Thermodynamics. Question 1. Question 2. Question 3 3/10/2010. Practice Questions PV TR PV T R

Save this PDF as:
 WORD  PNG  TXT  JPG

Size: px
Start display at page:

Download "Thermodynamics. Question 1. Question 2. Question 3 3/10/2010. Practice Questions PV TR PV T R"

Transcription

1 /10/010 Question 1 1 mole of idel gs is rought to finl stte F y one of three proesses tht hve different initil sttes s shown in the figure. Wht is true for the temperture hnge etween initil nd finl sttes? Thermodynmis rtie Questions ) It s the sme for ll proesses. ) It s the smllest for proess 1. ) It s the smllest for proess. ) It s the smllest for proess. E) It s the sme for proesses 1 nd. The temperture for gs t ny point is given y: TR T R Sine ll of the three proesses end t the sme temperture, we need only look t the proess tht strts losest to: 0 0 T R R p 0 1 p 0 p roess 1. R R tully egins t this temperture,. so its hnge is 0 R R. R R F Question 1 mole of idel gs is rought to finl stte F y one of three proesses tht hve different initil sttes s shown in the figure. Wht is true for the work done y the gs? ) It s positive for proess 1 nd, ut negtive for proess. ) It s the smllest for proess 1. p 0 ) It s the smllest for proess. 1 ) It s the smllest for proess. p E) Zero work is done long proess. 0 p F Question n idel gs is rought from S -> F y three different pths: SRF, SF, STF. The temperture t S is the sme s the temperture t F. Whih of the following is true? ) roess SRF ours t onstnt temperrure ) The work done y the gs long SRF is the sme s the work done y the gs long STF. ) Net het into the gs long STF is greter thn the work done y the gs long this pth. ) The hnge in the internl energy is the sme for ll three pths. E) Work done y the gs long STF is greter thn the work done y the gs long SF. The internl energy of n idel gs depends only on temperture S R The internl energy of n idel gs depends only on temperture. If the initil nd finl temp re the sme, there is no hnge in internl energy. U=0 for ll three pths. W= - Q for eh proess, whih elimintes Work done y the gs is equl to the re under the urve T F re onsidertions eliminte nd E Temperture t R is greter thn the temperture t S (gs lw) elimintes 1

2 /10/010 Wht is true for the proess ->? Question 4 Question 5 Wht is true for the two step proess -> ->? ) U=0 Q>0. ) U=0 Q<0 ) W=0 U>0 ) W=0 U<0 E) W=0 Q=0 p 0 ) U=0 Q=0 ) U=0 Q>0 ) W=0 Q>0 ) W=0 Q<0 E) W>0 Q<0 p 0 Sine the re (work) under -> is zero, this elimintes nd p 0 Sine the re (work) under -> -> is not zero, this elimintes nd p 0 E is eliminted euse if Q = 0 nd W=0, then U would hve to e zero, thus no hnge in ould our. The internl energy must hve inresed to doule the pressure, therefore U> is eliminted euse if U=0 nd Q=0 then W must e 0, ut it is not. The idel gs lw tells us tht the temperture does not hnge etween nd, so U=0. nd sine expnding Q> The work done on the gs is negtive (the gs does positive work when expnding), therefore Q>0, this elimintes nd E Question 6 00 enters rnot engine from the hot reservoir, held t 400 K. uring the entire engine yle, 50 of useful work is performed y the engine. Wht is the temperture of the old reservoir? ) 100 K ) 00 K ) 00 K ) 50 K E) 150 K From the definition of effiieny: e W y 1 Q H Wy 50 e 0.5 Q 00 H T e T H Question 7 For n idel gs in ontiner with fixed volume nd onstnt temperture, whih of the following is true? I. ressure results from moleulr ollisions with the wlls. II. The moleules ll hve the sme speed. III. The verge kineti energy is diretly proportionl to the temperture ) I, II, nd III ) I nd II only ) II nd III only ) II only E) I nd III only Sine this is rnot engine, we hve T e 1 T H T T 00K The distriution of speeds mong the moleules, leding to n verge vlue, ut the moleules ertinly do not ll hve the sme speed.

3 /10/010 Question 8 1 mole of idel gs is in ontiner of volume, temperture T, nd pressure. If the volume is hlved nd the temperture douled, the pressure will e: ) ) ) 4 ) ½ E) ¼ Sine n idel gs: nrt TR T R 1 TR 4 4 Question 9 Two identil ontiners ontin 1 mole eh of two different montomi idel gses, gs nd gs, with the mss of gs 4 times the mss of gs. oth gses re t the sme temperture, nd 10 of het is dded to gs, resulting in temperture hnge T. How muh het must e dded to gs to use the sme T? ) 10 ) 100 ) 40 ).5 E) 1600 For the internl Energy of Idel Gs U nrt For the sme numer of moles, T depends only on the energy dded, not the mss of the gs moleules Question 10 The entropy of losed mrosopi system will never derese euse ) Energy wouldn t e onserved if entropy deresed ) For lrge systems, the proility of suh derese is negligile ) Mehnil equilirium ouldn t e sustined with derese in entropy ) Het n never e mde to flow from old ojet to hot ojet E) Moleulr motions reh their minimum only t solute 0 Question 11 uring ertin proess, 600 of het is dded to system. While this ours, the system performs 00 of work on the surroundings. The hnge in internl energy of the system is: ) 00 ) 800 ) 400 ) E) Impossile to determine without knowing the temperture The proility of suh mro stte ourring is essentilly 0 The First Lw of Thermodynmis U U Q W

4 /10/010 Question 1 Whih of the following is not true out therml rdition? ) It s mehnism of energy exhnge etween systems t different tempertures ) It involves tomi exittions ) Ie t will emit therml rdition ) It requires some mteril sustne to trvel through E) It s y produt of the food we et Eletromgneti rdition doesn t need mteril medium to trvel through. In E, we et to mintin ody temperture, nd we emit lot of energy s therml rdition Question 1 To inrese the dimeter of n luminium ring from 50.0 mm to 50.1 mm, the temperture of the ring must inrese y 8. Wht temperture hnge would e neessry to inrese the dimeter of n luminium ring from mm to mm? ) ) 4 ) 8 ) 11 E) 16 We pply the formul for length expnsion: 0.1mm mm 0.1mm mm Therefore 0.1mm T mm 40 L T L o Question 14 gs is enlosed in metl ontiner with movele piston on top. Het is dded to the gs y pling ndle flme in ontt with the ontiner's ottom. Wht of the following is true out the temperture of the gs? ) The temperture must go up if the piston remins sttionry. ) The temperture must go up if the piston is pulled out drmtilly. ) The temperture must go up no mtter wht hppens to the piston. ) The temperture must go down no mtter wht hppens to the piston. E) The temperture must go down if the piston is ompressed drmtilly. Use the First Lw of Thermodynmis, U=Q+W. The temperture dds het to the gs, so Q is positive. Internl energy is diretly relted to temperture, so if U is positive, then the temperture goes up (nd vie vers). Here we n e sure tht U is positive if the work done on the gs is either positive or zero. The only possile nswer is. For, the work done on the gs is negtive euse the gs expnds (Note: euse we dd het to gs does NOT men the temperture utomtilly goes up) Question 15 smll het engine opertes using pn of 10 oiling wter s the high temperture reservoir nd the tmosphere s low temperture reservoir. ssuming idel ehviour, how muh more effiient is the engine on old, dy thn on wrm, dy? ) out 1. times s effiient. ) times s effiient.. ) 0 times s effiient. ) Infinitely more effiient. E) ust s effiient. T Using the Engine effiieny eqution: e 1 TH 0 7 Hot y: e old y: 0 7 e Rtio:

5 /10/010 Question 16 In three seprte experiments, gs is trnsformed from i, i to stte f, f long pths (1,, nd ) illustrted elow. The work done on the gs is? Question 17 1 m ontiner ontins 10 moles of idel gs t room temperture. t wht frtion of tmospheri pressure is the gs inside the ontiner? ) Gretest for pth 1 ) Lest for pth ) The sme for pths 1 nd ) The gretest for pth E) The sme for ll three pths i 1 ) 1/40 tm ) 1/0 tm ) 1/10 tm ) ¼ tm E) ½ tm p 0 f Using the Idel Gs Lw nrt p 0 The work done on the gs during thermodynmi proess is equl to the re of the region in the - digrm ove the xis nd elow the pth the system mkes from its initil to its finl stte. f i nrt 10 mol K mol K 1m N 490 m 0 Rtio: 0 N 490 m 0.5 N m Question 18 n idel gs is ompressed isothermlly from 0 L to 10 L. uring this proess, 5 of work done to ompress the gs. Wht is the hnge of internl energy for this gs? ) -10 ) - 5 ) 0 ) 5 E) 10 Question 19 Through series of thermodynmi proesses, the internl energy of smple of onfined gs is inresed y 560. If the net mount of work done on the smple y its surroundings is 0, how muh het ws trnsferred etween the gs nd its environment? ) 40 sored ) 40 dissipted ) 880 sored ) 880 dissipted E) None of the ove The First Lw of Thermodynmis U Q W The First Lw of Thermodynmis U Q W uring n isotherml proess, U is LWYS ZERO 560 Q 0 Q

6 /10/010 Question 0 In one of the steps of the rnot yle, the gs undergoes n isotherml expnsion. Whih of the following sttements is true onerning this step? ) No het is exhnged etween the gs nd its surrounding, euse the proess is isotherml. ) The temperture dereses euse the gs expnds. ) This steps violtes the Seond Lw of Thermodynmis euse ll the het sored is trnsferred into work. ) The internl energy of the gs remins onstnt. E) The internl energy of the gs dereses due to the expnsion. Sttement is wrong euse no het exhnged etween the gs nd its surroundings is the definition of diti, not isotherml. Sttement nnot e orret sine the step desried in question is isotherml; y definition, the temperture does not hnge. Sttement is flse, euse lthough the het sored is onverted ompletely to work, it does not inlude eing returned k to it is initil stte nd hving 100% onversion to het ( dereses nd inreses). Isotherml y definition sttes internl energy remins onstnt Question 1 rnot engine using montomi idel gs s working sustne opertes etween two reservoirs held t 00 K nd 00 K, respetively. Strting t point with pressure nd volume s indited on the grph, the gs expnds isothermlly to point, where the volume is 4 x 10 - m. 500 of het energy is sored y the gs in this proess. ) How mny moles of gs re present? ) Find the work done on the gs during -> proess. ) Find the work done on the gs during the proess -> diti expnsion. ) How muh het is expelled in the proess ->? E) etermine the hnge in internl energy in the proess ->. x10 5 Isotherm t T H dit: Q=0 dit: Q=0 Isotherm t T x10-4x10 - Question 1 rnot engine using montomi idel gs s working sustne opertes etween two reservoirs held t 00. K nd 00. K, respetively. Strting t point with pressure nd volume s indited on the grph, the gs expnds isothermlly to point, where the volume is 4.0 x 10 - m of het energy is sored y the gs in this proess. ) How mny moles of gs re present? ) Find the work done on the gs during -> proess. ) Find the work done on the gs during the proess -> diti expnsion. ) How muh het is expelled in the proess ->? E) etermine the hnge in internl energy in the proess ->. Using the gs lw nd the given,, nd T.0x10 5 Isotherm t T H R n nt RT dit: Q=0 dit: Q=0 5 N m m Isotherm t T K mol K.0x10-4.0x10-0.4moles Question 1 rnot engine using montomi idel gs s working sustne opertes etween two reservoirs held t 00 K nd 00 K, respetively. Strting t point with pressure nd volume s indited on the grph, the gs expnds isothermlly to point, where the volume is 4.0 x 10 - m. 500 of het energy is sored y the gs in this proess. ) How mny moles of gs re present? ) Find the work done on the gs during -> proess. ) Find the work done on the gs during the proess -> diti expnsion. ) How muh het is expelled in the proess ->? E) etermine the hnge in internl energy in the proess ->. Rell: if the temperture is onstnt, then so is the internl energy. long -> isotherm, there is no hnge in internl energy, so the first lw gives us U = 0 = Q + W. ut Q is given s 500, so W = -500 is done ON the gs..0x10 5 dit: Q=0 QIsotherm H 500 t T H Isotherm t T.0x10-4.0x10 - dit: Q=0 6

7 /10/010 Question 1 rnot engine using montomi idel gs s working sustne opertes etween two reservoirs held t 00 K nd 00 K, respetively. Strting t point with pressure nd volume s indited on the grph, the gs expnds isothermlly to point, where the volume is 4.0 x 10 - m. 500 of het energy is sored y the gs in this proess. ) How mny moles of gs re present? ) Find the work done on the gs during -> proess. ) Find the work done on the gs during the proess -> diti expnsion. ) How muh het is expelled in the proess ->? E) etermine the hnge in internl energy in the proess ->. Rell: if the proess is diti, there is no het trnsfer (Q=0). Therefore U=W..0x10 5 Now we know the dit onnets the two isotherms where we know the temperture, nd sine the idel gs internl energy depends only upon temperture. U nrt 0.4 moles K 00 K mol K 99 W= - 99 dit: Q=0 Isotherm t T H Isotherm t T.0x10-4.0x10 - dit: Q=0 Question 1 rnot engine using montomi idel gs s working sustne opertes etween two reservoirs held t 00 K nd 00 K, respetively. Strting t point with pressure nd volume s indited on the grph, the gs expnds isothermlly to point, where the volume is 4.0 x 10 - m. 500 of het energy is sored y the gs in this proess. ) How mny moles of gs re present? ) Find the work done on the gs during -> proess. ) Find the work done on the gs during the proess -> diti expnsion. ) How muh het is expelled in the proess ->? E) etermine the hnge in internl energy in the proess ->. We re looking for Q.0x10 5 T 00K For the rnot yle: e T H 00K Isotherm t T H dit: Q=0 Q dit: Q=0 lso: e 1 QH Q Isotherm t T 0. 1 Therefore: 500.0x10-4.0x10 - Q Question 1 rnot engine using montomi idel gs s working sustne opertes etween two reservoirs held t 00 K nd 00 K, respetively. Strting t point with pressure nd volume s indited on the grph, the gs expnds isothermlly to point, where the volume is 4.0 x 10 - m. 500 of het energy is sored y the gs in this proess. ) How mny moles of gs re present? ) Find the work done on the gs during -> proess. ) Find the work done on the gs during the proess -> diti expnsion. ) How muh het is expelled in the proess ->? E) etermine the hnge in internl energy in the proess ->. One gin we re onneting two isotherms. Sine the internl energy depends only on temperture, we hve U nrt 0.4 moles K 00 K mol K 99 The opposite of ->.0x10 5 dit: Q=0 Isotherm t T H Isotherm t T.0x10-4.0x10 - dit: Q=0 Question When system is tken from stte to stte long the pth shown in the figure elow, 70 of het flows into the system, nd the system does 0 of work. ) When the system is returned from stte to stte long the urved pth shown, 60 of het flows out of the system. oes the system perform work on its surroundings or do the surroundings perform work on the system? How muh work is done? ) If the system does 10 of work in trnsforming from stte to stte long pth d, does the system sor or does it emit het? How muh het is trnsferred? ) If U =0 nd U d =0, determine the het sored in the proess d nd d. ) For the proess d, identify eh of the following quntities s positive, negtive, or zero. W = Q = U = d 7

8 /10/010 Question When system is tken from stte to stte long the pth shown in the figure elow, 70 of het flows into the system, nd the system does 0 of work. ) When the system is returned from stte to stte long the urved pth shown, 60 of het flows out of the system. oes the system perform work on its surroundings or do the surroundings perform work on the system? How muh work is done? We require the vlue of W First let s lulte U U Q W 70 0 U 40 euse U -> does not depend on pth tken from to, then sine U = 40 then U = -U = 40-0 euse system does work U Q W W W 0 Sine the vlue is + 0, the surroundings does the work. d Question When system is tken from stte to stte long the pth shown in the figure elow, 70 of het flows into the system, nd the system does 0 of work.. If the system does 10 of work in trnsforming from stte to stte long pth d, does the system sor or does it emit het? How muh het is trnsferred? gin, using the ft tht U -> does not depend on the pth tken, we know tht U d = 40 Now we require Q d U Qd Wd d 40 Qd 10 Q 50 d Therefore the system sors 50 of het euse the system does 10 of work d Question When system is tken from stte to stte long the pth shown in the figure elow, 70 of het flows into the system, nd the system does 0 of work. ) If U =0 nd U d =0, determine the het sored in the proess d nd d. For the proess d, there is no hnge in volume, therefore W d = 0. U Q W U d d d d Q Now, U = 40, nd U = 0, tells us tht U = 40, therefore: U U U d d d Q 10 d For d: W d =W d +W d =W d +0=W d Sine W d =-10, then W d =-10 U Qd Wd d U U Qd 10 d 0 0 Qd 10 Q 40 d d Question When system is tken from stte to stte long the pth shown in the figure elow, 70 of het flows into the system, nd the system does 0 of work.. For the proess d, identify eh of the following quntities s positive, negtive, or zero. W = Q = U = The proess d is yli, so U = zero euse the proess is trversed ounterlokwise in the digrm, we know tht W is positive. U=Q+W therefore Q must e negtive The system is het engine, emitting het on eh yle d 8

9 /10/010 Question 1 mole of montomi idel gs is rought through yle ->->->-> s depited elow. ll proesses re performed slowly. Respond to the following in terms of 0, 0, nd R. ) Find the temperture t eh vertex. ) Find the het dded to the gs for the proess ->. ) Find the work done on the gs for the proess ->. ) Find the het dded to the gs fro the proess -> E) Find the hnge in internl energy for i. roess -> ii. The entire yle Question 1 mole of montomi idel gs is rought through yle ->->->-> s depited elow. ll proesses re performed slowly. Respond to the following in terms of 0, 0, nd R. ) Find the temperture t eh vertex. ) Find the het dded to the gs for the proess ->. ) Find the work done on the gs for the proess ->. ) Find the het dded to the gs fro the proess -> E) Find the hnge in internl energy for i. roess -> ii. The entire yle 0 From the Idel Gs Lw: T nr T R T R R T R R T R Question 1 mole of montomi idel gs is rought through yle ->->->-> s depited elow. ll proesses re performed slowly. Respond to the following in terms of 0, 0, nd R. ) Find the temperture t eh vertex. ) Find the het dded to the gs for the proess ->. ) Find the work done on the gs for the proess ->. ) Find the het dded to the gs fro the proess -> E) Find the hnge in internl energy for i. roess -> ii. The entire yle 0 Question 1 mole of montomi idel gs is rought through yle ->->->-> s depited elow. ll proesses re performed slowly. Respond to the following in terms of 0, 0, nd R. ) Find the temperture t eh vertex. ) Find the het dded to the gs for the proess ->. ) Find the work done on the gs for the proess ->. ) Find the het dded to the gs fro the proess -> E) Find the hnge in internl energy for i. roess -> ii. The entire yle 0 roess -> ours t onstnt pressure. So we n use the First Lw U Q W Q U W nrt W nrt T W 4 R W R R 0 Sine Work done on the gs is negtive of the re under grph R R 5 0 Q R The re under the proess -> W 0 The derese in volume orresponds to positive work done on the gs The sign of the net work is negtive for yles tht run lokwise

10 /10/010 Question 1 mole of montomi idel gs is rought through yle ->->->-> s depited elow. ll proesses re performed slowly. Respond to the following in terms of 0, 0, nd R. ) Find the temperture t eh vertex. ) Find the het dded to the gs for the proess ->. ) Find the work done on the gs for the proess ->. ) Find the het dded to the gs fro the proess -> E) Find the hnge in internl energy for i. roess -> ii. The entire yle 0 roess -> ours t onstnt volume. So we n use the First Lw U Q W QU nrt nrt T R R R Sine t onstnt volume, no work is done 0 Q Question 1 mole of montomi idel gs is rought through yle ->->->-> s depited elow. ll proesses re performed slowly. Respond to the following in terms of 0, 0, nd R. ) Find the temperture t eh vertex. ) Find the het dded to the gs for the proess ->. ) Find the work done on the gs for the proess ->. ) Find the het dded to the gs fro the proess ->. E) Find the hnge in internl energy for i. roess -> ii. The entire yle 0 i) Using: U nrt U RT T 4 R R R ii) For the entire yle, you end up k in the sme stte, so U = 0 0 Question mol smple of n idel ditomi gs undergoes slow hnges from stte to stte to stte nd k to long the yle shown. th is n isotherm nd the work done y the gs s it hnges isothermlly from stte to stte is given y the eqution: (x10 5 ) W nrt ln 9.1 mol K ) Wht is the temperture of: mol K i. Stte ii. Stte iii. Stte ) etermine the hnge in internl energy of the gs for: 0.6 i. Step. ii. Step. iii. Step ) How muh work, W is done y the gs during step? (x10 - m ) ) Wht is the totl work done over the yle? E) Is het sored or disrded step? F) If so how muh? G) Wht is the mximum possile effiieny (without violting the Seond Lw of Thermodynmis) for ylil het engine tht opertes etween the tempertures of stte nd? Question mol smple of n idel ditomi gs undergoes slow hnges from stte to stte to stte nd k to long the yle shown. th is n isotherm nd the work done y the gs s it hnges isothermlly from stte to stte is given y the eqution: (x10 5 ) W nrt ln.4 ) Wht is the temperture of: i. Stte ii. Stte iii. Stte 0.6 From the Idel Gs Lw: T nr m T 870K 0.4mol 8.1 mol K T 870K Stte is on the Isotherm with stte m T 0K 0.4mol 8.1 mol K 1 48 (x10 - m ) 10

11 /10/010 Question mol smple of n idel ditomi gs undergoes slow hnges from stte to stte to stte nd k to long the yle shown. th is n isotherm nd the work done y the gs s it hnges isothermlly from stte to stte is given y the eqution: (x10 5 ) W nrt ln 9.1 mol K ) etermine the hnge in internl energy of the gs for: mol K i. Step. ii. Step. iii. Step. 0.6 Sine step tkes ple long n isotherm, U = 0 U Q W npt 1 48 (x10 - m ) mol K U U nr n vt T U U U U U or mol mol K K 0 0 K K mol mol K K U 5400 U U mol 9.1 0K 870K m m Question mol smple of n idel ditomi gs undergoes slow hnges from stte to stte to stte nd k to long the yle shown. th is n isotherm nd the work done y the gs s it hnges isothermlly from stte to stte is given y the eqution: (x10 5 ) W nrt ln 9.1 mol K ) How muh work, W is done y the gs during step? mol K W nrt ln 4810 m 0.4mol K ln 1 10 mol K m (x10 - m ) Question mol smple of n idel ditomi gs undergoes slow hnges from stte to stte to stte nd k to long the yle shown. th is n isotherm nd the work done y the gs s it hnges isothermlly from stte to stte is given y the eqution: (x10 5 ) W nrt ln 9.1 mol K ) Wht is the totl work done over the yle? mol K Question mol smple of n idel ditomi gs undergoes slow hnges from stte to stte to stte nd k to long the yle shown. th is n isotherm nd the work done y the gs s it hnges isothermlly from stte to stte is given y the eqution: (x10 5 ) W nrt ln 9.1 mol K E) Is het sored or disrded step? mol K F) If so how muh? The totl work done over yle is equl to the sum of the vlues of the work done over eh step. W W W W yle (x10 - m ) y the First Lw of Thermodynmis U Q W Q 0 Q W W 4000 U Q W 0.6 Sine Q is positive, this represents het sored y the gs (x10 - m ) 11

12 /10/010 Question mol smple of n idel ditomi gs undergoes slow hnges from stte to stte to stte nd k to long the yle shown. th is n isotherm nd the work done y the gs s it hnges isothermlly from stte to stte is given y the eqution: W nrt ln G) Wht is the mximum possile effiieny (without violting the Seond Lw of Thermodynmis) for ylil het engine tht opertes etween the tempertures of stte nd? (x10 5 ) The mximum possile effiieny is the effiieny of the rnot engine..4 T e 1 T H 0K K 75% (x10 - m ) Question 5 You hve just purhsed up of offee (45 g), ut you notie tht the offee is old, so you ll over your host. It s 45 o, you sy, ut I like it 65 o.. The offee in the pot is 95 o, how muh offee should the host pour to rise the temperture of the 45g of offee in your up? Look t the het lost y the pot s offee nd the het gined y your offee offee in up: offee in pot: Q m T T f i Q m T T f Q mt i m Now, Q1 Q m g0 m 0 m 0g 1

a) Read over steps (1)- (4) below and sketch the path of the cycle on a P V plot on the graph below. Label all appropriate points.

a) Read over steps (1)- (4) below and sketch the path of the cycle on a P V plot on the graph below. Label all appropriate points. Prole 3: Crnot Cyle of n Idel Gs In this prole, the strting pressure P nd volue of n idel gs in stte, re given he rtio R = / > of the volues of the sttes nd is given Finlly onstnt γ = 5/3 is given You

More information

Thermal energy 2 U Q W. 23 April The First Law of Thermodynamics. Or, if we want to obtain external work: The trick of using steam

Thermal energy 2 U Q W. 23 April The First Law of Thermodynamics. Or, if we want to obtain external work: The trick of using steam April 08 Therml energy Soures of het Trnsport of het How to use het The First Lw of Thermoynmis U W Or, if we wnt to otin externl work: U W 009 vrije Universiteit msterm Close yle stem power plnt The trik

More information

Chapter 19: The Second Law of Thermodynamics

Chapter 19: The Second Law of Thermodynamics hpter 9: he Seon Lw of hermoynmis Diretions of thermoynmi proesses Irreversile n reversile proesses hermoynmi proesses tht our in nture re ll irreversile proesses whih proee spontneously in one iretion

More information

U Q W The First Law of Thermodynamics. Efficiency. Closed cycle steam power plant. First page of S. Carnot s paper. Sadi Carnot ( )

U Q W The First Law of Thermodynamics. Efficiency. Closed cycle steam power plant. First page of S. Carnot s paper. Sadi Carnot ( ) 0-9-0 he First Lw of hermoynmis Effiieny When severl lterntive proesses involving het n work re ville to hnge system from n initil stte hrterize y given vlues of the mrosopi prmeters (pressure p i, temperture

More information

QUADRATIC EQUATION. Contents

QUADRATIC EQUATION. Contents QUADRATIC EQUATION Contents Topi Pge No. Theory 0-04 Exerise - 05-09 Exerise - 09-3 Exerise - 3 4-5 Exerise - 4 6 Answer Key 7-8 Syllus Qudrti equtions with rel oeffiients, reltions etween roots nd oeffiients,

More information

AP Physics - Heat Engines

AP Physics - Heat Engines A hysis - et Engines et engines operte y onverting het into work. Exmples of het engines ound - ound! Sys the hysis Khun - ll round us. Gsoline engines, jet engines, diesel engines, stem engines, Stirling

More information

Project 6: Minigoals Towards Simplifying and Rewriting Expressions

Project 6: Minigoals Towards Simplifying and Rewriting Expressions MAT 51 Wldis Projet 6: Minigols Towrds Simplifying nd Rewriting Expressions The distriutive property nd like terms You hve proly lerned in previous lsses out dding like terms ut one prolem with the wy

More information

6.5 Improper integrals

6.5 Improper integrals Eerpt from "Clulus" 3 AoPS In. www.rtofprolemsolving.om 6.5. IMPROPER INTEGRALS 6.5 Improper integrls As we ve seen, we use the definite integrl R f to ompute the re of the region under the grph of y =

More information

Numbers and indices. 1.1 Fractions. GCSE C Example 1. Handy hint. Key point

Numbers and indices. 1.1 Fractions. GCSE C Example 1. Handy hint. Key point GCSE C Emple 7 Work out 9 Give your nswer in its simplest form Numers n inies Reiprote mens invert or turn upsie own The reiprol of is 9 9 Mke sure you only invert the frtion you re iviing y 7 You multiply

More information

Chapter 4 The second law of thermodynamics

Chapter 4 The second law of thermodynamics hpter 4 he second lw of thermodynmics Directions of thermodynmic processes et engines Internl-combustion engines Refrigertors he second lw of thermodynmics he rnotcycle Entropy Directions of thermodynmic

More information

NON-DETERMINISTIC FSA

NON-DETERMINISTIC FSA Tw o types of non-determinism: NON-DETERMINISTIC FS () Multiple strt-sttes; strt-sttes S Q. The lnguge L(M) ={x:x tkes M from some strt-stte to some finl-stte nd ll of x is proessed}. The string x = is

More information

6.3.2 Spectroscopy. N Goalby chemrevise.org 1 NO 2 CH 3. CH 3 C a. NMR spectroscopy. Different types of NMR

6.3.2 Spectroscopy. N Goalby chemrevise.org 1 NO 2 CH 3. CH 3 C a. NMR spectroscopy. Different types of NMR 6.. Spetrosopy NMR spetrosopy Different types of NMR NMR spetrosopy involves intertion of mterils with the lowenergy rdiowve region of the eletromgneti spetrum NMR spetrosopy is the sme tehnology s tht

More information

Introduction to Olympiad Inequalities

Introduction to Olympiad Inequalities Introdution to Olympid Inequlities Edutionl Studies Progrm HSSP Msshusetts Institute of Tehnology Snj Simonovikj Spring 207 Contents Wrm up nd Am-Gm inequlity 2. Elementry inequlities......................

More information

The Stirling Engine: The Heat Engine

The Stirling Engine: The Heat Engine Memoril University of Newfounln Deprtment of Physis n Physil Oenogrphy Physis 2053 Lortory he Stirling Engine: he Het Engine Do not ttempt to operte the engine without supervision. Introution Het engines

More information

for all x in [a,b], then the area of the region bounded by the graphs of f and g and the vertical lines x = a and x = b is b [ ( ) ( )] A= f x g x dx

for all x in [a,b], then the area of the region bounded by the graphs of f and g and the vertical lines x = a and x = b is b [ ( ) ( )] A= f x g x dx Applitions of Integrtion Are of Region Between Two Curves Ojetive: Fin the re of region etween two urves using integrtion. Fin the re of region etween interseting urves using integrtion. Desrie integrtion

More information

Solutions for HW9. Bipartite: put the red vertices in V 1 and the black in V 2. Not bipartite!

Solutions for HW9. Bipartite: put the red vertices in V 1 and the black in V 2. Not bipartite! Solutions for HW9 Exerise 28. () Drw C 6, W 6 K 6, n K 5,3. C 6 : W 6 : K 6 : K 5,3 : () Whih of the following re iprtite? Justify your nswer. Biprtite: put the re verties in V 1 n the lk in V 2. Biprtite:

More information

QUADRATIC EQUATION EXERCISE - 01 CHECK YOUR GRASP

QUADRATIC EQUATION EXERCISE - 01 CHECK YOUR GRASP QUADRATIC EQUATION EXERCISE - 0 CHECK YOUR GRASP. Sine sum of oeffiients 0. Hint : It's one root is nd other root is 8 nd 5 5. tn other root 9. q 4p 0 q p q p, q 4 p,,, 4 Hene 7 vlues of (p, q) 7 equtions

More information

Math 32B Discussion Session Week 8 Notes February 28 and March 2, f(b) f(a) = f (t)dt (1)

Math 32B Discussion Session Week 8 Notes February 28 and March 2, f(b) f(a) = f (t)dt (1) Green s Theorem Mth 3B isussion Session Week 8 Notes Februry 8 nd Mrh, 7 Very shortly fter you lerned how to integrte single-vrible funtions, you lerned the Fundmentl Theorem of lulus the wy most integrtion

More information

6.3.2 Spectroscopy. N Goalby chemrevise.org 1 NO 2 H 3 CH3 C. NMR spectroscopy. Different types of NMR

6.3.2 Spectroscopy. N Goalby chemrevise.org 1 NO 2 H 3 CH3 C. NMR spectroscopy. Different types of NMR 6.. Spetrosopy NMR spetrosopy Different types of NMR NMR spetrosopy involves intertion of mterils with the lowenergy rdiowve region of the eletromgneti spetrum NMR spetrosopy is the sme tehnology s tht

More information

1.3 SCALARS AND VECTORS

1.3 SCALARS AND VECTORS Bridge Course Phy I PUC 24 1.3 SCLRS ND VECTORS Introdution: Physis is the study of nturl phenomen. The study of ny nturl phenomenon involves mesurements. For exmple, the distne etween the plnet erth nd

More information

10. AREAS BETWEEN CURVES

10. AREAS BETWEEN CURVES . AREAS BETWEEN CURVES.. Ares etween curves So res ove the x-xis re positive nd res elow re negtive, right? Wrong! We lied! Well, when you first lern out integrtion it s convenient fiction tht s true in

More information

Chapter 9 Definite Integrals

Chapter 9 Definite Integrals Chpter 9 Definite Integrls In the previous chpter we found how to tke n ntiderivtive nd investigted the indefinite integrl. In this chpter the connection etween ntiderivtives nd definite integrls is estlished

More information

Section 4.4. Green s Theorem

Section 4.4. Green s Theorem The Clulus of Funtions of Severl Vriles Setion 4.4 Green s Theorem Green s theorem is n exmple from fmily of theorems whih onnet line integrls (nd their higher-dimensionl nlogues) with the definite integrls

More information

Lesson 2: The Pythagorean Theorem and Similar Triangles. A Brief Review of the Pythagorean Theorem.

Lesson 2: The Pythagorean Theorem and Similar Triangles. A Brief Review of the Pythagorean Theorem. 27 Lesson 2: The Pythgoren Theorem nd Similr Tringles A Brief Review of the Pythgoren Theorem. Rell tht n ngle whih mesures 90º is lled right ngle. If one of the ngles of tringle is right ngle, then we

More information

Global alignment. Genome Rearrangements Finding preserved genes. Lecture 18

Global alignment. Genome Rearrangements Finding preserved genes. Lecture 18 Computt onl Biology Leture 18 Genome Rerrngements Finding preserved genes We hve seen before how to rerrnge genome to obtin nother one bsed on: Reversls Knowledge of preserved bloks (or genes) Now we re

More information

CS 310 (sec 20) - Winter Final Exam (solutions) SOLUTIONS

CS 310 (sec 20) - Winter Final Exam (solutions) SOLUTIONS CS 310 (sec 20) - Winter 2003 - Finl Exm (solutions) SOLUTIONS 1. (Logic) Use truth tles to prove the following logicl equivlences: () p q (p p) (q q) () p q (p q) (p q) () p q p q p p q q (q q) (p p)

More information

Symmetrical Components 1

Symmetrical Components 1 Symmetril Components. Introdution These notes should e red together with Setion. of your text. When performing stedy-stte nlysis of high voltge trnsmission systems, we mke use of the per-phse equivlent

More information

Continuous Random Variables Class 5, Jeremy Orloff and Jonathan Bloom

Continuous Random Variables Class 5, Jeremy Orloff and Jonathan Bloom Lerning Gols Continuous Rndom Vriles Clss 5, 8.05 Jeremy Orloff nd Jonthn Bloom. Know the definition of continuous rndom vrile. 2. Know the definition of the proility density function (pdf) nd cumultive

More information

CALCULATING REACTING QUANTITIES

CALCULATING REACTING QUANTITIES MODULE 2 14 WORKSHEET WORKSHEET For multiple-hoie questions 1 5 irle the letter orresponding to the most orret nswer. 1 The lned eqution for the urning of utnol (C 4 H 9 OH) is given elow: C 4 H 9 OH(l)

More information

Non Right Angled Triangles

Non Right Angled Triangles Non Right ngled Tringles Non Right ngled Tringles urriulum Redy www.mthletis.om Non Right ngled Tringles NON RIGHT NGLED TRINGLES sin i, os i nd tn i re lso useful in non-right ngled tringles. This unit

More information

CS 373, Spring Solutions to Mock midterm 1 (Based on first midterm in CS 273, Fall 2008.)

CS 373, Spring Solutions to Mock midterm 1 (Based on first midterm in CS 273, Fall 2008.) CS 373, Spring 29. Solutions to Mock midterm (sed on first midterm in CS 273, Fll 28.) Prolem : Short nswer (8 points) The nswers to these prolems should e short nd not complicted. () If n NF M ccepts

More information

QUADRATIC EQUATIONS OBJECTIVE PROBLEMS

QUADRATIC EQUATIONS OBJECTIVE PROBLEMS QUADRATIC EQUATIONS OBJECTIVE PROBLEMS +. The solution of the eqution will e (), () 0,, 5, 5. The roots of the given eqution ( p q) ( q r) ( r p) 0 + + re p q r p (), r p p q, q r p q (), (d), q r p q.

More information

Algorithm Design and Analysis

Algorithm Design and Analysis Algorithm Design nd Anlysis LECTURE 8 Mx. lteness ont d Optiml Ching Adm Smith 9/12/2008 A. Smith; sed on slides y E. Demine, C. Leiserson, S. Rskhodnikov, K. Wyne Sheduling to Minimizing Lteness Minimizing

More information

Solutions to Assignment 1

Solutions to Assignment 1 MTHE 237 Fll 2015 Solutions to Assignment 1 Problem 1 Find the order of the differentil eqution: t d3 y dt 3 +t2 y = os(t. Is the differentil eqution liner? Is the eqution homogeneous? b Repet the bove

More information

I 3 2 = I I 4 = 2A

I 3 2 = I I 4 = 2A ECE 210 Eletril Ciruit Anlysis University of llinois t Chigo 2.13 We re ske to use KCL to fin urrents 1 4. The key point in pplying KCL in this prolem is to strt with noe where only one of the urrents

More information

1 Nondeterministic Finite Automata

1 Nondeterministic Finite Automata 1 Nondeterministic Finite Automt Suppose in life, whenever you hd choice, you could try oth possiilities nd live your life. At the end, you would go ck nd choose the one tht worked out the est. Then you

More information

(a) A partition P of [a, b] is a finite subset of [a, b] containing a and b. If Q is another partition and P Q, then Q is a refinement of P.

(a) A partition P of [a, b] is a finite subset of [a, b] containing a and b. If Q is another partition and P Q, then Q is a refinement of P. Chpter 7: The Riemnn Integrl When the derivtive is introdued, it is not hrd to see tht the it of the differene quotient should be equl to the slope of the tngent line, or when the horizontl xis is time

More information

Algebra 2 Semester 1 Practice Final

Algebra 2 Semester 1 Practice Final Alger 2 Semester Prtie Finl Multiple Choie Ientify the hoie tht est ompletes the sttement or nswers the question. To whih set of numers oes the numer elong?. 2 5 integers rtionl numers irrtionl numers

More information

SOLUTIONS FOR ADMISSIONS TEST IN MATHEMATICS, COMPUTER SCIENCE AND JOINT SCHOOLS WEDNESDAY 5 NOVEMBER 2014

SOLUTIONS FOR ADMISSIONS TEST IN MATHEMATICS, COMPUTER SCIENCE AND JOINT SCHOOLS WEDNESDAY 5 NOVEMBER 2014 SOLUTIONS FOR ADMISSIONS TEST IN MATHEMATICS, COMPUTER SCIENCE AND JOINT SCHOOLS WEDNESDAY 5 NOVEMBER 014 Mrk Scheme: Ech prt of Question 1 is worth four mrks which re wrded solely for the correct nswer.

More information

Chem Homework 11 due Monday, Apr. 28, 2014, 2 PM

Chem Homework 11 due Monday, Apr. 28, 2014, 2 PM Chem 44 - Homework due ondy, pr. 8, 4, P.. . Put this in eq 8.4 terms: E m = m h /m e L for L=d The degenery in the ring system nd the inresed sping per level (4x bigger) mkes the sping between the HOO

More information

Arrow s Impossibility Theorem

Arrow s Impossibility Theorem Rep Voting Prdoxes Properties Arrow s Theorem Arrow s Impossiility Theorem Leture 12 Arrow s Impossiility Theorem Leture 12, Slide 1 Rep Voting Prdoxes Properties Arrow s Theorem Leture Overview 1 Rep

More information

How do we solve these things, especially when they get complicated? How do we know when a system has a solution, and when is it unique?

How do we solve these things, especially when they get complicated? How do we know when a system has a solution, and when is it unique? XII. LINEAR ALGEBRA: SOLVING SYSTEMS OF EQUATIONS Tody we re going to tlk out solving systems of liner equtions. These re prolems tht give couple of equtions with couple of unknowns, like: 6= x + x 7=

More information

Matrices SCHOOL OF ENGINEERING & BUILT ENVIRONMENT. Mathematics (c) 1. Definition of a Matrix

Matrices SCHOOL OF ENGINEERING & BUILT ENVIRONMENT. Mathematics (c) 1. Definition of a Matrix tries Definition of tri mtri is regulr rry of numers enlosed inside rkets SCHOOL OF ENGINEERING & UIL ENVIRONEN Emple he following re ll mtries: ), ) 9, themtis ), d) tries Definition of tri Size of tri

More information

DIRECT CURRENT CIRCUITS

DIRECT CURRENT CIRCUITS DRECT CURRENT CUTS ELECTRC POWER Consider the circuit shown in the Figure where bttery is connected to resistor R. A positive chrge dq will gin potentil energy s it moves from point to point b through

More information

Core 2 Logarithms and exponentials. Section 1: Introduction to logarithms

Core 2 Logarithms and exponentials. Section 1: Introduction to logarithms Core Logrithms nd eponentils Setion : Introdution to logrithms Notes nd Emples These notes ontin subsetions on Indies nd logrithms The lws of logrithms Eponentil funtions This is n emple resoure from MEI

More information

u( t) + K 2 ( ) = 1 t > 0 Analyzing Damped Oscillations Problem (Meador, example 2-18, pp 44-48): Determine the equation of the following graph.

u( t) + K 2 ( ) = 1 t > 0 Analyzing Damped Oscillations Problem (Meador, example 2-18, pp 44-48): Determine the equation of the following graph. nlyzing Dmped Oscilltions Prolem (Medor, exmple 2-18, pp 44-48): Determine the eqution of the following grph. The eqution is ssumed to e of the following form f ( t) = K 1 u( t) + K 2 e!"t sin (#t + $

More information

Lecture 27: Diffusion of Ions: Part 2: coupled diffusion of cations and

Lecture 27: Diffusion of Ions: Part 2: coupled diffusion of cations and Leture 7: iffusion of Ions: Prt : oupled diffusion of tions nd nions s desried y Nernst-Plnk Eqution Tody s topis Continue to understnd the fundmentl kinetis prmeters of diffusion of ions within n eletrilly

More information

m A 1 1 A ! and AC 6

m A 1 1 A ! and AC 6 REVIEW SET A Using sle of m represents units, sketh vetor to represent: NON-CALCULATOR n eroplne tking off t n ngle of 8 ± to runw with speed of 6 ms displement of m in north-esterl diretion. Simplif:

More information

= state, a = reading and q j

= state, a = reading and q j 4 Finite Automt CHAPTER 2 Finite Automt (FA) (i) Derterministi Finite Automt (DFA) A DFA, M Q, q,, F, Where, Q = set of sttes (finite) q Q = the strt/initil stte = input lphet (finite) (use only those

More information

Nondeterministic Finite Automata

Nondeterministic Finite Automata Nondeterministi Finite utomt The Power of Guessing Tuesdy, Otoer 4, 2 Reding: Sipser.2 (first prt); Stoughton 3.3 3.5 S235 Lnguges nd utomt eprtment of omputer Siene Wellesley ollege Finite utomton (F)

More information

Physics 505 Homework No. 11 Solutions S11-1

Physics 505 Homework No. 11 Solutions S11-1 Physis 55 Homework No 11 s S11-1 1 This problem is from the My, 24 Prelims Hydrogen moleule Consider the neutrl hydrogen moleule, H 2 Write down the Hmiltonin keeping only the kineti energy terms nd the

More information

Calculus Module C21. Areas by Integration. Copyright This publication The Northern Alberta Institute of Technology All Rights Reserved.

Calculus Module C21. Areas by Integration. Copyright This publication The Northern Alberta Institute of Technology All Rights Reserved. Clculus Module C Ares Integrtion Copright This puliction The Northern Alert Institute of Technolog 7. All Rights Reserved. LAST REVISED Mrch, 9 Introduction to Ares Integrtion Sttement of Prerequisite

More information

Physics 9 Fall 2011 Homework 2 - Solutions Friday September 2, 2011

Physics 9 Fall 2011 Homework 2 - Solutions Friday September 2, 2011 Physics 9 Fll 0 Homework - s Fridy September, 0 Mke sure your nme is on your homework, nd plese box your finl nswer. Becuse we will be giving prtil credit, be sure to ttempt ll the problems, even if you

More information

PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

PAIR OF LINEAR EQUATIONS IN TWO VARIABLES PAIR OF LINEAR EQUATIONS IN TWO VARIABLES. Two liner equtions in the sme two vriles re lled pir of liner equtions in two vriles. The most generl form of pir of liner equtions is x + y + 0 x + y + 0 where,,,,,,

More information

2.1 ANGLES AND THEIR MEASURE. y I

2.1 ANGLES AND THEIR MEASURE. y I .1 ANGLES AND THEIR MEASURE Given two interseting lines or line segments, the mount of rottion out the point of intersetion (the vertex) required to ring one into orrespondene with the other is lled the

More information

Chapter 6 Continuous Random Variables and Distributions

Chapter 6 Continuous Random Variables and Distributions Chpter 6 Continuous Rndom Vriles nd Distriutions Mny economic nd usiness mesures such s sles investment consumption nd cost cn hve the continuous numericl vlues so tht they cn not e represented y discrete

More information

Reflection Property of a Hyperbola

Reflection Property of a Hyperbola Refletion Propert of Hperol Prefe The purpose of this pper is to prove nltill nd to illustrte geometrill the propert of hperol wherein r whih emntes outside the onvit of the hperol, tht is, etween the

More information

Calculus Cheat Sheet. Integrals Definitions. where F( x ) is an anti-derivative of f ( x ). Fundamental Theorem of Calculus. dx = f x dx g x dx

Calculus Cheat Sheet. Integrals Definitions. where F( x ) is an anti-derivative of f ( x ). Fundamental Theorem of Calculus. dx = f x dx g x dx Clulus Chet Sheet Integrls Definitions Definite Integrl: Suppose f ( ) is ontinuous Anti-Derivtive : An nti-derivtive of f ( ) on [, ]. Divide [, ] into n suintervls of is funtion, F( ), suh tht F = f.

More information

Exam 2 Solutions ECE 221 Electric Circuits

Exam 2 Solutions ECE 221 Electric Circuits Nme: PSU Student ID Numer: Exm 2 Solutions ECE 221 Electric Circuits Novemer 12, 2008 Dr. Jmes McNmes Keep your exm flt during the entire exm If you hve to leve the exm temporrily, close the exm nd leve

More information

Interpreting Integrals and the Fundamental Theorem

Interpreting Integrals and the Fundamental Theorem Interpreting Integrls nd the Fundmentl Theorem Tody, we go further in interpreting the mening of the definite integrl. Using Units to Aid Interprettion We lredy know tht if f(t) is the rte of chnge of

More information

Section 7.1 Area of a Region Between Two Curves

Section 7.1 Area of a Region Between Two Curves Section 7.1 Are of Region Between Two Curves White Bord Chllenge The circle elow is inscried into squre: Clcultor 0 cm Wht is the shded re? 400 100 85.841cm White Bord Chllenge Find the re of the region

More information

Section 4: Integration ECO4112F 2011

Section 4: Integration ECO4112F 2011 Reding: Ching Chpter Section : Integrtion ECOF Note: These notes do not fully cover the mteril in Ching, ut re ment to supplement your reding in Ching. Thus fr the optimistion you hve covered hs een sttic

More information

PART 1 MULTIPLE CHOICE Circle the appropriate response to each of the questions below. Each question has a value of 1 point.

PART 1 MULTIPLE CHOICE Circle the appropriate response to each of the questions below. Each question has a value of 1 point. PART MULTIPLE CHOICE Circle the pproprite response to ech of the questions below. Ech question hs vlue of point.. If in sequence the second level difference is constnt, thn the sequence is:. rithmetic

More information

CIT 596 Theory of Computation 1. Graphs and Digraphs

CIT 596 Theory of Computation 1. Graphs and Digraphs CIT 596 Theory of Computtion 1 A grph G = (V (G), E(G)) onsists of two finite sets: V (G), the vertex set of the grph, often enote y just V, whih is nonempty set of elements lle verties, n E(G), the ege

More information

Fundamentals of Analytical Chemistry

Fundamentals of Analytical Chemistry Homework Fundmentls of nlyticl hemistry hpter 9 0, 1, 5, 7, 9 cids, Bses, nd hpter 9(b) Definitions cid Releses H ions in wter (rrhenius) Proton donor (Bronsted( Lowry) Electron-pir cceptor (Lewis) hrcteristic

More information

Pythagoras theorem and surds

Pythagoras theorem and surds HPTER Mesurement nd Geometry Pythgors theorem nd surds In IE-EM Mthemtis Yer 8, you lernt out the remrkle reltionship etween the lengths of the sides of right-ngled tringle. This result is known s Pythgors

More information

200 points 5 Problems on 4 Pages and 20 Multiple Choice/Short Answer Questions on 5 pages 1 hour, 48 minutes

200 points 5 Problems on 4 Pages and 20 Multiple Choice/Short Answer Questions on 5 pages 1 hour, 48 minutes PHYSICS 132 Smple Finl 200 points 5 Problems on 4 Pges nd 20 Multiple Choice/Short Answer Questions on 5 pges 1 hour, 48 minutes Student Nme: Recittion Instructor (circle one): nme1 nme2 nme3 nme4 Write

More information

Section 4.8. D v(t j 1 ) t. (4.8.1) j=1

Section 4.8. D v(t j 1 ) t. (4.8.1) j=1 Difference Equtions to Differentil Equtions Section.8 Distnce, Position, nd the Length of Curves Although we motivted the definition of the definite integrl with the notion of re, there re mny pplictions

More information

PDE Notes. Paul Carnig. January ODE s vs PDE s 1

PDE Notes. Paul Carnig. January ODE s vs PDE s 1 PDE Notes Pul Crnig Jnury 2014 Contents 1 ODE s vs PDE s 1 2 Section 1.2 Het diffusion Eqution 1 2.1 Fourier s w of Het Conduction............................. 2 2.2 Energy Conservtion.....................................

More information

The Ellipse. is larger than the other.

The Ellipse. is larger than the other. The Ellipse Appolonius of Perg (5 B.C.) disovered tht interseting right irulr one ll the w through with plne slnted ut is not perpendiulr to the is, the intersetion provides resulting urve (oni setion)

More information

where the box contains a finite number of gates from the given collection. Examples of gates that are commonly used are the following: a b

where the box contains a finite number of gates from the given collection. Examples of gates that are commonly used are the following: a b CS 294-2 9/11/04 Quntum Ciruit Model, Solovy-Kitev Theorem, BQP Fll 2004 Leture 4 1 Quntum Ciruit Model 1.1 Clssil Ciruits - Universl Gte Sets A lssil iruit implements multi-output oolen funtion f : {0,1}

More information

Functions. mjarrar Watch this lecture and download the slides

Functions. mjarrar Watch this lecture and download the slides 9/6/7 Mustf Jrrr: Leture Notes in Disrete Mthemtis. Birzeit University Plestine 05 Funtions 7.. Introdution to Funtions 7. One-to-One Onto Inverse funtions mjrrr 05 Wth this leture nd downlod the slides

More information

Proving the Pythagorean Theorem

Proving the Pythagorean Theorem Proving the Pythgoren Theorem W. Bline Dowler June 30, 2010 Astrt Most people re fmilir with the formul 2 + 2 = 2. However, in most ses, this ws presented in lssroom s n solute with no ttempt t proof or

More information

MA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp.

MA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp. MA123, Chpter 1: Formuls for integrls: integrls, ntiderivtives, nd the Fundmentl Theorem of Clculus (pp. 27-233, Gootmn) Chpter Gols: Assignments: Understnd the sttement of the Fundmentl Theorem of Clculus.

More information

STRAND J: TRANSFORMATIONS, VECTORS and MATRICES

STRAND J: TRANSFORMATIONS, VECTORS and MATRICES Mthemtics SKE: STRN J STRN J: TRNSFORMTIONS, VETORS nd MTRIES J3 Vectors Text ontents Section J3.1 Vectors nd Sclrs * J3. Vectors nd Geometry Mthemtics SKE: STRN J J3 Vectors J3.1 Vectors nd Sclrs Vectors

More information

Strategy: Use the Gibbs phase rule (Equation 5.3). How many components are present?

Strategy: Use the Gibbs phase rule (Equation 5.3). How many components are present? University Chemistry Quiz 4 2014/12/11 1. (5%) Wht is the dimensionlity of the three-phse coexistence region in mixture of Al, Ni, nd Cu? Wht type of geometricl region dose this define? Strtegy: Use the

More information

dy ky, dt where proportionality constant k may be positive or negative

dy ky, dt where proportionality constant k may be positive or negative Section 1.2 Autonomous DEs of the form 0 The DE y is mthemticl model for wide vriety of pplictions. Some of the pplictions re descried y sying the rte of chnge of y(t) is proportionl to the mount present.

More information

Geometry of the Circle - Chords and Angles. Geometry of the Circle. Chord and Angles. Curriculum Ready ACMMG: 272.

Geometry of the Circle - Chords and Angles. Geometry of the Circle. Chord and Angles. Curriculum Ready ACMMG: 272. Geometry of the irle - hords nd ngles Geometry of the irle hord nd ngles urriulum Redy MMG: 272 www.mthletis.om hords nd ngles HRS N NGLES The irle is si shpe nd so it n e found lmost nywhere. This setion

More information

Basic Angle Rules 5. A Short Hand Geometric Reasons. B Two Reasons. 1 Write in full the meaning of these short hand geometric reasons.

Basic Angle Rules 5. A Short Hand Geometric Reasons. B Two Reasons. 1 Write in full the meaning of these short hand geometric reasons. si ngle Rules 5 6 Short Hnd Geometri Resons 1 Write in full the mening of these short hnd geometri resons. Short Hnd Reson Full Mening ) se s isos Δ re =. ) orr s // lines re =. ) sum s t pt = 360. d)

More information

Thermal Diffusivity. Paul Hughes. Department of Physics and Astronomy The University of Manchester Manchester M13 9PL. Second Year Laboratory Report

Thermal Diffusivity. Paul Hughes. Department of Physics and Astronomy The University of Manchester Manchester M13 9PL. Second Year Laboratory Report Therml iffusivity Pul Hughes eprtment of Physics nd Astronomy The University of nchester nchester 3 9PL Second Yer Lbortory Report Nov 4 Abstrct We investigted the therml diffusivity of cylindricl block

More information

Math 8 Winter 2015 Applications of Integration

Math 8 Winter 2015 Applications of Integration Mth 8 Winter 205 Applictions of Integrtion Here re few importnt pplictions of integrtion. The pplictions you my see on n exm in this course include only the Net Chnge Theorem (which is relly just the Fundmentl

More information

Improper Integrals. The First Fundamental Theorem of Calculus, as we ve discussed in class, goes as follows:

Improper Integrals. The First Fundamental Theorem of Calculus, as we ve discussed in class, goes as follows: Improper Integrls The First Fundmentl Theorem of Clculus, s we ve discussed in clss, goes s follows: If f is continuous on the intervl [, ] nd F is function for which F t = ft, then ftdt = F F. An integrl

More information

ILLUSTRATING THE EXTENSION OF A SPECIAL PROPERTY OF CUBIC POLYNOMIALS TO NTH DEGREE POLYNOMIALS

ILLUSTRATING THE EXTENSION OF A SPECIAL PROPERTY OF CUBIC POLYNOMIALS TO NTH DEGREE POLYNOMIALS ILLUSTRATING THE EXTENSION OF A SPECIAL PROPERTY OF CUBIC POLYNOMIALS TO NTH DEGREE POLYNOMIALS Dvid Miller West Virgini University P.O. BOX 6310 30 Armstrong Hll Morgntown, WV 6506 millerd@mth.wvu.edu

More information

Unit #10 De+inite Integration & The Fundamental Theorem Of Calculus

Unit #10 De+inite Integration & The Fundamental Theorem Of Calculus Unit # De+inite Integrtion & The Fundmentl Theorem Of Clculus. Find the re of the shded region ove nd explin the mening of your nswer. (squres re y units) ) The grph to the right is f(x) = -x + 8x )Use

More information

H 4 H 8 N 2. Example 1 A compound is found to have an accurate relative formula mass of It is thought to be either CH 3.

H 4 H 8 N 2. Example 1 A compound is found to have an accurate relative formula mass of It is thought to be either CH 3. . Spetrosopy Mss spetrosopy igh resolution mss spetrometry n e used to determine the moleulr formul of ompound from the urte mss of the moleulr ion For exmple, the following moleulr formuls ll hve rough

More information

Naming the sides of a right-angled triangle

Naming the sides of a right-angled triangle 6.2 Wht is trigonometry? The word trigonometry is derived from the Greek words trigonon (tringle) nd metron (mesurement). Thus, it literlly mens to mesure tringle. Trigonometry dels with the reltionship

More information

Homework Assignment 3 Solution Set

Homework Assignment 3 Solution Set Homework Assignment 3 Solution Set PHYCS 44 6 Ferury, 4 Prolem 1 (Griffiths.5(c The potentil due to ny continuous chrge distriution is the sum of the contriutions from ech infinitesiml chrge in the distriution.

More information

set is not closed under matrix [ multiplication, ] and does not form a group.

set is not closed under matrix [ multiplication, ] and does not form a group. Prolem 2.3: Which of the following collections of 2 2 mtrices with rel entries form groups under [ mtrix ] multipliction? i) Those of the form for which c d 2 Answer: The set of such mtrices is not closed

More information

The DOACROSS statement

The DOACROSS statement The DOACROSS sttement Is prllel loop similr to DOALL, ut it llows prouer-onsumer type of synhroniztion. Synhroniztion is llowe from lower to higher itertions sine it is ssume tht lower itertions re selete

More information

Section 5.1 #7, 10, 16, 21, 25; Section 5.2 #8, 9, 15, 20, 27, 30; Section 5.3 #4, 6, 9, 13, 16, 28, 31; Section 5.4 #7, 18, 21, 23, 25, 29, 40

Section 5.1 #7, 10, 16, 21, 25; Section 5.2 #8, 9, 15, 20, 27, 30; Section 5.3 #4, 6, 9, 13, 16, 28, 31; Section 5.4 #7, 18, 21, 23, 25, 29, 40 Mth B Prof. Audrey Terrs HW # Solutions by Alex Eustis Due Tuesdy, Oct. 9 Section 5. #7,, 6,, 5; Section 5. #8, 9, 5,, 7, 3; Section 5.3 #4, 6, 9, 3, 6, 8, 3; Section 5.4 #7, 8,, 3, 5, 9, 4 5..7 Since

More information

Chapter Gauss Quadrature Rule of Integration

Chapter Gauss Quadrature Rule of Integration Chpter 7. Guss Qudrture Rule o Integrtion Ater reding this hpter, you should e le to:. derive the Guss qudrture method or integrtion nd e le to use it to solve prolems, nd. use Guss qudrture method to

More information

Farey Fractions. Rickard Fernström. U.U.D.M. Project Report 2017:24. Department of Mathematics Uppsala University

Farey Fractions. Rickard Fernström. U.U.D.M. Project Report 2017:24. Department of Mathematics Uppsala University U.U.D.M. Project Report 07:4 Frey Frctions Rickrd Fernström Exmensrete i mtemtik, 5 hp Hledre: Andres Strömergsson Exmintor: Jörgen Östensson Juni 07 Deprtment of Mthemtics Uppsl University Frey Frctions

More information

f (x)dx = f(b) f(a). a b f (x)dx is the limit of sums

f (x)dx = f(b) f(a). a b f (x)dx is the limit of sums Green s Theorem If f is funtion of one vrible x with derivtive f x) or df dx to the Fundmentl Theorem of lulus, nd [, b] is given intervl then, ording This is not trivil result, onsidering tht b b f x)dx

More information

2.4 Linear Inequalities and Problem Solving

2.4 Linear Inequalities and Problem Solving Section.4 Liner Inequlities nd Problem Solving 77.4 Liner Inequlities nd Problem Solving S 1 Use Intervl Nottion. Solve Liner Inequlities Using the Addition Property of Inequlity. 3 Solve Liner Inequlities

More information

Department of Mechanical Engineering ME 322 Mechanical Engineering Thermodynamics. Lecture 33. Psychrometric Properties of Moist Air

Department of Mechanical Engineering ME 322 Mechanical Engineering Thermodynamics. Lecture 33. Psychrometric Properties of Moist Air Deprtment of Mechnicl Engineering ME 3 Mechnicl Engineering hermodynmics Lecture 33 sychrometric roperties of Moist Air Air-Wter Vpor Mixtures Atmospheric ir A binry mixture of dry ir () + ter vpor ()

More information

( ) as a fraction. Determine location of the highest

( ) as a fraction. Determine location of the highest AB/ Clulus Exm Review Sheet Solutions A Prelulus Type prolems A1 A A3 A4 A5 A6 A7 This is wht you think of doing Find the zeros of f( x) Set funtion equl to Ftor or use qudrti eqution if qudrti Grph to

More information

Alg. Sheet (1) Department : Math Form : 3 rd prep. Sheet

Alg. Sheet (1) Department : Math Form : 3 rd prep. Sheet Ciro Governorte Nozh Directorte of Eduction Nozh Lnguge Schools Ismili Rod Deprtment : Mth Form : rd prep. Sheet Alg. Sheet () [] Find the vlues of nd in ech of the following if : ) (, ) ( -5, 9 ) ) (,

More information

MATH FIELD DAY Contestants Insructions Team Essay. 1. Your team has forty minutes to answer this set of questions.

MATH FIELD DAY Contestants Insructions Team Essay. 1. Your team has forty minutes to answer this set of questions. MATH FIELD DAY 2012 Contestnts Insructions Tem Essy 1. Your tem hs forty minutes to nswer this set of questions. 2. All nswers must be justified with complete explntions. Your nswers should be cler, grmmticlly

More information

CS12N: The Coming Revolution in Computer Architecture Laboratory 2 Preparation

CS12N: The Coming Revolution in Computer Architecture Laboratory 2 Preparation CS2N: The Coming Revolution in Computer Architecture Lortory 2 Preprtion Ojectives:. Understnd the principle of sttic CMOS gte circuits 2. Build simple logic gtes from MOS trnsistors 3. Evlute these gtes

More information

CS 311 Homework 3 due 16:30, Thursday, 14 th October 2010

CS 311 Homework 3 due 16:30, Thursday, 14 th October 2010 CS 311 Homework 3 due 16:30, Thursdy, 14 th Octoer 2010 Homework must e sumitted on pper, in clss. Question 1. [15 pts.; 5 pts. ech] Drw stte digrms for NFAs recognizing the following lnguges:. L = {w

More information