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FIRST PAGES Chapter 5 5-1 MSS: σ1 − σ3 = Sy/n ⇒ n = Sy σ1 − σ3 DE: n = Sy σ ′ σ ′ = (σ 2A − σAσB + σ 2B)1/2 = (σ 2x − σxσy + σ 2y + 3τ 2xy)1/2…
Chapter 9, Problem 1. Given the sinusoidal voltage v(t) = 50 cos (30t + 10 o ) V, find: (a) the amplitude V m ,(b) the period T, (c) the frequency f, and (d) v(t) at t =…
Chapter 13 13.1 Show that the 1948 Hill criterion and flow rules predict an angular variation of R as: € Rθ= H+(2N−F−G−4H)sin2θcos2θ Fsin2θ+Gcos2θ . Solution:…
1 Problem Solutions for Chapter 2 2-1. E · 100cos 2π10 8 t + 30° ( ) e x + 20 cos 2π10 8 t − 50° ( ) e y + 40cos 2π10 8 t + 210° ( ) e z 2-2. The general form is:…
Concise Complex Analysis Solution of Exercise Problems Ai Shu Xue March 9, 2008 1 Contents 1 Calculus 3 2 Cauchy Integral Theorem and Cauchy Integral Formula 14 3 Theory…
1 Problem Solutions for Chapter 2 2-1. E · 100cos 2π10 8 t + 30° ( ) e x + 20 cos 2π10 8 t − 50° ( ) e y + 40cos 2π10 8 t + 210° ( ) e z 2-2. The general form is:…
C H A P T E R 1 Cameras PROBLEMS 1.1. Derive the perspective equation projections for a virtual image located at a distance f in front of the pinhole. Solution We write again…
Gerald W. Recktenwald Overview • Preliminary considerations and bracketing. • Roots of Polynomials Example: Picnic Table Leg Example: Picnic Table Leg Computing
Convex Analysis and Optimization Chapter 1 Solutions Dimitri P. Bertsekas with Angelia Nedi´c and Asuman E. Ozdaglar Massachusetts Institute of Technology Athena Scientific,…
FIRST PAGES Chapter 7 7-1 a DE-Gerber Eq 7-10: A = {4226002 + 3184002}12 = 2920 lbf · in B = {4225002 + 3183002}12 = 2391 lbf · in d = 822920π30 000 1…
APPLIED ANALYSES IN GEOTECHNICS: SOLUTIONS MANUAL 1 APPLIED ANALYSES IN GEOTECHNICS Fethi Azizi SOLUTIONS MANUAL [email protected] Non fissured rock h = 14m 2m 2 m 2…
14-1 Chapter 14 14.1 Let GOL(jωc) = R + jI where ωc is the critical frequency. Then, according to the Bode stability criterion | GOL(jωc)| = 1 = 22 IR + ∠GOL(jωc) =…
1 © John Bird & Carl Ross Published by Taylor and Francis CHAPTER 1 REVISIONARY MATHEMATICS EXERCISE 1, Page 4 1. Convert the following angles to degrees correct to…
1 Problem Solutions for Chapter 2 2-1. E · 100cos 2π10 8 t + 30° ( ) e x + 20 cos 2π10 8 t − 50° ( ) e y + 40cos 2π10 8 t + 210° ( ) e z 2-2. The general form is:…
5.3 (a) Looking into the base of Q 1 we see an equivalent resistance of r π1 , so we can draw the following equivalent circuit for finding R in : R in R 1 R 2 r π1 R in…
3.1 (a) IX = VX R1 0 VX < 0 VX > 0 IX VX (V) Slope = 1/R1 3.2 IX = Plotting IX (t), we have VX R1 0 VX < 0 VX > 0 V0 IX (t) for VB = 1 V (Solid) 0 0 −V0 /R1…
4.4 According to Equation (4.8), we have IC = AE qDn n2 VBE /VT i −1 e NB WB 1 ∝ WB We can see that if WB increases by a factor of two, then IC decreases by a factor…
3.1 (a) IX = { VX R1 VX < 0 0 VX > 0 VX (V) IX Slope = 1/R1 3.2 IX = { VX R1 VX < 0 0 VX > 0 Plotting IX(t), we have 0 −V0/R1I X (t ) fo r V B = 1 V (S ol id…
Chapter 3, Solution 1. 8 Ω v2 2 Ω 40 Ω v1 6 A 10 A At node 1, 6 = v1/(8) + (v1 - v2)/4 48 = 3v1 - 2v2 (1) At node 2, v1 - v2/4 = v2/2 + 10 40 = v1 - 3v2 (2) Solving…
Chapter 2 Density, Specific Gravity, Specific Weight 1. What is the specific gravity of 38◦API oil? 38◦API oil sp.gr. = 141.5 131.5+◦API = 141.5 131.5 + 38 sp. gr.…