Chap06-Solution1
5 Pages

Chap06-Solution1

Course Number: PHYS 213, Spring 2008

College/University: Kansas State University

Word Count: 1705

Rating:

Document Preview

Chap.6 HW1 Questions v horizontal fS 1. In three experiments, three different horizontal force F forces are applied to the same block lying on the same countertop. The force magnitudes are F1 = 12 N, F2 = 8 N, and F3 = 4 N. In each experiment, the block remains stationary in spite of the applied force. Rank the forces according to (a) the magnitude fs of the static frictional force on the block from the countertop...

Unformatted Document Excerpt
Coursehero >> Kansas >> Kansas State University >> PHYS 213

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.

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.

HW1 Chap.6 Questions v horizontal fS 1. In three experiments, three different horizontal force F forces are applied to the same block lying on the same countertop. The force magnitudes are F1 = 12 N, F2 = 8 N, and F3 = 4 N. In each experiment, the block remains stationary in spite of the applied force. Rank the forces according to (a) the magnitude fs of the static frictional force on the block from the countertop and (b) the maximum value fs,max of that force, greatest first. a) Fnet = ma Since the block is stationary v = constant and a = 0. Therefore F- fs = 0 F = fs. and the ranking is F1, F2, and F3 b) fs,max = sFN = smg in this problem and does not depend on the horizontal force. Therefore they all tie. v FN v 2. In Figure 6-14, horizontal force F1 of magnitude 10 N is v fS applied to a box on a floor, but the v does not slide. Then, box as the magnitude of vertical force F2 is increased from zero, mg do the following quantities increase, decrease, or stay the v same: (a) the magnitude of the frictional force f s on the v y box; (b) the magnitude of the normal force FN on the box from the floor; (c) the maximum value fs,max of the magnitude of the static frictional force on the box? (d) Does the box x eventually slide? Hint: Draw a free-body diagram of the mass. There will be a total of four forces. Now apply Newton's 2nd law. Note that in a), b), and c) the box has constant velocity (actually v = 0)so a = 0. a) Applying Newton's 2nd law along x: F1 fs = 0 F1 = fs F2 does not affect fS and so fS stays the same. b) Applying Newton's 2nd law along y: FN F2 mg = 0 FN = F2 + mg Thus as F2 increases FN also increases. c) Since fs,max = sFN, fs,max increases d) Since fs,max increases and fS is constant, the box will not move. 6. In Figure 6-16, a block of mass m is held v stationary on a ramp by the frictional force on it from the ramp. A force F , directed up the ramp, is then applied to the block and gradually increased in magnitude from zero. During the increase, what happens to the direction and magnitude of the frictional force on the block? v Initially (when F is zero) the friction force points up the slope (opposite the component of mg down the slope). 1 As F increases, the total force on the mass must remain zero (provided the mass does not move) so the friction force must decrease. Eventually the friction force comes zero v y As F is further increased the friction force reverses direction and increases. v Finally when F becomes large enough the block abruptly moves. Problems x 6. The floor of a railroad flatcar is loaded with FN v loose crates having a coefficient of static friction of fs,max 0.25 with the floor. If the train is initially moving at a speed of 48 km/h, in how short a distance can the train be stopped at constant acceleration Fg without causing the crates to slide over the floor? nd Hint: Draw a free-body diagram. Apply Newton's 2 law. As the train slows down, the box slows down because static friction holds it in place on the flatcar. The maximum value of this force (which will produce the maximum deceleration) is fs,max = sFN = smg Then Newton's law along x gives -fs,max = ma. Therefore -smg = ma a = -sg Therefore the distance the train can stop in is x - x0 = - v2/2a = - v2/(-2 sg) = (13 m/s)2/(2 x 0.25 x 9.8 m/s2) = 36 m 7. A 3.5 kg block is pushed along a horizontal floor by a force v F of magnitude 15 N at an angle = 40 with the horizontal (Figure 6-20). The coefficient of kinetic friction between the block and the floor is 0.25. Calculate the magnitudes of (a) the frictional force on the block from the floor and (b) the block's acceleration. Hint: Draw a free-body diagram for the block. The forces on it are the applied force, the force of gravity, the normal force of the floor, and the frictional force of the floor. Write the Newton's second law equation in component form, taking the axis to be horizontal and the y axis to be vertical. a) The F = 15 N force has components Fx = F cos and Fy = F sin . Applying Newton's 2nd law along y: FN - F sin - mg = 0 FN = (15) sin 40 + (3.5) (9.8) = 44 N. f k = k FN = 11 N b) Apply Newton's second law to the x axis: F cos - f k = ma a = v y x (15) cos 40 - 11 3.5 = 0.14 m/s 2 . Since the result is positive-valued, then the block is accelerating in the +x direction. 2 S 18. You testify as an expert witness in a case y involving an accident in which car A slid into the rear of car B, which was stopped at a red light along a road headed down a (Figure hill x 6-27). You find that the slope of the hill is = 12.0, that the cars were separated by distance d = 24.0 m when the driver of car A put the car into a slide (it lacked any automatic anti-brake-lock system), and that the speed of car A at the onset of braking was v0 = 18.0 m/s. With what speed did car A hit car B if the coefficient of kinetic friction was (a) 0.60 (dry road surface) and (b) 0.10 (road surface covered with wet leaves)? (a) applying Newton's 2nd law along x to car A mg sin - f = ma Applying Newton's 2nd law along y to car A x v FN - mg cos = 0 FN f k = k FN = mg cos Therefore car mg sin - f mg sin - k mg cos a= = = mg (sin - k cos ) v m m y v fk 3.72 m/s2 The minus sign means a points uphill and so the car is decelerating. 2 Now the final velocity v is given by v 2 = v0 + 2ad . Therefore 2 v = v0 + 2ad = 12.1 m/s. mg b) If k = 0.1 then a = +1.1 m/s2, and the speed (when impact occurs) is 19.4 m/s. 20. A loaded penguin sled weighing 80 N rests on a plane inclined at angle = 20 to the horizontal (Figure 6-29). Between the sled and the plane, the coefficient of static friction is 0.25, and the coefficient of kinetic friction is 0.15. (a) What is v the least magnitude of the force F , parallel to the plane, that will prevent the sled from slipping down the plane? (b) What is the minimum magnitude F that will start the sled moving up the plane? (c) What value of F is required to move the sled up the plane at constant velocity? Note: m = W/g = 8.2 kg v (a) If F is at its minimum value to stop the box r sliding then we must choose f s to point up the slope and be at its max value. i.e. fs = fs, max = s FN . 3 v FN v F v fs v mg y x Applying Newton's 2nd law to the block along the x directions Fmin + f s ,max - mg sin = 0 Applying Newton's 2nd law to the block along the x directions FN - mg cos = 0 f s ,max = s FN = s mg cos Therefore Fmin = - f s , max + mg sin = - s mg cos + mg sin = mg (- s cos + sin ) = 8.57 N (b)Now F is increased and just before the block starts moving uphill, fs = fs, max = s FN and points downhill. Repeating the calculation above Fmin = f s ,max + mg sin = s mg cos + mg sin = mg ( s cos + sin ) = 46 N (c) The block is actually moving and so the friction is kinetic, , fk = k FN F = f kx + mg sin = k mg cos + mg sin = mg ( s cos + sin ) = 39 N v 21. Block B in Figure 6-30 weighs 711 N. The coefficient of static friction between block and table is 0.25; angle is 30; assume that the cord between B and the knot is horizontal. Find the maximum weight of block A for which the system will be stationary. Hint: Draw a free body diagram for the knot. Use the info given about mass B to find the maximum possible tension in the horizontal string. Use this to find the weight of A Solution When mA becomes too large it causes mB to slide. This occurs when the tension in the horizontal string becomes large enough that it exceeds fs,max for B. y Step 1: Calculate fs,max for block B. Applying Newton's 2nd law Summing forces along y and noting that ay = 0): FN - mBg= may = 0 So FN = mBg and fs,max= s FN = smBg = (0.25)(711 N) = 178 N x mB 4 Summing forces along x, assuming fs is at its maximum value and noting that ax = 0: T1 - fs,max = max = 0 T1 = fs,max = 178 N Step 2. Will now find weight of block A if T1 has the above value. Appling Newton's second law (and noting that the acceleration of block B = 0) along x: T2 cos - T1 = 0 along y: T2 sin - mAg = 0 T2 can be found from the first equation and plugged into the 2nd equation: (T1/cos )sin - mAg = 0 mAg = T1tan = (178 N)tan 300 = 103 N 22. In Figure 6-31, a box of Cheerios (mass mC = 1.0 kg) and a box of Wheaties (mass mw = 3.0 kg) are accelerated v across a horizontal surface by a horizontal force F applied to the Cheerios box. The magnitude of the frictional force on the Cheerios box is 2.0 N, and the magnitude of the v frictional force on the Wheaties box is 4.0 N. If the magnitude of F is 12 N, what is the magnitude of the force on the Wheaties box from the Cheerios box? Solution To find the acceleration, we treat the two boxes (total mass mC + mW ) as our system. Newton's 2nd law along x (x is to the right): F fsC - fsw = (mC + mw)a a = (F fsC - fsw)/ (mc + mw) = (12 2 4)/(1 + 3) = 1.5 m/s2. Now consider the wheaties box only A free-body diagrams for the block A+C together, FN y fS FWC x Applying Newton's 2nd law along x: Fwc fs = mwa Fwc = fs + mwa = 4 + 3 x 1.5 = 8.5 N 5

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:

Kansas State University - PHYS - 213
Chap.6 HW2 Questions 9. In Figure 6-17, a horizontal force of 100 N is to be applied to a 10 kg slab that is initially stationary on a frictionless floor, to accelerate the slab. A 10 kg block lies on top of the slab; the coefficient of friction bet
Kansas State University - PHYS - 213
Chap. 7. HWQuestions4. Figure 7-18a shows two horizontal forces that act on a block that is sliding to the right across a frictionless floor. Figure 7-18b shows three plots of the block's kinetic energy K versus time t. Which of the plots best cor
Kansas State University - PHYS - 213
Chap. 8 HW1Questions 2. In Figure 8-21, a horizontally moving block can take three frictionless routes, differing only in elevation, to reach the dashed finish line. Rank the routes according to (a) the speed of the block at the finish line In (1) g
Temple - BIO - 2204
Probability What is the chance that a given event will occur? what is the chance that a child, or a family of children, will have a given phenotype? Probability - expressed in numbers between 0 and 1. Probability = 0 means the event never happens
Temple - BIO - 2204
InheritanceFor a long time, general ideas of inheritance were known ie. cows give birth to cows, not horses brown dogs usually gave birth to brown dogsWhat was really lacking was a quantitative understanding of how particular traits were passed do
Temple - BIO - 2204
Chapter 7Sex Determination and Sex ChromosomesPages 169 - 182Copyright 2006 Pearson Prentice Hall, Inc.Embryos created with DNA from 3 people February 5 , 2008 LONDON - British scientists We are not trying to alter genes, we're just trying
Temple - BIO - 2204
Chapter 8Chromosome Mutations: Variation in Chromosome Number and ArrangementMost diploid species normally contain precisely two haploid chromosome setsHuman Female KaryotypeEvery organism contain a specific number of chromosomes in the genom
Temple - BIO - 2204
Chromosome Mapping in Eukaryotes Chapter 5Copyright 2006 Pearson Prentice Hall, Inc.Early ObservationsWalter Sutton By 1903 Sutton pointed out likelihood that there were many more "unit factors" than chromosomes in most species Shortly, observ
Temple - BIO - 2204
Gene Interactions and Modified Mendelian RatiosEpigenesisMany phenotypes affected/controlled by more than one gene "gene interaction" (occurs at many levels for many reasons)Development is a cascade of eventsEach ensuing step of development inc
Temple - BIO - 2204
Genetic Analysis in HumansAfter rediscovery of Mendel's principles, an early task was to show that they were true for animals And especially in humansResistance to this from English geneticistsI am a geneticist currently I am taking Bio.203 Gene
Temple - BIO - 2204
Lecture 9 Gene MappingChapter 5Life Cycle of ChlamydomonasForm diploid cells (zygotes) by the fusion of reproductive cells (gametes or isogametes when "look" identical)Life Cycle of Saccharaomyces cerevisiaeLife Cycle of NeurosporaLife Cyc
Temple - MATH - 1022
Math 1022: Review, Test 1 Spring 2008The time allotted to the test is 1 hour 15 minutes. NO CALCULATORS OR SCRAP PAPER are allowed on the test. MSRC Reviews Thursday, February 14; 4:10pm - 6:00pm Anderson Lecture 14 Monday, February 18; 4:40pm - 6
Temple - BIO - 2204
DNA ReplicationDNA Is Reproduced by Semiconservative Replication Replication is essential function of the genetic material and must be executed precisely Based on their DNA model, Watson and Crick proposed how replication occurs, termed Semiconser
Temple - BIO - 2204
The Genetic CodeThe Genetic Code Exhibits a Number of Characteristics The genetic code is written in linear form, using the ribonucleotide bases that compose mRNA molecules as "letters." The sequence of RNA is derived from the complementary bases
Temple - MATH - 1022
Math 1022: Review, Test 2 Spring 2008 Thursday, March 20 and Friday, March 21The time allotted to the test is 1 hour 15 minutes. Please bring a SCIENTIFIC CALCULATOR to the test. NO GRAPHING CALCULATORS are allowed. No scrap paper is allowed. MSRC
Temple - MATH - 1022
Math 1022 Test 1 Review Answer Key Spring 2008 Solutions for odd problems and chapter 3 review exercises can be found in the textbook. 3.1 32. (a) - 36 (b) 4 - 3 2 - u 4 (c) 4 + 6 - 4 2 (d) - 8 + 6 + 2 2 u u u u u 9 9 60. - ,
Temple - MATH - 1022
Math 1022 Test 2 Review Answer Key Spring 2008 Solutions for odd problems can be found in the textbook. 5.1 14. 1 54. x = 3 118. a) $11,871.65 b) $20,427.93 5.3 24. 1 40. u - 31 ) 88. 6 log b x - log b ( x + 4 2 ) 94. log b (
Temple - BIO - 2204
Gene Mutation, DNA repair and TranspositionBasics on mutations A mutation is an alteration in DNA sequence. Very important process, without the variation that arises from changes in DNA sequence, there would be no phenotypic variability, no adapta
Temple - BIO - 2204
DNA and RNA structureInformation flow: The central dogma of molecular genetics is that DNA makes RNA, which makes proteinsAlthough it was know that chromosomes are transmitted through gametes from parents to offspring and this contributed to the
Temple - BIO - 2204
Genetic Basis of CancerCancer Is a Genetic Disease Cancer cells share two properties: uncontrolled cell proliferation and metastatic spread. All cancer cells in primary and secondary tumors are clonal, meaning that they originated from a common an
Temple - BIO - 2204
DNA Organization in ChromosomesViral and Bacterial Chromosomes Are Relatively Simple DNA Molecules Bacterial and viral chromosomes are usually a single nucleic acid molecule, are largely devoid of associated proteins, and are much smaller than euk
Temple - BIO - 2204
Transcription and RNA processingHow does DNA, a nucleic acid, specify a protein composed of amino acids? There must be an intermediateEvidence for RNA as the intermediate: DNA is nuclear. RNA is localized in both nucleus and cytoplasm and ribosome
Montgomery CC - SOC - 1376/1396
ORDER MODEL VS. CONFLICT MODEL: THEORIES OF MAJORITY-MINORITY RELATIONSThe "Issue" Order/Functionalist Perspective Either the itself meets a specific social need OR the is the result of another social condition that is useful to overall societySt
Temple - BIO - 2204
Gene Mutation, DNA repair and TranspositionGenetic Techniques, Cell Culture, and Pedigree Analysis Are All Used to Detect Mutations Nutritional mutations in Neurospora crassa are detected by growth on complete medium or supplemental minimal medium
Temple - BIO - 2204
NMT - PHYS - 122
Physics 122: Introductory Physicssec.#03 - SPRING 2008 - CRN. 37815"Eighty percent of success is showing up." - Woody Allen (1935 - ) Class Location: Class Meeting Times: Recitation Meeting: Instructor: Office hours: Workman Center 109 11:00 am - 1
Penn State - COMM - 409
Comm 409 Exam 1. In the Citizen Jayne case, Miller's board memberships could be questioned because as the board member she was dealing with officials that could have an influence on whether or not the groups she was involved in received public funds.
Rutgers - CHEM - 210
Chem 210Training Set IOrganic Chemistry: Chem 210Training Set IThe questions are arranged approximately in the order of the coverage of material in the lecture. This set covers the material of the first exam.1. A bond is made by the overlap
Rutgers - CHEM - 210
Chem 210Training Set IIOrganic Chemistry: Chem 210Training Set IIThe questions are arranged approximately in the order of the coverage of material in the lecture. This set covers the material of the second exam.1. The following molecule is f
Rutgers - CHEM - 210
Chem 210Training Set IIIOrganic Chemistry: Chem 210 Training Set IIIThe questions are arranged approximately in the order of the coverage of material in the lecture. This set covers the material of the third exam.Notes: I. When listing reagent
Rutgers - CHEM - 210
Chem 210Training Set IVOrganic Chemistry: Chem 210 Training Set IVThe questions are arranged approximately in the order of the coverage of material in the lecture. This set covers the material of the final exam. Questions 1-50 cover the material
Rutgers - CHEM - 307
Chemistry 307 Final Exam Review Answers will be provided at the review session on Wednesday night, December 12, at 8:00-11:00 PM in SERC 209 (Busch Campus) DO NOT EMAIL PROF. POWELL FOR DETAILED ANSWERS. 1. Assign the following sets of frequencies to
Rutgers - CHEM - 307
Chemistry 307 Chapter 7 When you learned about the SN2 reaction, you had considered two possible mechanisms, involving either a single step (nucleophilic substitution with second order kinetics, SN2) or two steps, i.e., 1. 2.+ + RLG R + NuR + LG
Rutgers - CHEM - 307
Chemistry 307 Chapter 7 Miscellaneous SN2, SN1, E2, E1 (with some interesting previews) Divergent rates of the E2 reactions of cis- and trans-1-bromo4-tert-butylcyclohexane; one of the isomers reacts significantly faster than the other.Would you l
Rutgers - CHEM - 307
Chemistry 307+o+QuickTimeTM and a Photo - JPEG decompressor are needed to see this picture.QuickTimeTM and a Photo - JPEG decompressor are needed to see this picture.
Rutgers - CHEM - 307
Chemistry 307 Chapter 8 AlcoholsAlcohols are alkanes containing the OH function; they have the general composition CnH2n+2O. They are ingredients of many natural products. 1. IUPAC Nomenclature: change ane to anol or to alkyl alcohol the "longest c
Rutgers - CHEM - 307
Chapter 8 ctd9. Reaction of oxirane with organometallic reagents Just as H can serve as a nucleophile in reactions with oxirane organometalic reagents also react with these substrates:orThe strength of the Grignard or organo lithium "carbon nuc
Rutgers - CHEM - 307
Chemistry 307 Chapter 9 Alcohols EthersYou are familiar with the ability of alcohols to form either alkoxide or alkoxonium ions and with some of the reactions that may occur.1.Alkoxide ions a) Reaction with strong bases generate alkoxide ions.
Rutgers - CHEM - 307
Chemistry 307 Chapter 10 Nuclear Magnetic ResonanceNuclear magnetic resonance (NMR) spectroscopy is one of three spectroscopic techniques that are useful tools for determining the structures of organic compounds. [We will learn about infrared (IR)
Rutgers - CHEM - 307
Chemistry 307 Chapter 11 Alkenes InfraRed Spectroscopy Mass Spectroscopy1. IUPAC nomenclature Rule 1: Find the longest chain containing the alkene function Rule 2: Number the ene function low Rule 3: Substituents are named as "prefixes" in alphab
Rutgers - CHEM - 307
ORGANIC CHEMISTRY 307LECTURE NOTES II R. BoikessA. Overall Organization and Systematization There are16 million organic compounds and more every day. We must have a way to systematize so that we can learn and communicate. a. Remember why there are
Rutgers - CHEM - 307
ORGANIC CHEMISTRY 307LECTURE NOTES I R. BoikessWelcome to Chemistry 307, Organic Chemistry: Be sure you understand the organizational and administrative aspects of the course. How to Succeed in Organic Chemistry 1. The first few weeks are critical
Rutgers - CHEM - 307
ORGANIC CHEMISTRY 307LECTURE NOTES III R. BoikessII. Principles of Organic Reactions1. Chemical reactions are the result of bond breaking and bond making. a. Most (but not all) bond making and bond breaking tends to be associated with a functiona
Rutgers - CHEM - 307
ORGANIC CHEMISTRY 307LECTURE NOTES IV R. BoikessCYCLOALKANES Consider an imaginary process in which a homolytic cleavage of a C-H bond at each end of a chain occurs and then the two ends come together to form a new C-C bond. Such a compound has tw
Rutgers - CHEM - 307
ORGANIC CHEMISTRY 307 LECTURE V Stereoisomerism: Already encountered. Two compounds are stereoisomers when they differ only in the spatial relationship of their parts. The composition is the same, the connectivity is the same. It's only when we consi
Rutgers - CHEM - 307
LECTURE VII.Haloalkanes Let's begin our survey of functional groups with a class of compounds that we have already considered from the point of view of nomenclature. A. Nomenclature 1. Recall that the halogen functional group is treated by the IU
Rutgers - CHEM - 308
Lecture I, Chapter 13, Chem 308ORGANIC CHEMISTRY 308 LECTURE ICHAPTER 13 Alkynes: Continuing our survey of functional groups. 1. An alkyne is a hydrocarbon with a triple bond. A triple bond introduces two degrees of unsaturation. Therefore the gen
Rutgers - CHEM - 308
Organic Chemistry 308, Lecture II, Chapter 14 ORGANIC CHEMISTRY 308 LECTURE IICHAPTER 14 So far, in approaching the chemistry of functional groups, we have focused directly on the functional group. But in many cases the chemistry of the functional g
Rutgers - CHEM - 308
ORGANIC CHEMISTRY 308 LECTURE III CHAPTER 15 ORGANIC CHEMISTRY 308 LECTURE IIICHAPTER 15I. Benzene Our starting point. Where we see further consequences of conjugationA. Historical Perspective. In the mid 19th century Benzene was a well known com
Rutgers - CHEM - 308
Lecture IV, Chapter 16, Chem 308ORGANIC CHEMISTRY 308 LECTURE IVCHAPTER 16Let us continue to explore electrophilic aromatic substitution, a major area of synthetic and mechanistic organic chemistry. Review the general mechanism of this reaction.
Rutgers - CHEM - 308
ORGANIC CHEMISTRY 308 LECTURE V CHAPTER 17 ORGANIC CHEMISTRY 308 LECTURE VCHAPTER 17I.The Carbonyl Group: -C=OVery important and widespread. Found in a number of functional groups. Strong bond because it is a multiple bond between two atoms of
Rutgers - CHEM - 308
ORGANIC CHEMISTRY 308 LECTURE VI CHAPTER 18 ORGANIC CHEMISTRY 308 LECTURE VICHAPTER 18We have seen that there is a vast array of reactions that are additions to the polarized C=O of ketones and aldehydes. These reactions are very useful synthetica
Rutgers - CHEM - 308
Organic Chemistry II with Dr RocheLecture Notes Email Room Phonehttp:/crab.rutgers.edu/~alroche alroche@crab.rutgers.edu Sci 311 225-6166 (labs: Sci 328F/309/304A/318/319)Text(a) "Organic Chemistry" Wade, 4th Edition (b) Solution Manual to Or
ASU - POS - 210
POS 210 Political Ideologies Fall 2007 SimhonyLattie Coor Hall, #6762 (480) 727- 8189 Error! Hyperlink reference not valid.Course TA: Ms Tracy Munsil Lattie Coor Hall, #tracy.munsil@asu.edu Office hours: Tuesday: 12:30-1:30 Office hours: Thursd
ASU - ECN - 212
Homework Set #2 ECN 212 Dr. RobertsHomework Set #2 is due no later than Monday, October 15, 2007. Answers must be turned in on a NCS Pearson Scantron, form # 229633, available at the ASU Bookstore. Answer sheets must be marked in pencil and contai
ASU - ECN - 212
ECN 212 - SLN 75851 Microeconomic PrinciplesOffice: Phone: E-mail: Hours:SyllabusDr. Nancy Roberts Fall 2007Website:BAC 650 (480) 965-4683 nancy.roberts@asu.edu TTH 8:00 a.m. - 9:00 a.m. TTH 3:00 p.m. - 4:00 p.m. Other times by appointment.
ASU - ECN - 212
ASU - ECN - 212
Homework Set #3 - ECN 212 - Dr. RobertsDue in the lab no later than Wednesday, November 7, 2007 or may be turned in at the end of a regularly scheduled class. Answers must be submitted on a NCS Pearson Scantron, Form 229633, available at the ASU Boo
Rutgers - CHEM - 309
Rutgers - CHEM - 309
Rutgers - CHEM - 309
ORGANIC CHJ~MISTRYLAB 309EXAM I, 2007NAME:Please wriite clearly and answer the questions directly and concisely. Partial credit will be given where appropriate. The total number of points in this exam is so. Total scores will be multiplied by 2