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### Lecture12

Course: ECON 398, Spring 2009
School: Michigan
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Word Count: 1034

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Strategies 3/3/2009 Mixed Matching Pennies We denote the strategies on the game bi-matrix Guildenstern q Heads Rosencrantz R t p 1-p Heads Hd Tails 1 , -1 1 -1 , 1 1-q Tails -1 , 1 1 1 , -1 Preferences Involving Gambles We revisit our description of payoffs in the game bimatrix. Consider Tom, who is faced with two outcomes A: get \$1 B: get \$4 Guildensterns payoff to H: (-1)p + 1(1-p) = 1-2p T: (1)p + (-1)(1-p)...

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Michigan - ECON - 398
3/3/2009Mixed Strategies Matching PenniesWe denote the strategies on the game bi-matrixGuildenstern q Heads Rosencrantz R t p 1-p Heads Hd Tails 1 , -1 1 -1 , 1 1-q Tails -1 , 1 1 1 , -1Mixed Strategies Matching PenniesRosencrantzs &amp; Guilden
Michigan - ECON - 398
3/5/2009Your Choice of GamblesA1= A2=.2 .8 .8 .2 \$4000 \$0 \$4000 \$0Your Choice of Gambles\$3000 \$0 \$3000 \$0B1= B2=.25 .75 1 0A1= A2=.2 .8 .8 .2\$4000 \$0 \$4000 \$0B1= B2=.25 .75 1 0\$3000 \$0 \$3000 \$0Suppose you prefer A1 to B1, but
Michigan - ECON - 398
3/10/2009Expert AdviceWe often rely on professionals to diagnose problems and recommend and carry out fixesDoctors Auto Mechanics and Repair Technicians p Lawyers Financial Advisors and Real Estate Agents Teachers and ResearchersExpert AdviceA
Michigan - ECON - 398
1/13/2009Strategic SituationsEcon 398: StrategyA Study of Strategic Interaction in Games Chris Proulx Lorch Hall M107 rabbit@umich.edu Office Hours: Tu, Th 4:15-5:15Many (2) people Each has many possible actions Different actions yield differe
Michigan - ECON - 398
1/22/2009Homework Due Friday 4:15PMHand homework in to your GSI during discussion section OR Turn in homework to the specially marked box outside Lorch Hall 109 by 4:15PM Today: Chapter 4, Next Week: Chapter 5Homework Due Friday 4:15PMDixit &amp; S
Michigan - ECON - 398
Multiple-Choice Questions: Select the best answer. Each correct answer is worth 5 points. Each wrong answer is worth zero points. Answers left blank are worth 1 point.Use the following to answer questions 1-2:Michigan and OSU are playing a footbal
University of Toronto - PHY - PHYA22
28.57. Model: The currents r the two segments of the wire are the same. in rVisualize: The electric fields E1 and E2 point in the direction of the current. Establish a cylindrical Gaussian surface with end area a that extends into both segments of
University of Toronto - PHY - PHYA22
28.2. Solve: We estimate a distance of 5 ft from the wall switch to the ceiling and then a distance of 8 ft to thecenter of the room. This yields a total length of approximately L 4.0 m that an electron will travel in time t. The drift speed of ele
University of Toronto - PHY - PHYA22
current through this wire is 2.0 1019 s 1 . Using Table 28.1 for the electron density of iron and Equation 28.3, the drift velocity is28.1. Solve: The wires cross-sectional area is A = r 2 = (1.0 10 3 m ) = 3.1415 10 6 m 2 , and the electron2
University of Toronto - PHY - PHYA22
28.6. Solve: For L = 1.0 cm = 1.0 10 2 m , the surface area of the wire isA = (2r)L = DL = (1.0 103 m)(1.0 102 m) = (1.0 105 m2) The surface charge density of the wire is1 19 Q (1000 cm 1 cm )1.60 10 C = = 5.1 10 12 C / m 2 A 1.0 10 5 m 2
University of Toronto - PHY - PHYA22
28.10. Visualize:Solve: (a) The charge density is not uniform along the wire. If it was, there would be no electric field inside the wire. The charge density is most positive near the positive terminal of the battery. It gradually decreases until b
University of Toronto - PHY - PHYA22
28.11. Model: A battery is a charge escalator.Solve: When a wire is connected to a battery, there is a sustained motion of electrons. A current of 1.5 A means that a charge of 1.5 C flows through a cross section of the wire per second. Because Q = N
University of Toronto - PHY - PHYA22
28.17. Solve: (a) The current density isJ= I I 0.85 A 7 2 = = 2 = 1.73 10 A / m A R 2 [ 1 (0.00025 m )] 2(b) The electron current, or number of electrons per second, isNe I 0.85 A 0.85 C / s = = = 5.31 1018 s 1 19 t e 1.60 10 C 1.60 10 19 C
University of Toronto - PHY - PHYA22
28.18. Solve: (a) From Equation 28.13 and Table 28.1, the current density in the gold wire isJ = nevd = (5.9 10 28 m 3 )(1.60 10 19 C)(3.0 10 4 m / s) = 2.83 10 6 A / m 2(b) The current isI = JA = (2.83 10 6 A / m 2 ) (0.25 10 3 m ) = 0.556
University of Toronto - PHY - PHYA22
28.25. Solve: From Equations 28.18 and 28.19, the resistivity is3 E E EA Er 2 (0.085 N / C) (1.5 10 m ) = = = = = = 5.01 10 8 m 12 A J IA I I 2
University of Toronto - PHY - PHYA22
28.24. Solve: From Equation 28.18 and Table 28.2, the electric field isE= J I 5.0 A = = = 0.159 N / C A (1.0 10 7 1 m 1 ) (1.0 10 3 m ) 2
University of Toronto - PHY - PHYA22
28.28. Visualize:Solve: The current density through the cube is J = E, and the actual current is I = AJ. Combining these equations, the conductivity is I 9.0 A = = = 1.8 10 7 1 m 1 AE 10 4 m 2 5.0 10 3 N / C If all quantities entering the calcul
University of Toronto - PHY - PHYA22
28.29. Solve: The density of aluminum is 2700 kg/m3, so 1.0 m3 of aluminum has a mass of 2700 kg. Theconduction-electron density is the number of electrons in 1.0 m3. Because the atomic mass of aluminum is 27 u, 27 g of aluminum contains NA = 6.02
University of Toronto - PHY - PHYA22
28.39. Model: We assume that the charge carriers are uniformly distributed throughout the wire.Solve: Using Equation 28.16, we can write the current density asJ= ne 2 I ne 2 E A = E I= m A mFor a given wire, the current is thus proportional
University of Toronto - PHY - PHYA22
28.38. Solve: The total charge in the battery isQ = It = (90 A)(3600 s) = 3.2 105 C
University of Toronto - PHY - PHYA22
28.9. Model: We will use the model of conduction to relate the electric field strength to the mean free timebetween collisions. Solve: From Equation 28.8, the electric field isE=(9.11 10 31 kg)(5.0 1019 s 1 ) mi = 2 = 0.31 N / C neA (8.5 10 2
University of Toronto - PHY - PHYA22
28.37. Solve: (a) Current is defined as I = Q/t, so the charge delivered in time t isQ = It = (150 A)(0.80 s) = 120 C (b) The drift speed is J IA I 150 A vd = = 5.617 10 4 m / s = =2= 2 ne ne r ne (0.0025 m ) 8.5 10 28 m 3 1.60 10 19 C()()
University of Toronto - PHY - PHYA22
28.19. Solve: From Equation 28.13, the current in the wire isI = JA = 7.50 10 5 A / m 2 (2.5 10 6 m 75 10 6 m ) = 0.141 mA()
University of Toronto - PHY - PHYA22
28.26. Solve: From Equations 28.18 and 28.19, the resistivity is3 E E EA Er 2 (0.0075 N / C) (0.5 10 m ) = = = = = = 1.51 10 6 m 3 3.9 10 A J IA I I 2From Table 8.2, we see that the wire is made of nichrome.
University of Toronto - PHY - PHYA22
28.16. Visualize:The current density J in a wire, as given by Equation 28.13, does not depend on the thickness of the wire. Solve: (a) The current in the wire isI wire = J wire Awire = ( 450,000 A / m 2 )[ (1.5 10 m)]1 2 32= 0.795 ABeca
University of Toronto - PHY - PHYA22
28.15. Solve: (a) The current density isJ= I 2.5 A = = 6.25 10 5 A / m 2 A 4.0 10 6 m 2(b) Using Equation 28.13 and Table 28.1, the drift speed isvd =J 6.25 10 5 A / m 2 = 6.51 10 5 m / s = ne (6.0 10 28 m 3 )(1.6 10 19 C)
University of Toronto - PHY - PHYA22
28.3. Solve: Using Equation 28.3 and Table 28.1, the electron current isi = nAvd = (5.9 10 28 m 3 ) (0.5 10 3 m ) 5.0 10 5 m / s = 2.3 1018 s 12()The time for 1 mole of electrons to pass through a cross section of the wire ist=N A 1
University of Toronto - PHY - PHYA22
28.40. Solve: Equation 28.13 defines the current density as J = I A . This meansA= I D 2 = D= J 4 4I = J 4(1.0 A ) = 0.050 cm = 0.50 mm (500 A / cm 2 )Assess: Fuse wires are usually thin.
University of Toronto - PHY - PHYA22
28.21. Model: Use the model of conduction to relate the mean time between collisions to conductivity.Solve: From Equation 28.17, Table 28.1, and Table 28.2, the mean time between collisions for aluminum is Al =7 31 1 1 m Al (9.11 10 kg)(3.5 10
University of Toronto - PHY - PHYA22
28.20. Visualize:Solve:The current-carrying cross section of the wire isA = r12 r22 = (0.0010 m ) (0.0005 m )2[2] = 2.356 106m2The current density isJ=10 A = 4.24 10 6 A / m 2 2.356 10 6 m 2
University of Toronto - PHY - PHYA22
28.22. Model: We will use the model of conduction to relate the mean time between collisions to conductivity.Solve: From Equation 28.17, Table 28.1, and Table 28.2, the mean time between collisions for silver is silver =7 31 1 1 m silver (9.11
University of Toronto - PHY - PHYA22
28.23. Solve: The current density is J = E. Using Equation 28.18 and Table 28.2, the current in the wire isI = EA = (3.5 10 7 1 m 1 )(0.012 N / C)( 4 10 6 m 2 ) = 1.68 A
University of Toronto - PHY - PHYA22
28.4. Solve: Equation 28.2 is N e = nAvd t . Using Table 28.1 for the electron density, we getA= Ne D 2 = nvd t 4D=4(1.0 1016 ) 4 Ne = = 9.26 10 4 m = 0.926 mm nvd t (5.8 10 28 m 3 )(8.0 10 4 m / s)(320 10 6 s)
University of Toronto - PHY - PHYA22
28.7. Solve: (a) Each gold atom has one conduction electron. Using Avogadros number and n as the number of moles, the number of atoms is r 2 L m V N = nN A = NA = NA = NA MA MA MA The density of gold is = 19,300 kg/m3, the atomic mass is MA = 197 g
University of Toronto - PHY - PHYA22
28.14. Visualize:r The direction of the current I in a material is opposite to the direction of motion of the negative charges and is the same as the direction of motion of positive charges. Solve: The charge due to positive ions moving to the righ
University of Toronto - PHY - PHYA22
28.27. Solve: (a) Since J = E and J = I A , the electric field isE= I I 0.020 A = = = 1.64 10 3 N / C A r 2 (0.25 10 3 m ) 2 (6.2 10 7 1 m 1 )(b) Since the current density is related to vd by J = I A = nevd , the drift speed isvd =I 0.020
University of Toronto - PHY - PHYA22
28.12. Solve: Equation 28.10 is Q = It. The amount of charge delivered is60 s Q = (10.0 A ) 5.0 min = 3000 C 1 min The number of electrons that flow through the hair dryer is Q 3000 C N= = = 1.88 10 22 e 1.60 10 19 C
University of Toronto - PHY - PHYA22
28.13. Solve: From Equation 28.10,I=13 19 Q Ne (2.0 10 )(1.60 10 C) = = = 0.0032 C / s = 0.0032 A = 3.2 mA 3 t t 1.0 10 s
University of Toronto - PHY - PHYA22
28.36. Solve: (a) A current of 1.8 pA for the potassium ions means that a charge of 1.8 pC flows through the potassium ion channel per second. The number of potassium ions that pass through the ion channel per second is1.8 10 12 C / s = 1.125 10 7
University of Toronto - PHY - PHYA22
28.31. Solve: (a) Let the new current be I such that I = 2I where I is the original current. The current density is J = I A . Since the area of cross section of the wire remains the same, we haveI I I J = J = 2J = I J J That is, doubling the c
UC Davis - HIS - 17B
February 10, 2009 Great Depression, First and Second New Deal I. Stock Market Crash a. Why did it crash? Why did this trigger a great depression? b. 1920s: stock market kept increasing c. economy was booming in 1920s d. a lot of this boom was due to
UC Davis - HIS - 17B
i. Civilian Conservation Corps 1. Paid 30 bucks a week 2. Worked in parks 3. Purpose was to give jobs who cant find one in private sectors 4. Since it pays so little, people will never turn down a private job when offered 5. Pays enough to survive th
UC Davis - HIS - 17B
February 12, 2009 Depression and War, 1936-45 I. Roosevelt targeted reform, relief and recovery a. Relief: low interest and cash payments b. Reforms: make sure crisis like this doesnt happen again i. Regulate banks and stock market ii. Also started g
UC Davis - HIS - 17B
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UC Davis - HIS - 17B
February 19, 2009 I. Second Red Scare a. Thought Soviets nuclear bomb, North Koreas invasion, Chinas conversion to communism b. Senator McCarthy listed the blasphemy of the world. c. How these happen?! d. He said that there were spies and conspiracie
UC Davis - HIS - 17B
II.III.a. King found it difficult also b. He was jailed! c. He wrote a letter in jail i. response to whites who say he was Wanting change too fast d. he said he refused to Wait e. when he was out of jail, he came up with new strategies f. since a
UC Davis - HIS - 17B
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UC Davis - HIS - 17B
i. Also support psychological warfares ii. Also draw the line in overthrowing government and assassinations iii. But the ext president does not feel that way! b. Dwight Eisenhower i. CIA should overthrow government that seem threatening ii. one of su
UC Davis - HIS - 17B
II.a. She though she needed to get to the root of the problem and dedicated her life to finding birth control! 2. No woman can be free without the freedom of choosing. 3. In Holland a. talked to doctors who were in fortfront of birth control b. dia
UC Davis - HIS - 17B
The Progressive Spirit: Jane Addams, Margaret Sanger, and the Suffragist 1. 2. 3. 4. 5. I. Progressive concerns and progressive reforms Margaret Sanger: from social justice to social control Jane Addams: from social control to social justice Woman su
UC Davis - HIS - 17B
February 26, 2009 Korean war, Vietnam war I. Japan had to give up all its companies to the winners of the war a. Such were China, and Korea! b. So what to do with these new territories? c. Victorious powers divided Korea to Soviet and U.S. d. Didnt g
UC Davis - HIS - 17B
February 26, 2009 Korean war, Vietnam war I. Japan had to give up all its companies to the winners of the war a. Such were China, and Korea! b. So what to do with these new territories? c. Victorious powers divided Korea to Soviet and U.S. d. Didnt g
UC Davis - HIS - 17B
i. With colonies, they can sell and even get raw materials and labor. ii. Industrialization provide motive iii. In technology 1. Industrialized vs. non-industrialized army 2. Accurate rifles iv. Ideaological 1. Imperialists said: People in that count
UC Davis - HIS - 17B
II.III.a. Britain was only one holding out against Nazis b. So Hitler started blitz of bombing on Britain as a threat before officially invading c. Opinions begin to change in U.S., but still majority is the same d. We shoud support Britain! e. S
UC Irvine - ECON - 161B
MikeLiao 83381880 InthearticlewrittenbyOlivierBlanchard,TheCrisis: BasicMechanisms,andAppropriatePolicies,theideaofthe paperistoexpressandlookbeyondthecomplexityofthe globalfinancialandeconomiccrisis,buttoidentifythe basicmechanismsandpoliciesneeded
UC Irvine - ECON - 161B
MikeLiaoJr. 83381880 InthearticleAnIntroductiontoCapitalControls, ChristopherNeelydefinesacapitalcontrolasanypolicythat isdesignedtolimitorredirectcapitalaccounttransactions. Thesecontrolscouldtaketheformoftaxesorquantity controls.AccordingtoNeely(1
Cornell - ANTHR - 2432
Cornell - ANTHR - 1400
Cornell - ANTHR - 1400
Cornell - ANTHR - 1400
Cornell - ANTHR - 1400