HW #1 Solutions. Lewis Structures: Using the above rules, determine the molecular structure for Cl2CO. Hint: C is at the center.

Size: px
Start display at page:

Download "HW #1 Solutions. Lewis Structures: Using the above rules, determine the molecular structure for Cl2CO. Hint: C is at the center."

Transcription

1 HW # Soluions Cron Mss Prolem: ssuming n erge surfce pressure of m, n erge ropospheric emperure of 55 K, n glol CO mixing rio of 385 ppm, wh is he curren mospheric Cron reseroir (in unis of g m -? Compre your pproximion wih Tle.3 in W+H. Surfce pressure of m cn e use o esime he mss of ir oer column, p mg/, yieling m/ g m -. We wn o now how much of his mss is cron. The rio of CO molecules o ir molecules is gien y he mixing rio of 385 ppm. Th CO is well mixe mens h his lue hols hroughou he mosphere. The rio of cron mss o ir mss is he mixing rio imes he cron mss per CO molecule ( MU iie y he ir mss per ir molecule (9 MU. 385 ppm x (/9.6x -4. This is he frcion of he g m - column mss which is cron. Therefore he cron column uren is.6 g m -. This grees spo on wih W+H. Lewis Srucures: Using he oe rules, eermine he moleculr srucure for ClCO. Hin: C is he cener. Cl Cl C O Henry s Lw: Wh is he equilirium concenrion of HCO3 in freshwer le, neglecing oher processes, gien moern-y CO mixing rio of 385 ppm? H CO.34 M m - p CO ξp 3.85x -4 m H CO 3 (.34 M m - (3.85x -4 m.3x -5 M Dissoluion of SO ino clou rops is imporn o he sulfur cycle, ecuse oxiion o sulfe occurs more efficienly in he queous phse hn in he gseous phse. The Nionl mien ir Quliy Snrs he se nnul erge limis of.3 ppm for SO. Using he Henry s Lw coefficien from S+P Tle 7., fin he pproxime equilirium concenrions (in M of he soluion species, HSO3 in clou rops his mixing rio. ssume emperure of 5 C, n pressure of 95 m. H SO.3 M m - p SO ξp.8x -8 m H CO 3 (.3 M m - (3x -8 m 3.4x -8 M

2 S+P prolem 3.: This prolem uses wh we lerne ou ermoleculr recions n pplies i o he conersion of NO o niric ci i NO + OH + M HNO 3 + M eing ermoleculr recion, we now h his is he sum of hree recions, where we he roppe he inermeiry prouc HNO 3 * in he oe recion re (see S+P. Thus his ne recion oesn follow he rrhenius form. Inse, he recion re is oine from Tle. R (T,zNO OH ( ( T, z + ( ( T.x ( T.5x 3 ( T M.6 ( T M/ ( T 3 T 6 cm molecule 3 3 cm molecule s { + log ( ( T M / ( T } s (,,c Where (TM is he re for he low-pressure limi (i.e. mping collisions of HNO 3 * wih M re rre enough o e limiing fcor, n (T is he re for he high pressure limi (he re of collisions eween NO n OH is he limiing fcor. You cn see from he form of ( h he slower re omines. The concenrion for he rnsiion eween low n high pressure limis, M, is foun where (TM (T, his mes he enominor in (, n hus M.5x 9 molecules cm -3. Using p MT, we he p.5 m. We e T 3 K for lc of guince from he prolem. Here I plo he soluion wih IDL.

3 S+P prolem 3.4: Collision heory preics R C π ( r + r 8T π m OHCHF + m 3 Plugging n chugging yiels R C 3.6x -6 m 3 molecule - s - OHCHF 3 3.6x - cm 3 molecule - s - OHCHF 3 Thus he frcion of collisions leing o recion, γ /3.6x - 7.7x -7, or less hn one in million. This is rre recion! Firs, he O sie of OH hs o hi he H sie of he HCF 3. Furhermore, he collision hs o e priculrly energeic n he on so he OH cn ig ino he elecron shell n pull ou he H om wihou eing eflece firs noe he reliely srong energy rrier (he -E/T erm in seen in Tle. for his recion (This ws iscusse in clss on Mony. See S+P iscussion in secion 3..3 if you woner wh I m ling ou. Furhermore, he O sie of OH is he more negie sie, n hus he elecron shell roun he fluorocron will e rcing he H in he OH, furher complicing he recion S+P Prolem 3.6: This prolem els wih heerogeneous recion (no iscusse in clss, u foun in S+P secion 3.7. I follows irecly from collision heory wih hree simplificions: he spee of he erosol pricle will e negligile compre o h of he gs molecule, so g ; he rius of he gs molecule will e negligile compre o h of he erosol, so r ; n 3 inse of n rrhenius form for he recion efficiency, we jus use fixe frcion, γ. (Don ge confuse h S+P hs ¼ ou fron of (3.36. Th s jus ecuse hey use p pricle surfce re, n he o pu he ¼ ou fron o coner c o cross-secionl re, which is more relen for collision heory. Using hese simplificions, we ge R γr g γπr C 8T πm γπ g n ( r + r N Where he erm in fron of he N O 5 is he firs orer re coefficien 7.5x -4 s -. I es n N O 5 molecule ou minues o fin n rec on he surfce of n erosol pricle O 5 n N O 5

4 S+P prolem 3.8: C ( is equilen o, ( C ( So he concenrion equions re C (,,c. pr -- Sole explicily ( is simple o sole, -. (3 Using his in (, we ge (4 Ting our lesson from he eigenecor/eigenlue nlysis, we ll see h his prolem hs soluions of he form. + (4 Which yiels ( + Now we jus group erms wih he sme eigenlue (ecy re, n use he iniil coniion h o ge ( (5, n

5 (6 ( ( Finlly, we cn sole for C y inegring (c wih (6 n using he iniil coniion h C C ( (7 You cn chec h his srs zero n hs ime eriie of zero, consisen wih he iniil coniions for C h we impose, n he fc h C s source, which is, is lso zero. You cn lso see h s >> / + /, hen C, which is wh we expec h eeryhing sring in eenully ens up s C.. pr jusify PSS for PSS for ses h he source n sin erms for re fr lrger hn heir ifference, / fer some iniil equilirium recion ime. Using scle nlysis on (, we cn formlly express his coniion s, << ~ (8 T PSS is concerne wih he longer imescles of se of recions fer pseuo sey se hs een reche. If we se T o e he lrger of / or /, hen we ge << (9 smller of, The only wy for his o wor is if <<. Here s n lerne rgumen using he eigenlue/eigenfuncion nlysis. PSS is jusifie when he wo eigenlues (i.e. jusmen res re well sepre. In his cse, he eigenlues re simply n (you cn o he mh. Plugging hese in o sole for he eigenecors, we see h he eigenlue correspons o n eigenecor where /( - n he eigenlue correspons o ecy purely in. If >>, we he ery oring soluion h he rpi eigenecor poins long he / - line, which simply mens h coners ino ery rpily hrough ( unil i reches se prllel o he secon eigenecor, which poins long. You will noe h his is he sme s PSS prolem where ( is se o zero. Since he prolem se ou seing ( o zero (no (, we he o moe on.

6 In he oher cse, where.>>, quicly equilires (inepenenly of unil i is on he line efine y he s eigenecor. The firs eigenecor in his limi preics h / / <<. fer reching his pseuo-equilirium, n ecy ogeher owrs zero he slower re. This is more meningful use of PSS, hn he.>> cse, simply ecuse here is some se where n re oh nonzero n in pseuo-equilirium.. Use PSS o sole prolem Seing ( yiels The soluion for ( is sill (3. Comining (3 n ( yiels Inegring (c using ( yiels / ( -. (3 ( / - ( C -- ( You coul he lso goen ( y noing h + + C, n <<, yieling C. You cn esily reconcile ( n ( wih (6 n (7 in he limi h <<.

e t dt e t dt = lim e t dt T (1 e T ) = 1

e t dt e t dt = lim e t dt T (1 e T ) = 1 Improper Inegrls There re wo ypes of improper inegrls - hose wih infinie limis of inegrion, nd hose wih inegrnds h pproch some poin wihin he limis of inegrion. Firs we will consider inegrls wih infinie

More information

(b) 10 yr. (b) 13 m. 1.6 m s, m s m s (c) 13.1 s. 32. (a) 20.0 s (b) No, the minimum distance to stop = 1.00 km. 1.

(b) 10 yr. (b) 13 m. 1.6 m s, m s m s (c) 13.1 s. 32. (a) 20.0 s (b) No, the minimum distance to stop = 1.00 km. 1. Answers o Een Numbered Problems Chper. () 7 m s, 6 m s (b) 8 5 yr 4.. m ih 6. () 5. m s (b).5 m s (c).5 m s (d) 3.33 m s (e) 8. ().3 min (b) 64 mi..3 h. ().3 s (b) 3 m 4..8 mi wes of he flgpole 6. (b)

More information

An object moving with speed v around a point at distance r, has an angular velocity. m/s m

An object moving with speed v around a point at distance r, has an angular velocity. m/s m Roion The mosphere roes wih he erh n moions wihin he mosphere clerly follow cure phs (cyclones, nicyclones, hurricnes, ornoes ec.) We nee o epress roion quniiely. For soli objec or ny mss h oes no isor

More information

More on Magnetically C Coupled Coils and Ideal Transformers

More on Magnetically C Coupled Coils and Ideal Transformers Appenix ore on gneiclly C Couple Coils Iel Trnsformers C. Equivlen Circuis for gneiclly Couple Coils A imes, i is convenien o moel mgneiclly couple coils wih n equivlen circui h oes no involve mgneic coupling.

More information

A Kalman filtering simulation

A Kalman filtering simulation A Klmn filering simulion The performnce of Klmn filering hs been esed on he bsis of wo differen dynmicl models, ssuming eiher moion wih consn elociy or wih consn ccelerion. The former is epeced o beer

More information

4.8 Improper Integrals

4.8 Improper Integrals 4.8 Improper Inegrls Well you ve mde i hrough ll he inegrion echniques. Congrs! Unforunely for us, we sill need o cover one more inegrl. They re clled Improper Inegrls. A his poin, we ve only del wih inegrls

More information

ENGR 1990 Engineering Mathematics The Integral of a Function as a Function

ENGR 1990 Engineering Mathematics The Integral of a Function as a Function ENGR 1990 Engineering Mhemics The Inegrl of Funcion s Funcion Previously, we lerned how o esime he inegrl of funcion f( ) over some inervl y dding he res of finie se of rpezoids h represen he re under

More information

Average & instantaneous velocity and acceleration Motion with constant acceleration

Average & instantaneous velocity and acceleration Motion with constant acceleration Physics 7: Lecure Reminders Discussion nd Lb secions sr meeing ne week Fill ou Pink dd/drop form if you need o swich o differen secion h is FULL. Do i TODAY. Homework Ch. : 5, 7,, 3,, nd 6 Ch.: 6,, 3 Submission

More information

The order of reaction is defined as the number of atoms or molecules whose concentration change during the chemical reaction.

The order of reaction is defined as the number of atoms or molecules whose concentration change during the chemical reaction. www.hechemisryguru.com Re Lw Expression Order of Recion The order of recion is defined s he number of oms or molecules whose concenrion chnge during he chemicl recion. Or The ol number of molecules or

More information

3 Motion with constant acceleration: Linear and projectile motion

3 Motion with constant acceleration: Linear and projectile motion 3 Moion wih consn ccelerion: Liner nd projecile moion cons, In he precedin Lecure we he considered moion wih consn ccelerion lon he is: Noe h,, cn be posiie nd neie h leds o rie of behiors. Clerl similr

More information

Minimum Squared Error

Minimum Squared Error Minimum Squred Error LDF: Minimum Squred-Error Procedures Ide: conver o esier nd eer undersood prolem Percepron y i > for ll smples y i solve sysem of liner inequliies MSE procedure y i = i for ll smples

More information

Minimum Squared Error

Minimum Squared Error Minimum Squred Error LDF: Minimum Squred-Error Procedures Ide: conver o esier nd eer undersood prolem Percepron y i > 0 for ll smples y i solve sysem of liner inequliies MSE procedure y i i for ll smples

More information

MTH 146 Class 11 Notes

MTH 146 Class 11 Notes 8.- Are of Surfce of Revoluion MTH 6 Clss Noes Suppose we wish o revolve curve C round n is nd find he surfce re of he resuling solid. Suppose f( ) is nonnegive funcion wih coninuous firs derivive on he

More information

Physic 231 Lecture 4. Mi it ftd l t. Main points of today s lecture: Example: addition of velocities Trajectories of objects in 2 = =

Physic 231 Lecture 4. Mi it ftd l t. Main points of today s lecture: Example: addition of velocities Trajectories of objects in 2 = = Mi i fd l Phsic 3 Lecure 4 Min poins of od s lecure: Emple: ddiion of elociies Trjecories of objecs in dimensions: dimensions: g 9.8m/s downwrds ( ) g o g g Emple: A foobll pler runs he pern gien in he

More information

Contraction Mapping Principle Approach to Differential Equations

Contraction Mapping Principle Approach to Differential Equations epl Journl of Science echnology 0 (009) 49-53 Conrcion pping Principle pproch o Differenil Equions Bishnu P. Dhungn Deprmen of hemics, hendr Rn Cmpus ribhuvn Universiy, Khmu epl bsrc Using n eension of

More information

Physics Worksheet Lesson 4: Linear Motion Section: Name:

Physics Worksheet Lesson 4: Linear Motion Section: Name: Physics Workshee Lesson 4: Liner Moion Secion: Nme: 1. Relie Moion:. All moion is. b. is n rbirry coorine sysem wih reference o which he posiion or moion of somehing is escribe or physicl lws re formule.

More information

1.0 Electrical Systems

1.0 Electrical Systems . Elecricl Sysems The ypes of dynmicl sysems we will e sudying cn e modeled in erms of lgeric equions, differenil equions, or inegrl equions. We will egin y looking fmilir mhemicl models of idel resisors,

More information

Chapter Three Systems of Linear Differential Equations

Chapter Three Systems of Linear Differential Equations Chaper Three Sysems of Linear Differenial Equaions In his chaper we are going o consier sysems of firs orer orinary ifferenial equaions. These are sysems of he form x a x a x a n x n x a x a x a n x n

More information

0 for t < 0 1 for t > 0

0 for t < 0 1 for t > 0 8.0 Sep nd del funcions Auhor: Jeremy Orloff The uni Sep Funcion We define he uni sep funcion by u() = 0 for < 0 for > 0 I is clled he uni sep funcion becuse i kes uni sep = 0. I is someimes clled he Heviside

More information

Motion. Part 2: Constant Acceleration. Acceleration. October Lab Physics. Ms. Levine 1. Acceleration. Acceleration. Units for Acceleration.

Motion. Part 2: Constant Acceleration. Acceleration. October Lab Physics. Ms. Levine 1. Acceleration. Acceleration. Units for Acceleration. Moion Accelerion Pr : Consn Accelerion Accelerion Accelerion Accelerion is he re of chnge of velociy. = v - vo = Δv Δ ccelerion = = v - vo chnge of velociy elpsed ime Accelerion is vecor, lhough in one-dimensionl

More information

PHYSICS 1210 Exam 1 University of Wyoming 14 February points

PHYSICS 1210 Exam 1 University of Wyoming 14 February points PHYSICS 1210 Em 1 Uniersiy of Wyoming 14 Februry 2013 150 poins This es is open-noe nd closed-book. Clculors re permied bu compuers re no. No collborion, consulion, or communicion wih oher people (oher

More information

Physics 2A HW #3 Solutions

Physics 2A HW #3 Solutions Chper 3 Focus on Conceps: 3, 4, 6, 9 Problems: 9, 9, 3, 41, 66, 7, 75, 77 Phsics A HW #3 Soluions Focus On Conceps 3-3 (c) The ccelerion due o grvi is he sme for boh blls, despie he fc h he hve differen

More information

1. Consider a PSA initially at rest in the beginning of the left-hand end of a long ISS corridor. Assume xo = 0 on the left end of the ISS corridor.

1. Consider a PSA initially at rest in the beginning of the left-hand end of a long ISS corridor. Assume xo = 0 on the left end of the ISS corridor. In Eercise 1, use sndrd recngulr Cresin coordine sysem. Le ime be represened long he horizonl is. Assume ll ccelerions nd decelerions re consn. 1. Consider PSA iniilly res in he beginning of he lef-hnd

More information

Chapter 2. Motion along a straight line. 9/9/2015 Physics 218

Chapter 2. Motion along a straight line. 9/9/2015 Physics 218 Chper Moion long srigh line 9/9/05 Physics 8 Gols for Chper How o describe srigh line moion in erms of displcemen nd erge elociy. The mening of insnneous elociy nd speed. Aerge elociy/insnneous elociy

More information

A 1.3 m 2.5 m 2.8 m. x = m m = 8400 m. y = 4900 m 3200 m = 1700 m

A 1.3 m 2.5 m 2.8 m. x = m m = 8400 m. y = 4900 m 3200 m = 1700 m PHYS : Soluions o Chper 3 Home Work. SSM REASONING The displcemen is ecor drwn from he iniil posiion o he finl posiion. The mgniude of he displcemen is he shores disnce beween he posiions. Noe h i is onl

More information

Chapter Direct Method of Interpolation

Chapter Direct Method of Interpolation Chper 5. Direc Mehod of Inerpolion Afer reding his chper, you should be ble o:. pply he direc mehod of inerpolion,. sole problems using he direc mehod of inerpolion, nd. use he direc mehod inerpolns o

More information

September 20 Homework Solutions

September 20 Homework Solutions College of Engineering nd Compuer Science Mechnicl Engineering Deprmen Mechnicl Engineering A Seminr in Engineering Anlysis Fll 7 Number 66 Insrucor: Lrry Creo Sepember Homework Soluions Find he specrum

More information

Physics 100: Lecture 1

Physics 100: Lecture 1 Physics : Lecure Agen for Toy Aice Scope of his course Mesuremen n Unis Funmenl unis Sysems of unis Conering beween sysems of unis Dimensionl Anlysis -D Kinemics (reiew) Aerge & insnneous elociy n ccelerion

More information

2D Motion WS. A horizontally launched projectile s initial vertical velocity is zero. Solve the following problems with this information.

2D Motion WS. A horizontally launched projectile s initial vertical velocity is zero. Solve the following problems with this information. Nme D Moion WS The equions of moion h rele o projeciles were discussed in he Projecile Moion Anlsis Acii. ou found h projecile moes wih consn eloci in he horizonl direcion nd consn ccelerion in he ericl

More information

PHYSICS 211 MIDTERM I 21 April 2004

PHYSICS 211 MIDTERM I 21 April 2004 PHYSICS MIDERM I April 004 Exm is closed book, closed notes. Use only your formul sheet. Write ll work nd nswers in exm booklets. he bcks of pges will not be grded unless you so request on the front of

More information

Homework 2 Solutions

Homework 2 Solutions Mah 308 Differenial Equaions Fall 2002 & 2. See he las page. Hoework 2 Soluions 3a). Newon s secon law of oion says ha a = F, an we know a =, so we have = F. One par of he force is graviy, g. However,

More information

( ) ( ) ( ) ( ) ( ) ( y )

( ) ( ) ( ) ( ) ( ) ( y ) 8. Lengh of Plne Curve The mos fmous heorem in ll of mhemics is he Pyhgoren Theorem. I s formulion s he disnce formul is used o find he lenghs of line segmens in he coordine plne. In his secion you ll

More information

1. (16 points) Answer the following derivative-related questions. dx tan sec x. dx tan u = du d. dx du tan u. du tan u d v.

1. (16 points) Answer the following derivative-related questions. dx tan sec x. dx tan u = du d. dx du tan u. du tan u d v. Exam #2 Soluions. (6 poins) Answer he following eriaie-relae quesions. (a) (8 poins) If y an sec x, fin. This is an applicaion of he chain rule in wo sages, in which we shall le u sec x, an sec x: an sec

More information

Mathematics 805 Final Examination Answers

Mathematics 805 Final Examination Answers . 5 poins Se he Weiersrss M-es. Mhemics 85 Finl Eminion Answers Answer: Suppose h A R, nd f n : A R. Suppose furher h f n M n for ll A, nd h Mn converges. Then f n converges uniformly on A.. 5 poins Se

More information

Properties of Logarithms. Solving Exponential and Logarithmic Equations. Properties of Logarithms. Properties of Logarithms. ( x)

Properties of Logarithms. Solving Exponential and Logarithmic Equations. Properties of Logarithms. Properties of Logarithms. ( x) Properies of Logrihms Solving Eponenil nd Logrihmic Equions Properies of Logrihms Produc Rule ( ) log mn = log m + log n ( ) log = log + log Properies of Logrihms Quoien Rule log m = logm logn n log7 =

More information

FM Applications of Integration 1.Centroid of Area

FM Applications of Integration 1.Centroid of Area FM Applicions of Inegrion.Cenroid of Are The cenroid of ody is is geomeric cenre. For n ojec mde of uniform meril, he cenroid coincides wih he poin which he ody cn e suppored in perfecly lnced se ie, is

More information

SOME USEFUL MATHEMATICS

SOME USEFUL MATHEMATICS SOME USEFU MAHEMAICS SOME USEFU MAHEMAICS I is esy o mesure n preic he behvior of n elecricl circui h conins only c volges n currens. However, mos useful elecricl signls h crry informion vry wih ime. Since

More information

The Fundamental Theorems of Calculus

The Fundamental Theorems of Calculus FunamenalTheorems.nb 1 The Funamenal Theorems of Calculus You have now been inrouce o he wo main branches of calculus: ifferenial calculus (which we inrouce wih he angen line problem) an inegral calculus

More information

Chapter 2: Evaluative Feedback

Chapter 2: Evaluative Feedback Chper 2: Evluive Feedbck Evluing cions vs. insrucing by giving correc cions Pure evluive feedbck depends olly on he cion ken. Pure insrucive feedbck depends no ll on he cion ken. Supervised lerning is

More information

Phys 110. Answers to even numbered problems on Midterm Map

Phys 110. Answers to even numbered problems on Midterm Map Phys Answers o een numbered problems on Miderm Mp. REASONING The word per indices rio, so.35 mm per dy mens.35 mm/d, which is o be epressed s re in f/cenury. These unis differ from he gien unis in boh

More information

Motion in a Straight Line

Motion in a Straight Line Moion in Srigh Line. Preei reched he mero sion nd found h he esclor ws no working. She wlked up he sionry esclor in ime. On oher dys, if she remins sionry on he moing esclor, hen he esclor kes her up in

More information

Section 5: Chain Rule

Section 5: Chain Rule Chaper The Derivaive Applie Calculus 11 Secion 5: Chain Rule There is one more ype of complicae funcion ha we will wan o know how o iffereniae: composiion. The Chain Rule will le us fin he erivaive of

More information

Solutions to assignment 3

Solutions to assignment 3 D Sruure n Algorihm FR 6. Informik Sner, Telikeplli WS 03/04 hp://www.mpi-.mpg.e/~ner/oure/lg03/inex.hml Soluion o ignmen 3 Exerie Arirge i he ue of irepnie in urreny exhnge re o rnform one uni of urreny

More information

Chapter 2 The Derivative Applied Calculus 107. We ll need a rule for finding the derivative of a product so we don t have to multiply everything out.

Chapter 2 The Derivative Applied Calculus 107. We ll need a rule for finding the derivative of a product so we don t have to multiply everything out. Chaper The Derivaive Applie Calculus 107 Secion 4: Prouc an Quoien Rules The basic rules will le us ackle simple funcions. Bu wha happens if we nee he erivaive of a combinaion of hese funcions? Eample

More information

INTEGRALS. Exercise 1. Let f : [a, b] R be bounded, and let P and Q be partitions of [a, b]. Prove that if P Q then U(P ) U(Q) and L(P ) L(Q).

INTEGRALS. Exercise 1. Let f : [a, b] R be bounded, and let P and Q be partitions of [a, b]. Prove that if P Q then U(P ) U(Q) and L(P ) L(Q). INTEGRALS JOHN QUIGG Eercise. Le f : [, b] R be bounded, nd le P nd Q be priions of [, b]. Prove h if P Q hen U(P ) U(Q) nd L(P ) L(Q). Soluion: Le P = {,..., n }. Since Q is obined from P by dding finiely

More information

M r. d 2. R t a M. Structural Mechanics Section. Exam CT5141 Theory of Elasticity Friday 31 October 2003, 9:00 12:00 hours. Problem 1 (3 points)

M r. d 2. R t a M. Structural Mechanics Section. Exam CT5141 Theory of Elasticity Friday 31 October 2003, 9:00 12:00 hours. Problem 1 (3 points) Delf Universiy of Technology Fculy of Civil Engineering nd Geosciences Srucurl echnics Secion Wrie your nme nd sudy numer he op righ-hnd of your work. Exm CT5 Theory of Elsiciy Fridy Ocoer 00, 9:00 :00

More information

5.1-The Initial-Value Problems For Ordinary Differential Equations

5.1-The Initial-Value Problems For Ordinary Differential Equations 5.-The Iniil-Vlue Problems For Ordinry Differenil Equions Consider solving iniil-vlue problems for ordinry differenil equions: (*) y f, y, b, y. If we know he generl soluion y of he ordinry differenil

More information

Y 0.4Y 0.45Y Y to a proper ARMA specification.

Y 0.4Y 0.45Y Y to a proper ARMA specification. HG Jan 04 ECON 50 Exercises II - 0 Feb 04 (wih answers Exercise. Read secion 8 in lecure noes 3 (LN3 on he common facor problem in ARMA-processes. Consider he following process Y 0.4Y 0.45Y 0.5 ( where

More information

Chapter 2 PROBLEM SOLUTIONS

Chapter 2 PROBLEM SOLUTIONS Chper PROBLEM SOLUTIONS. We ssume h you re pproximely m ll nd h he nere impulse rels uniform speed. The elpsed ime is hen Δ x m Δ = m s s. s.3 Disnces reled beween pirs of ciies re ( ) Δx = Δ = 8. km h.5

More information

PHY2048 Exam 1 Formula Sheet Vectors. Motion. v ave (3 dim) ( (1 dim) dt. ( (3 dim) Equations of Motion (Constant Acceleration)

PHY2048 Exam 1 Formula Sheet Vectors. Motion. v ave (3 dim) ( (1 dim) dt. ( (3 dim) Equations of Motion (Constant Acceleration) Insrucors: Field/Mche PHYSICS DEPATMENT PHY 48 Em Ferur, 5 Nme prin, ls firs: Signure: On m honor, I he neiher gien nor receied unuhoried id on his eminion. YOU TEST NUMBE IS THE 5-DIGIT NUMBE AT THE TOP

More information

Solutions to Problems from Chapter 2

Solutions to Problems from Chapter 2 Soluions o Problems rom Chper Problem. The signls u() :5sgn(), u () :5sgn(), nd u h () :5sgn() re ploed respecively in Figures.,b,c. Noe h u h () :5sgn() :5; 8 including, bu u () :5sgn() is undeined..5

More information

dt = C exp (3 ln t 4 ). t 4 W = C exp ( ln(4 t) 3) = C(4 t) 3.

dt = C exp (3 ln t 4 ). t 4 W = C exp ( ln(4 t) 3) = C(4 t) 3. Mah Rahman Exam Review Soluions () Consider he IVP: ( 4)y 3y + 4y = ; y(3) = 0, y (3) =. (a) Please deermine he longes inerval for which he IVP is guaraneed o have a unique soluion. Soluion: The disconinuiies

More information

For the reaction, R P, the is given by,

For the reaction, R P, the is given by, Dr JADU SAMUEL CHEMICAL KINETICS Inroducion Chemicl ineics is brnch of physicl chemisry, which dels wih he sudy of he re of chemicl recions nd he vrious fcors ffecing i Such sudies lso enble us o elucide

More information

CHAPTER 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES

CHAPTER 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES CHAPTER PARAMETRIC EQUATIONS AND POLAR COORDINATES. PARAMETRIZATIONS OF PLANE CURVES., 9, _ _ Ê.,, Ê or, Ÿ. 5, 7, _ _.,, Ÿ Ÿ Ê Ê 5 Ê ( 5) Ê ˆ Ê 6 Ê ( 5) 7 Ê Ê, Ÿ Ÿ $ 5. cos, sin, Ÿ Ÿ 6. cos ( ), sin (

More information

Some Inequalities variations on a common theme Lecture I, UL 2007

Some Inequalities variations on a common theme Lecture I, UL 2007 Some Inequliies vriions on common heme Lecure I, UL 2007 Finbrr Hollnd, Deprmen of Mhemics, Universiy College Cork, fhollnd@uccie; July 2, 2007 Three Problems Problem Assume i, b i, c i, i =, 2, 3 re rel

More information

CHAPTER 2: Describing Motion: Kinematics in One Dimension

CHAPTER 2: Describing Motion: Kinematics in One Dimension CHAPTER : Describing Moion: Kinemics in One Dimension Answers o Quesions A cr speeomeer mesures only spee I oes no gie ny informion bou he irecion, n so oes no mesure elociy By efiniion, if n objec hs

More information

3. Renewal Limit Theorems

3. Renewal Limit Theorems Virul Lborories > 14. Renewl Processes > 1 2 3 3. Renewl Limi Theorems In he inroducion o renewl processes, we noed h he rrivl ime process nd he couning process re inverses, in sens The rrivl ime process

More information

The solution is often represented as a vector: 2xI + 4X2 + 2X3 + 4X4 + 2X5 = 4 2xI + 4X2 + 3X3 + 3X4 + 3X5 = 4. 3xI + 6X2 + 6X3 + 3X4 + 6X5 = 6.

The solution is often represented as a vector: 2xI + 4X2 + 2X3 + 4X4 + 2X5 = 4 2xI + 4X2 + 3X3 + 3X4 + 3X5 = 4. 3xI + 6X2 + 6X3 + 3X4 + 6X5 = 6. [~ o o :- o o ill] i 1. Mrices, Vecors, nd Guss-Jordn Eliminion 1 x y = = - z= The soluion is ofen represened s vecor: n his exmple, he process of eliminion works very smoohly. We cn elimine ll enries

More information

Characteristic Function for the Truncated Triangular Distribution., Myron Katzoff and Rahul A. Parsa

Characteristic Function for the Truncated Triangular Distribution., Myron Katzoff and Rahul A. Parsa Secion on Survey Reserch Mehos JSM 009 Chrcerisic Funcion for he Trunce Tringulr Disriuion Jy J. Kim 1 1, Myron Kzoff n Rhul A. Prs 1 Nionl Cener for Helh Sisics, 11Toleo Ro, Hysville, MD. 078 College

More information

23.5. Half-Range Series. Introduction. Prerequisites. Learning Outcomes

23.5. Half-Range Series. Introduction. Prerequisites. Learning Outcomes Half-Range Series 2.5 Inroducion In his Secion we address he following problem: Can we find a Fourier series expansion of a funcion defined over a finie inerval? Of course we recognise ha such a funcion

More information

Practice Problems - Week #4 Higher-Order DEs, Applications Solutions

Practice Problems - Week #4 Higher-Order DEs, Applications Solutions Pracice Probles - Wee #4 Higher-Orer DEs, Applicaions Soluions 1. Solve he iniial value proble where y y = 0, y0 = 0, y 0 = 1, an y 0 =. r r = rr 1 = rr 1r + 1, so he general soluion is C 1 + C e x + C

More information

Magnetostatics Bar Magnet. Magnetostatics Oersted s Experiment

Magnetostatics Bar Magnet. Magnetostatics Oersted s Experiment Mgneosics Br Mgne As fr bck s 4500 yers go, he Chinese discovered h cerin ypes of iron ore could rc ech oher nd cerin mels. Iron filings "mp" of br mgne s field Crefully suspended slivers of his mel were

More information

Module II, Part C. More Insight into Fiber Dispersion

Module II, Part C. More Insight into Fiber Dispersion Moule II Par C More Insigh ino Fiber Dispersion . Polariaion Moe Dispersion Fiber Birefringence: Imperfec cylinrical symmery leas o wha is known as birefringence. Recall he HE moe an is E x componen which

More information

( ) ( ) ( ) ( u) ( u) = are shown in Figure =, it is reasonable to speculate that. = cos u ) and the inside function ( ( t) du

( ) ( ) ( ) ( u) ( u) = are shown in Figure =, it is reasonable to speculate that. = cos u ) and the inside function ( ( t) du Porlan Communiy College MTH 51 Lab Manual The Chain Rule Aciviy 38 The funcions f ( = sin ( an k( sin( 3 38.1. Since f ( cos( k ( = cos( 3. Bu his woul imply ha k ( f ( = are shown in Figure =, i is reasonable

More information

Version 001 test-1 swinney (57010) 1. is constant at m/s.

Version 001 test-1 swinney (57010) 1. is constant at m/s. Version 001 es-1 swinne (57010) 1 This prin-ou should hve 20 quesions. Muliple-choice quesions m coninue on he nex column or pge find ll choices before nswering. CubeUniVec1x76 001 10.0 poins Acubeis1.4fee

More information

6. Gas dynamics. Ideal gases Speed of infinitesimal disturbances in still gas

6. Gas dynamics. Ideal gases Speed of infinitesimal disturbances in still gas 6. Gs dynmics Dr. Gergely Krisóf De. of Fluid echnics, BE Februry, 009. Seed of infiniesiml disurbnces in sill gs dv d, dv d, Coninuiy: ( dv)( ) dv omenum r r heorem: ( ( dv) ) d 3443 4 q m dv d dv llievi

More information

Two Coupled Oscillators / Normal Modes

Two Coupled Oscillators / Normal Modes Lecure 3 Phys 3750 Two Coupled Oscillaors / Normal Modes Overview and Moivaion: Today we ake a small, bu significan, sep owards wave moion. We will no ye observe waves, bu his sep is imporan in is own

More information

COMPETITIVE GROWTH MODEL

COMPETITIVE GROWTH MODEL COMPETITIVE GROWTH MODEL I Assumpions We are going o now solve he compeiive version of he opimal growh moel. Alhough he allocaions are he same as in he social planning problem, i will be useful o compare

More information

Collision Detection and Bouncing

Collision Detection and Bouncing Collision Deecion nd Bouncing Collisions re Hndled in Two Prs. Deecing he collision Mike Biley mj@cs.oregonse.edu. Hndling he physics of he collision collision-ouncing.ppx If You re Lucky, You Cn Deec

More information

A new model for limit order book dynamics

A new model for limit order book dynamics Anewmodelforlimiorderbookdynmics JeffreyR.Russell UniversiyofChicgo,GrdueSchoolofBusiness TejinKim UniversiyofChicgo,DeprmenofSisics Absrc:Thispperproposesnewmodelforlimiorderbookdynmics.Thelimiorderbookconsiss

More information

Chapter 2. First Order Scalar Equations

Chapter 2. First Order Scalar Equations Chaper. Firs Order Scalar Equaions We sar our sudy of differenial equaions in he same way he pioneers in his field did. We show paricular echniques o solve paricular ypes of firs order differenial equaions.

More information

Chapter 15: Phenomena. Chapter 15 Chemical Kinetics. Reaction Rates. Reaction Rates R P. Reaction Rates. Rate Laws

Chapter 15: Phenomena. Chapter 15 Chemical Kinetics. Reaction Rates. Reaction Rates R P. Reaction Rates. Rate Laws Chaper 5: Phenomena Phenomena: The reacion (aq) + B(aq) C(aq) was sudied a wo differen emperaures (98 K and 35 K). For each emperaure he reacion was sared by puing differen concenraions of he 3 species

More information

RESPONSE UNDER A GENERAL PERIODIC FORCE. When the external force F(t) is periodic with periodτ = 2π

RESPONSE UNDER A GENERAL PERIODIC FORCE. When the external force F(t) is periodic with periodτ = 2π RESPONSE UNDER A GENERAL PERIODIC FORCE When he exernl force F() is periodic wih periodτ / ω,i cn be expnded in Fourier series F( ) o α ω α b ω () where τ F( ) ω d, τ,,,... () nd b τ F( ) ω d, τ,,... (3)

More information

Chapter Introduction. 2. Linear Combinations [4.1]

Chapter Introduction. 2. Linear Combinations [4.1] Chper 4 Inrouion Thi hper i ou generlizing he onep you lerne in hper o pe oher n hn R Mny opi in hi hper re heoreil n MATLAB will no e le o help you ou You will ee where MATLAB i ueful in hper 4 n how

More information

can be viewed as a generalized product, and one for which the product of f and g. That is, does

can be viewed as a generalized product, and one for which the product of f and g. That is, does Boyce/DiPrim 9 h e, Ch 6.6: The Convoluion Inegrl Elemenry Differenil Equion n Bounry Vlue Problem, 9 h eiion, by Willim E. Boyce n Richr C. DiPrim, 9 by John Wiley & Son, Inc. Someime i i poible o wrie

More information

Math 2142 Exam 1 Review Problems. x 2 + f (0) 3! for the 3rd Taylor polynomial at x = 0. To calculate the various quantities:

Math 2142 Exam 1 Review Problems. x 2 + f (0) 3! for the 3rd Taylor polynomial at x = 0. To calculate the various quantities: Mah 4 Eam Review Problems Problem. Calculae he 3rd Taylor polynomial for arcsin a =. Soluion. Le f() = arcsin. For his problem, we use he formula f() + f () + f ()! + f () 3! for he 3rd Taylor polynomial

More information

22.615, MHD Theory of Fusion Systems Prof. Freidberg Lecture 9: The High Beta Tokamak

22.615, MHD Theory of Fusion Systems Prof. Freidberg Lecture 9: The High Beta Tokamak .65, MHD Theory of Fusion Sysems Prof. Freidberg Lecure 9: The High e Tokmk Summry of he Properies of n Ohmic Tokmk. Advnges:. good euilibrium (smll shif) b. good sbiliy ( ) c. good confinemen ( τ nr )

More information

Physics 101 Lecture 4 Motion in 2D and 3D

Physics 101 Lecture 4 Motion in 2D and 3D Phsics 11 Lecure 4 Moion in D nd 3D Dr. Ali ÖVGÜN EMU Phsics Deprmen www.ogun.com Vecor nd is componens The componens re he legs of he righ ringle whose hpoenuse is A A A A A n ( θ ) A Acos( θ) A A A nd

More information

WEEK-3 Recitation PHYS 131. of the projectile s velocity remains constant throughout the motion, since the acceleration a x

WEEK-3 Recitation PHYS 131. of the projectile s velocity remains constant throughout the motion, since the acceleration a x WEEK-3 Reciaion PHYS 131 Ch. 3: FOC 1, 3, 4, 6, 14. Problems 9, 37, 41 & 71 and Ch. 4: FOC 1, 3, 5, 8. Problems 3, 5 & 16. Feb 8, 018 Ch. 3: FOC 1, 3, 4, 6, 14. 1. (a) The horizonal componen of he projecile

More information

Laplace Examples, Inverse, Rational Form

Laplace Examples, Inverse, Rational Form Lecure 3 Ouline: Lplce Exple, Invere, Rionl For Announceen: Rein: 6: Lplce Trnfor pp. 3-33, 55.5-56.5, 7 HW 8 poe, ue nex We. Free -y exenion OcenOne Roo Tour will e fer cl y 7 (:3-:) Lunch provie ferwr.

More information

Physics 201, Lecture 5

Physics 201, Lecture 5 Phsics 1 Lecue 5 Tod s Topics n Moion in D (Chp 4.1-4.3): n D Kinemicl Quniies (sec. 4.1) n D Kinemics wih Consn Acceleion (sec. 4.) n D Pojecile (Sec 4.3) n Epeced fom Peiew: n Displcemen eloci cceleion

More information

Section P.1 Notes Page 1 Section P.1 Precalculus and Trigonometry Review

Section P.1 Notes Page 1 Section P.1 Precalculus and Trigonometry Review Secion P Noe Pge Secion P Preclculu nd Trigonomer Review ALGEBRA AND PRECALCULUS Eponen Lw: Emple: 8 Emple: Emple: Emple: b b Emple: 9 EXAMPLE: Simplif: nd wrie wi poiive eponen Fir I will flip e frcion

More information

LAPLACE TRANSFORMS. 1. Basic transforms

LAPLACE TRANSFORMS. 1. Basic transforms LAPLACE TRANSFORMS. Bic rnform In hi coure, Lplce Trnform will be inroduced nd heir properie exmined; ble of common rnform will be buil up; nd rnform will be ued o olve ome dierenil equion by rnforming

More information

Seminar 5 Sustainability

Seminar 5 Sustainability Seminar 5 Susainabiliy Soluions Quesion : Hyperbolic Discouning -. Suppose a faher inheris a family forune of 0 million NOK an he wans o use some of i for himself (o be precise, he share ) bu also o beques

More information

f t f a f x dx By Lin McMullin f x dx= f b f a. 2

f t f a f x dx By Lin McMullin f x dx= f b f a. 2 Accumulion: Thoughs On () By Lin McMullin f f f d = + The gols of he AP* Clculus progrm include he semen, Sudens should undersnd he definie inegrl s he ne ccumulion of chnge. 1 The Topicl Ouline includes

More information

KEY. Math 334 Midterm III Winter 2008 section 002 Instructor: Scott Glasgow

KEY. Math 334 Midterm III Winter 2008 section 002 Instructor: Scott Glasgow KEY Mah 334 Miderm III Winer 008 secion 00 Insrucor: Sco Glasgow Please do NOT wrie on his exam. No credi will be given for such work. Raher wrie in a blue book, or on your own paper, preferably engineering

More information

Topics in Combinatorial Optimization May 11, Lecture 22

Topics in Combinatorial Optimization May 11, Lecture 22 8.997 Topics in Combinaorial Opimizaion May, 004 Lecure Lecurer: Michel X. Goemans Scribe: Alanha Newman Muliflows an Disjoin Pahs Le G = (V,E) be a graph an le s,,s,,...s, V be erminals. Our goal is o

More information

total distance cov ered time int erval v = average speed (m/s)

total distance cov ered time int erval v = average speed (m/s) Physics Suy Noes Lesson 4 Linear Moion 1 Change an Moion a. A propery common o eeryhing in he unierse is change. b. Change is so imporan ha he funamenal concep of ime woul be meaningless wihou i. c. Since

More information

Temperature Rise of the Earth

Temperature Rise of the Earth Avilble online www.sciencedirec.com ScienceDirec Procedi - Socil nd Behviorl Scien ce s 88 ( 2013 ) 220 224 Socil nd Behviorl Sciences Symposium, 4 h Inernionl Science, Socil Science, Engineering nd Energy

More information

EXERCISE - 01 CHECK YOUR GRASP

EXERCISE - 01 CHECK YOUR GRASP UNIT # 09 PARABOLA, ELLIPSE & HYPERBOLA PARABOLA EXERCISE - 0 CHECK YOUR GRASP. Hin : Disnce beween direcri nd focus is 5. Given (, be one end of focl chord hen oher end be, lengh of focl chord 6. Focus

More information

Mon Apr 9 EP 7.6 Convolutions and Laplace transforms. Announcements: Warm-up Exercise:

Mon Apr 9 EP 7.6 Convolutions and Laplace transforms. Announcements: Warm-up Exercise: Mah 225-4 Week 3 April 9-3 EP 7.6 - convoluions; 6.-6.2 - eigenvalues, eigenvecors and diagonalizabiliy; 7. - sysems of differenial equaions. Mon Apr 9 EP 7.6 Convoluions and Laplace ransforms. Announcemens:

More information

Linear Response Theory: The connection between QFT and experiments

Linear Response Theory: The connection between QFT and experiments Phys540.nb 39 3 Linear Response Theory: The connecion beween QFT and experimens 3.1. Basic conceps and ideas Q: How do we measure he conduciviy of a meal? A: we firs inroduce a weak elecric field E, and

More information

Chapter 7: Solving Trig Equations

Chapter 7: Solving Trig Equations Haberman MTH Secion I: The Trigonomeric Funcions Chaper 7: Solving Trig Equaions Le s sar by solving a couple of equaions ha involve he sine funcion EXAMPLE a: Solve he equaion sin( ) The inverse funcions

More information

KEY. Math 334 Midterm I Fall 2008 sections 001 and 003 Instructor: Scott Glasgow

KEY. Math 334 Midterm I Fall 2008 sections 001 and 003 Instructor: Scott Glasgow 1 KEY Mah 4 Miderm I Fall 8 secions 1 and Insrucor: Sco Glasgow Please do NOT wrie on his eam. No credi will be given for such work. Raher wrie in a blue book, or on our own paper, preferabl engineering

More information

ECE Microwave Engineering. Fall Prof. David R. Jackson Dept. of ECE. Notes 10. Waveguides Part 7: Transverse Equivalent Network (TEN)

ECE Microwave Engineering. Fall Prof. David R. Jackson Dept. of ECE. Notes 10. Waveguides Part 7: Transverse Equivalent Network (TEN) EE 537-635 Microwve Engineering Fll 7 Prof. Dvid R. Jcson Dep. of EE Noes Wveguides Pr 7: Trnsverse Equivlen Newor (N) Wveguide Trnsmission Line Model Our gol is o come up wih rnsmission line model for

More information

Lecture 6 - Testing Restrictions on the Disturbance Process (References Sections 2.7 and 2.10, Hayashi)

Lecture 6 - Testing Restrictions on the Disturbance Process (References Sections 2.7 and 2.10, Hayashi) Lecure 6 - esing Resricions on he Disurbance Process (References Secions 2.7 an 2.0, Hayashi) We have eveloe sufficien coniions for he consisency an asymoic normaliy of he OLS esimaor ha allow for coniionally

More information

MATH 4330/5330, Fourier Analysis Section 6, Proof of Fourier s Theorem for Pointwise Convergence

MATH 4330/5330, Fourier Analysis Section 6, Proof of Fourier s Theorem for Pointwise Convergence MATH 433/533, Fourier Analysis Secion 6, Proof of Fourier s Theorem for Poinwise Convergence Firs, some commens abou inegraing periodic funcions. If g is a periodic funcion, g(x + ) g(x) for all real x,

More information

V L. DT s D T s t. Figure 1: Buck-boost converter: inductor current i(t) in the continuous conduction mode.

V L. DT s D T s t. Figure 1: Buck-boost converter: inductor current i(t) in the continuous conduction mode. ECE 445 Analysis and Design of Power Elecronic Circuis Problem Se 7 Soluions Problem PS7.1 Erickson, Problem 5.1 Soluion (a) Firs, recall he operaion of he buck-boos converer in he coninuous conducion

More information

THE 2-BODY PROBLEM. FIGURE 1. A pair of ellipses sharing a common focus. (c,b) c+a ROBERT J. VANDERBEI

THE 2-BODY PROBLEM. FIGURE 1. A pair of ellipses sharing a common focus. (c,b) c+a ROBERT J. VANDERBEI THE 2-BODY PROBLEM ROBERT J. VANDERBEI ABSTRACT. In his shor noe, we show ha a pair of ellipses wih a common focus is a soluion o he 2-body problem. INTRODUCTION. Solving he 2-body problem from scrach

More information

EXISTENCE AND UNIQUENESS OF SOLUTIONS FOR A SECOND-ORDER ITERATIVE BOUNDARY-VALUE PROBLEM

EXISTENCE AND UNIQUENESS OF SOLUTIONS FOR A SECOND-ORDER ITERATIVE BOUNDARY-VALUE PROBLEM Elecronic Journl of Differenil Equions, Vol. 208 (208), No. 50, pp. 6. ISSN: 072-669. URL: hp://ejde.mh.xse.edu or hp://ejde.mh.un.edu EXISTENCE AND UNIQUENESS OF SOLUTIONS FOR A SECOND-ORDER ITERATIVE

More information