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UCSD - ECON - 1
January 07, 2009 We learn economic due to scarcity (limited resources) need to make choices and have trade-offs Scarcity principle: theres boundless needs and wants, resources are limited having move of one thing means having less of another t
UCSD - ECON - 1
January 9, 2009 Economist agents are rational: their objective is to make benefits through certain goals Example: cost-benefit principle One listens to a CD, but there are 2 songs that one hates. However, one is comfortably seated and does not wa
UC Irvine - BIOCHEM - 98
Bio 98A Midterm Exam 2/11/08ID # _TA Use OnlyA term used to describe a molecule that contains both a hydrophobic and a hydrophilic region . a b c d e clathrate amphipathic micelle bipolar amphoteric(1) _The predominate form of an important
UC Irvine - BIOCHEM - 98
AB
UC Irvine - BIOCHEM - 98
Bio. 98 Midterm ReviewInstructors: Poulos/Cumsky Peer Tutor: Michael ChauFormat of this sessionThis review session:Focuses on: A few important concepts per lecture Common questions associated with topics Does NOT focus on: Memorization-type
UC Irvine - BIOCHEM - 98
2/9/07Bio 98A MidtermName _score)If one combines 0.1 moles of NaH2PO4 and 0.2 moles of Na2HPO4 in 900 ml of water, the final pH will be _7.16 or 7.2_. -log R = D, so log (0.2/0.1) = 0.301, The basic form prevails around the pKa of 6.86 So
UC Irvine - BIOCHEM - 98
2/1/06 )Bio 98A Midterm ExamName _ (4) _For the following two ligands, what type of forces could be involved in their non-covalent binding to a protein's active site? (do not include van der Waals forces) leucine ionic or charge-charge or elect
Arizona - CHEM - CHEM 151
General Chemistry CHEM 151Week 2UA GenChemWeek 2 Reading AssignmentWeek 2 Section 5.8 Kinetic Molecular theory : a Model for Gases 5.9 Mean Free Path Tools of Chemistry equations and graphs for gases Section 5.1 pressure and Units 5.2 Pressure
Arizona - CHEM - CHEM 151
WELCOME to the UNIVERSITY OF ARIZONA FUNDAMENTALS OF CHEMISTRYCHEM 151 and 103AUA GenChemDr. Wayne WesolowskiDoc Weso Dr. W ProfessorUA GenChemMeet Someone !Share with someone you DONT know. Your name; Your major; Why you are taking
Arizona - CHEM - CHEM 151
The Following statement is from Dr. Anne Padias, Director of Academic Services for the Chemistry Department. The exams for all sections of Chem 151, including the students registered for Chem 103a, are scheduled for the following Mondays at 5 5:50 p
Arizona - CHEM - CHEM 151
General Chemistry CHEM 151Week 4UA GenChemWeek 4 Reading Assignment See D2LUA GenChemCounting ParticlesIs it practical, in a lab situation, to physically COUNT the particles? NO! The particles are too small! So how do we typically measure
Arizona - CHEM - Chem 151
General Chemistry CHEM 151Week 5UA GenChemWeek 5 Reading Assignment See D2L CONTENTUA GenChemUnit 2: Zooming InWhat are they made of? Why are they joined in this way?They could be atoms, molecules, ionsWhat holds them together? Why are
Arizona - CHEM - Chem 151
General Chemistry CHEM 151Week 6UA GenChemPhotoelectron SpectroscopyAnother spectroscopic technique can be used to determine number of electrons with a determined energy in an atom. KEhIonization Energy = h - KE The nth Ionization Energy is
Arizona - CHEM - Chem 151
General Chemistry CHEM 151Week 7UA GenChemElectrons in AtomsElectrons in an atom can be classified into two main groups:CORE electrons: Strong attraction to protons in the nucleus (closer) Unavailable for exchange or bonds. VALENCE electrons:
Arizona - CHEM - Chem 151
General Chemistry CHEM 151Week 8UA GenChemIonic Compounds: PropertiesHard High Melting Very High Boiling Do not Conduct Electricity as solids, but conduct when melted ( ions then can move) Shatter when struck. UA GenChemA Powerful Model
Arizona - CHEM - Chem 151
General Chemistry CHEM 151Week 9UA GenChemREVIEWING CHEMICAL BONDINGIn a chemical reaction between two atoms, their valence electrons are reorganized. Atoms lose or share electrons because that leads to a more stable state (full electron shells
Arizona - CHEM - Chem 151
General Chemistry CHEM 151Week 10UA GenChemMolecular GeometryValence Shell Electron Pair Repulsion Theory The most important factor in determining geometry is the relative repulsion between electron groups around the central atom Molecules adop
Arizona - CHEM - Chem 151
General Chemistry CHEM 151Week 11UA GenChemForces and Particles INTRAMOLECULAR forces WITHIN the molecule (particle) which hold atoms together, determine the electron and molecular geometries. NN has strong intramolecular forces INTERMOLECULA
Arizona - CHEM - Chem 151
General Chemistry CHEM 151Week 12UA GenChemAqueous SolutionsA variety of molecular and ionic compounds are soluble in water. Their solubility depends on the strength of the intermolecular forces between particles in the system. solventA soluti
Cornell - NTRES - 2010
Feeding rate attack rate handling time prey encounter encounters per min0.2 0.5 1 0.08 Change this one, this equals to 5 sec handling time in minutes. 5 seconds is 5/60 minut FR Functional response 1 0.2 0.48 2 0.39 0.92 3 0.57 1.33 4 0.75 1.71 5 0
N.C. State - ECE - 220
iECE220 Homework Assignment #5 SolutionsProblem 7.5 Consider the differential equation dv(t) + 2.0 106 v(t) = 10u(t) dt (0.1)with the initial condition v(0) = 0. (a) Find vc (t). (b) Find vp (t). (c) Find the total solution v(t). (d) Plot all
N.C. State - ECE - 220
iECE220 Homework Assignment #1 SolutionsProblem 2.15. Sketch the following signals v1 (t) = (t)[et + cos(2106 t)] v2 (t) = (t 5) + (t + 5) v3 (t) = (t)e3t [u(t) u(t + 2)] sin(21000t) Solution See gures 0.1 through 0.3. Note the value of the sig
N.C. State - ECE - 220
iECE220 Homework Assignment #2 SolutionsProblem 3.18. For safety reasons, electrical devices limit the output voltage vout (t) to a maximum allowed value A, regardless of the value of the input voltage vin (t). Mathematically, we can represent su
N.C. State - ECE - 220
iECE220 Homework Assignment #3 SolutionsProblem 6.21. Let a and b be real-valued constants. Let 2 x1 2 3a 1 2 2 x2 = b 1 x3 1 34 Determine, if possible, values for a and b that will make the system have no solution (i.e., inconsistent),
N.C. State - ECE - 220
iECE220 Homework Assignment #4 SolutionsProblem 4.33 Write the following signals in phasor notation: v1 (t) = 4 cos(23000t + 4/5) v2 (t) = 60 cos(2400t ) v3 (t) = Re[10ej(21000t/3) ] v4 (t) = 3.4 cos(500t) v5 (t) = 3.5 sin(400t) v6 (t) = 100 v7
N.C. State - ECE - 220
iECE220 Homework Assignment #6 SolutionsProblem 7.10 Consider the differential equation obtained from an RLC circuit d2 v(t) R dv(t) 1 1 + + v(t) = vs (t) (0.1) 2 dt L dt LC LC where R = 10, L = 1, C = 2 and the initial conditions are v(0) = 0, v
N.C. State - ECE - 220
iECE220 Homework Assignment #7 SolutionsProblem 8.1 Find the Laplace transform of v(t) = 4e6t cos(7t)u(t) by direct integration. (Hint: using Eulers identity may be helpful.) Solution We haveV (s) =04e6t cos(7t)u(t)est dt e(s+6)t cos(7t)dt
N.C. State - ECE - 220
iECE220 Homework Assignment #8 SolutionsProblem 9.11 Without using any integrals, determine the Fourier series coefcients, ak , bk and k , of the following functions. Note the frequency that corresponds to each index k. The time variable is measu
N.C. State - ECE - 220
iECE220 Homework Assignment #9 SolutionsProblem 10.1 Using direct integration, nd the Fourier transform of the signal s(t) = e25t cos(2100t + /4)u(t) You can check your result by using the Laplace transform table in Chapter 8. Solution s(t) = e25
N.C. State - ECE - 220
ECE 220 Spring 2008 Problem Lab #08 Solutions* Note, all gures are at the end of the document. I. Laplace transforms, use of MATLAB laplace() functionRead the help le for the MATLAB function laplace(). Question 1. Use laplace() to evaluate the La
N.C. State - ECE - 220
ECE220 Problem Lab #09 Review for Final Exam Date: Week of April 21, 2008The goal of this lab is to review material in preparation for the nal exam. Make sure you also review the problems we covered in the preparation labs for tests 1, 2, and 3. Wo
N.C. State - ECE - 220
ECE220 Problem Lab #3 Getting Practice complex numbers and functions Date: Week of February 5, 2007I. Basic concepts in complex numbers 1. 2+j2Magnitude: Phase:2 + j 2 = 2 2 + 2 2 = 2 2 = 2.82842 + j 2 = / 4 = 0.785Q1 - Show Coordinates2
N.C. State - ECE - 220
Solutions, ECE220 Problem Lab #1 Getting practice with MATLAB, periodic signals and mathematical background skills Date: Week of January 21, 2008The goals of this lab are to: (a) make sure you are up to speed with the required/assumed math backgrou
N.C. State - ECE - 220
Solutions, ECE220 Problem Lab #2 Signal Plots, Vectors, Matrices, and Linear Equations Date: Week of January 28, 2008The main goal of this lab is to give you practice on: (a) signal calculation, processing and plotting, and, (b) vectors, matrices a
N.C. State - ECE - 220
ECE220 Problem Lab #5 Solutions Dierential equations and transfer functions Date: Week of February 25, 2008The goal of this lab is to (a) summarize and reinforce all the methods we presented in chapter 7, about solving rst order dierential equation
N.C. State - ECE - 220
ECE220 Solution Lab #6 Review for Test 2 Date: Week of March 10, 2008The goal of this lab is to review material in preparation for Test 2. Work out as many problems as you can. If you have worked on a problem before, you may skip it. Do not worry i
N.C. State - ECE - 220
ECE 220 Test 3 Solutions14 April 2008Problem 1. (20 points) A system is described by the dierential equation given below: dv(t) + 4v(t) = vs (t) dt where vs (t) is the input. Let vs (t) = 10 cos(5t) and v(0) = 0. Determine the solution in the Lapla
N.C. State - ECE - 220
ECE 220 Lab9 SolutionMade by Rong GuoProblems 7.20 myfunction.m -%Refer to matlab script7.3 on Page381. function yd=myfunction(t,v) A=[3 2;6 0]; b=[0;cos(2*pi*60*t)]; yd=A*v+b; return --pro7x20.m -clear all [t,v]=ode23('myfunction',[0,1],[-1/6;0
N.C. State - ECE - 220
4.384.394.404.42The phase and magnitude plots obtained are as shown:5.105.11C1 = [1 5 6] C2 = Not possible. Matrix dimensions do not agree. Atranspose= [1 02 10 C3= [-2 6 1] C4 = Matrix dimensions do not agree. -1 3 1 1]5.125.205.
N.C. State - ECE - 220
Plot for Problem 7.56 x 10-6Vc(t) Vp(t) V(t) 420-2-4-600.20.40.60.811.2 x 10-5Plots for Problem 7.63 Driving force Particular solution 210-1-2-3012345 Time (t)6789103 yc (t) 2.5 yp(t)
N.C. State - ECE - 220
Name:ID#4Signature2 2_This Exam is "Closed-Book." "Notes," in any form, are NOT allowed. Calculators cannot be used in "complex number mode." Calculators can ONLY be used to perform "real number arithmetic and trig" operations. At the end o
N.C. State - ECE - 220
ECE 220, Section 001 Final Exam Solutions 13 December 2005Problem 1. (10 points) Consider the pulse, p(t), in Equation 1: p(t) = 1, 0 t < 10, 0, otherwise. (1)Express the signal s(t) in Equation 2 in terms of p(t). t < 0, 2, 0 t < 5, s(t) = 1,
N.C. State - ECE - 220
ECE 220, Sections 001, 002, Test 1 Solutions 22 September 2005Problem 1. (15 points) Consider the graph of the signal s(t) shown in gure 1. The graph is accurate. Write down a mathematical expression for this signal, using (time-shifted and scaled,
N.C. State - ECE - 220
ECE 220, section 1 Test 3 Solutions18 November 2004Problem 1. (10 points) Consider the dierential equation 1: d2 v(t) dv(t) d3 v(t) +2 +3 + 4v(t) = cos(2100t)u(t). 3 2 dt dt dt(1)Let v(0) = 0, v(0) = 0, v (0) = 0. Determine the total solution i
N.C. State - ECE - 220
Lab Test Fall 2007I. Complex numbers(25 Points)- jp / 9 1. Let z = 2.75e and w = 1.23 - j 3.62 . Use Matlab to compute 4.01z + w e - j 9002.5 z2 in both Cartesian and exponential formats. l MATLAB program instructions: (4 points) z=2.75*exp(-j*pi/9
N.C. State - ECE - 220
Fall 2007 Lab2 SolutionCreated by Rong Guo4.1> z1=3+5j z1 = 3.0000 + 5.0000i > compass(z1) > z2=10+2j z2 = 10.0000 + 2.0000i > compass(z2) > z3=-3+8j z3 = -3.0000 + 8.0000i > compass(z3) > z4=-1+4j z4 = -1.0000 + 4.0000i > compass(z4)4.2%Four
N.C. State - ECE - 220
ECE220P Fall2007 Lab3 solutionCreated by Rong GuoProblem1 part.1clear all R=250 C=1.5e-6 L=0.65 E=120 w0=1/sqrt(L*C) w=0.1*w0:0.1:10*w0; I=E./(R+j*(w*L-1./(w*C); %Both of magnitude, and frequency axis are in linear scales. subplot(2,1,1) plot(w,a
N.C. State - ECE - 220
10/6/2007Page 1 of 6ECE220 Lab4 SolutionCreated by Rong Guo6.35.Part.1. I1-I4-I5=0 I2+I5-I6=0Use KCL at nodes, and use KVL at Loop1, Loop2, andLoop3, we can get six equations.I3+I6+I4=0 7*I4-5*I6-8*I5=0 4*I2-8*I5-2*I1 =20 4*I2+5*I6-3
N.C. State - ECE - 220
ECE220 LAB5 SolutionsCreated by Rong Guo%Set up the step size h to 0.001, the simulation time now is t=h*N=0.5s clear all v(1)=1; t(1)=0; h=0.001; N=500; for k=1:N t(k+1)=t(k)+h; v(k+1)=v(k)-h*t(k)^3*v(k)+h*10*t(k); end plot(t,v); grid on xlabel('t
N.C. State - ECE - 220
ECE220 LAB6 Solution%Example 7.30 %The array v contains the solution %Set up the initial conditions v(1,1)=1; v(2,1)=0; t(1)=0; %Setup the step size h=0.001; %Set up the recursion for 10000 steps N=10000; %Set up the A matrix. A=[0 1;-7/4 -1/2]; for