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Johns Hopkins - PHY - 171.101
1. (a) The amplitude is half the range of the displacement, or xm = 1.0 mm. (b) The maximum speed vm is related to the amplitude xm by vm = xm, where is the angular frequency. Since = 2f, where f is the frequency,vm = 2 fxm = 2 (120 Hz ) (1.0 10 -
Johns Hopkins - PHY - 171.101
1. (a) The motion from maximum displacement to zero is one-fourth of a cycle so 0.170 s is one-fourth of a period. The period is T = 4(0.170 s) = 0.680 s. (b) The frequency is the reciprocal of the period:f = 1 1 = = 1.47 Hz. T 0.680 s(c) A sinuso
Johns Hopkins - PHY - 171.101
1. The time it takes for a soldier in the rear end of the column to switch from the left to the right foot to stride forward is t = 1 min/120 = 1/120 min = 0.50 s. This is also the time for the sound of the music to reach from the musicians (who are
Johns Hopkins - PHY - 171.101
1. We take p3 to be 80 kPa for both thermometers. According to Fig. 18-6, the nitrogen thermometer gives 373.35 K for the boiling point of water. Use Eq. 18-5 to compute the pressure:pN =T 373.35 K p3 = 273.16 K 273.16 K(80 kPa) = 109.343kPa.
Johns Hopkins - PHY - 171.101
1. Each atom has a mass of m = M/NA, where M is the molar mass and NA is the Avogadro constant. The molar mass of arsenic is 74.9 g/mol or 74.9 103 kg/mol. 7.50 1024 arsenic atoms have a total mass of (7.50 1024) (74.9 103 kg/mol)/(6.02 1023 mol
Johns Hopkins - PHY - 171.101
1. (a) Since the gas is ideal, its pressure p is given in terms of the number of moles n, the volume V, and the temperature T by p = nRT/V. The work done by the gas during the isothermal expansion isW=V2 V1p dV = n RTV2 V1dV V = n RT ln 2 . V
Johns Hopkins - PHY - 171.101
1. (a) With a understood to mean the magnitude of acceleration, Newton's second and third laws lead to m2 a2 = m1a1c6.3 10 kghc7.0 m s h = 4.9 10 m =-7 2 2-79.0 m s2kg.(b) The magnitude of the (only) force on particle 1 isq q q F = m1
Johns Hopkins - PHY - 171.101
1. We note that the symbol q2 is used in the problem statement to mean the absolute value of the negative charge which resides on the larger shell. The following sketch is for q1 = q2 .The following two sketches are for the cases q1 > q2 (left figu
Johns Hopkins - PHY - 171.101
1. The vector area A and the electric field E are shown on the diagram below. The angle between them is 180 35 = 145, so the electric flux through the area is = E A = EA cos = (1800 N C ) 3.2 10-3 m cos145 = -1.5 10-2 N m 2 C.2()2. We u
Johns Hopkins - PHY - 171.101
1. (a) An Ampere is a Coulomb per second, so84 A h = 84FG HCh sIJ FG 3600 s IJ = 3.0 10 K H hK5C.(b) The change in potential energy is U = qV = (3.0 105 C)(12 V) = 3.6 106 J.2. The magnitude is U = eV = 1.2 109 eV = 1.2 GeV.3. T
Johns Hopkins - PHY - 171.101
1. Charge flows until the potential difference across the capacitor is the same as the potential difference across the battery. The charge on the capacitor is then q = CV, and this is the same as the total charge that has passed through the battery.
Johns Hopkins - PHY - 171.101
1. (a) The charge that passes through any cross section is the product of the current and time. Since 4.0 min = (4.0 min)(60 s/min) = 240 s, q = it = (5.0 A)(240 s) = 1.2 103 C. (b) The number of electrons N is given by q = Ne, where e is the magnitu
Johns Hopkins - PHY - 171.101
1. (a) The cost is (100 W 8.0 h/2.0 W h) ($0.80) = $3.2 102. (b) The cost is (100 W 8.0 h/103 W h) ($0.06) = $0.048 = 4.8 cents.2. The chemical energy of the battery is reduced by E = q, where q is the charge that passes through in time t = 6.
Ohio State - ART ED - 160
Kristen DibleyI looked at "Luxury Goods" by Thomas Frontini. It is a larger painting, about 4 feet by 5 feet. The main focus is on a girl standing, facing the viewer straight ahead, wearing a white sundress and a straw hat. She has brown hair, bare
Ohio State - HIST - 398
1 Kristen Dibley History 398 1/23/08Film, Public Opinion, and the Cold WarFilm from a certain time period helps us understand a public's opinion, or how the public felt about a certain topic. It may not encompass every attitude that the public ho
Union - ASL - 101
Laura Dunn 10/26/06 ASL The book Inside Deaf Culture by Carol Padden and Tom Humphries was, for me, an eye-opening book. There were many things included that caught my attention and were very interesting and informative. The confusion I had before ab
UNL - ECON - 422
Margaret Dempsey ECON 422 Assignment 1 January 22, 20081.The first benefit from dealing with strangers outside your immediate clan is the increasein productivity due to the ability of specialization, as described in the text. Upon reviewing exa
Johns Hopkins - PHY - 171.101
1. (a) Eq. 28-3 leads to 6.50 10-17 N FB v= = = 4.00 105 m s . -19 -3 eB sin 160 10 C 2.60 10 T sin 23.0 .chch(b) The kinetic energy of the proton isK=2 1 2 1 mv = 167 10-27 kg 4.00 105 m s = 134 10-16 J. . . 2 2chchThis is
Johns Hopkins - PHY - 171.101
1. (a) The magnitude of the magnetic field due to the current in the wire, at a point a distance r from the wire, is given byB= 0i2r.With r = 20 ft = 6.10 m, we havec4 10 B=hb 2 b6.10 mg-7T m A 100 Ag = 3.3 10-6T = 3.3 T.(b
University of Toronto - ECE - 110
1. The vector area A and the electric field E are shown on the diagram below. The angle between them is 180 35 = 145, so the electric flux through the area is = E A = EA cos = (1800 N C ) 3.2 10-3 m cos145 = -1.5 10-2 N m 2 C.2()2. We u
Johns Hopkins - PHY - 171.101
1. The amplitude of the induced emf in the loop is m = A 0 ni0 = (6.8 10-6 m 2 )(4 10 -7 T m A)(85400 / m)(1.28 A)(212 rad/s)= 1.98 10-4 V.2. (a) =d B d = 6.0t 2 + 7.0t = 12t + 7.0 = 12 2.0 + 7.0 = 31 mV. dt dtchb g(b) Appealing to
University of Toronto - ECE - 231
University of Toronto - ECE - 231
University of Toronto - ECE - 231
University of Toronto - ECE - 231
University of Toronto - ECE - 231
University of Toronto - ECE - 243
UNIVERSITY OF TORONTO FACULTY OF APPLIED SCIENCE AND ENGINEERING FINAL EXAMINATION, APRIL 2006 Second Year ECE243H1 S COMPUTER ORGANIZATION Exam Type: D Examiners A. Moshovos, G. SteffanInstructionsThis is a type D exam. You are allowed to use th
University of Toronto - ECE - 243
Student # (use if pages get separated) _ UNIVERSITY OF TORONTO FACULTY OF APPLIED SCIENCE AND ENGINEERING FINAL EXAMINATION, APRIL 2007 Second Year ECE243H1 S COMPUTER ORGANIZATION Exam Type: D Examiners A. Moshovos, G. Steffan. H. TimorabadiInstr
Johns Hopkins - PHY - 171.101
1. (a) All the energy in the circuit resides in the capacitor when it has its maximum charge. The current is then zero. If Q is the maximum charge on the capacitor, then the total energy is2.90 10-6 C Q2 U= = = 117 10-6 J. . -6 2C 2 3.60 10 Fc
Baylor - PSC - 101
Test Four: Chapter 14: The First Amendment in a Democratic Society: Political Speech and National Security I. First Amendment- Free Speech- Constitutional Analysis a. Analytical pathway to determine what test to use; how do they analyze certain types
Johns Hopkins - PHY - 171.101
1. (a) The flux through the top is +(0.30 T)r2 where r = 0.020 m. The flux through the bottom is +0.70 mWb as given in the problem statement. Since the net flux must be zero then the flux through the sides must be negative and exactly cancel the tota
UMass (Amherst) - MKTG - 301
Marketing 301 Exam 1 Review Sheet Spring 2008 The exam consists of 50 multiple choice questions. I certainly recommend reading all the chapter material in Ch. 1 and Ch.7 through bases for segmentation. Additionally, I strongly encourage you to review
Johns Hopkins - PHY - 171.101
1. In air, light travels at roughly c = 3.0 108 m/s. Therefore, for t = 1.0 ns, we have a distance of d = ct = (3.0 108 m / s) (1.0 10-9 s) = 0.30 m.2. (a) From Fig. 33-2 we find the smaller wavelength in question to be about 515 nm, (b) and the
Baylor - PSC - 101
Test Three: Chapter Nine: Racial Equality (Equal Protection Clause of the 14 th Amendment) I. Two views of racial equality under the constitution (What the government can do) a. Anti-Discrimination Principle i. Allied with individualism school of tho
Johns Hopkins - PHY - 171.101
1. The image is 10 cm behind the mirror and you are 30 cm in front of the mirror. You must focus your eyes for a distance of 10 cm + 30 cm = 40 cm.2. The bird is a distance d2 in front of the mirror; the plane of its image is that same distance d2
Union - HUM - 101
Card #1 Governer Smith: The people need to be educated and aware of the issues existing in their regions. This includes addressing the problems and being totally honest about existing and proposed solutions-their positives, negatives, and possible co
NYU - MATH - Linear Alg
1.1SOLUTIONSNotes: The key exercises are 7 (or 11 or 12), 1922, and 25. For brevity, the symbols R1, R2,., stand for row 1 (or equation 1), row 2 (or equation 2), and so on. Additional notes are at the end of the section.1.x1 + 5 x2 = 7 -2 x1
N.C. State - MA - 341
HW-1 SOLUTION 2. Second-order ordinary differential equation(ODE); y is the dependent variable and x is the independent variable; linear 3. ODE; first-order; y is the dependent variable and x is the independent variable; nonlinear 5. ODE; first-order
N.C. State - MA - 341
MA341Homework 11Solutions7.6 0, 0 < t < 1; 2, 1 < t < 2; 5. g(t) = 1, 2 < t < 3; 3, 3 < t. g(t) = 2u(t - 1) - u(t - 2) + 2u(t - 3). The Laplace transform of u(t - a) is e-as . L{u(t - a)}(s) = s Thus the Laplace transform of g(t) is e-
N.C. State - MA - 341
MA341Homework 9Solution7.25. cos 2t. Using the definition of the Laplace transform, we have for s > 0:F (s) =0e-st cos 2tdt 1 -st = e-st sin 2tdt e sin 2t 0 + s 2 0 1 = - s e-st d(cos 2t) 4 0 1 = - s e-st cos 2t 0 + s e-st cos 2tdt
N.C. State - MA - 341
MA341Homework 10Solution7.59. z + 5z - 6z = 21et-1 ; z(1) = -1, z (1) = 9 (1) To use the method of Laplace transforms, we first move the initial conditions to t = 0. This can be done by setting y(t) = z(t + 1). Then, y (t) = z (t + 1), y (t) =
N.C. State - MA - 341
MA341Homework 12Solutions9.3 1 1 1 [A|I] = 1 2 1 2 3 2 1 1 1 0 1 0 0 1 0 1 1 1 0 1 0 0 0 0 The matrix A is singular, hence the inverse of A 13. [A|I] = 11. | 1 0 0 | 0 1 0 | 0 0 1 | 1 0 0 | -1 1 0 | -2 0 1 | 1 0 0 | -1 1 0 | -1 -1
N.C. State - MA - 341
MA341Homework 13Solutions9.51 3 3 x(t), x(0) = . 3 1 1 The characteristic equation for this matrix is 1-r 3 = r2 - 2r - 8 = (r + 2)(r - 4) = 0. A - rI = 3 1-r Hence, the eigenvalues of this matrix are r1 = -2 and r2 = 4. To find the eigenvecto
N.C. State - MA - 341
Sec 3.2 12. First, we equate the expression (21) for p0 at time t a and t b , we getp0Since tbpa p1e Ap1ta and p0 p1 pa (1 e Ap1ta )eAp1tapb p1e Ap1tb . p1 pb (1 e Ap1tb )e2Ap1tb2ta , we setand2, hencepa ( p1 pb ) pa p1 pb p1 .
N.C. State - MA - 341
HW-3 SOLUTION Sec 2.22. No 3. Yes 6. Yes 8. We separate the variables and rewrite the equation in the form: 1 y 3dy dx . x Integrating both sides, we have: 1 y 3dy dx , x 1 4 y ln x C . 4 Then, solving the last equation for y we obtain:1y(4 ln
Baylor - PSC - 101
Test Two: Chapter Four: I. Judicial Activism vs. Judicial Restraint a. Judicial Activism- a judge is likely to overturn (hold unconstitutional) a law passed by congress b. Judicial Restraint- opposite of activism; tendency to uphold (find constitutio
N.C. State - MA - 341
MA341Homework 54.13. The derivatives of y are: y = 6 cos 3t - 3 sin 3t, y = -18 sin 3t - 9 cos 3t. Thus, 2y + 18y = 2(-18 sin 3t - 9 cos 3t) + 18(2 sin 3t + cos 3t) = -36 sin 3t - 18 cos 3t + 36 sin 3t + 18 cos 3t = 0. Substitution into the init
N.C. State - MA - 341
MA341Homework 8Solution 3. This equation is already in the standard form, hence p1 (x) = -1, p2 (x) = x - 1, g(x) = tan x. Now p1 (x) is continuous on (-, +), p2 (x) is continuous on [1, +), and g(x) is defined on (- + k, + k), thus p1 , p2 ,
N.C. State - MA - 341
MA341Homework 7Solution4.65. y - 2y + y = t-1 et . The characteristic equation for the corresponding homogeneous equation is r2 - 2r + 1 = 0; it has a double root: r = 1. Thus, et and tet are two linearly independent solutions to the homogeneo
N.C. State - MA - 341
HW-2 SOLUTION Sec 1.2 2. (a) We differentiate the implicit function with respect to x, which yields: dy 2y 1 0. dx dy 1 This is equivalent to , where y 0 . dx 2y Furthermore, y 2 3 x 0 x ( ,3) . (b) The definition of the implicit function reduces to
N.C. State - MA - 341
MA341003Homework 6Solution4.41. No. t-1 et is not a product of polynomials and exponentials. 3. Yes. Since 3t = eln 3 = et ln 3 , then we can apply the method of undetermined coefficients to this equation. 5. No. sec = 1/ cos it not a sine/c
Baylor - PSC - 101
CHAPTER 9 PROFIT PLANNING I. The Budget A. DefinitionB. Planning vs. controlC. Advantages of budgeting (pg. 379) D. Responsibility accountingE. Budget periodF. Success of budgeting 1. self-imposed2. Problems with self-imposed budgets a. b.
Baylor - PSC - 101
BUS 3315 Research AssignmentOverview Your purpose in completing the business writing report project is to analyze a company's response to a recent issue or problem. The issue can be one that affects only your company, or it can be one that impacts t
Baylor - PSC - 101
Case Law/Concept Review Sheet: PSC 2302, 2nd ExamNote: Cases listed below are the ones most likely to be on the exam. The material on this sheet is not the only material that will be on the exam; however, most of the case law questions will come fro
Baylor - PSC - 101
Practice Problems (Ch 16) Problem 1: Financial Statement Analysis The balance sheet and income statements for Waverly Inc. are presented below. Waverly Inc. Balance Sheet December 31, 20X4 and 20X3 Dec. 31, 20X4 Cash Trading securities Accounts recei
Baylor - PSC - 101
Review Sheet for the Final ExamChapter 14: Freedom of Speech Political speech & Natl SecuritySymbolic Speech [421]Laws found to be regulating the symbolic speech for content-neutral reasonsUnited States v. O' Brien (1968) [371]: Establishes tes
Baylor - PSC - 101
Chapter 9 Racial Equality and Equal Protection PRE-BROWN ERA [1868-1954] I. Traditional/Intentional/Explicit Use of Race [to harm minorities] a. Early analysis of Equal Protection concerning race[1880-1950s] 1) Dred Scott v. Sanford (1857), p.224, a
Baylor - PSC - 101
MC-1Practice Multiple Choice Chapter 16 1. The following data on sales has been provided: 20x5 20x4 20x3 20x2 Sales $4,203 $3,632 $3,827 $3,504 Using 20x1 as the base year, the sales trend is a. 100%, 103%, 112%, 107%, 123%. b. 97%, 100%, 109%, 104%
Baylor - PSC - 101
Poly Science Test ReviewNotes-On TestMarbury v. Madison 1. Marbury goes to Supreme Court to and asks them to issue a Writ of Mandamus claiming SC can issue Writ of M in ANY case based on section 13 of the judiciary act of 1789. 2. John Marshall, n
UCSC - CHEM - 1M
Titration Curve12 10 8pH6 4 2 0 1 3 5 7 9 11 13 15 17 19 21 23 25 27 Volume NaOHSeries2
UCSC - AMST - 1
D'Sousa: "Pericles' Funeral Oration" "Our system of government does not copy the institution of its neighbors, it is more of a case of us being models to others, and then imitating them" Private lives but public law. Tocqueville: "Democracy in Americ