hwsol9
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hwsol9

Course Number: PHYS 2A, Winter 2008

College/University: UCSD

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Homework Solutions #9 Spinning the Wheels: An Introduction to Angular Momentum Part A: Which of the following is the SI unit of angular momentum? :r.::. i"~ m 'J. W.::: .J Y.s 1" h Since.. L-= TiJJ e S.:r:. V n it--.5 C\...r e )S~m:; Part B: An object has rotational inertia 1. The object, initially at rest, begins to rotate with a constant angular acceleration of magnitude (}.What is the...

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Solutions Homework #9 Spinning the Wheels: An Introduction to Angular Momentum Part A: Which of the following is the SI unit of angular momentum? :r.::. i"~ m 'J. W.::: .J Y.s 1" h Since.. L-= TiJJ e S.:r:. V n it--.5 C\...r e )S~m:; Part B: An object has rotational inertia 1. The object, initially at rest, begins to rotate with a constant angular acceleration of magnitude (}.What is the magnitude of the angular momentum L of the object after time t? Jj n 0{ C. e.... w(t).::: cA-t L ;: I. W(t) [1: t 1 Part C: A rigid, uniform bar with mass mand length b rotates about the axis passing through the midpoint of the bar perpendicular to the bar. The linear speed of the end points of the bar isu What is the magnitude of the angular momentum Lof the bar? L :;.J:. u-> \.V e. Ji r. (3 Vv" ::c:: I;), m,-J ...L. L .2.. c~ ~ e.e. iDa (j /"') Lu:::- ~ r ::. .JL.. P/~ ~ +-Au:5: L-=..Lmb~.~ )),. J9 1+ 1 ~vb b V Part D: The uniform bar shown in the diagram has a length of 0.80 m. The bar begins to rotate from rest in the horizontal plane about the axis passing through its left end. What will be the magnitude of the angular momentum IJ of the bar 6.0 s after the motion has begun? The forces acting on the bar are shown. , ') Torl1 lJ e.S --<. L> [J, is bo..r IS ~xpe,..ie.VlGlnj [, =- J;)'N(O.brr'l) T;).. =- -8 tv ( 0 . go 0( yY\) Ll=- O~gNVV\=01.= (l N''!' I g -r Q.-ho u t- ,-'45 ( +-s f' vo-r- 13 M.L~ L=-Io<.t- =- O_~tJ'm (65);;:; '-/. I", J 8 (Y\ 'J. J Part E: Each of the four bars shown can rotate freely in the horizontal which diagrams is the net torque equal to zero? plane about its left end. For For - B <t'(. :2- r :::.0 Part F: Consider the figures for Part E. For which diagrams is the angular momentum constant? B C}c_ rf- ZL +-he-n L :;.aY\S 1-o-n c Y Part G: Each of the disks in the figure has radius r. Each disk can rotate freely about the axis passing through the center of the disk perpendicular to the plane of the figure, as shown. For which diagrams is the angular momentum constant? In your calculations, use the information provided in the diagrams. For A Z r:::::: /P- r - 4r ;l g r =- 0 J Lj For 8 For L 2. f:::: rX ~ -r tr: +- 4 r;;2.. L==. L(r r -g', - l')...r -:;;::..;:l.. IA vr Fo-r 0 ~r:;;; ?$r +-b r +<tr-~r-6 ';).. - Part H: Three disks are spinning independently on the same axle without friction. Their respective 4I . . . ') rotatlOna I' lOertms an did angu ar spee s are l,w( c Ioc k wise ) ; 2ft&"';( counterc Ioc k wise ; an d ,W/2 (clockwise). The disks then slide together and stick together, forming one piece with a single angular velocity. What will be the direction and the rate of rotation wDdofthe single piece? The- -rofO..l yv\ 0 tn en -ru yY'\ J'Yl uS -r be co f} Se r veol ::.. 3 I {)..J L.::: - I I Lf w +- b Xw -;l r w t- li :r:. := (I + ;).I ) UJ J- L i =- L.(!. 37w_ ~ 7/- UJ+ lw 7 Problem 10.86 A unifornl rod of mass 111 land length rotates in a horizontal plane about afixed (L'(isthrough its center and perpendicular to the rod. Two small rings, each with massm'2, are mounted so that they can slide along the rod. They are initially held by catches at positions a distance ron each side from the center of the rod, and the system is rotating at an angular velocity w. Without otherwise changing the system, the catches are released, and the rings slide outward along the rod and fly off at the ends. Part A: What is the angular speed of the system at the instant when the rings reach the ends of the rod? Lj:::L~ (1:1 In L I 'l..j.- ?.( fYl;). rl) fI\ I ) W:::' (Ii m, L ?, +- 2( yY',,-L W ,;)j) w {Jjf ( L. '). +. IJ... ' rY\ILa. ( )').. +-:t :> TV' r~) 0---. .L L'J...) yY\:l Part B: What is the angular speed of the rod after the rings leave it? -r), e.. Ctn3 u lOf 5 DfYl ~ t-() Sj'J e..ed- I::> +Ae Ih .s 6.111 e .. Ie 0( e. o -f +he t-CA..-} yY\ 0 J?1 e-11 r lJ /..:s ca.-I" f' o If vJ; 1-' h +--A e. r; /7f) S ()... . S . Wf - -- (~ ):2 " .J-Jm~r 1 ~ _ W (. yn,~ ~J~ +- Lm d- L <) ~ The Center of it All Part A: Two particles of masses fTlland m2('111 < m2) are located 10 meters apart. Where is the center of mass of the system located? 0 fYl, CvY' -rhe cwre( of yY);). } Yno.Ss IS G) Gloser ne.o..vier +0 fh ~ par-hele- "I h P art B : F or t h e sys t em 0 f th ree partlc es sown, where is the center of mass located? "d" w h" h h ave masses j\il , 2M , an d 3M as III Ica t e d , IC 0 Ih e- c.e...n+e r wj/} f Iv{~.s.s I,M. o ).1"\. eM. o o 31'\ be. Lt nd be-rwe.eY) ;2.JVl 3 M . Part C: For the system of particles described in the Part B, find the x coordinate :r.o'lofthe center of mass" Assume that the particle of mass At is at the origin and the positive x axis is directed to the right. Xcm ::: (tM)[O) !.(;)./.J.)( L) +- (3 M) (.:<.,-) (IMt:2M-r3!'-'t) --- -- :lML-t-bN.L bt1. Xc.. W\ :::- 8~L bj1 -- case. The system includes three particles of equal Part D: Let us now consider a two-dimensional mass M located at the vertices of an isosceles triangle as shown in the figure. Which arrow best shows the location of the center of mass of the system? Do not calculate. lhe. ceJllfe.r \\ CJ{- VVL~5 i5 Clr Joca.+~OI1 [j}e +-h~ . Jocot--ed. (j) b r:- 1- CC-t Y't' t b e... Q -t (J) b e..,e-a. u .s e..vY1Gc,<,S \'~ d--is (pUtlvJ\1 f\ e-Io.>e+o (0 cf~' ()e> or't)' hy -rAe- .yA (l.r~. Part E: What is the x coordinate xCfriofhe center of mass of the system described in Part D? t /rem':::' (:1M)(O) +- Lf-t 3M -~ I - L ~ '- Part F: A system of four buckets forms a square as shown in the figure. Initially, the buckets have different masses (it is not known how these masses are related). A student begins to add water gradually to the bucket located at the origin. As a result, what happens to the coordinates of the center of mass of the system of buckets? G I .9 co ~ ~ em j' " I' 4J I~ i +-i Gt. Jl Y '~I I !C em 0 .{-;(lO-I/i ! () fl e A5 e p a-r uf ./-( / L c rnCLS~') b e- <:'0 rYl e5 mOr e MO-SS i ve...+-6 -the center MOVe-3 C/6<5e..- pur +Ao.-+tie-Ie Part G: Find the x coordinate figure. xc:mofthe center of mass of the system of particles shown in the Xc,," - (30 hc:J) CO) +-' /00 (70 K J) (6 - 3 )N\) /~ Ae.- YYl ~ (). 56 YY) Part H: Find the y coordinate previous part. of the center of mass of the system of particles described in the "Y eM:;::" (10 h~ +-'10 ;"'y() I DO +-- 50 V,:] K) (o-Z;'>'\) Y C Y)'\':::'- O,~YYl Part I: Let us now consider an extended object. A wire is bent at its midpoint through a right angle as shown in the figure. Which arrow best shows the location of the center of mass of the system? Do not calculate. 1 h~ -rhe- c-- C)" -I- e yo+- (Y\U-<S~<) i j'e.s U?Otl \ " ~ 3 ~ " of H ne co (\" ect; ~ +A e- c1 e-e-/l-re (/'..5 yYlCLS-~ lor -r-h e. sep erv.-fe.. hCA z<)' !II if.. 0 Part J: A straight rod has one end at the origin and the other end at the point (L, 0)and a linear A = aJ? density given by , where (lis a known constant and Xis the x coordinate. Since this wire is not JM __ dm M uniform, you will have to use integration Find xCtllforthis rod. to solve this part. Use to find the total mass . XcI{ =- MJ I~( ;q::::; d:rf\ X J /\ cA o V\...--'" 11 L X ::. _ ---3 0.... L 3 .'V 1\('"m .- I r:t ~x j L :l ;X 0\ X :::;; oM. ~J o f tjL -X '1 o 'xc Nt --- _L Lj a... l/ " CtL 3 ----.3 =- Tipping Crane Part A: While watching the crane in operation, an observer mentions to you that for a given load there is a maximum angle Olll""thatthe crane arm can make with the horizontal without tipping the crane over. Is this correct? NOI, Part B: Later that week, while watching the same crane in operation, a different observer mentions to you that there is a maximum load the crane can lift without tipping, and you can find that maximum load by observing the minimum angle Omlnthat he crane arm makes with the horizontal. t Is this correct? Yes) Part C: Select the correct explanation for why you can determine the maximum load given that 8ml:n is the minimum angle the crane arm can make with the horizontal. ~ +~ <9 Yl'\ iY\ +h.e lev'er ()..rfYI , 0- r'\ ct (IGose) +0 r~ is ~t)'& mo.-X'; ~ )V\Ol,xfmUyY'I -r he. y-(l. -ore / u e. n, j z. eo{ . tc...osB Part D: You know that the torques must sum to zero about equilibrium. if an object is in static a Y f 0 Y\t J'\ j 0 11 0 r 0 +F rhe ho cA'I Part E: This implies that you can pick the point about which to sum the torques to simplify the calculation. Often it is best to pick the point where an unknown force acts, so that the torque due to that force is zero. In this problem the simplest equations result if you take torques about fh. e roi '"+- vJ A ere. + A e. f'h e. .(!ro n .J.- -/-t re..5 1-0 v ~ A 5'0 IJ h(;)... . Part F: Given the angle Olllill, what is the maximum weight (or load) WL,ml'Uthathis crane can lift t without tipping forward? (Recall that weight has units of force.) Sa I Ve. \tV&.. ) 1-0 r W L} M~)c' mox w(%) W~ (0. cos G~i1\) t \NL- w~ aX.COj (e-mil"l) Part G: What is W~llr", the largest weight of the load that is safe to lift regardless of the angle of the crane's arm? fA e.. I'YI/).)(; m Ii M +0 r ~u e rs e.-xe. (' ted.-. w he 11 B ~O \tv l%: )::0 \,Alb R w:::. S w~ ...., :lQ. Part H: Notice that we have the weight of the crane exerting a torque about the front wheels of the same crane. To create a torque, a force must be present, so it would seem that somehow the weight of the crane is exerting a force upon its front wheels. However, the crane is one object, and it follows from Newton's laws that an object cannot exert a net force upon itself. This crane seems to be defying Newton's laws. What's going on here? Th.e- e..o.. r- T h. . ( j e. 3ro..Il ; +/) e-x e. ,rf-s }-n e- c r (A "e a ol -r It e. 10 o..cJ.. .. Oil Y\ {orc-.5 Part I: Assume you get a summer job as a crane operator. On the first day you are lifting a heavy piece of machinery. Even though you have the arm at 7(}" above the horizontal, the crane begins to tip slowly forward. Consider the following possible actions: A}} 0 f +It e. (Y1 l. 1. Release the brake on the lifting cable so that the load accelerates downward. As j-he 100-0... Clc.c.eleraJ-es forc.. a.ownw~ro{) I~-j- exe,fs Je.s~ 2. Decrease Or) f-he cra.n so that the load accelerates downward. e... A~o.it'\} rAe. }oo.d.. OLe c elera..f-i"<j (o..nd... }or~ue) 011 ti~w n wa..r-ol lesselt...S +Ae lever. vhe, +orc.e fJ 3. Increase while simultaneously letting out the lifting cable so that the load accelerates downward. , +J1e.. n 0. e.lerQ..h'o n 1\ ~ CL-~ As 10 ~> r e} eJ- cc w;l} be. +-0 wa.,r tk. tA L /00. t() r olA- nd. .. 4. Put the crane wheels in gear and accelerate the crane forward. Ye.s I f he- et SYtI i Y\~S bo.. c K (I ess e tl i n3 .J-Ae- I-or~(J(?) None of these solutions is ideal, but which will have the short-term effect of restoring contact of the crane's rear wheels with the ground? r~is ;05 ex. ba.ct. VV } id-elAJ howeve.r he-co..use .fAe.. crane.. CAe. n c. f ash W hen you .s 1-0p e. e 1 e ,a. t ;1'\ 3. Problem 11.60 Part A: In the figure a uniform beam of length Lis hanging from a point a distance x to the right of its center. The beam weighs and makes an angle of 1 with vertical. the At the right-hand end of the beam a concrete block weighing w1ishung; an unknown weight l/Jhangs at the other end. If the system is in equilibrium, what is w? You can ignore the thickness of the beam. !::..-?t ;;). 0 ll'l ~v Fo r r his .s ys l-e"" -to he. j n .s 1-0.. +i G e-~ u Ii bri U M OY) z...,-;::. 0 le..f+-Sicl~ rne ~he W' ~ (L- a -I- x ) +- ~ W.( ?<' ) ==" w~ J}n e ( 1;- + x) +- WeIw e:x sit1 On riCj"t siote. GJ, the- rwo 1 Sine (~-x) W, <S j 1\ sid. ..s ty\ IJS t e..~uo../ : C Q.,n C e. /.5 0 U 'jJ VV~ - e {~ - ~ ):::..W (~. s} n B( ~ :r~) The 5 j n , d. ep e n.d en c e. +: .}- W<si"lt (?() 'AI 1 ...;t').:::: w{ -.i +-:t)+ W;x \,4/, ( ; 02 Part B: If the beam makes, instead, an angle of with the vertical, what i x) -- W'X __ ( '-;:1. + ;%) -}-h (). r n f 0..r+- A) dep enol~f'c.e 0.( we, f 0 V flrA YltJf fA It -r +- A e (;}- LCLYlCe-ls ou+) wi Jl t-nefe lore.. o..oti~efn+ Vo./ve.. f) ec+- -rJ.. e Ct. n.s 141 e 1'. \AI;::::.. W, (~ -X) - \VA (~+x) Problem 11.72 You are trying to raise a bicycle wheel of mass Tnand radius Rup over a curb of height h. To do this, you apply a horizontal force F Part A: What is the least magnitude of the force that will succeed in raising the wheel onto the curb when the force is applied at the center of the wheel? F Ii \AI e- TYl ust {,'nd- fh e Components 0f F q.. tnj fntAt ().re fer pen oI..icu Io..r J;U~. h '), fo ~A e. Ie va/' a f' /Y\ n .: e I t'\ F.L = FSin e F o.,rm 'elle" p.-h m3.L.= m3cos8 s,n8 FJ. z. m3J- F ~ fYlj c~se _> F > rn ')jJ.ll.h-h' Part B: What is the least magnitude of the force that will succeed in raising the wheel onto the curb when the force is applied at the top of the wheel? F ~ F J~-h FJ..;; F s i rY\3 +- ::::- tY':J F .2 g COS ~ J- 8 ~1., ::::. 1Yl M 0\ C 0 s~ ~") J s,'n9 r h'lj--j ~ R. h - h 'J. ::1.R.-), JAt{ 'h ~h),... e J.f\-~ Part C: In which case is less force required? fJ fJ./+ B - where- -role.e i's aft /,'e-d 1-0 +he ~ fop, A Matter of Some Gravity Part A: Two particles are separated by a certain distance. The force of gravitational interaction between them is Fh. Now the separation between the particles is tripled. Find the new force of gravitational interaction Fl. Ie..r f;: I be. +he iYlJrio.' 3'0 ts ~Ae. fiYlO;..I r6 sefo..l'o.-fjoY\, se.pa./'CA..fiOT\ r. -= 0 """l1y r().:l. mt4.c;. ~ (3 ro) 1 J mMC-r _ - Po -;;;- r o J. Part B: A satellite revolves around a planet at an altitude equal to the radius of the planet. The force of gravitational interaction between the satellite and the planet is Fo Then the satellite moves to a different orbit, so that its altitude is tripled. Find the new force of gravitational interaction F2 F :::. o mj.{ G; _ (d. rJ'J.. yVJ ( - ~ Lj rn f{ r~~ Yl1!1 G J ~ F =;l. t1 r e);l. f{ G_ =- -l /6 - rt ,. - G - if Part C: A satellite revolves around a planet at an altitude equal to the radius of the planet. The force of gravitational interaction between the satellite and the planet is Fo Then the satellite is brought back to the surface of the planet. Find the new force of gravitational interaction F4 F ::.. o ~J{G (~1'e.)~ ~ If ::: !'" f{ C?. ::::- (rJ ~ 1tjF.l Part D: Two satellites revolve around the Earth. Satellite A has mass 'mand has an orbit of radius r. Satellite B has mass 6mand an orbit of unknown radius ft,. The forces of gravitational attraction between each satellite and the Earth is the same. Findt'I,. r ~ mMG r:t PI (8; ~m r M Cit :l.. b o ,'n C e. Fit ==- '8 ,.b .;). br~ rb'" ='/ ~Jb rJ Part E: An adult elephant has a mass of about 5.0 tons. An adult elephant shrew has a mass of about 50 grams. How far rfrom the center of the Earth should an elephant be placed so that its weight equals that of the elephant shrew on the surface of the Earth? The radius of the Earth is 6400 kill .( 1 tOll = 103 kg .) m.shre,w ~~ ---------f'~ e - m ele.pna.nf- ~j( :XI0 X'?; 3 r~ (0.05 (,. )1~) <1,x ItJ6m)?. _..s - r~ --------{!-:::.:1 ~ J 0 6 /A,w} Part F: Find the net gravitational force F;letacting on the Earth in the Sun-Earth-Moon system during the new moon (when the moon is located directly between the Earth and the Sun). -(he- f.orc.t. he-tween +Ile e.CA.J"fJ.. ancJ ,"oon:::-;> PM::" M~.MM r a !3 :l. :20 (Yl FM.=-1.9'6;</O rJ the fot'ce betlN'eetJ F =-F", T Fs::: r rite. etA.rfil ).l1d S.UI1 7 r- _ r 5 - --:.----::::- .M.,s Me ~ 5 'f )rIO :;J~N) 3.5 D )( /D '1:>, rs ~ Fs ~ 3. 521 g' NO :J'LN Part G: Find the net gravitational force Ft;'alactingon the Earth in the Sun-Earth-Moon system during the full moon (when the Earth is located directly between the moon and the sun). F.:::. Ps - F M ~( tJ 1 Problem 12.24 When in orbit, a communication satellite attracts the earth with a force of F and the earth-satellite gravitational potential energy (relative to zero at infinite separation) is _ U. Part A: Find the satellite's altitude above the earth's surface. r he. p (' 0 b Ie fY\ U;::. Sf e.c;~ie5 U Q..- P mM.G, F.:::- mM~ r r~ e. wnere. r:::. o./+i-rucJe -r r ~ u r= :::r.:=. r fhvs ,-' --- + ~ If,' tVcJ.~ exff-i tUc{e. .=: Y,:: -re., Part B: Find the mass of the satellite. tJ 0te. + h oJ' for U A.:::. yY) '1. W\ ~ M 'J. G :l. r'l. U ~r~ - SotVit\J m?.. M :J. q:l. - u ~ lU) M:I."G.~ 'J. _ U &j 1, lp -:::; F f1"i:L fYl;:; MGrF u'J. - Problem 12.58 Planet X rotates in the same manner as the earth, around an axis through its north and south poles, and is perfectly spherical. An astronaut who weighs IVon the earth weighs Wlat the north pole of Planet X and only W"latits equator. The distance from the north pole to the equator is L, measured along the surface of Planet X. Part A: How long is the day on Planet X? A+ +ht- pDIe.s) your At o...ccel~ro.flon f.s cJ.ve. sl1Ylply -to ~rt>-vd .. . y . +}o,e e~ vlAtor) fJ,c. ro.cAja/ w- m~ R I accelero.-tiol1 Y0l) e)(pel"ieYlce (fne..-l'la.ne+-w~v/~ co. u.s~ you +-0 w~}~n S/ljh+ly less. _ w~ WI yo!-afjJ1;jJ _ mv R m:::. ":l" -/ -~ w, _ w~ - ,., v_ .N\ ~ R No+' c e.. A.. .:::.. ;).,L rr ::::;.) ~\I Lj ~= ;?1fR we ~now V Jhus =- I;)'~L -:) (w la/ ~ ~ (on ~().("fh) rr lAnl"A ~L TT ~ (W, -W.1 ) J oJ. bur w.e.. ~ J~no }V r-(Ic1o.y);::; ttL - ~x.::-vtD..X -=- ~ 7fr for Id-o..v Part B: If a msatellite is placed in a circular orb.t orbital period? /:~L(W/-V~)h C1~1Y\x r?" Ci1Y\.x _ =-'tL ove the surface of Planet X, what will be its / We, l~now: ---r ~V2. vJ he,re..- v=~ r fl1 US: S6Jve. -For r I: (o/) G yY\x. _ r {~n)J.r~ ~ -rJ.. T'l.-:::: (lTT)'},r3 - Gmy T.:: /'trT'>-r' &m.x - Problem 12.43 At the Galaxy's Core. Astronomers have observed a small, massive object at the center of our Milky Way galaxy (Section 12.8). A ring of material orbits this massive object; the ring has a diameter of about 12.0 kmfE light years and an orbital speed of about 160 . Part A: Determine the mass of the massive object at the center of the Milky Way galaxy. Give your answer in kilograms. for Circ.vlo.f" orb'/t Mp(G. _ rX - rv~ r 11\:;0.; r v~ r:;:. }'l li~bt ')... yeo-cs Q 1- '16/ ,X/I) ?.. /5 r(\ V;:::. :Zoo J{~;:::.. ~x/o5~ M-:::- ('1.7305)(/0 ;[ J!J ) ( Wi '1X/O I0 ?) ~ b.6 73 N m;, h o..ct v ---:, -:::- 2..836)< 10 36 ~ /1.=1 /'K" ~ /' /0 Part B: Give your answer in solar masses (one solar mass is the mass of the sun). 1 m o.s s ; V ~ 0 b j e. c:+ he., M~ J. } Sol ().T 0... m~.s S 0 r 836)(/656 X /0 s6 /-< 5 Nt. (). s s:::' /. 90/ J~) 1e y t A +0 rf2.36 'J . 8 3 (, X10 'j =:=- / /. '1 J. 5 / . '! Cf )(/0 36 1'$9- I,) it'S rs ,x 1() 6 S61a.r ;J.. a.5 S e..5 Part C: Observations of stars, as well as theories of the structure of stars, suggest that it is impossible for a single star to have a mass of more than about 50 solar masses. Can this massive object be a single, ordinary star? No ... /\"jOb .f o..r e.-x c e. d.. Q. 5 6() I Part D: Many astronomers believe that the massive object at the center of the Milky Way galaxy is a black hole. If so, what must the Schwarzschild radius of this black hole be? 1he- 5CAWCArz,sch;/et TCAditJS is. cAe..~"Y1eo{ o-.s . As::" -- 16tH. '),v~(\ c. ~ -----C). Part E: Would a black hole of this size fit inside the earth's orbit around the sun? V-e. Sl , I Problem 12.66 Part A: Calculate how much work is required to launch a spacecraft of mass 71lfromthe surface of the earth (mass mE, radius RE) and place it in a circular low earth orbit, that is, an orbit whose altitude above the earth's surface is much less than Re. (As an example, the International Space Station is in low earth orbit at an altitude of about 400 km, much less than RE = 6380 km.) You can ignore the kinetic energy that the spacecraft has on the ground due to the earth's rotation. E bE;: j c::: -~ m M. r~ Ef Gml'\ . re., ~ (' e.- Gfn/i ;l r~ Ef - i - (- ; 1-1) Part B: Calculate the minimum amount of additional work required to move the spacecraft from low earth orbit to a very great distance from the earth. You can ignore the gravitational effects of the sun, the moon, and the other planets. Ihe. e..1\ e.r~y , In orb,. f- J'j -0 ~ Gm~ ~r~ F -00 ~E. =- 0 a c- = E&::J - Eo f"' Problem 12.70 If a satellite is in a sufficiently low orbit, it will encounter air drag from the earth's atmosphere. Since air drag does negative work (the force of air drag is directed opposite the motion), the mechanical energy will decrease. According to Eq. (12.15), if E decreases (becomes more negative), the radius Tofthe orbit will decrease. If air drag is relatively small, the satellite can be considered to be in a circular orbit of continually decreasing radius. Part A: A satellite with mass mis initially in a circular orbit a distance hlabove the earth's surface. Due to air drag, the satellite's altitude decreases to It.!. Calculate the initial orbital speed. Vlt Hno \AI Jliv').. _ ~f"'Y At 11 - - 1'1 ,;y( C, -) (Re.+h }~ v =- L.ME. Gi 'liEt' h I Part B: Calculate the increase in orbital speed. h<!.i'Jl-tt h"J ~;;:jt1G I? of- h J... 50 wL Lo../CU/Ulre f -Vi .:::. fhe At thJ ChO-fl e {'11 lie I oc/ fy .. LW ~ V /j1)e. g:' _ ":3. (jte. GJ Jf.f,.:. "/ Part C: Calculate the initial mechanical ~ ,; ::; J". E r P j;. = Tf\.V hI G. tt, 1Yt h , J. lie) but wLjrJ,s+Ii. =- - \}IrAn+ whl-cA I~ fhe (Y1 e chan" QJ I e. l1e/'(j Y : -k m tI ~ ;:::: m V.f. ~ ) /~ I m M G ~ {AT:; +-h2 Part D: Calculate the change in kinetic energy. 6 KE ;;. ,/;A I rn. v., ~ J ---- ---- ;;: ME 1 ;;l rn bi ( fi.E.-fh} Part E: Calculate the change in gravitational potential energy. rl1;~io.lly: u- :::: I j m f{r; G - h, tfE Fi Jl\(X lIy ~ -VMMEG J hJ.-rf' - -U + 6. u ~ Uf - U' '=- JN1 ffE G h,tA[ yY1 /VIE G hJ.+-(\E Part F: Calculate the change in mechanical energy. fr 0 I'" rhUS f 0.'+ C ~ E-, ;;.. - 0 M ~ 2eAr- I'" +h,) I; 1-\ e vV"ISC [ f - ".'-.1 - - G I,t jV\ ------ :J. (f{Er J,,:z.) 6.. E =. - G /1E -:t. M ( I Aef- /" 1. _, I_ ) At ..,.. },,\ Part G: Calculate the work done by the force of air drag. the VJor-l'-, i J1 d..Dne- bl f,ic+ion caI e)1 "I) t..Ltu(Li -1-0 +he- e, hCA "'c:J e- rn Q..Ch Cl () ; e,rS y-, vJ::- - G ME. ~ (' :l. I Ae+hL _ 1) I\e,+h l

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UCSD - PHYS - 2A
Homework 10 Good Vibes: Introduction to OscillationConsider a block of massl/1.attached to a springwith force constant k, as shown in the figure. The spring can be either stretched or compressed. The block slides on a frictionless horizontal
UCSD - PHYS - 2A
Physics 2A Quiz 1 answer keyQuiz Form Question1 2 3 4 5 6 7AC B E D A D BBC D B D C E CCD E B B D B DDA A E B E C AEB C C E C A EFA B E B A A AGE E A D D A DHB E D B E C DIC A A C E B CJA B A E E D BKE D D E D D C
UCSD - PHYS - 2A
Physics 2A Quiz 2 answer keyQuiz Form Question1 2 3 4 5 6 7AC E D E C D ABE A D B E D ECD A C D C D EDB D D E B B EED B D C E E BFD E D A E E AGA E D A B C DHB E D A E C AIA C E A A B CJA B B B E E AKC A C C B D A
UCSD - PHYS - 2A
Physics 2A Quiz 3 answer keyQuiz Form Question1 2 3 4 5 6 7AB D B A D A ABB B A C B B DCA B B C B D CDC C D C E C AEC A C A C B AFD A A B C D BGC B C B C B CHB A D D C B BIA B D B D B CJD C A B A C CKA A B C C C A
UCSD - PHYS - 2A
Physics 2A Quiz 4 answer keyQuiz Form Question1 2 3 4 5 6 7AB D C A D E DBE C D C D E ACD D D C B E BDA D D B E C CEB D D C B E AFA B D C D E BGD A C A C B BHC E B A A B DIB E E B B D BJA A C B E C CKA E B C D B E
UCSD - PHYS - 2A
Physics 2A Quiz 5 answer keyQuiz Form Question1 2 3 4 5 6 7AA A B A A A EBA A E A D A ACC D C A D D BDE A B A A D BEE B D B A D BFB C A E A E AGA D D B B B AHB B C D C E AID B A A B B CJA B A B C E DKC B D A A E B
UCSD - PHYS - 2A
Physics 2A Quiz 6 answer keyQuiz Form Question1 2 3 4 5 6 7AA A E A A A ABE C D B A B CCB E B B A C DDC C A A E C AEC C A E D B CFA A C A A E BGB A B A E D CHD A A C D E AIC C A B E A CJB C C A A E AKA C C C A A E
UCSD - PHYS - 2A
Physics 2A Quiz 7 answer keyQuiz Form Question1 2 3 4 5 6 7ABCDEB B A D C E DFC C E D C D AGA E D B B E BHA B D E E C CIB E C D D B CJE B B E A D CKA D E D A E ALA B E D A C AUCSD Physics 2A Winter 2007 Instru
UCSD - PHYS - 2A
Physics 2A Quiz 8 answer keyQuiz Form Question1 2 3 4 5 6 7AA D D D B A ABD A C B A B CCB B D D B D DDA A B D A D AED E E A A C CFA C A A D D EGA A B C E A BHB C E E C E BIE C D C E B BJB C C D C D AKA E A E E D D
UCSD - PHYS - 2A
Physics 2A Quiz 9 answer keyQuiz Form Question1 2 3 4 5 6 7AB A C E D E EBC D B B E D BCC C A C C C ADD C D C A B DEB C C D E C AFD B C A C A DGC A A B A B BHA E E A B E DID D C A C D DJD D B E E A DKD A D A B C B
UCSD - PHYS - 2C
Physics 2C Summer Session I Solutions to Quiz #11. The increase in pressure is found from P = gh so that h= P= g = (1:03 105 P a = 9:9m ! C: 103 kg=m3 ) 9:8m=s22. Balancing torques we .nd PA w = F (1) ; 2where w is the width of the door. This le
UCSD - PHYS - 2C
Physics 2C Summer Session I Quiz #21. From the ideal gas law PV NO2 NO2 : 2. From kinetic theory of ideal gasses we have 3 1 m v2 = kT ! v 2 = 3kT =m 2 2 r r p 3NA kT 3RT 2i = v hv = rms = NA m mmole r 3 8:314 473 = 1717m= sec (B) vrms = :004 3. The
UCSD - PHYS - 2C
Physics 2C Summer Session I Quiz #31. The speed of sound is given by vs = the molar density is n=V = P=RT: Hence the mass density on Titan is = nmmol =V = = :387kg=m3 (ideal gas law). For a pressure of P = 1:5 105 P a the speed of sound is p vs = 1:
UCSD - PHYS - 2C
Physics 2C Summer Session I Quiz #4Multiple Choice. Choose the answer that best completes the statement or answers the question.1. From the law of Malus S=S0 = cos2 : Hence for four polarizers each of which has an angle of = =8 we have S=S0 = cos2
UCSD - CENG - 100
y2)t ru Basis: n:o - . _lb_. , _ r lr t b.lrl ft)' l(s ' x l b ,) = 0 .Ia (ffx tb , )Let A = arca of the Pipe md v = water velocity. The flow rate isc =nv =[1.;tv)(o'),gal nl(2 inll (l ft)' 1 3Rl60 s 17. 48 gal 4l 129.37 / min| l(12 in
UCSD - CENG - 100
UCSD - CENG - 100
6.5Based on the process measurements, there are 5,000 Ib/h more flow for the process leaving the heat exchanger than the feed rate to the heat exchanger; therefore, the material balance for the process fluid does not close. The reason for this discr
UCSD - CENG - 100
9.7Basis: Data given in the problemMassof hydrate 10.407g Massof dry sample .9.520 g Mass of water 0.887g9.520 g Bal,ll g mol Bal, 391 g Bal,= 0.0243g molBal,0.887 g H,O 1 g mol H,O = 0.0493 g mol H,O\ 18.02g H,O0.0243 = 0.49. or 2 H,O
UCSD - CENG - 100
Problem 11.4 Since the problem statement said there are n components in the feed, apparently the &quot;two components&quot; in the column refer to the vapor and liquid streams inside the column. Since they are inside the column and, thus, inside the system bou
UCSD - CENG - 100
Solutions Chapter 24 Since Pfi~I'V,noI' nd liHfi~' are in the tables, you a(8)have to get a liH.noI' and Vrm.,vaporgiven an assumed Pfi&quot;,'.that satisfY the equation. liHfi&quot;,' wilJ be composed of saturatedj(7b)liquid and vapor. liHL(8-m)
UCSD - PHYS - 2CL
Physics 2CLSummer II 2007Name: Student ID: Section: A0Quiz 1 Physics 2CL: Electricity, Magnetism, Waves, &amp; Optics Laboratory 8 August 2007Problem 1 [2 points]For an RC circuit having component values of R = 500 and C = 0.01 F, compute the
UCSD - PHYS - 2CL
Physics 2CLSummer II 2007Name: Student ID: Section: A0Quiz 2 Physics 2CL: Electricity, Magnetism, Waves, &amp; Optics Laboratory 13 August 2007Problem 1 [2 points]For the amplifier and resistor network depicted on the left, you must achieve a ga
UCSD - PHYS - 2CL
Physics 2CLSummer II 2007Name: Student ID: Section: Day_ Time _Quiz 3 Physics 2CL: Electricity, Magnetism, Waves, &amp; Optics Laboratory 15 August 2007Problem 1 [2 points]A flip coil is wound on a 3 cm diameter and has 10 turns. It would be con
UCSD - PHYS - 2CL
Physics 2CLSummer II 2007Name: Student ID: Section: Day_ Time _Quiz 4 Physics 2CL: Electricity, Magnetism, Waves, &amp; Optics Laboratory 20 August 2007Problem 1 [2 points]For the circuit depicted on the left, write the expression for gain (clos
UCSD - PHYS - 2CL
Physics 2CLSummer II 2007Name: Student ID: Section: Day_ Time _Quiz 5 Physics 2CL: Electricity, Magnetism, Waves, &amp; Optics Laboratory 22 August 2007Problem 1 [2 points]The speed of sound in water is 1493 m/s. What will be the lowest frequenc
UCSD - PHYS - 2CL
Physics 2CLSummer II 2007Name: Student ID: Section: Day_ Time _Quiz 6 Physics 2CL: Electricity, Magnetism, Waves, &amp; Optics Laboratory 27 August 2007Problem 1 [4 points]1 1 1 + = and a table of data for u -1 and v -1 on can determine u v f th
UCSD - PHYS - 2CL
Physics 2CLSummer II 2007Name: Student ID: Section: Day_ Time _Quiz 7 (Last pre-laboratory quiz!) Physics 2CL: Electricity, Magnetism, Waves, &amp; Optics Laboratory 27 August 2007Problem 1 [2 points]Suppose that you conduct a single-slit diffra
UCSD - CENG - 100
On a/&lt;3-.at^-,a-+/1.?:crl4, o/ /v ( c, t/ro / ) H, . y ( ) &quot; , , i l ,.)Ca&quot;*72-r.r.,-J2,.6LLrr_-*-c * t.f, a, ), td.*-J _( n/t-fI , c o _ , t + _ , &gt; ) e , w. 2 ( . , Fd-.^j-o ,*.|;.-.^ :(.0, ( . l O(o { -&gt; (-l^ / + -11 |t-LO du.or&lt;L,Vr
UCSD - CHEM - 131
CHEM 131K. Lindenberg Solutions to Homework No. 1 Posted Monday October 8 Fall 2007These solutions to Homework No. 1 were written partly by Dario and partly by Jim. Note: they chose how to divide up the task, so the solutions are not in the order
UCSD - CHEM - 131
CHEM 131K. Lindenberg Solutions to Homework No. 2 Posted Monday October 15 Fall 2007These solutions to Homework No. 2 were written partly by Dario and partly by Jim. Note: they chose how to divide up the task, so the solutions are not in the order
UCSD - CHEM - 131
Chemistry 131, Fall 2007 Solutions to Homework No. 3Problem 1.The latter expression is equal to H - T S only if T is constant (Levine eq. 4.64).(c) True. Follows from the definitions of enthalpy, H U + P V (Levine eq. 2.45), Gibbs energy, G H
UCSD - CHEM - 131
Chemistry 131, Fall 2007 Solutions to Homework No. 4Problem 1.See Levine, problem 4.39.(b) True. At constant pressure, H = U +P V . Since the reaction halves the number of moles of the gas, the volume of the system decreases and the term P V must
UCSD - CHEM - 131
(h) True. The ideal-gas equilibrium constant depends on temperature only (Levine p. 182). The temperature dependence of KP is described by the van't Hoff equation (Levine eq. 6.36). Although each equilibrium partial pressure Pi,eq deProblem 1. pends
UCSD - CHEM - 131
Chemistry 131, Fall 2007 Solutions to Homework No. 6Problem 1.See Levine, problem 7.2. Use the phase rule for systems with no reactions, f = 2 - p + c (Levine eq. 7.7).Problem 5.See Levine, problem 7.21. (a) True (Levine eq. 3.15). (b) True. The
UCSD - CHEM - 131
Chemistry 131, Fall 2007 Solutions to Homework No. 7Problem 1.See Levine, problem 8.4. Let x = 1/Vm and solve Levine eq. 8.4 to get P = RT x + Bx2 + Cx3 + . . . . Next write an expression for Z with eq. 8.5 and substitute the expression for P found
UCSD - CHEM - 131
Chemistry 131, Fall 2007 Solutions to Homework No. 8Problem 1.See Levine, problem 9.33. (a) True. Since xi &lt; 1 for each component i in solution and since ln xi &lt; 0 when xi &lt; 1, then mix G = RT i ni ln xi &lt; 0, at constant T and P (Levine eq. 9.44).
UCSD - CHEM - 132
Chemistry 132, Winter 2008 Solutions to Homework No. 1Problem 1.See Levine problem 11.1. (a) True. The activity of component i in a reaction mixture is defined in terms of an exponential function, which is di mensionless: ai = e(i -i )/RT (Levine e
UCSD - CHEM - 132
Chemistry 132, Winter 2008 Solutions to Homework No. 2Problem 1.See Levine problem 12.1. (a) True. Addition of a solute B at constant T and P to pure solvent A decreases the mole fraction xA of the solvent. A Since A &gt; 0 (Levine eq. 4.90), then add
UCSD - CHEM - 132
Chemistry 132, Winter 2008 Solutions to Homework No. 3Problem 1.See Levine problem 14.12. (a) True. Taking the derivative of E (Levine eq. 14.41) with respect to the activity aj of substance j gives E /aj = -j RT /nF aj , which is negative when j &gt;
UCSD - CHEM - 132
(1.15) True. The distribution function for vx , g(vx ), has a maximum at vx = 0 (Levine p. 469) independently of the type of gas. Since all velocity components have the same distribution function (Levine p. 463), the most probable Problem 1. value of
UCSD - CHEM - 132
Chemistry 132, Winter 2008 Solutions to Homework No. 5Problem 1.Given n possible values xi , i = 1 n, of a discrete random variable X, and given a function f (X) of the variable X, the average value of f (X) is (Levine eq. 15.40)nFigure 1: Ex
UCSD - CHEM - 132
Chemistry 132, Winter 2008 Solutions to Homework No. 6Problem 1.See Levine problem 16.6. (a) False. By integrating Newton's viscosity law (Levine eq. 16.15) it can be shown that for a layer with radius s the velocity vy along the pipe is a paraboli
UCSD - CHEM - 132
Chemistry 132, Winter 2008 Solutions to Homework No. 7Problem 1.See Levine problem 17.1.products. Neither an intermediate nor a catalyst appears in the overall reaction. A reaction intermediate first appears as the product of a step of the mechan
UCSD - CHEM - 132
Chemistry 132, Winter 2008 Solutions to Homework No. 8Problem 1.See Levine problem 17.44. (a) Assume a rate law of the form r = k[A] [B] [E] (Levine eq. 17.5), where A, B, and E represent ClO- , Cl- , and OH- , respectively. Let ri denote the initi
UCSD - CHEM - 132
Chemistry 132, Winter 2008 Solutions to Homework No. 9Problem 1.In a chain reaction, the chain carriers are reactive intermediates that occur in chain propagation steps (Levine p. 565). A step that generates chain carriers from relatively unreactiv
UCSD - CHEM - 132
Chemistry 132, Winter 2008 Solutions to Homework No. 10Problem 1.See Levine problem 22.54. Solve for N2 , at T = 300 K with rot = 2.86 K and zrot = 51.0. (a) The fraction of molecules in the rotational energy level corresponding to quantum number J
UCSD - CHEM - 132
UCSD - CHEM - 132
CHEM 132MIDTERM EXAMK. Lindenberg Winter 2007STUDENT ID_1./8 /16 /7 /21 /10 /27 /105TAL MIDTERMTOTAL NUMBER OF PAGES INCLUDING THIS COVER PAGE: 13. NOTE: As announced, this exam is based on homeworks and book examples, with details chan
UCSD - CHEM - 131
UCSD - CHEM - 131
SUNY Geneseo - PHYS - 125
PHYS 125Formula SheetSpring 2008Units and Constants e = 1.6 10-19 C me = 9.11 10-31 kg k = 8.99 109 Nm2/C2 -12 C2 /Nm2 0 = 8.85 10 0 = 4 10-7 Tm/A c = 2.998 108 m/s v = 343 m/sELECTROSTATICS F = E= 1 q1 q2 kq1q2 = 2 r 4 0 r2 F = q0 E E=
SUNY Geneseo - PHYS - 125
zzMcLean, James GPhysics 125, Analytical Physics II Date Sat, Apr 12, 2008 at 11:00 New Material 1. [4pt] A conducting loop of wire is placed in a magnetic field that is normal to the plane of the loop, as shown. Section 99 4. [5pt] For each stateme
SUNY Geneseo - PHYS - 125
SUNY Geneseo - PHYS - 125
SUNY Geneseo - PHYS - 125
SUNY Geneseo - PHYS - 125
SUNY Geneseo - PHYS - 125
SUNY Geneseo - PHYS - 125
SUNY Geneseo - PHYS - 125
SUNY Geneseo - PHYS - 125
zzIyer, Savi V Physics 125, Analytical Physics IIDate Thr, Mar 13, 2008 at 21:00Section 99 Exam 2 4. [6pt] The figure shows the equipotential contours due tothree point charges. The labels on the contours are in Volts. Indicate T-True or F-False
SUNY Geneseo - PHYS - 125
pl{~S \~5 -Q&amp;.ef ~Ot\k~ F1n'\e) ~ t~i\1.E&quot;~.~Mtv\~\s~\) g.JAE) 1=:~ vJvi~.e)C&quot;~fO(eI\~(M.II.e)G~\oiW~~f~rt? 't~~&lt;;~1~~-k ~~Ap~n1L~ WovJj.~.e-.&gt;r ~Q~rVll1'l\t.eA -*~. (~.t&lt; etA'(ec-~\?il