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Course: HIS 103, Spring 2008School: University of Toronto
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of Sources British Strength and StaminaPoliticalConstitutional monarchy More difficult during the wartime Voters are merchants Internal stability Raw EconomicColonies materials Taxation Slavery Monopoly Industrial Revolution MilitaryThe Navy Mercenaries ***BRITAIN WAS ON AN ISLAND
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of Sources British Strength and StaminaPoliticalConstitutional monarchy More difficult during the wartime Voters are merchants Internal stability Raw EconomicColonies materials Taxation Slavery Monopoly Industrial Revolution MilitaryThe Navy Mercenaries
***BRITAIN WAS ON AN ISLAND
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University of Toronto - HIS - 103
Charlie Kramer, Student ENG100H1 University of Toronto Toronto, Ontario M1K 2W3 December 6, 2007 Dr. Deirdre Flynn Department of English University of Toronto Toronto, Ontario M4K 1G2 Course Diagnostic Letter:Dear Professor Flynn, My objective in t
University of Toronto - HIS - 103
Office hours: 12-1 SS 0593 New forms of warfare as result of French Revolution:b Total National mobilization Complete destruction of enemy Levee en masse Conscription Warfare Supplies Artillery Aggression 1 veteran with 2 rookies
University of Toronto - HIS - 103
---------20 September 1792 at Valmy field in Eastern France, the artillery in the invading Prussian army, opened a blistering artillery barrage against the The Austro-Prussian invasion was meant to re-establish the King and Queen of
University of Toronto - HIS - 103
History Lecture 12- The War of Austrian Succession and the Diplomatic Revolution - French and British relationship demanded psychological restraint and commitment. - Great powers were inclined to not go to war and towards peace. - Outsiders (countrie
University of Toronto - HIS - 103
------Scaffoldarian party were opposed by the Republicans who opposed war because the SP would gain seats in the Dutch Republic. The internal debate stopped immediately because the policy of neutrality was popular. All the Dutch governmen
University of Toronto - HIS - 103
Bismarck and the States System- 1880-90 Bismarck wanted to make a good treaty with Russia. There was a national community of interest between Prussia and Russia. Both countries had conservative political elites. Bismarck argued the natural congruence
University of Toronto - HIS - 103
History Lecture 13 One permanent feature of the international order was the Hapsburg (Spain and Austrian monarchies) and French confrontations. It was a permanent standing confrontation. Between 1755-56 the permanent feature of the international cons
University of Toronto - HIS - 103
The Origins of the First World War: Part One July 1891- The French naval squadrons sailed into the naval based in St Petersburg, called the Kronstadt Affair. It electrified much of the continent. Charles Freycinet and Ribot sat down and drafted an al
University of Toronto - HIS - 103
History Lecture 17 4 July, an American colonel made a speech on Independence Day, in Paris, on the Champs D'Elysee (main boulevard in Paris). He said very little other than a few niceties. He did say however, "Lafayette was here." It was the French's
University of Toronto - HIS - 103
Britain provided monetary support to their allies. The British lent troops to their allies as well. Industrial Revolution going on in Britain at the same time. Napoleon Bonaparte was a terrible diplomat. Russia was more a rallying point as an ally fo
N. Arizona - SPA - 102
REPASO: CAPITULO 4VOCABULARIONOMBRE_Mariana thinks about mornings in her house. Fill in the blanks with the correct room according to what the person is doing. (pg. 130) Mis padres siempre se levantan temprano. Mam prepara el desayuno en la _1 m
N. Arizona - SPA - 101
Espaol 101 Notas de la clase 28 de agosto de 2007 P.4 Saludos (greetings) Espaol English Equivelant Hola Hello Cmo estas? Cmo estas? How are you? Cmo se llama usted? Cmo te llamas? What is your name? Buenos das Good morning Buenas tardes Good afterno
Temple - ECO - 101
1. Gilligan and Robinson are stranded on a desert island. To feed themselves each day they can either catch fish or pick fruit as specified in the table below. a. Gilligan's opportunity cost of catching/producing a fish is foregone fruit. 2 b. Gillig
Virginia Tech - STAT - 4105
STAT 4105 Exam 1 Study SheetExamplesThere are 8 women and 11 men in a graduating class. Three are chosen to give speeches. What is the probability that all three are women?Urn Problems 1) Ordered Lists with Replacement # Lists = n n n . nk
Embry-Riddle FL/AZ - PS - 150
2.54 cm in . 1 km 10 5 cm 1.61 km 1.1: 1 mi 5280 ft mi 12 in. ft Although rounded to three figures, this conversion is exact because the given conversion from inches to centimeters defines the inch.
Penn State - STAT - 418
Chapter 2Problems1. 2. (a) S = {(r, r), (r, g), (r, b), (g, r), (g, g), (g, b), (b, r), b, g), (b, b)} (b) S = {(r, g), (r, b), (g, r), (g, b), (b, r), (b, g)} S = {(n, x1, ., xn-1), n 1, xi 6, i = 1, ., n - 1}, with the interpretation that the o
N.C. State - CE - 313
1Tension, Compression, and ShearNormal Stress and StrainProblem 1.2-1 A solid circular post ABC (see figure) supports a load P1 2500 lb acting at the top. A second load P2 is uniformly distributed around the shelf at B. The diameters of the upper
N.C. State - CE - 313
SECTION 12.6Polar Moments of Inertia741Polar Moments of InertiaProblem 12.6-1 Determine the polar moment of inertia IP of an isosceles triangle of base b and altitude h with respect to its apex (see Case 5, Appendix D) Solution 12.6-1 Polar mo
N.C. State - CE - 313
32CHAPTER 1Tension, Compression, and ShearProblem 1.6-10 A flexible connection consisting of rubber pads (thickness t 9 mm) bonded to steel plates is shown in the figure. The pads are 160 mm long and 80 mm wide. (a) Find the average shear strai
N.C. State - CE - 313
2Axially Loaded NumbersChanges in Lengths of Axially Loaded MembersProblem 2.2-1 The T-shaped arm ABC shown in the figure lies in a vertical plane and pivots about a horizontal pin at A. The arm has constant cross-sectional area and total weight
N.C. State - CE - 313
80CHAPTER 2Axially Loaded NumbersProblem 2.3-8 A bar ABC of length L consists of two parts of equal lengths but different diameters (see figure). Segment AB has diameter d1 100 mm and segment BC has diameter d2 60 mm. Both segments have length
N.C. State - CE - 313
106CHAPTER 2Axially Loaded NumbersProblem 2.5-3 A rigid bar of weight W 750 lb hangs from three equally spaced wires, two of steel and one of aluminum (see figure). The diameter of the wires is 1/8 in. Before they were loaded, all three wires h
N.C. State - CE - 313
122CHAPTER 2Axially Loaded NumbersStresses on Inclined SectionsProblem 2.6-1 A steel bar of rectangular cross section (1.5 in. 2.0 in.) carries a tensile load P (see figure). The allowable stresses in tension and shear are 15,000 psi and 7,000
Delaware - MATH - 241
Delaware - MATH - 241
Delaware - MATH - 241
Virginia Tech - ESM - 2204
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Virginia Tech - ACIS - 2115
Chapter 3Systems Design: Job-Order CostingSolutions to Questions3-1 By definition, overhead consists of costs that cannot practically be traced to products or jobs. Therefore, if they are to be assigned to products or jobs, overhead costs must be
Ohio State - EE - 300
G. Rizzoni, Principles and Applications of Electrical EngineeringProblem solutions, Chapter 2Chapter 2 Instructor NotesChapter 2 develops the foundations for the first part of the book. Coverage of the entire Chapter would be typical in an intro
Ohio State - EE - 300
G. Rizzoni, Principles and Applications of Electrical EngineeringProblem solutions, Chapter 3Chapter 3 Instructor NotesChapter 3 presents the principal topics in the analysis of resistive (DC) circuits. The presentation of node voltage and mesh
Ohio State - EE - 300
G. Rizzoni, Principles and Applications of Electrical EngineeringProblem solutions, Chapter 4Chapter 4 Instructor NotesThe chapter starts by developing the dynamic equations for energy storage elements. The analogy between electrical and hydraul
Ohio State - EE - 300
G. Rizzoni, Principles and Applications of Electrical EngineeringProblem solutions, Chapter 5Chapter 5 Instructor NotesChapter 5 has been reorganized in response to a number of suggestions forwarded by users of the third edition of this book. Th
Virginia Tech - CEE - 3104
INSTRUCTOR'S MANUAL To Accompany ENGINEERING MECHANICS - DYNAMICS Volume 2 Fifth Edition, 2002 J. L. Meriam and L. G. Kraige Copyright 2002 by John Wiley & Sons, Inc.USE OF THE INSTRUCTOR'S MANUAL The problem solution portion of this manual has be
Penn State - LING - 102
DANTE: Catalyst to Italian StandardizationLING 102 Final - Fall, 2007Alex Caire-Broudy December 21, 20070DANTE: Catalyst to Italian StandardizationDante Alighieri was the main propulsor of Italian standardization in Late Latin's decay and i
UCLA - MATH - 32B
CHAPTER 1Section 1. Differential Equation Models1.1. Let y(t) be the number of bacteria at time t. The rate of change of the number of bacteria is y (t). Since this rate of change is given to be proportional to y(t), the resulting differential equ
University of Texas - EE - 411
Chapter 1, Solution 1 (a) q = 6.482x1017 x [-1.602x10-19 C] = -0.10384 C (b) q = 1. 24x1018 x [-1.602x10-19 C] = -0.19865 C (c) q = 2.46x1019 x [-1.602x10-19 C] = -3.941 C (d) q = 1.628x1020 x [-1.602x10-19 C] = -26.08 C Chapter 1, Solution 2 (a) (b)
University of Texas - EE - 411
Chapter 2, Solution 1 v = iR i = v/R = (16/5) mA = 3.2 mAChapter 2, Solution 2 p = v2/R Chapter 2, Solution 3 R = v/i = 120/(2.5x10-3) = 48k ohms Chapter 2, Solution 4 (a) (b) i = 3/100 = 30 mA i = 3/150 = 20 mA R = v2/p = 14400/60 = 240 ohmsCha
University of Texas - EE - 411
Chapter 3, Solution 1. v18 2 40 v26A10 AAt node 1, 6 = v1/(8) + (v1 - v2)/4 At node 2, v1 - v2/4 = v2/2 + 10 Solving (1) and (2), v1 = 9.143V, v2 = -10.286 V2 v1 (9.143)2 P8 = = = 10.45 W 8 848 = 3v1 - 2v2(1)40 = v1 - 3v2(2)P4 =
University of Texas - EE - 411
Chapter 4, Solution 1. 1 i 5 io1V+ -838 (5 + 3) = 4 , i =1 1 = 1+ 4 5io =1 1 i= = 0.1A 2 10Chapter 4, Solution 2.6 (4 + 2) = 3, i1 = i 2 =1 A 25 4io =1 1 i1 = , v o = 2i o = 0.5V 2 4i1ioi2 1A 8 6 2If is = 1A, then vo
University of Texas - EE - 411
Chapter 5, Solution 1. (a) (b) (c) Rin = 1.5 M Rout = 60 A = 8x104 Therefore AdB = 20 log 8x104 = 98.0 dBChapter 5, Solution 2. v0 = Avd = A(v2 - v1) = 105 (20-10) x 10-6 = 0.1V Chapter 5, Solution 3. v0 = Avd = A(v2 - v1) = 2 x 105 (30 + 20) x 10
University of Texas - EE - 411
Chapter 6, Solution 1.i=Cdv = 5 2e -3t - 6 + e -3 t = 10(1 - 3t)e-3t A dt()p = vi = 10(1-3t)e-3t 2t e-3t = 20t(1 - 3t)e-6t W Chapter 6, Solution 2.1 2 1 Cv1 = (40)(120) 2 2 2 1 2 1 w2 = Cv1 = (40)(80) 2 2 2 w1 = w = w 1 - w 2 = 20(120 2
University of Texas - EE - 411
Chapter 7, Solution 1. Applying KVL to Fig. 7.1. 1 t i dt + Ri = 0 C - Taking the derivative of each term, i di +R =0 C dt di dt or =- i RCIntegrating, i( t ) - t = ln I 0 RCi( t ) = I 0 e - t RC v( t ) = Ri( t ) = RI 0 e - t RC or v(t ) = V0
University of Texas - EE - 411
Chapter 8, Solution 1. (a) At t = 0-, the circuit has reached steady state so that the equivalent circuit is shown in Figure (a).6 VS+ -6 +6+ vL 10 H (b) v -10 F -(a)i(0-) = 12/6 = 2A, v(0-) = 12V At t = 0+, i(0+) = i(0-) = 2A, v(0+) =
University of Texas - EE - 411
Chapter 9, Solution 1. (a) (b) angular frequency frequency f = = 103 rad/s = 159.2 Hz 2(c)periodT =1 = 6.283 ms f(d)Since sin(A) = cos(A 90), vs = 12 sin(103t + 24) = 12 cos(103t + 24 90) vs in cosine form is vs = 12 cos(103t 66) V
University of Texas - EE - 411
Chapter 10, Solution 1.=110 cos( t - 45) 10 - 45 5 sin( t + 30) 5 - 601H 1F jL = j 1 = -j jCThe circuit becomes as shown below.3 Vo j10-45 V+ -2 Io+ -5-60 VApplying nodal analysis, (10 - 45) - Vo (5 - 60) - Vo Vo + = 3 j
University of Texas - EE - 411
Chapter 11, Solution 1.v( t ) = 160 cos(50t ) i( t ) = -20 sin(50t - 30) = 2 cos(50t - 30 + 180 - 90) i( t ) = 20 cos(50t + 60) p( t ) = v( t ) i( t ) = (160)(20) cos(50t ) cos(50t + 60) p( t ) = 1600 [ cos(100 t + 60) + cos(60) ] W p( t ) = 800 +
University of Texas - EE - 411
Chapter 12, Solution 1. (a) If Vab = 400 , then Van = 400 3 - 30 = 231 - 30 VVbn = 231 - 150 V Vcn = 231 - 270 V(b) For the acb sequence, Vab = Van - Vbn = Vp 0 - Vp 120 1 3 Vab = Vp 1 + - j = Vp 3 - 30 2 2 i.e. in the acb sequence, Vab la
University of Texas - EE - 411
Chapter 14, Solution 1.Vo R jRC = = Vi R + 1 jC 1 + jRCj 0 , 1 + j 0H () =H () =where 0 =1 RCH = H () = 0 1 + ( 0 ) 2 = H () = - tan -1 2 0 This is a highpass filter. The frequency response is the same as that for P.P.14.1
University of Texas - EE - 411
Chapter 15, Solution 1.e at + e - at 2 1 1 1 s L [ cosh(at ) ] = + = s2 - a2 2s - a s + a (a)cosh(at ) =(b)sinh(at ) =e at - e - at 2 1 1 1 a L [ sinh(at ) ] = - = s2 - a2 2s - a s + a Chapter 15, Solution 2.(a)f ( t ) = cos(t
University of Texas - EE - 411
Chapter 16, Solution 1. Consider the s-domain form of the circuit which is shown below. 1 I(s)1/s+ -1/ssI(s) = 1s 1 1 = 2 = 1 + s + 1 s s + s + 1 (s + 1 2) 2 + ( 3 2) 2i( t ) =2 3 e - t 2 sin 2 t 3 i( t ) = 1.155 e -0.5t sin (0.
University of Texas - EE - 411
Chapter 17, Solution 1. (a) This is periodic with = which leads to T = 2/ = 2.(b) y(t) is not periodic although sin t and 4 cos 2t are independently periodic. (c) Since sin A cos B = 0.5[sin(A + B) + sin(A B)], g(t) = sin 3t cos 4t = 0.5[sin 7t