Course Hero has millions of student submitted documents similar to the one
below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.
Find millions of documents on Course Hero - Study Guides, Lecture Notes, Reference Materials, Practice Exams and more.
Course Hero has millions of course specific materials providing students with the best way to expand
their education.
Below is a small sample set of documents:
HKU - BIOC - BIOC2603
Genetic Engineering and Applications (BIOC 2603)Dr. Danny Chan Department of BiochemistryEmail: chand@hkusua.hku.hk Phone: 28199842Topics in year 1 Restriction enzymes. DNA modifying enzymes. Cloning using plasmids as vectors. cDNA library. Screening
HKU - BIOC - BIOC2603
BIOC2603 (2007) BIOC2603 (2007)Genomes and GenomicsBrian Wong bcwwong@hkucc.hku.hkReferencesBrown TA (2007) Genomes 3rd ed. Chapters 7-8 TA (2007) Genomes 3rd ed Chapters (see http:/www.ncbi.nlm.nih.gov/books/ for 2nd ed.) International Human Genome S
HKU - BIOC - BIOC2603
BIOC2603 Principles of Molecular GeneticsRecombination Brian Wong bcwwong@hkucc.hku.hkGenetic recombination: reassortment of genetic material(A) Independent assortment (for genes on different chromosomes) (B) crossing over (for genes on the same chromo
HKU - BIOC - BIOC2603
BIOC2603 Principles of Molecular GeneticsRecombinationBrian Wong bcwwong@hkucc.hku.hkPart IISite-specific recombinationSite-specific recombination Reciprocal recombination between two short specific DNA sequences. No net synthesis or deletion of DNA
HKU - BIOC - BIOC2603
BIOC2603 Principles of Molecular GeneticsSplicing and RNA processing RNA editing Gene evolution: exons and introns Dr. Mai Har Sham Department of BiochemistryAdvanced ReferencesWatson et al. (2004) Molecular Biology of the Gene. 5th Ed. Pearson/ Benjam
Skyline College - PHYS - 121
12.) Which of the following are permissible amounts of charge for an object to have? Select all that are correct (and none that are incorrect) to receive credit. (You should assume the elementary charge value is exactly e=1.610-19 C.)A neutral object is
USC - BISC - 220
" \ r toN.t "I.:l 2 S pring 010 bisc220sileaders@gmail.com* " -l'3ls l t oSarah,SeanandYuree www.usc.edu/siBISC 220 SI- Exam #2 ReviewCHAPTER 8: ENERGY AND METABOLISM p 1.T he m olecule ictured n t he r ight i s A T F o ATP + H2O-+ DP + Ir A The bre
Abraham Baldwin Agricultural College - ACC - 11101
CHAPTER 7 DEDUCTIONS AND LOSSES: CERTAIN BUSINESS EXPENSES AND LOSSES SOLUTIONS TO PROBLEM MATERIALSQuestion/ Problem 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24Topic Bad debts: cash basis taxpayer Bad debts: business versus nonbusine
TAMU Kingsville - MGMT - 5337
Statistics Data Paper
The Petroleum Institute - PHYSICS - 191262
80. (a) Using Eq. 7-32, the work becomes W = 2 x2 x3 (SI units understood). The plot is shown below:9(b) We see from the graph that its peak value occurs at x = 3.00 m. This can be verified by taking the derivative of W and setting equal to zero, or sim
The Petroleum Institute - PHYSICS - 191262
51. (a) The objects displacement is d = d f di = (8.00 m) + (6.00 m) + (2.00 m) k . i j Thus, Eq. 7-8 gives W = F d = (3.00 N)(8.00 m) + (7.00 N)(6.00 m) + (7.00 N)(2.00 m) = 32.0 J. (b) The average power is given by Eq. 7-42: Pavg = W 32.0 = = 8.00 W. t
The Petroleum Institute - PHYSICS - 191262
52. According to the problem statement, the power of the car is P= dW d 1 2 dv = mv = mv = constant. dt dt 2 dtThe condition implies dt = mvdv / P , which can be integrated to giveT0dt =vT0mvdv PT=2 mvT 2Pwhere vT is the speed of the car at t =
The Petroleum Institute - PHYSICS - 191262
53. (a) We set up the ratio50 km E = 1 km 1 megaton and find E = 503 1 105 megatons of TNT.FG HIJ K1/ 3(b) We note that 15 kilotons is equivalent to 0.015 megatons. Dividing the result from part (a) by 0.013 yields about ten million bombs.
The Petroleum Institute - PHYSICS - 191262
54. (a) The compression of the spring is d = 0.12 m. The work done by the force of gravity (acting on the block) is, by Eq. 7-12, W1 = mgd = (0.25 kg) 9.8 m / s2 (0.12 m) = 0.29 J. (b) The work done by the spring is, by Eq. 7-26, 1 1 W2 = kd 2 = (250 N /
The Petroleum Institute - PHYSICS - 191262
55. One approach is to assume a path from ri to rf and do the line-integral accordingly. Another approach is to simply use Eq. 7-36, which we demonstrate: W=xf xiFx dx +yf yiFy dy =4 2(2x)dx +3 3(3) dywith SI units understood. Thus, we obtain W =
The Petroleum Institute - PHYSICS - 191262
56. (a) The force of the worker on the crate is constant, so the work it does is given by WF = F d = Fd cos , where F is the force, d is the displacement of the crate, and is the angle between the force and the displacement. Here F = 210 N, d = 3.0 m, and
The Petroleum Institute - PHYSICS - 191262
57. There is no acceleration, so the lifting force is equal to the weight of the object. We note that the persons pull F is equal (in magnitude) to the tension in the cord. (a) As indicated in the hint, tension contributes twice to the lifting of the cani
The Petroleum Institute - PHYSICS - 191262
58. With SI units understood, Eq. 7-8 leads to W = (4.0)(3.0) c(2.0) = 12 2c. (a) If W = 0, then c = 6.0 N. (b) If W = 17 J, then c = 2.5 N. (c) If W = 18 J, then c = 15 N.
The Petroleum Institute - PHYSICS - 191262
59. Using Eq. 7-8, we find W = F d = ( F cos sin (x + y = Fx cos + Fy sin i+F j) i j) where x = 2.0 m, y = 4.0 m, F = 10 N, and = 150 . Thus, we obtain W = 37 J. Note that the given mass value (2.0 kg) is not used in the computation.
The Petroleum Institute - PHYSICS - 191262
60. The acceleration is constant, so we may use the equations in Table 2-1. We choose the direction of motion as +x and note that the displacement is the same as the distance traveled, in this problem. We designate the force (assumed singular) along the x
The Petroleum Institute - PHYSICS - 191262
61. The total weight is (100)(660 N) = 6.60 104 N, and the words raises at constant speed imply zero acceleration, so the lift-force is equal to the total weight. Thus P = Fv = (6.60 104)(150 m/60.0 s) = 1.65 105 W.
The Petroleum Institute - PHYSICS - 191262
62. (a) The force F of the incline is a combination of normal and friction force which is serving to cancel the tendency of the box to fall downward (due to its 19.6 N weight). Thus, F = mg upward. In this part of the problem, the angle between the belt a
The Petroleum Institute - PHYSICS - 191262
63. (a) In 10 min the cart moves d = 6.0 so that Eq. 7-7 yields W = Fdcos = (40 lb)(5280 ft) cos 30 = 1.8 105 ft lb. (b) The average power is given by Eq. 7-42, and the conversion to horsepower (hp) can be found on the inside back cover. We note that 10 m
The Petroleum Institute - PHYSICS - 191262
64. Using Eq. 7-7, we have W = Fd cos = 1504 J . Then, by the work-kinetic energy theorem, we find the kinetic energy Kf = Ki + W = 0 + 1504 J. The answer is therefore 1 .5 kJ .
The Petroleum Institute - PHYSICS - 191262
65. (a) To hold the crate at equilibrium in the final situation, F must have the same magnitude as the horizontal component of the ropes tension T sin , where is the angle between the rope (in the final position) and vertical: = sin 1FG 4.00IJ = 19.5 .
The Petroleum Institute - PHYSICS - 191262
66. From Eq. 7-32, we see that the area in the graph is equivalent to the work done. We 1 find the area in terms of rectangular [length width] and triangular [ 2 base height] areas and use the work-kinetic energy theorem appropriately. The initial point i
The Petroleum Institute - PHYSICS - 191262
67. (a) Noting that the x component of the third force is F3x = (4.00 N)cos(60), we apply Eq. 7-8 to the problem: W = [5.00 N 1.00 N + (4.00 N)cos 60](0.20 m) = 1.20 J. (b) Eq. 7-10 (along with Eq. 7-1) then yields v = 2W/m = 1.10 m/s.
The Petroleum Institute - PHYSICS - 191262
68. (a) In the work-kinetic energy theorem, we include both the work due to an applied force Wa and work done by gravity Wg in order to find the latter quantity. K = Wa + Wg leading to Wg = 2.1 102 J . (b) The value of Wg obtained in part (a) still applie
The Petroleum Institute - PHYSICS - 191262
69. (a) Eq. 7-6 gives Wa = Fd = (209 N)(1.50 m) 314 J. (b) Eq. 7-12 leads to Wg = (25.0 kg)(9.80 m/s2)(1.50 m)cos(115) 155 J. (c) The angle between the normal force and the direction of motion remains 90 at all times, so the work it does is zero. (d) The
The Petroleum Institute - PHYSICS - 191262
70. After converting the speed to meters-per-second, we find K = 2 mv2 = 667 kJ.1
The Petroleum Institute - PHYSICS - 191262
71. (a) Hookes law and the work done by a spring is discussed in the chapter. Taking absolute values, and writing that law in terms of differences F and x , we analyze the first two pictures as follows: | F | = k | x| 240 N 110 N = k (60 mm 40 mm) which y
The Petroleum Institute - PHYSICS - 191262
72. (a) Using Eq. 7-8 and SI units, we find W = F d = (2 4 (8 + c = 16 4c i j) i j) which, if equal zero, implies c = 16/4 = 4 m. (b) If W > 0 then 16 > 4c, which implies c < 4 m. (c) If W < 0 then 16 < 4c, which implies c > 4 m.
The Petroleum Institute - PHYSICS - 191262
73. A convenient approach is provided by Eq. 7-48. P = F v = (1800 kg + 4500 kg)(9.8 m/s2)(3.80 m/s) = 235 kW. Note that we have set the applied force equal to the weight in order to maintain constant velocity (zero acceleration).
The Petroleum Institute - PHYSICS - 191262
74. (a) The component of the force of gravity exerted on the ice block (of mass m) along the incline is mg sin , where = sin 1 0.91 15 gives the angle of inclination for the . inclined plane. Since the ice block slides down with uniform velocity, the work
The Petroleum Institute - PHYSICS - 191262
75. (a) The plot of the function (with SI units understood) is shown below.Estimating the area under the curve allows for a range of answers. Estimates from 11 J to 14 J are typical. (b) Evaluating the work analytically (using Eq. 7-32), we have W=2 01
The Petroleum Institute - PHYSICS - 191262
76. (a) Eq. 7-10 (along with Eq. 7-1 and Eq. 7-7) leads to vf = (2 m F cos )1/2= (cos )1/2, where we have substituted F = 2.0 N, m = 4.0 kg and d = 1.0 m. (b) With vi = 1, those same steps lead to vf = (1 + cos )1/2. (c) Replacing with 180 , and still usi
The Petroleum Institute - PHYSICS - 191262
which (by taking two derivatives) we find the acceleration to be a = 0.20 m/s2. The (constant) force is therefore F = ma = 0.40 N, with a corresponding work given by W = 2 Fx = 50 t(t 10). It also follows from the x expression that vo = 1.0 m/s. This mean
The Petroleum Institute - PHYSICS - 191262
78. The problem indicates that SI units are understood, so the result (of Eq. 7-23) is in Joules. Done numerically, using features available on many modern calculators, the result is roughly 0.47 J. For the interested student it might be worthwhile to quo
The Petroleum Institute - PHYSICS - 191262
79. (a) To estimate the area under the curve between x = 1 m and x = 3 m (which should yield the value for the work done), one can try counting squares (or half-squares or thirds of squares) between the curve and the axis. Estimates between 5 J and 8 J ar
National University of Singapore - CVE - 3132
AY200910CE3132WaterResourcesEngineering DepartmentofCivilEngineering NationalUniversityofSingaporeHomeworkNo.5 Questions#2and#4tobediscussed Questions#1and#3tobesubmittedon12thMar2010 TominimizedistractionstotheresearchersinE10822,pleasesubmityourhomew
Waterloo - CS - 447
Software Testing, Quality Assurance and MaintenanceWinter 2010Lecture 9 January 22, 2010Patrick Lam version 1So far, weve seen a number of coverage criteria for graphs, but Ive been vague about how to actually construct graphs. For the most part, its
Waterloo - CS - 447
Software Testing, Quality Assurance and MaintenanceWinter 2010Lecture 8 January 20, 2010Patrick Lam version 1Building on the notion of a def-clear path: Denition 1 A du-path with respect to v is a simple path that is def-clear with respect to v from a
Waterloo - CS - 447
Software Testing, Quality Assurance and MaintenanceWinter 2010Lecture 7 January 18, 2010Patrick Lam version 2Input: Directed graph G Output: List of prime paths in G, primePaths nonextendablePaths ; primePaths ; worklist all paths of length 0, i.e. no
Waterloo - CS - 447
Software Testing, Quality Assurance & Maintenance (ECE453/CS447/SE465): Assignment 3 v1Patrick LamDue: March 3, 2010You may discuss the assignment with others, but I expect each of you to do the assignment independently. I will follow UWs Policy 71 if
Waterloo - CS - 447
/ d d d d d d d d d dWWWdddddddddd W WW//h/hWapplyFactorToTextSize, since there only exist one big nested if-statement instead of many if statements. The choice for paths are limited, thus to cover ADC, AUC and ADUPC are simple(because you cant ha
Waterloo - CS - 447
Software Testing, Quality Assurance & Maintenance (ECE453/CS447/SE465): Assignment 2 SolutionsPatrick LamQuestion 2 (12 points)Predicate 1 a (b c) Determination analysis: a determines p i [true (b c)] [false (b c)] (b c) false (b c) i.e. b : false, c :
Waterloo - CS - 447
Software Testing, Quality Assurance & Maintenance (ECE453/CS447/CS647/SE465): Midterm SolutionsFebruary 10, 2009This open-book midterm has 5 questions and 90 points. Answer the questions in your answer book. You may consult any printed material (books,
Waterloo - CS - 447
Software Testing, Quality Assurance & Maintenance (ECE453/CS447/CS647/SE465): Midterm Practice QuestionsHere are a couple of midterm practice questions. These questions may be slightly ambiguous; Ive taken more care in drafting the actual midterm questio
Waterloo - CS - 447
Software Testing, Quality Assurance & Maintenance (ECE453/CS447/SE465): Assignment 1 SolutionsPatrick LamIve excluded solutions for questions 2, 3 and 7. Ill try to make them available later, but they wont help you with midterm preparation anyway. The C
Waterloo - CS - 447
Software Testing, Quality Assurance and MaintenanceWinter 2010Lecture 12 January 29, 2010Patrick Lam version 2Graph Coverage for SpecicationsWell move further up the abstraction chain now and talk about testing based on specications. Specication-base
Waterloo - CS - 447
Software Testing, Quality Assurance and MaintenanceWinter 2010Lecture 11 January 27, 2010Patrick Lam version 1Data Flow Graph Coverage for Design ElementsThe structural coverage criteria for design elements were not very satisfying: basically we only
Waterloo - CS - 447
Software Testing, Quality Assurance and MaintenanceWinter 2010Lecture 10 January 25, 2010Patrick Lam version 1Dataow Graph Coverage for Source CodeLast time, we saw how to construct graphs which summarized a control-ow graphs structure. Lets enrich o
Waterloo - CS - 447
Software Testing, Quality Assurance and MaintenanceWinter 2010Lecture 6 January 15, 2010Patrick Lam version 3Prime Path Coverage versus Complete Path Coverage. n0 n1 n3 n2 Prime paths: path(t1 ) = path(t2 ) = T1 = cfw_t1 , t2 satises both PPC and CP
Waterloo - CS - 447
Software Testing, Quality Assurance and MaintenanceWinter 2010Lecture 5 January 13, 2010Patrick Lam version 2Lets consider an example of a test set which satises node coverage on D, the double-diamond graph from last time. Start with a test case t1 ;
Waterloo - CS - 447
Software Testing, Quality Assurance and MaintenanceWinter 2010Lecture 4 January 11, 2010Patrick Lam version 2Some binary distinctionsLets digress for a bit and dene some older terms which we wont use much in this course, but which we should discuss b
Waterloo - CS - 447
Software Testing, Quality Assurance and MaintenanceWinter 2010Lecture 3 January 8, 2010Patrick Lam version 2SubsumptionSometimes one coverage criterion is strictly more powerful than another one: any test set that satises C1 might automatically satis
Waterloo - CS - 447
Software Testing, Quality Assurance and MaintenanceWinter 2010Lecture 2 January 6, 2010Patrick Lam version 1RIP Fault ModelRecall that a fault is something thats latent or hiding, while a failure is visible (e.g. EPIC FAIL). To get from a fault to a
Waterloo - CS - 447
Software Testing, Quality Assurance & MaintenanceLecture 1Patrick Lam University of WaterlooJanuary 4, 2010The Testing CourseCourse mechanicsTextbook: I will be (somewhat loosely) following Ammann and Offutt. Website: http:/patricklam.ca/stqam Grace
Waterloo - REC - 280
The Impacts of Tourism Paper REC 280 Introduction to Tourism Percent of Final Grade: 25% Due Date: Wednesday March 17, 2010 (Hard-Copy Submissions Only)Assignment Description: Imagine that you will be traveling in the near future. Describe the place(s) t
Waterloo - REC - 280
REC 280 Impacts of Tourism Rubric (W2010) CONTENT: Responses to Assignment QuestionsExcellent Discussed in detail all the aspects of their trip (transport, accommodations, environment, geography, culture and economy) Identification of motivation for tra
Waterloo - REC - 280
Assignment 1 Test Questions Grading Weight 5% of final course grade Due at the beginning of class on Wednesday, February 3, 2010 Question Criteria: Create 5 multiple choice questions using your notes from class to guide your development of the questions.