/s strikes a wall at a 45° angle tennis ball and rebounds with the same speed at 45° (Fig. 7-29). What is the impulse given the wall? Magnitude “FEE N-s t Direction ' a C (_) in the direction of the ball's original motion :E r L) opposite the direction of the ball's original motion r (0) normal and into the wall “ (_) in the direction of the ball's final motion r (_) opposite the direction of the ball's final motion X .3332 Imludse = FAI‘: = AF- Sn i had velocity Vl' = V {1E m/S flamed) ve'w cl'hj Vf i V; Lg] MLS y—Lom‘wheyxfi' of \I and V3C (imagm’l-uol—L B [VL-xi' [\QX‘ 7— Vsr'n 9 y_t°m¥when} Os; Vt. anal V5 (mm3,n\'MLL ) V £03 @- Wit/Viki: lfi mamenl-um "7 ‘I’d;mkw Chang). E e -vsme) = (QWS‘W " ’ — ' e h, l' Mame;an m x A“ Ab‘x; m‘gfmviz .. m( Vsm \// 4; avg; n ' o 2.5— N [AH :[wlmvst'ne]z+z([email protected]%)(gfim/S)Smog ., s T: Peybeflcll'cuiaY/ X—O‘l‘Y (hm/mo) l-DUOOJJ.) and e (:0le rs in ImFuJSL fil‘ffi‘fl *0 1*“ ihl‘o the WW’ 12. Giancoli6 7.P.018. [3529141 0/2 points Show Details You are the design engineer in charge of the crashworthiness of new automobile models. Cars are tested by smashing them into fixed, massive barriers at 50 km/h (30 mph). A new model of massll OOOlkg takes|0.12ls from the time of impact until it is 3 brought to rest. (Take the positive direction to be the original direction of motion.) (a) Calculatethe average force exerted on the car by the barrier. l Xe- N (b) Calculate the average deceleration of the car. I 76m m/s2 800m RMYW Wu F ; AE ; akaan l‘h momfrdrum At Ckcmfik r‘h h'mt L U3) F : maA ~llEsj§°JN> = (@ttl“ H6 “ISL ; -—— “5.17? “7/9. 1 '1: a 3 5V / decele'rodw'oh : ’m [S L ...
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HW-7-sol - PHYS 34M Hw ’7 SoLUTtONg 1 Gianco|i6 6.P.060[3531051 0/1 points Show Details A 1100 kg sports car accelerates from rest tokm/h in 7.4

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