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Chapter 12

Course: MGMT 370, Spring 2008
School: Iowa State
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12 Chapter 1. The nature of Organization Design a. Organization design- The overall set of structural elements and the relationships among those elements used to manage the total organization. 2. Universal Perspectives on Organization Design a. Bureaucratic Model i. Bureaucracy- A model of organization design based on a legitimate and formal system of authority ii. Examples 1. Government Agencies 2. Universities...

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12 Chapter 1. The nature of Organization Design a. Organization design- The overall set of structural elements and the relationships among those elements used to manage the total organization. 2. Universal Perspectives on Organization Design a. Bureaucratic Model i. Bureaucracy- A model of organization design based on a legitimate and formal system of authority ii. Examples 1. Government Agencies 2. Universities 3. Situational Influences on organization design a. Behavioral Model i. Behavioral Model- A model of organization design consistent with the human relations movement and stressing attention to developing work groups and concern with interpersonal processes ii. System 1 design1. Predetermined information 2. Follow a formal prescribed conduit Thou 3. shalt not deviate iii. System 4 design 1. Tell me exactly what you are trying to do 2. Here's what I think you need and why you need it 3. Let's try this 4. So what!?!? b. Environment i. Mechanist organization- Similar to the bureaucratic or system 1 mode, most frequently found in stable environments ii. Organic Organization- Very flexible and informal model of organizational design, most often found in unstable and unpredictable environments 1. Very hard to determine chart, tons of things going on. c. The learning organization i. Learning organization- One that works to facilitate the lifelong learning and personal development of all its employees while continually transforming itself to respond to changing demands and needs.
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Iowa State - MGMT - 370
Chapter 13 1. The nature of organization change i. Organization change- Any substantive modification to some part of the organization b. Changing Technology and operations i. Enterprise resource planning- A large-scale information system for integrat
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1.4:g 11.3 cm31 kg 1000 g100 cm 1m31.13 10 4kg . m3
Texas A&M - PHYS - 218
1.5:327 in 32.54 cm in31 L 1000 cm 35.36 L.
Texas A&M - PHYS - 218
1.6:1 m31000 L 1 m3 2111.9 bottles1 gal 128 oz. 3.788 L 1 gal 2112 bottles1 bottle . 16 oz.The daily consumption must then be 1 yr bottles 2.11 103 yr 365.24 da5.78bottles . da
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1.10: a) 60mi hr ft s21h 3600 s 30.48 cm 1ft5280 ft 1 mi 1m 100 cm88ft s m s2b) 329.8g c) 1.0 3 cm100 cm 1m31 kg 1000 g103kg m3
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TEXAS A&M UNIVERSITY ENGR221/502 Introduction to Engineering Mechanics Fall 2005Final ExamDecember 9, 3-5 pm, 2005Closed books and closed notes. Needed formula sheets are provided in the exam. This exam has 5 questions for a total of 100 points.
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Chapter 2CHAPTER 2 - Describing Motion: Kinematics in One Dimension 1. We find the time from average speed = d/t; 15 km/h = (75 km)/t , which givest = 5.0 h. 88 km/h.2. 3.We find the average speed from average speed = d/t = (280 km)/(3.2 h) =
Texas A&M - PHYS - 208
Chapter 3CHAPTER 3 - Kinematics in Two Dimensions; Vectors 1. We choose the west and south coordinate system shown. D 2x For the components of the resultant we have W RW = D1 + D2 cos 45 D 2y D 2 = (200 km) + (80 km) cos 45 = 257 km; RS = 0 + D2 =
Texas A&M - PHYS - 208
Chapter 41CHAPTER 4 - Dynamics: Newton's Laws of Motion 1. We convert the units: # lb = (0.25 lb)(4.45 N/lb) ~ 1 N. If we select the bike and rider as the object, we apply Newton's second law to find the mass: ?F = ma; 255 N = m(2.20 m/s2), which
Texas A&M - PHYS - 208
Chapter 5CHAPTER 5 - Further Applications of Newton's Laws 1. The friction is kinetic, so Ffr = kFN. With constant velocity, the acceleration is zero. Using the force diagram for the crate, we can write ?F = ma: x-component: F kFN = 0; y-component
Texas A&M - PHYS - 208
Chapter 6CHAPTER 6 - Gravitation and Newton's Synthesis 1. Because the spacecraft is 2 Earth radii above the surface, it is 3 Earth radii from the center. The gravitational force on the spacecraft is F = GMEM/r2 = (6.67 1011 N m2/kg2)(5.98 1024 kg
Texas A&M - PHYS - 208
Chapter 7CHAPTER 7 - Work and Energy 1. 2. 3. The displacement is in the direction of the gravitational force, thus W = Fh cos 0 = mgh = (250 kg)(9.80 m/s2)(2.80 m) = 6.86 103 J. The displacement is opposite to the direction of the retarding force,
Texas A&M - PHYS - 208
Chapter 8CHAPTER 8 - Conservation of Energy 1. The potential energy of the spring is zero when the spring is not compressed (x= 0). For the stored potential energy, we have U = ! kxf2 0; 35.0 J = ! (82.0 N/m)xf2 0, which gives xf = 2. 3. 4. For t
Texas A&M - PHYS - 208
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Texas A&M - PHYS - 208
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PROBLEM 1.1 KNOWN: Heat rate, q, through one-dimensional wall of area A, thickness L, thermalconductivity k and inner temperature, T1. FIND: The outer temperature of the wall, T2. SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction in the x-dire
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PROBLEM 1.41KNOWN: Hot plate-type wafer thermal processing tool based upon heat transfer modes by conduction through gas within the gap and by radiation exchange across gap. FIND: (a) Radiative and conduction heat fluxes across gap for specified hot
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Texas A&M - ELEN - 214
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Texas A&M - ELEN - 214
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7 Response of First-Order RL and RCCircuitsAssessment ProblemsAP 7.1 [a] The circuit for t < 0 is shown below. Note that the inductor behaves like a short circuit, effectively eliminating the 2 resistor from the circuit.First combine the 30 an
Texas A&M - ELEN - 214
8 Natural and Step Responses of RLCCircuitsAssessment ProblemsAP 8.1 [a] 1 1 = , (2RC)2 LC 1 , [b] = 5000 = 2RC s1,2 = -5000 [c] 1 = 20,000, LC therefore C = 500 nF C = 1 Ftherefore 25 106 - therefore(103 )(106 ) = (-5000 j5000) rad/s 20
Texas A&M - ELEN - 214
Sinusoidal Steady State Analysis9Assessment ProblemsAP 9.1 [a] V = 170/-40 V [b] 10 sin(1000t + 20 ) = 10 cos(1000t - 70 ) . I = 10/-70 A[c] I = 5/36.87 + 10/-53.13 = 4 + j3 + 6 - j8 = 10 - j5 = 11.18/-26.57 A [d] sin(20,000t + 30 ) = cos(20,0
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11 Balanced Three-Phase CircuitsAssessment ProblemsAP 11.1 Make a sketch:We know VAN and wish to find VBC . To do this, write a KVL equation to find VAB , and use the known phase angle relationship between VAB and VBC to find VBC . VAB = VAN + VN
Texas A&M - ELEN - 214
12 Introduction to the Laplace TransformAssessment ProblemsAP 12.1 [a] cosh t = et + e-t 2 Therefore, 1 -(s-)t [e + e-(s+)t ]dt L{cosh t} = 2 0- = = [b] sinh t = 1 e-(s-)t 2 -(s - ) 1 2 0-+e-(s+)t -(s + ) = s2 0-1 1 + s- s+s - 2et - e
Texas A&M - ELEN - 214
13 The Laplace Transform in CircuitAnalysisAssessment ProblemsAP 13.1 [a] Y = 1 1 C[s2 + (1/RC)s + (1/LC) + + sC = R sL s 1 = 25 108 LC106 1 = = 80,000; RC (500)(0.025) Therefore Y = [b] -z1,225 10-9 (s2 + 80,000s + 25 108 ) s = -40,000 1
Texas A&M - ELEN - 214
14Introduction to Frequency-Selective CircuitsAssessment ProblemsAP 14.1 fc = 8 kHz, c = 1 ; RC c = 2fc = 16 krad/s R = 10 k;. C =1 1 = = 1.99 nF c R (16 103 )(104 )AP 14.2 [a] c = 2fc = 2(2000) = 4 krad/s L= 5000 R = = 0.40 H c 4000 4000
Texas A&M - ELEN - 214
Active Filter Circuits15Assessment ProblemsAP 15.1 H(s) = -(R2 /R1 )s s + (1/R1 C) R1 = 1 , . C = 1 F1 = 1 rad/s; R1 C R2 = 1, R1 .. R2 = R1 = 1 -s s+1Hprototype (s) =AP 15.2 H(s) =-20,000 -(1/R1 C) = s + (1/R2 C) s + 5000 C = 5 F1
Texas A&M - ELEN - 214
Fourier Series16Assessment ProblemsAP 16.1 av = 1 T 2 T2T /3 0Vm dt +1 TT 2T /3Vm 3T7 dt = Vm = 7 V 9 Vm cos k0 t dt 3ak = = bk = =2T /3 0Vm cos k0 t dt + sin 4k 3 =2T /34Vm 3k0 T 2 T2T /3 06 4k sin k 3T 2T /3Vm sin k
Texas A&M - ELEN - 214
The Fourier Transform17Assessment ProblemsAP 17.1 [a] F () = =0 - /2(-Ae-jt ) dt + /2 0Ae-jt dtA [2 - ej /2 - e-j /2 ] j ej /2 + e-j /2 2A 1- = j 2 -j2A [1 - cos ] = 2 [b] F () = AP 17.2 f (t) = = = = 1 2 0te-at e-jt dt = 4ejt d +