...Trial Examination 2014 VCE Physics Units 3&4 Written Examination Suggested Solutions Neap Trial Exams are licensed to be photocopied or placed on the school intranet and used only within the confines of the school purchasing them, for the purpose of examining that school’s students only. They may not be otherwise reproduced or distributed. The copyright of Neap Trial Exams remains with Neap. No Neap Trial Exam or any part thereof is to be issued or passed on by any person to any party inclusive of other schools, non-practising teachers, coaching colleges, tutors, parents, students, publishing agencies or websites without the express written consent of Neap. Copyright © 2014 Neap ABN 49 910 906 643 96–106 Pelham St Carlton VIC 3053 Tel: (03) 8341 8341 Fax: (03) 8341 8300 TEVPHYU34_SS_2014.FM VCE Physics Units 3&4 Trial Examination Suggested Solutions SECTION A – CORE Area of study – Motion in one and two dimensions Question 1 (10 marks) a. v v = 12 sin 60° = 10.4 m s b. t air = 2 × t top 0 – 10.4 t top = ------------------ = 1.04 – 10 t air = 2 × 1.04 = 2.1 1 2 s = -- at 2 2 1 = -- ( 10 ) ( 1.04 ) 2 –1 1 mark 1 mark 1 mark Note: Consequential on part a. c. 1 mark 1 mark = 5.4 m d. Gravity is 10 m s ∴ 10 m s e. –2 –2 down 1 mark 1 2 KE = -- mv 2 v h = 12 cos 60° =6ms –1 1 mark 2 1 KE = -- ( 80 ) ( 6 ) 2 = 1440 J f. R = v h t air = ( 6 ) ( 2.08 ) = 12.5 m 1 mark 1 mark 1 mark Note: Consequential on part e. 2 ...
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...Certificate of Education May–June Summer 2016 Examination Timetable – FINAL For more information on Edexcel qualifications please visit http://qualifications.pearson.com Notes Pearson Edexcel General Certificate of Education May–June Summer 2016 Examination Timetable – FINAL Home Notes Pearson Edexcel GCE Summer 2016 Examination View by Week Week Week 2 Week 3 Week 4 Week 5 Week 6 Monday 16 May Monday 23 May Monday 6 June Monday 13 June Monday 20 June Monday 27 June Tuesday 17 May Tuesday 24 May Tuesday 7 June Tuesday 14 June Tuesday 21 June Tuesday 28 June Wednesday 18 May Wednesday 25 May Wednesday 8 June Wednesday 15 June Wednesday 22 June Wednesday 29 June Thursday 19 May Thursday 26 May Thursday 9 June Thursday 16 June Thursday 23 June Thursday 30 June Friday 20 May Day Week 1 Friday 27 May Friday 10 June Friday 17 June Friday 24 June Friday 1 July View by Subject Subject A B C D E F G H I J K L M N O P Q R S T U V W X Y Z Pearson Edexcel General Certificate of Education May–June Summer 2016 Examination Timetable – FINAL Notes Home Notes 1. Conduct of Examinations 2. Key Dates • Each examination must be taken on the day and at the time as shown on the timetable. • The date for the restricted release of results to centres is Wednesday 17 August 2016. • The published starting time of all examinations for UK centres is either 9.00 a.m. or 1.30 p.m. Candidates with more than one examination in a session...
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...MANAGEMENT Question 1: A) 7 A + 10 B = 73 4 53 3 15 2 4 + 101 5 40 1 33 2 2 = 10 35 4221 21 735 14 28 + 10 50 400 10 3030 20 20 = 24 85 8221 31 3765 34 48 B) I. Matrix inverse method: 2x1 + 3x2 + 4x3 = 29 X1 + x2 + 2x3 = 13 3x1 + 2x2 + x3 = 16 1) Matrix form ,A: A =211 312 413121 122 223331 232 133 XX1 X2 X3 b291316 Ax = b ; x = A-1b A-1 =1A - A djoint A 2) Determine ( Row 2 as sample) A = (1) (-1)2+13421 + (1) (-1)2+2 2431 + 2(1)2+32332 = (-1) [(3)(1) – (4)(2)] + (1) [(2)(1)-(4)(3)] + (-2) [(2)(2)-(3)(3)] = (-1)(3-8) + (1)(2-12) + (-2)(4-9) = 5 – 10 + 10 A=5 3) Minor A A11 = 1221 = 1(1) – (2)(2) = 1 – 4 = -3 A12 = 1231 = 1(1) – (2)(3) = 1 – 6 = -5 A13 = 1232 = 1(1) – (1)(3) = 2 – 3 = -1 A21 = 3421 = 3(1) – (4)(2) = 3 – 8 = -5 A22 = 2431 = 2(1) – (4)(3) = 2 – 12 = -10 A23 = 2332 = 2(2) –(3)(3) = 4 – 9 = -5 A31 = 3412 = 3(2) – (4)(1) = 6 – 4 = 2 A33 = 2311 = 2(1) – (3)(1) = 2 – 3 = -1 Minor A = -3-5-1-5-10-520-1 4) Cofactor A = +-3--5+-1--5+-10--5+2-0+-1 = -35-15-10520-1 5) Ad-joint A = [Cofactor A] T = -35-15-10520-1T = -3525-100-15-1 6) Inverse A, A-1 A-1 = 1A – Ad-Joint A = 15-3525-100-15-1 = -3/512/51-20-1/55-1/5 7) Find X1 , X2 , X3 I) X1X2 X3 = -3/512/51-20-1/51-1/5291316 = -...
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...Chapter 4. Job Costing 4-16 Job costing a c e g h k l n o p q r s u Process costing b d f i j m t 4-17 1. B 180% A 190% 2. A $127,000 N $124,000 3. N underallocated $145,000 4-18 1. B $50 A $47 2a. L $188,926 M $221,720 2b. L $186,166 M $218,600 4-19 1. $24 per mh 2. $4,080,000 3. overallocated $30,000 4-20 1. M $36 A 180% 2. $99,000 3. M $120,000 under; A $(260,000) over 4-21 1. 260% 2. 400% 3. TC $32,400; BP $36,000 4-22 1. 2Q $74; 3Q $99 2. 2Q $84; 3Q $84. 3. 2Q $109.20; 3Q $109.20 4-23 1. $22 per dlh 2. dr. WIP Control $4,664,000 (this would be the amount for the year) 3. cr. COGS $14,000 4-24 (Note: skip part 1) 2. (11) dr. MOH Allocated $2,080; cr. MOH Control $1,950; cr. COGS $130 4-25 8. dr. MOH Allocated $215,800; cr. MOH Control $202,900; cr. COGS $12,900 4-26 Skip part 1. 3. dr. MOH Allocated $93; dr. CGS $1; cr. MOH Control $94 4. $115 GM 4-27 1. M1 $561,000 M2 $419,000 2. $510 3. dr. FG Control $561,000 cr. WIP Control $561,000 4. $419,000 dr. balance 4-28 1. (a) $53; $48 (b) $53; $45 (c) $60; $45 3. (a) $18,685 (b) $18,130 (c) $19,425 4-29 1a. D: $25 I: $14 1b. D: $25 I: $15 1c. D: $26 I: $15 3a. $3,081 2b. $3,160 2c. $3,239 4-30 1. 50% 2. underallocated $3,000 3. COGS a. $552,250 b. $551,200 c. $551,200 4-31 2. M $49; F...
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...GRE MATH FORMULAE Mixtures first.... 1. when you mix different quantities (say n1 and n2) of A and B, with different strengths or values v1 and v2 then their mean value vm after mixing will be: Vm*(n1 + n2) = (v1.n1 + v2.n2) you can use this to find the final price of say two types of rice being mixed or final strength of acids of different concentration being mixed etc.... the ratio in which they have to be mixed in order to get a mean value of vm can be given as: n1/n2 = (v2 - vm)/(vm - v1) When three different ingredients are mixed then the ratio in which they have to be mixed in order to get a final strength of vm is: n1 : n2 : n3 = (v2 - vm)(v3 - vm) : (vm - v1)(v3 - vm) : (vm - v2)(vm - v1) 2. If from a vessel containing M units of mixtures of A & B, x units of the mixture is taken out & replaced by an equal amount of B only .And If this process of taking out & replacement by B is repeated n times , then after n operations, Amount of A left/ Amount of A originally present = (1-x/M)^n 3. If the vessel contains M units of A only and from this x units of A is taken out and replaced by x units of B. if this process is repeated n times, then: Amount of A left = M [(1 - x/M)^n] This formula can be applied to problem involving dilution of milk with water, etc... EXPLAINATION TO THE ABOVE FORMULA when you mix different quantities (say n1 and n2) of A and B, with different strengths or values v1 and v2 then their mean value vm after mixing...
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...Assignment 1 for All Units Assignment 1 for All Units: Chapter Exercises (NT1430.GA1) Learning Objectives and Outcomes Each unit exercise covers the Learning Objectives and Outcomes for that unit. You can review those objectives and outcomes in your syllabus. Assignment Requirements Type out answers for the end-of-chapter exercise questions indicated in the list below for each unit. You will be graded on accuracy and completeness. Required Resources Sobell, M. G. (2012). A practical guide to Fedora and Red Hat Enterprise Linux. (6 ed.). Upper Saddle River, NJ: Prentice Hall. th Submission Requirements Submit your word-processed answers to your instructor at the beginning of the next class after the assignment. Units 1-10 Chapter Exercises Unit 1 § § § § Unit 2 § § Unit 3 § § § Unit 4 § § Sobell, Chapter 14, p. 582, Exercises 1-5 Sobell, Chapter 10, p. 403, Exercise 1 Sobell, Chapter 7, pp. 251, Exercises 1, 3, 4, 8 Sobell, Chapter 9, pp. 356, Exercises 1, 2, 3, 4, 5, 6, 8 Sobell, Chapter 9, pp. 357, Advanced Exercises 10, 11 Sobell, Chapter 5, p. 182, Exercises 1, 3, 4, 8, 10 Sobell, Chapter 6, p. 221, Exercises 1, 2, 3, 4, 6 Chapter 1, pp. 20-21, Exercises 1, 4 Chapter 2, p. 53, Exercise 1 Chapter 3, p. 86, Exercises 1, 4, 8 Chapter 11, p. 498, Exercises 1, 3, 7 © ITT Educational Services, Inc. All Rights Reserved. -118- Change Date: 05/30/2012 NT1430 Linux Networking STUDENT COPY: Assignment 1 for All Units Unit 5 § § Unit 6 § § Unit 7 § § Unit...
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...Newsvendor Model Chapter 11 1 utdallas.edu/~metin Learning Goals Determine the optimal level of product availability – Demand forecasting – Profit maximization Service measures such as a fill rate utdallas.edu/~metin 2 Motivation Determining optimal levels (purchase orders) – Single order (purchase) in a season – Short lifecycle items 1 month: Printed Calendars, Rediform 6 months: Seasonal Camera, Panasonic 18 months, Cell phone, Nokia Motivating Newspaper Article for toy manufacturer Mattel Mattel [who introduced Barbie in 1959 and run a stock out for several years then on] was hurt last year by inventory cutbacks at Toys “R” Us, and officials are also eager to avoid a repeat of the 1998 Thanksgiving weekend. Mattel had expected to ship a lot of merchandise after the weekend, but retailers, wary of excess inventory, stopped ordering from Mattel. That led the company to report a $500 million sales shortfall in the last weeks of the year ... For the crucial holiday selling season this year, Mattel said it will require retailers to place their full orders before Thanksgiving. And, for the first time, the company will no longer take reorders in December, Ms. Barad said. This will enable Mattel to tailor production more closely to demand and avoid building inventory for orders that don't come. - Wall Street Journal, Feb. 18, 1999 utdallas.edu/~metin For tax (in accounting), option pricing (in finance)...
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...Name: __________________________ Date: _____________ 1. The production quantity model, a variation of the basic EOQ model, assumes noninstantaneous replenishment. A) True B) False 2. The order quantity cannot be calculated for a periodic inventory system that experiences variable demand. A) True B) False sh is ar stu ed d v i y re aC s ou o u rc rs e eH w er as o. co m 3. The periodic inventory system is often preferred for high quantity, low value items. A) True B) False Th 4. The order quantity for a periodic inventory system remains constant. A) True B) False https://www.coursehero.com/file/11816292/8/ Page 1 Answer Key A B A B Th sh is ar stu ed d v i y re aC s ou o u rc rs e eH w er as o. co m 1. 2. 3. 4. https://www.coursehero.com/file/11816292/8/ Page 2 Numerical Question ar stu ed d vi y re aC s ou ou rc rs e eH w er as o. co m 1. The Ambrosia Bakery makes cakes for freezing and subsequent sale. The bakery, which operates five days a week, 52 weeks a year, can produce cakes at the rate of 116 cakes per day. The bakery sets up the cake- production operation and produces until a predetermined number (Q) has been produced. When not producing cakes, the bakery uses its personnel and facilities for producing other bakery items. The setup cost for a production run of cakes is $700. The cost of holding frozen cakes in storage is $9 per cake per year. The annual demand for frozen cakes, which is constant over time, is 6000 cakes. Determine the following: a. Optimal...
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...Mathematics Extension 1 Preliminary Course Answers 8. o oo oo o (a) 0.83 (b) 0.07 (c) 0.13 (d) 0.16 o oo o or 0. 142857 (h) 1.18 (g) 0.142857 9. (a) 8 9 (h) Chapter 1: Basic arithmetic 13 60 Problem 5 Exercises 1.1 1. 2. (a) Rational (b) Rational (e) Rational (f) Irrational (i) Rational (j) Irrational (e) - 4.3 (a) 18 (b) 11 (c) 6 (d) 11 (h) 1 3. (c) Rational (g) Irrational 19 20 (i) 2 (j) 3 (d) Irrational (h) Rational (f) −1 (g) 2 7 15 1 3 (f) 0.17 (g) 0.36 (h) 1.20 (i) - 4.27 1300 8. 5. 950 600 16. 1.7 6. 3000 (j) 8.16 1. 7. 11 000 8. 17. 79 cents 18. 2.73 19. 1.1 20. 3.6 m 21. $281.93 22. 1.8 g (b) 2 3 20 (d) 12. 0.73 13. 33 14. 3.248 15. 4.21 Exercises 1.2 6. - 1.2 10. - 2 15. 5 3. - 56 4. 10 (a) 7. - 7.51 8. - 35.52 11. - 7 12. −23 13. 10 (b) 17. 1 14. 1 18. 60 19. −20 20. 9 51 1000 (c) 5 1 20 (d) 11 7 20 3 (e) 5 (a) 4. 7 18 (d) 2 6 11 (g) 7 45 oo (e) 1.72 4 45 (e) 14. 17.5% 3 28 17 20 3. (a) 15. 41.7% (b) 7 10 (c) 1 7. $65 179 cm 9. (a) 11.9 (b) 5.3 (c) 19 (d) 3.2 (e) 3.5 (f) 0.24 (g) 0.000 18 (h) 5720 (i) 0.0874 (j) 0.376 15. 402.5 g 19. 573 12. 1152.125 g 16. 41.175 m 13. $10.71 17. $30.92 20. $2898 5 minutes after 1 o’clock. 11 Exercises 1.5 1 64 1. (a) 500 (b)...
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...CHAPTER = 11 Q-1 | | |Factor |Factor Score for Each Location | | |Location Factor |Weight |A |B |C |D | | |1. Labor climate | 5 |5 | | |Location Factor |Weight |A |B |C |D | | |1. Rent |25 |3 |75 |1 |25 |2 |50 |5 |125 | | |2. Quality of life |20 |2 |40 |5 |100 |5 |100 |4 |80 | | |3. Schools |5 |3 |15 |5 |25 |3 |15 |1 |5 | | |4. Proximity to work |10 |5 |50 |3 |30 |4 |40 |3 |30 | | |5. Proximity to recreation |15 |4 |60 |4 |60 |5 |75 |2 |30 | | |6. Neighborhood security |15 |2 |30 |4 |60 |4 |60 |4 |60 | | |7. Utilities |10 |4 |40 |2 |20 |3 |30 |5 |50 | | |Total ...
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...ME 2330 By K. L. Ting 5. Tangential and Normal components Let R(s) = x(s) i + y(s) j represent a curve and s is the independent parameter representing the arc length of the curve measured from a reference point. At s and s’, a particle moves from P to P’ and the displacement is dR (Fig. 1). If P and P’ is infinitesimally separated, i.e. ds= s’-s → 0, l dR l = ds, dR/ds = ut. ut is a unit vector (called unit tangent) tangent to the curve at P (Fig. 2). At P and P’, let ut and ut‘ be the unit vectors in the tangent direction. The lines normal to ut and ut‘ intersect at O (Fig. 2). With ds → 0, point O is the center of curvature and ds/dθ = ρ where ρ, called radius of curvature, is the length OP. In Fig. 3, one may observe that dut is normal to ut and since ut and ut’ are unit vectors, dut/dθ is a unit vector, designated as un. un is the unit vector (called unit normal) in the direction from P to O. That is, dut/ds =(dut/dθ)(dθ/ds) = (1/ρ) un or dut/ds = k un where k (= 1/ρ = dθ/ds) is the curvature at P. Fig. 1 Fig. 2 Fig. 3 With a Cartesian coordinate system, the radius of curvature at (x, y) can be calculated as ρ = [1 + (dy/dx)2]3/2/ ld2y/dx2l or ρ= where the dots on x and y represent the first and second derivatives with respect to any independent parameter, such as t or s. In a circle, the center of curvature is the center of the circle and the radius of curvature is the radius of the circle. Let a particle move along a path R(s). Differentiating...
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...Mathematics Classes 9-10 Chapter One Real Number Mathematics is originated from the process of expressing quantities in symbols or numbers. The history of numbers is as ancient as the history of human civilization. Greek Philosopher Aristotle According to the formal inauguration of mathematics occurs in the practice of mathematics by the sect of priest in ancient Egypt. So, the number based mathematics is the creation of about two thousand years before the birth of Christ. After that, moving from many nations and civilization, numbers and principles of numbers have gained an universal form at present. The mathematicians in India first introduce zero (0) and 10 based place value system for counting natural numbers, which is considered a milestone in describing numbers. Chinese and Indian mathematicians extended the idea zero, real numbers, negative number, integer and fractional numbers which the Arabian mathematicians accepted in the middle age. But the credit of expressing number through decimal fraction is awarded to the Muslim Mathematicians. Again they introduce first the irrational numbers in square root form as a solution of the quadratic equation in algebra in the 11th century. According to the historians, very near to 50 BC the Greek Philosophers also felt the necessity of irrational number for drawing geometric figures, especially for the square root of 2. In the 19th century European Mathematicians gave the real numbers a complete shape...
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...Operations Management I 73-331 Winter 2012 Odette School of Business University of Windsor Midterm Exam II Solution Saturday, March 24 Education Building Room ED 1101 Aids Permitted: Calculator, straightedge. Time available: 1 hour 20 min Instructions: 0 This exam has 12 pages including this cover page and excluding 5 pages of formula and table. 1 Please be sure to put your name and student ID number on each odd numbered page. 2 Show your work. 3 State results up to four decimal places. 4 Do not return tables and formula. Grading: Question Marks: 1 /10 2 /15 3 /15 4 /10 5 /15 Total: /65 Question 1: (10 points) Circle the most appropriate answer 1.1 Which of the following is a type I error? a. Process is in-control, but the sample measurement lies outside limit b. Process is out-of-control, but the sample measurements lie within the limit c. Process is in control, and the sample measurements lie within the limit d. Process is out-of-control, and the sample measurement lies outside limit 1.2 Which of the following control charts is the cheapest a. X char b. R chart c. p chart d. 3 sigma chart 1.3 Type II error increases if a. sample size increases c. control limits are narrower d. control limits are wider 1.4 If EF=12 and LF=12 for the same activity, then a. Slack time of the activity is 12 b...
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...REVIEW ARTICLE H e p a t i t i s M o n t h l y 2 0 0 7 ; 7 ( 3 ) : 1 5 3 -1 6 2 1 Hepatitis C among Hemodialysis Patients: A Review on Epidemiologic, Diagnostic, and Therapeutic Features Seyed-Moayed Alavian 1, Seyed Mohammad-Mehdi Hosseini-Moghaddam 2*, Mohammad Rahnavardi 2 M M M 1 Baqiyatallah Research Center for Gastroenterology and Liver Diseases, Baqiyatallah University of Medical Sciences & Tehran Hepatitis Center, Tehran, Iran 2 Urology and Nephrology Research Center (UNRC), Shaheed Beheshti University of Medical Sciences, Tehran, Iran Hepatitis C virus (HCV) is a major public health problem and is the most common liver disease among hemodialysis (HD) patients. The seroprevalence of HCV infection among HD ranged from 1.9% to 80% in reports published since 1999. The main risk factor for HCV acquisition in HD patients seems the length of time on HD. Phylogenetic analysis of HCV viral isolates has suggested nosocomial patient-to-patient transmission of HCV infection among HD patients. Lack of strict adherence to universal precautions by staff and sharing of articles such as multidose drugs might be the main mode of nosocomial HCV spread among HD patients. Currently, there are several dilemmas on the management of these patients: should HCV-RNA testing be included in the routine screening of HD population for HCV infection?; does periodic serum alanine aminotransferase testing have a role in screening HD patients for HCV infection?; can dialysis really 'save'...
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...Table of Contents Section 1 – Motor and Load Basics.................................................................................................... 9 AC Motors........................................................................................................................................... 11 NEMA Design Types........................................................................................................................ 12 Motor Synchronous Speed............................................................................................................... 12 3-Phase Motor Connections - NEMA ............................................................................................... 13 Rotor Inertia – NEMA....................................................................................................................... 16 3-Phase Motor Connections – IEC Nomenclature............................................................................ 17 Rotor Inertia – IEC ........................................................................................................................... 18 AC Motor Operation above Base Speed .......................................................................................... 19 Synchronous Motors ........................................................................................................................ 20 Wound Rotor........................................................................................
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