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### Midterm1

Course: PHYS 47020, Fall 2007
School: UC Irvine
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UC Irvine - PHYS - 47020
Bucknell - PHYS - 212
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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
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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 &quot;bump&quot; between 0 and L. From this graph you see that the par
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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 &quot;distance = speed time&quot; (with c as a speed) to get an estimate
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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
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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 ) =
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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
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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
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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
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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
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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
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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
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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
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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
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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#
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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 &lt; 0 if x &gt; 0#19 f (x) = x(6 + 5x) = 6x1/2 + 5x3/2 f (x) = 4x3/2 + 2x5/2 + C f (1) = 6 + C
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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
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Function Templates1Function TemplatesRecall our discussion on function overloading. Consider the two overloaded versions of the swap function. / swaps int variables void swap(int &amp;x, int &amp;y) { int temp = x; x = y; y = temp; } / end function swap
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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
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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
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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
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