Psych Study Guide 3
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Psych Study Guide 3

Course Number: PSY 111, Fall 2009

College/University: Elon

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PSY 111 Exam 3 Study Guide Some tips for studying for the exam: Actively work at understanding the material in the lecture notes. Those are written in my words, not yours, and you will need to translate them into your words if you really want to understand the material. If you need help understanding something, ask me for clarification. Good luck! Social Psychology 1. What are social norms? The rules that a group...

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PSY 111 Exam 3 Study Guide Some tips for studying for the exam: Actively work at understanding the material in the lecture notes. Those are written in my words, not yours, and you will need to translate them into your words if you really want to understand the material. If you need help understanding something, ask me for clarification. Good luck! Social Psychology 1. What are social norms? The rules that a group uses for appropriate and inappropriate values, beliefs, attitudes and behaviors 2. What is normative social influence? What is informational social influence? Normative influence from persons desire to gain approval or avoid disapproval Informational influence resulting from ones willingness to accept others opinions about reality 3. Understand Aschs experiment as it relates to conformity Everyone would say the wrong answer and the person being tested would always conform and choose the obviously wrong answer as well. Even though they knew it was wrong. 4. How is conformity strengthened? Why do we conform? Strengthening Conformity 1. One is made to feel incompetent or insecure. 2. The group has at least three people. 3. The group is unanimous. 4. One admires the groups status and attractiveness. 5. One has no prior commitment to a response. 6. The group observes ones behavior. 7. Ones culture strongly encourages respect for a social standard. Why Do We Conform?- Gain social approval - Fear of being labeled negatively and being socially ostracized - The need for social respect - Helps avoid social chaos - Convenience allows us not to think - Groups provide us with valuable info 5. Understand Milgrams experiment as it relates to obedience Stanley Milgram designed a study that investigates the effects of authority on obedience.- Milgrams infamous experiment with electric shocks- Told to shock someone when they get answer wrong 6. What is social facilitation? In what circumstances is behavior enhanced/impaired?

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Elon - PSY - 111
PSY 111 Exam 1 Study Questions Some tips for studying for the exam: Actively work at understanding the material in the lecture notes. Those are written in my words, not yours, and you will need to translate them into your words if you really want to under
Elon - ANT - 112
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Grace NE - SOCY - 145
Last update fall 2008Style and Reference Guide for Undergraduate Essays Department of Sociology Queen's UniversityThe following style and reference guide is based on the American Sociological Association Style Guide (2nd Edition) (a copy of the ASA Styl
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COMM 376Module 1 and 2: Japan & Korea South Korea North Korea is bigger than South Korea (55% and 45%) Three kingdoms in control: Koguryo (north), Paekche (southwest), Silla (southeast) Silla conquered then Koryo then Mongols then Choson Dynesty Later, J
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COMM 324 INVESTMENTS AND PORTFOLIO MANAGEMENTASSIGNMENT 2 Due: October 18 1. (Spreadsheet question) The monthly return data are given in the spreadsheet that are available on the course website. a. Calculate their variance-covariance matrix b. Plot the m
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TARIQ ELHASSANI Homework #6 Problem 13.5Linregr (x, y) a= 0.3525 G the slope St = 55.6000 Sr = 9.0740 r= 0.9148 Intercept = 4.8515 Graph12 11 10 9 8 7 6 5 402468101214161820B) switching the variables we get Linregr (y, x) a= 2.3741 + the slo
Old Dominion - ME - ME335
Distance (m) 0.0160 0.0150 0.0140 0.0130 0.0120 0.0110 0.0100 0.0090 0.0080 0.0070 0.0060 0.0050 0.0040 0.0030 0.0020 0.0010 0.00000.00 2.03 3.02 4.74 5.88 7.45 8.12 9.48 10.47 11.30 11.98 13.07 13.80 14.42 14.53 14.84 14.740.00 1.66 3.33 4.53 6.09 7.13
Old Dominion - ME - ME335
water= 1000 g = 9.81Transducer (Volts) 2.23 2.38 2.53 2.68 2.83 2.97 3.12 3.26 3.41kg/m3 m/s2 Differentiel Pressure (KPa) 0.00 2.49 4.98 7.48 9.97 12.46 14.95 17.44 19.93Manometer (in) 0 10 20 30 40 50 60 70 80Manometer (m) 0 0.25 0.51 0.76 1.02 1.2
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calculation of the mole fraction0.20 0.40 0.60 0.80 1.00n total446.8 227 153.33 116.62 94.56n O274 28 12.33 4.62 0n N2347.8 174 116 87 69.54y CO20.0269 0.0529 0.0783 0.1029 0.1269y H2O0.0291 0.0573 0.0848 0.1115 0.1375y O20.1656 0.1233 0.0804
Old Dominion - ME - ME335
SOLUTION MANUAL SI UNIT PROBLEMS CHAPTER 2SONNTAG BORGNAKKE VAN WYLENFUNDAMENTALSofThermodynamicsSixth EditionCONTENTSUBSECTION Correspondence table Concept-Study Guide Problems Properties and Units Force and Energy Specific Volume Pressure Manomet
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SOLUTION MANUAL SI UNIT PROBLEMS CHAPTER 5SONNTAG BORGNAKKE VAN WYLENFUNDAMENTALSofThermodynamicsSixth EditionSonntag, Borgnakke and van WylenCONTENTSUBSECTION Correspondence table Concept-Study Guide Problems Kinetic and potential energy Properti
Old Dominion - ME - ME335
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62/87,21 0$18$/ (1*/,6+ 81,7 352%/(06 &+$37(5 SONNTAG BORGNAKKE VAN WYLENFUNDAMENTALSofThermodynamicsSixth Edition6RQQWDJ %RUJQDNNH DQG YDQ :\OHQ&+$37(5 &217(17 &+$37(5 68%6(&7,21 &RUUHVSRQGHQFH WDEOH (TXLOLEULXP &KHPLFDO HTXLOLEULXP (TXLOLEULXP
Old Dominion - ME - ME335
SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 16SONNTAG BORGNAKKE VAN WYLENFUNDAMENTALSofThermodynamicsSixth EditionSonntag, Borgnakke and van WylenCHAPTER 16CONTENT CHAPTER 16SUBSECTION Correspondence table Stagnation properties Momentum Equatio
Old Dominion - ME - ME335
TARIQ ELHASSANI Problem 2-9Su 60 64 65 82 101 119 120 130 134 145 180 195 205 207 210 213 225 225 227 230 238 242 265 280 295 325 325 355 SUM 5462S'e 30 48 29.5 45 51 50 48 67 60 64 84 78 96 87 87 75 99 87 116 105 109 106 105 96 99 114 117 122 2274.5Co
William & Mary - ACCT - 400
Financial Accounting Liza Semenova, Stephanie Schaeberle, Greg Frink, Mike Stack, Kyle Turcotte 09/20/10 CP2-3 CP3-3 1. Comparing Companies within an Industry Financial Leverage Ratio: Average Total Assets/Average Total Stockholders Equity a. American Eag
Old Dominion - ME - ME335
SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 3SONNTAG BORGNAKKE VAN WYLENFUNDAMENTALSofThermodynamicsSixth EditionSonntag, Borgnakke and van WylenCONTENT CHAPTER 3SUBSECTION Correspondence table Study guide problems Phase diagrams, triple and cri
Old Dominion - ME - ME335
SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 3SONNTAG BORGNAKKE VAN WYLENFUNDAMENTALSofThermodynamicsSixth EditionSonntag, Borgnakke and WylenCHAPTER 3SUBSECTION Concept-Study Guide Problems Phase diagrams General Tables Ideal Gas Compressibility
Old Dominion - ME - ME335
CHAPTER 4 SI UNIT PROBLEMS SOLUTION MANUALSONNTAG BORGNAKKE VAN WYLENFUNDAMENTALSofThermodynamicsSixth EditionSonntag, Borgnakke and van WylenCONTENTSUBSECTION Correspondence table Concept problems Force displacement work Boundary work: simple one
Old Dominion - ME - ME335
SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 4SONNTAG BORGNAKKE VAN WYLENFUNDAMENTALSofThermodynamicsSixth EditionSonntag, Borgnakke and WylenCHAPTER 4SUBSECTION Concept-Study Guide Problems Simple processes Review Problems Polytropic processes M
Old Dominion - ME - ME335
SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 5SONNTAG BORGNAKKE VAN WYLENFUNDAMENTALSofThermodynamicsSixth EditionSonntag, Borgnakke and WylenCHAPTER 5SUBSECTION Concept-Study Guide Problems Kinetic and Potential Energy Properties from General Ta
Old Dominion - ME - ME335
SOLUTION MANUAL SI UNIT PROBLEMS CHAPTER 6SONNTAG BORGNAKKE VAN WYLENFUNDAMENTALSofThermodynamicsSixth EditionSonntag, Borgnakke and van WylenCONTENTSUBSECTION Correspondence table Concept-Study guide problems Continuity equation and flow rates Si
Old Dominion - ME - ME335
SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 6SONNTAG BORGNAKKE VAN WYLENFUNDAMENTALSofThermodynamicsSixth EditionSonntag, Borgnakke and WylenCHAPTER 6SUBSECTION Concept-Study Guide Problems Continuity and Flow Rates Single Flow Devices Multiple
Old Dominion - ME - ME335
SOLUTION MANUAL SI UNIT PROBLEMS CHAPTER 7SONNTAG BORGNAKKE VAN WYLENFUNDAMENTALSofThermodynamicsSixth EditionSonntag, Borgnakke and van WylenCONTENT CHAPTER 7SUBSECTION Correspondence table Concept-Study guide problems Heat engines and refrigerat
Old Dominion - ME - ME335
SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 7SONNTAG BORGNAKKE VAN WYLENFUNDAMENTALSofThermodynamicsSixth EditionSonntag, Borgnakke and WylenCHAPTER 7SUBSECTION Concept-Study Guide Problems Heat Engines and Refrigerators Carnot Cycles and Absolu
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SOLUTION MANUAL SI UNIT PROBLEMS CHAPTER 8SONNTAG BORGNAKKE VAN WYLENFUNDAMENTALSofThermodynamicsSixth EditionSonntag, Borgnakke and van WylenCONTENTSUBSECTIONPROB NO.Correspondence table Concept-Study Guide problems 1-20 Inequality of Clausius
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SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 8SONNTAG BORGNAKKE VAN WYLENFUNDAMENTALSofThermodynamicsSixth EditionSonntag, Borgnakke and WylenCHAPTER 8SUBSECTION Concept-Study Guide Problems Entropy, Clausius Reversible Processes Entropy Generati
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SOLUTION MANUAL SI UNIT PROBLEMS CHAPTER 9SONNTAG BORGNAKKE VAN WYLENFUNDAMENTALSofThermodynamicsSixth EditionSonntag, Borgnakke and van WylenCONTENTSUBSECTION Correspondence table Concept-Study Guide Problems Steady State Reversible Processes Sin
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SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 9SONNTAG BORGNAKKE VAN WYLENFUNDAMENTALSofThermodynamicsSixth EditionSonntag, Borgnakke and WylenCONTENTSUBSECTION Concept-Study Guide Problems Steady Single Flow Devices Steady Irreversible Processes
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SOLUTION MANUAL SI UNIT PROBLEMS CHAPTER 10SONNTAG BORGNAKKE VAN WYLENFUNDAMENTALSofThermodynamicsSixth EditionSonntag, Borgnakke and van WylenCONTENTSUBSECTION PROB NO.Correspondence table Concept-Study Guide Problems Available energy, reversibl
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SOLUTION MANUAL SI UNIT PROBLEMS CHAPTER 11SONNTAG BORGNAKKE VAN WYLENFUNDAMENTALSofThermodynamicsSixth EditionCONTENTSUBSECTION PROB NO.Correspondence table Concept-Study guide problems 1-20 Rankine cycles, power plants Simple cycles 21-35 Reheat
Old Dominion - ME - ME335
SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 11SONNTAG BORGNAKKE VAN WYLENFUNDAMENTALSofThermodynamicsSixth EditionSonntag, Borgnakke and van WylenCHAPTER 11SUBSECTION Rankine Cycles Brayton Cycles Otto, Diesel, Stirling and Carnot Cycles Refrige
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Old Dominion - ME - ME335
SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 12SONNTAG BORGNAKKE VAN WYLENFUNDAMENTALSofThermodynamicsSixth EditionSonntag, Borgnakke and van WylenCHAPTER 12SUBSECTION Concept Problems Mixture Composition and Properties Simple Processes Entropy G
Old Dominion - ME - ME335
CHAPTER 13 SOLUTION MANUALSONNTAG BORGNAKKE VAN WYLENFUNDAMENTALSofThermodynamicsSixth EditionSonntag, Borgnakke and van WylenFundamentals of Thermodynamics 6 Edition Sonntag, Borgnakke and van Wylen CONTENT CHAPTER 13SUBSECTION Correspondence tab
Old Dominion - ME - ME335
62/87,21 0$18$/ (1*/,6+ 81,7 352%/(06 &+$37(5 SONNTAG BORGNAKKE VAN WYLENFUNDAMENTALSofThermodynamicsSixth Edition6RQQWDJ %RUJQDNNH DQG YDQ :\OHQ&+$37(5 &217(17 &+$37(5 68%6(&7,21 &RUUHVSRQGHQFH WDEOH &ODSH\URQ HTXDWLRQ 9ROXPH ([SDQVLYLW\ DQG &RP
Old Dominion - ME - ME335
SOLUTION MANUAL SI UNIT PROBLEMS CHAPTER 14SONNTAG BORGNAKKE VAN WYLENFUNDAMENTALSofThermodynamicsSixth EditionSonntag, Borgnakke and van WylenCONTENTSUBSECTION PROB NO.Correspondence table Concept-Study Guide Problems 1-20 Fuels and the Combusti
Old Dominion - ME - ME335
SOLUTION MANUAL SI UNIT PROBLEMS CHAPTER 15SONNTAG BORGNAKKE VAN WYLENFUNDAMENTALSofThermodynamicsSixth EditionSonntag, Borgnakke and van WylenCONTENTSUBSECTION PROB NO. 1-20 21-24 25-66 67-73 74-78 79-88 89-106Correspondence table Concept-Study
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Old Dominion - ME - ME335
Chapter 1Problems 1-1 through 1-4 are for student research. 1-5Impending motion to left E1 f A1 f B G Fcr FcrDCFaccConsider force F at G, reactions at B and D. Extend lines of action for fully-developed friction D E and B E to find the point of c
Old Dominion - ME - ME335
Chapter 22-1 (a)12 10 8 6 4 2 060708090 100 110 120 130 140 150 160 170 180 190 200 210(b) f/(N x) = f/(69 10) = f/690 x 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210 f 2 1 3 5 8 12 6 10 8 5 2 3 2 1 0 1 69 fx 120 70 240 450 800 1320 7
Old Dominion - ME - ME335
Chapter 33-1 From Table A-20 Sut = 470 MPa (68 kpsi), S y = 390 MPa (57 kpsi) 3-2 From Table A-20 Sut = 620 MPa (90 kpsi), S y = 340 MPa (49.5 kpsi) Ans. 3-3 Comparison of yield strengths: Sut of G10 500 HR is S yt of SAE1020 CD is 620 = 1.32 times large
Old Dominion - ME - ME335
Chapter 44-1W 2 A RA 1 RB B A RA 1 1 RB W 2 B(a)1 RD 3 RA RB 2 D(b)1 RC CABW(c)W 1 RC RB RA RB 2 RA W(d) (e)ARA12 W RBx B RBx 1 RB RBy RByScale of corner magnified(f)Chapter 4514-2 (a)2 kN 60 2 30 2 kN 60 RB RA 30 RA RB 90R A = 2 si
Old Dominion - ME - ME335
Chapter 55-1 (a)k1 k2 k3 F yk= so (b)k1 k2 k3 F yF ; yy=F F F + + k1 k2 k3 Ans.k=1 (1/k1 ) + (1/k2 ) + (1/k3 )F = k1 y + k2 y + k3 y k = F/y = k1 + k2 + k3 Ans.(c)k2 k1 k31 1 1 = + k k1 k2 + k3k=1 1 + k1 k2 + k3-15-2 For a torsion bar, k
Old Dominion - ME - ME335
Chapter 66-1 MSS: DE: 1 - 3 = S y /n n= Sy n= Sy 1 - 31/22 2 = A - AB + B1/22 2 2 = x - x y + y + 3x y(a) MSS:1 = 12, 2 = 6, 3 = 0 kpsi 50 n= = 4.17 Ans. 12 = (122 - 6(12) + 62 ) 1/2 = 10.39 kpsi, 12 22DE:n=50 = 4.81 10.39Ans.12 (b) A , B = 2
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Chapter 77-1 H B = 490 Eq. (3-17): Eq. (7-8): Table 7-4: Eq. (7-18): Eq. (7-19): Eq. (7-17): 7-2 (a) Sut = 68 kpsi, Se = 0.495(68) = 33.7 kpsi (b) Sut = 112 kpsi, Se = 0.495(112) = 55.4 kpsi (c) 2024T3 has no endurance limit (d) Eq. (3-17): Se = 107 kpsi
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Chapter 88-1 (a)2.5 mm 25 mm 5 mm 2.5Thread depth = 2.5 mmAns.Width = 2.5 mm Ans. dm = 25 - 1.25 - 1.25 = 22.5 mm dr = 25 - 5 = 20 mm l = p = 5 mm Ans. Thread depth = 2.5 mm Ans. Width at pitch line = 2.5 mm Ans. dm = 22.5 mm dr = 20 mm l = p = 5 mm
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Chapter 99-1 Eq. (9-3): F = 0.707hl = 0.707(5/16)(4)(20) = 17.7 kip Ans. 9-2 Table 9-6: all = 21.0 kpsi f = 14.85h kip/in = 14.85(5/16) = 4.64 kip/in F = f l = 4.64(4) = 18.56 kip Ans. 9-3 Table A-20: 1018 HR: Sut = 58 kpsi, S y = 32 kpsi 1018 CR: Sut =
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Chapter 1010-11" 4"1" 21" 4"1" 210-2A = Sd m dim( Auscu ) = dim(S) dim(d m ) = kpsi inm m dim( ASI ) = dim(S1 ) dim d1 = MPa mmm MPa mmm . m Auscu = 6.894 757(25.40) m Auscu = 6.895(25.4) m Auscu ASI = kpsi in For music wire, from Table 10-4: Auscu
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Chapter 1111-1 For the deep-groove 02-series ball bearing with R = 0.90, the design life x D , in multiples of rating life, is xD = The design radial load FD is FD = 1.2(1.898) = 2.278 kN From Eq. (11-6), C10 540 = 2.278 0.02 + 4.439[ln(1/0.9)]1/1.483 =
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Chapter 1212-1 Given dmax = 1.000 in and bmin = 1.0015 in, the minimum radial clearance is cmin = Also 1.0015 - 1.000 bmin - dmax = = 0.000 75 in 2 2l/d = 1 r = 1.000/2 = 0.500 r/c = 0.500/0.000 75 = 667 N = 1100/60 = 18.33 rev/s P = W/(ld) = 250/[(1)(1
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Chapter 1313-1 d P = 17/8 = 2.125 in dG = N2 1120 (2.125) = 4.375 in dP = N3 544NG = PdG = 8(4.375) = 35 teeth Ans. C = (2.125 + 4.375)/2 = 3.25 in Ans. 13-2 n G = 1600(15/60) = 400 rev/min Ans. p = m = 3 mm Ans. C = [3(15 + 60)]/2 = 112.5 mm Ans. 13-3
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Chapter 1414-1 22 N = = 3.667 in P 6 Y = 0.331 d= V = Eq. (14-4b): Kv = Wt = Eq. (14-7): = 14-2 1.96(429.7)(6) Kv W t P = = 7633 psi = 7.63 kpsi Ans. FY 2(0.331) dn (3.667)(1200) = = 1152 ft/min 12 12 1200 + 1152 = 1.96 1200 63 025H 63 025(15) T = = = 42
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Chapter 1515-1 Given: Uncrowned, through-hardened 300 Brinell core and case, Grade 1, NC = 109 rev of pinion at R = 0.999, N P = 20 teeth, NG = 60 teeth, Q v = 6, Pd = 6 teeth/in, shaft angle 90, n p = 900 rev/min, J P = 0.249 and JG = 0.216 (Fig. 15-7),