UC Irvine - PHYS - 47020

Bucknell - PHYS - 212

Solutions for Hand-In Set #1 PHYS 212 - Spring 2007Sketch Force Diagram for Left Ballon (Blue)Problem A2 - Repulsive BalloonsBlew up the balloons (i.e., inflated them - I didn't use any explosives or anything) and tied them to the ends of a piec

Bucknell - PHYS - 212

Solutions for Hand-In Set #2 PHYS 212 - Spring 2007Problem A5 (Problem 2 on E-fields program)The signs are (1) positive; (2) negative; (3) negative; (4) negative; (5) positive. The way to figure this out: bring the positive test charge very close

Bucknell - PHYS - 212

Solutions for Hand-In Set #3 PHYS 212 - Spring 2007Problem A15 - Mapping the field of a permanent magnetor vf = 2IHBL . m(a) Sketch of a bar magnet with direction of magnetic field around the magnet:Comment: The solution given above makes it

Bucknell - PHYS - 212

Solutions for Hand-In Set #4 PHYS 212 - Spring 2007Problem A21- An electromagnetTipler 27-54Tried this out with two batteries in series - it worked very well. I first tried touching the two nails together before applying the current - there was

Bucknell - PHYS - 212

Review Sheet for Hour-Test I - PHYS 212 NOTE: This review sheet is NOT a substitute for doing the assigned and hand-in problems and drills. You CAN'T expect to do well on the test unless you successfully complete all of the problems, including the EP

Bucknell - PHYS - 212

Solutions for Hand-In Set #6 PHYS 212 Spring 2007A32 - Traveling Waves on a Magic SpringA35 - Annoying your roommateFor the longitudinal wave, I found that the pulse did about 5 full round trips in about 4.4 seconds. (It was a bit difficult to

Bucknell - PHYS - 212

Review Sheet for Hour-Test II - PHYS 212, Spring 2007 NOTE: This review sheet is NOT a substitute for doing the assigned problems, hand-in problems, reading quizzes and drills. You CAN'T expect to do well on the test unless you successfully complete

Bucknell - PHYS - 212

Bucknell - PHYS - 212

Bucknell - PHYS - 212

Solutions for Hand-In Set #9 PHYS 212 Spring 2007A62 - Uncertainty(c) For the second trial solution we have 2 (x) = B sin kx d2 = Bk cos kx dx d 2 1 = -Bk 2 sin kx. dx2 Plugging these into the rearranged Schrdinger equation o gives -Bk 2 sin kx

Bucknell - PHYS - 212

Solutions for Hand-In Set #10 PHYS 212 Spring 2007Supplementary Reading 1.4Supplementary Reading 1.10When you plot the square of the wavefunction for the ground state you see a broad "bump" between 0 and L. From this graph you see that the par

Bucknell - PHYS - 212

Solutions for Hand-In Set #12 PHYS 212 Spring 2007Supplemental Reading - 5.4Supplementary Reading - 5.12To complete this problem, use the same steps as in 5.3, but in reverse: use "distance = speed time" (with c as a speed) to get an estimate

Bucknell - PHYS - 212

Solutions for Hand-In Set #12 PHYS 212 Spring 2007A81 - Flipping inside atomsThe energy of the photon is equal to the energy difference between the two energy levels of the antiparallel spin state and the parallel spin state. The energy of a pho

Bucknell - PHYS - 212

Solutions for Hand-In Set #12 PHYS 212 Spring 2007A81 - Flipping inside atomsThe energy of the photon is equal to the energy difference between the two energy levels of the antiparallel spin state and the parallel spin state. The energy of a pho

Pittsburgh - ENGR - 0031

ProblemsSection 1-2 Electric Circuits and Current Flow P1.2-1i (t ) =P1.2-2d 4 1 - e-5t = 20 e-5t A dt()t t t t 4 4 q ( t ) = i ( ) d + q ( 0 ) = 4 1 - e -5 d + 0 = 4 d - 4 e -5 d = 4 t + e -5t - C 0 0 0 0 5 5()P1.2-3 q ( t ) =

Pittsburgh - ENGR - 0031

Section 1-3 Systems of Units P1.3-1q = i t =( 4000 A )( 0.001 s )= 4CP1.3-2 i= q 45 10-9 = = 9 10-6 = 9 A t 5 10-3P1.3-3 electron C -19 i = 10 billion 1.602 10 electron = s C 9 electron -19 1010 1.602 10 electron s elect

Pittsburgh - ENGR - 0031

Section 1-5 Power and Energy P1.5-1 a.) q =b.) P = v i = (110 V )(10 A ) = 1100 W c.) Cost = P1.5-2 i dt = it = (10 A )( 2 hrs )( 3600s/hr ) = 7.2100.06$ 1.1kW 2 hrs = 0.132 $ kWhr4CP = ( 6 V )(10 mA ) = 0.06 W t = 200 Ws w = = 3.33103 s

Pittsburgh - ENGR - 0031

Section 1.8 How Can We Check.? P 1.8-1 Notice that the element voltage and current of each branch adhere to the passive convention. The sum of the powers absorbed by each branch are: (-2 V)(2 A)+(5 V)(2 A)+(3 V)(3 A)+(4 V)(-5 A)+(1 V)(5 A) = -4 W + 1

Pittsburgh - ENGR - 0031

ProblemsSection 2-2 Engineering and Linear Models P2.2-1 The element is not linear. For example, doubling the current from 2 A to 4 A does not double the voltage. Hence, the property of homogeneity is not satisfied. P2.2-2 (a) The data points do ind

Oregon State - ECON - 201

chapter1> FirstPrinciplesSection 1: Individual Choice: The Core of EconomicsIndividual choice is the decision by an individual of what to do, which necessarily involves a decision of what not to do.Every economic issue involves, on its most

Oregon State - ECON - 201

chapter1> FirstPrinciplesSection 2: Interaction: How Economies WorkAs we learned in the Introduction, an economy is a system for coordinating the productive activities of many people. In a market economy, such as the one we live in, that coor

Oregon State - PHL - 280

Writing Philosophy Papers: A Student GuideWelcome toWriting Philosophy Papers: A Student GuideDepartment of Philosophy Oregon State Universitycopyright 1997 by Department of Philosophy, Oregon State University. Reprinted with permission.file

Clarkson - ES - 220

Stevens - MA - 115

MA115 Mathematical Analysis I Test 1Instructions The test consists of two sections, Part 1: Questions 1 6 Part 2: Questions 7 10 You are to do all of Part 1. For Part 2, select and complete 2 of the 4 problems given. Please clearly indicate which

Stevens - MA - 115

MA115 Mathematical Analysis I Test 2Instructions The test consists of three sections, Part 1: Questions 1 6 Part 2: Questions 7 10 Part 3: Bonus Question You are to do all of Part 1. For Part 2, select and complete 2 of the 4 problems given. Plea

Stevens - MA - 115

MA115 Mathematical Analysis I Test 3 Answer KeyInstructions The test consists of three sections, Part 1: Questions 1 6 Part 2: Questions 7 10 You are to do all of Part 1. For Part 2, select and complete 2 of the 4 problems given. Please clearly i

Stevens - MA - 115

Ma 115 Homework Solutions for Week 1 1.1 #23f (x) = 4 + 3x - x2 f (3 + h) = 4 + 3(3 + h) - (3 + h)2 = 4 + 9 + 3h - (9 + 6h + h2 ) = 4 - 3h - h2 (4-3h-h2 -4) f (3+h)-f (3) = = h(-3-h) = -3 - h h h h 1.1 f (x) = #285x+4 x2 +3x+2is defined for al

Johns Hopkins - MATH - 110.109

Johns Hopkins - MATH - 110.109

Johns Hopkins - MATH - 110.109

Stevens - MA - 115

Ma 115 Homework Solutions for Week 2 2.2 #4x0(a) lim f (x) = 3 (c) lim f (x) = 2 +x3(b) lim f (x) = 4 -x3(d) lim f (x) does not exist because the limits in (b) and (c) are not equal.x3(e) f (3) = 32.2xa#6lim f (x) exists for all a

Stevens - CS - 181

Programmer Defined Types - Classes and Objects1Programmer Defined Types - Classes and ObjectsObject oriented programming (OOP) revolves around the notion of an object. Objects are created using programmer defined types, called classes. In C+, a

Stevens - MA - 115

MA 115 Homework Solutions for Week 3 2.7 #4(a) Since g(5) = -3, the point (5, -3) is on the graph of g. Since g (5) = 4, the slope of the tangent line at x = 5 is 4. Using the point-slope form of a line gives us y - (-3) = 4(x - 5), or y = 4x - 23

Stevens - MA - 115

MA 115 Homework Solutions for Week 43.5#2 u = u1/2 .Let u = g(x) = 4 + 3x and y = f (u) = Thendy dx=dy du du dx1 = 2 u-1/2 (3) =3 2 u=3 . 2 4+3x3.5#6Let u = g(x) = ex and y = f (u) = sin u. Thendy dx=dy du du dx= (co

Stevens - MA - 115

MA 115 Homework Solutions for Week 5 4.2 #6Absolute maximum value is f (8) = 5; absolute minimum value is f (2) = 0; local maximum values are f (1) = 2, f (4) = 4, and f (6) = 3; local minimum values are f (2) = 0, f (5) = 2, and f (7) = 1.4.2#

Stevens - MA - 115

MA 115 Homework Solutions for Week 6 4.9 #12 x +x+1 1 = x + 1 + F (x) = f (x) = x x 4.921 2 2x 1 2 2x+ x + ln |x| + C1 + x + ln |x| + C2if x < 0 if x > 0#19 f (x) = x(6 + 5x) = 6x1/2 + 5x3/2 f (x) = 4x3/2 + 2x5/2 + C f (1) = 6 + C

Stevens - MA - 115

MA 115 Homework Solutions for Week 7 5.5 #10 xex dx =2Let u = x2 . Then du = 2x, so 5.51 21 eu ( 2 du) = 1 eu + C = 1 ex + C 2 22#201 2 du, soLet u = 2. Then du = 2d and d = sec 2 + C #22sec 2 tan 2d =sec u tan u( 1 du) = 21 2sec

Stevens - MA - 115

MA 115 Homework Solutions for Week 8 5.7 #18 A B x-1 = + . Multiply both sides by (x + 1)(x + 2) to get x - 1 = A(x + 2) + B(x + 1). x2 + 3x + 2 x+1 x+2 Substituting -2 for x gives -3 = -B B = 3. Substituting -1 for x gives -2 = A. Thus,1 0x-1 dx

Stevens - MA - 115

C+ Basics - Part 11C+ BasicsInteger data types In our first program the data type of the variables we have used is int. Thus we can store whole numbers in these variables. In this part, we will discuss: two other integer types long and short fo

Stevens - CS - 181

C+ Basics - Part 21More C+ BasicsFloating-point types As mentioned before, the double data type provides 15 digits of precision. For even greater precision, upto 19 digits, the data type long double can be used. The float data type provides the

Stevens - CS - 181

Control Structures - Selection Statements1Control Structures: Selection StatementsAny C+ program can be written using only the following three control structures. (1) sequence (2) selection (3) repetition Sequential statements: These statements

Stevens - CS - 181

Control Structures - Repetition Statements1Control Structures: Repetition StatementsMost programs require some action that is repeated a number of times. For example, our rectangle area.cpp program computes the area once per one run of the progr

Stevens - CS - 181

Predefined Functions1Predefined FunctionsThe programs that we have seen so far are very simple ones. In larger and more complicated programs, we have to divide the problem into smaller subtasks and then implement each subtask as a function. Inst

Stevens - CS - 181

Basic File I/O1Basic File I/OWe discussed file output briefly before. Now we describe more C+ file input and output features that are very useful. To begin with, consider the following rectangle area program discussed before. The program sends s

Stevens - CS - 181

Value Returning Functions1Programmer Defined Functions - Part 1 Value Returning FunctionsThe C+ library provides a number of useful predefined functions. However, in many situations we may have to write our own functions. Writing a function invo

Stevens - CS - 181

Void Functions1Programmer Defined Functions - Part 2 Void FunctionsThe functions we discussed so far have returned a value. Some times, we would like to write functions that perform subtasks other than returning a value. In such cases, we can us

Stevens - CS - 181

Argument Passing by Reference1Programmer Defined Functions - Part 3 Argument Passing by ReferenceSo far in our example programs, when a function call is made, even if a variable is supplied as an argument, the function takes only the value of th

Stevens - CS - 181

Function Overloading & Default Parameters1Overloading Function Names & Functions with Default ParametersOverloading function names In C+, we can use the same function name for two or more functions. Using the the same function name for two (or

Stevens - CS - 181

Function Templates1Function TemplatesRecall our discussion on function overloading. Consider the two overloaded versions of the swap function. / swaps int variables void swap(int &x, int &y) { int temp = x; x = y; y = temp; } / end function swap

Stevens - CS - 181

Arrays - Declaration and Initialization1Arrays - Declaration and InitializationSuppose that we want to write a program to do the following. 1. Read in five exam scores (integer values). 2. Find the highest score among these five scores. 3. Outpu

Stevens - CS - 181

Arrays as Arguments in Functions1Arrays as Arguments in FunctionsAny individual element (indexed variable) or the full array itself can be used as an argument in function calls. Array elements as arguments Any array element, such as score[2], ca

Stevens - CS - 181

Programming With Arrays - Searching1Programming With Arrays - Searching Searching is the process of locating a particular item among a list of items. Sorting is the process of arranging a sequence of items (or values) in order. For example, we

Stevens - CS - 181

Programming With Arrays - Sorting1Programming With Arrays - SortingOne of the most commonly encountered programming tasks is sorting a list of items. An example is a list of exam scores that must be sorted from lowest to highest or from highest

Stevens - CS - 181

Programming With Arrays - Function Templates1Programming With Arrays - Function TemplatesFunction templates facilitate code reuse. They enable a programmer to write the template for a function only once. The task of producing an actual version o

Stevens - CS - 181

Pointer Types - Basics1Pointer Types - BasicsPointer type is one of the most powerful features of the C+ programming language. We begin with the basic pointer concepts. Pointer declaration Pointers are variables that contain memory addresses as

Stevens - CS - 181

Arrays and Pointers1Arrays and PointersIn this part we investigate the relationship between pointers and arrays. Using const in pointer declarations Consider the following declaration. int a = 10; / int variable a is modifiable const int b = 20;

Stevens - CS - 181

Pointers and Character Strings1Pointers and Character StringsA character enclosed between single quotes is a character constant. For example, 'A' represents the character A. We have seen before that each character constant is represented in the

Stevens - CS - 181

Pointers and Dynamic Memory Allocation1Pointers and Dynamic Memory AllocationSince a pointer can be used to access a variable, a program can access and modify variables even if the variables have no identifiers to name them. The operator new can

Stevens - CS - 181

Programmer Defined Types - Classes and Constructors1Programmer Defined Types - Classes and ConstructorsWhen we declare an object, we often want to initialize some or all the data members of the object. To do this, the class definition requires s

Stevens - CS - 181

Operator Functions as Nonmember Functions1Operator Overloading Operator Functions as Nonmember FunctionsWe have seen earlier that C+ supports function overloading; so we can use the same name for two or more functions in the same program. C+ lan