in the case of the Bacillis strain, why would a young culture be use.pdf

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in the case of the Bacillis strain, why would a young culture be used for gram staining while an old culture is used when spore staining? Solution Older clustures donot retain the staining due to weak cell walls hence for old bacillus cultures endospore staining is done as endospores are more stable structures. The young cultures wil retain the stain so gram negative staining is done..

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Day care centers expose children to a wider variety of germs than th.pdf

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Consider a two-state paramagnet with a very large number N of elemen.pdf

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An administrator wishes to force remote access users to connect using SSH. Which command must be entered to disable Telnet access and force SSH use? A. The no telnet Global Configuration command B. no telnet in vty Line Configuration mode C. The transport input ssh Global Configuration command D. transport input ssh in vty Line Configuration mode Solution D is correct. This is the correct command, and must be entered in Line Configuration mode. A, B, and C are incorrect. A and B are incorrect because no telnet is not a valid IOS command. C is incorrect because it is the correct command, but it’s in the wrong configuration mode..

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Below is a quote from Bobby Kennedy on what the Gross National Product means and more importantly what it does not mean. Too much and too long, we seem to have surrendered community excellence and community values in the mere accumulation of material things. Our gross national product ... if we should judge America by that - counts air pollution and cigarette advertising, and ambulances to clear our highways of carnage. It counts special locks for our doors and the jails for those who break them. It counts the destruction of our redwoods and the loss of our natural wonder in chaotic sprawl. It counts napalm and the cost of a nuclear warhead, and armored cars for police who fight riots in our streets. It counts Whitman\'s rifle and Speck\'s knife, and the television programs which glorify violence in order to sell toys to our children. \"Yet the gross national product does not allow for the (1) health of our children, (2) the quality of their education, or the joy of their play. It does not include the (3) beauty of our poetry or the strength of our marriages; (4) the intelligence of our public debate or (5) the integrity of our public officials. It measures neither our (6) wit nor our (7) courage; neither our (8) wisdom nor our learning; neither our (9) compassion nor our (10) devotion to our country; it measures everything, in short, except that which makes life worthwhile. And it tells us everything about America except why we are proud that we are Americans.\" Kennedy mentions at least 10 things that aren\'t measured by GDP. YOUR task is to argue that he is wrong. Find a way to prove that we can somewhat account for the factors he mentions in our GDP. Solution As long as these points is not subject to any productive work he is right but if these lap with productive goods and service programmes they can be included in GDP. For example, if devotion is spread through tv channels then, people get benefit and trp, satelight rights etc. yield a return. Hence, it can be possible to include in GDP. Similarly on several grounds we can do that too for the other items to recorded in the ten points..

Below is a quote from Bobby Kennedy on what the Gross National Produ.pdf

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Abstract Data Types (a) Explain briefly what is meant by the term abstract data type (ADT). Give two reasons why use of ADTs is good programming practice. (b) Write out a signature, or interface, that defines the operations of a stack ADT. (c) Consider a string of characters of the form ... (.( ... ).) ... where ... indicates an arbitrary sequence of characters (except for parentheses), (.( indicates an arbitrary number (one or more) of opening parentheses, and similarly ).) indicates an arbitrary number of closing parentheses. Using only the stack abstraction operations defined above, write pseudocode for an algorithm that determines, using a stack, whether or not the number of closing parentheses is the same as the number of opening parentheses. You may assume the existence of a function read(str,ch) that reads the next character of string str into ch. You may also assume that you can invoke a function reportFail, that will cause termination with failure, and similarly, reportSuccess causes termination with a success indication. Further, you may also assume that you can call a function newStack(S) to create a new empty stack S, and eos(str) that returns false when you reach the end of the string. Solution (a) Explain briefly what is meant by the term abstract data type (ADT). Give two reasons why use of ADTs is good programming practice. A data type is a collection of values and a set of operations on those values. That collection and these operations form a mathematical construct that may be implemented with the use of a particular hardware or software data structure. The term abstract data type (ADT) refers to the basic mathematical concept that defines the data type. We have discussed four different implementations of the list data structure. In case of implementation of the list with the use of an array, the size of the array gives difficulty if increased. To avoid this, we allocate memory dynamically for nodes before connecting these nodes with the help of pointers. For this purpose, we made a singly linked list and connected it with the next pointer to make a chain. Moving forward is easy but going back is a difficult task. To overcome this problem, we made a doubly linked list using prev andnext pointers. With the help of these pointers, we can move forward and backward very easily. Now we face another problem that the prev pointer of first node and the next pointer of the last node are NULL. Therefore, we have to be careful in case of NULL pointers. To remove the NULL pointers, we made the circular link list by connecting the first and last node. The program employing the list data structure is not concerned with its implementation. We do not care how the list is being implemented whether through an array, singly linked list, doubly linked list or circular linked list. It has been witnessed that in these four implementations of the list, the interface remained the same i.e. it implements the same methods like add, get, next, start a.

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Across the field of human development, “development” is defined as: Question 1 options: Change across the lifespan. Universal and sequential. There are multiple definitions of development. Gains and losses. Save Question 2 (3 points) Question 2 Saved The traditional view of human development states that development is: Question 2 options: The biological unfolding of genetic potential. Any physical changes that occur between birth and maturity. Positive, normative, universal growth that occurs from conception to maturity. Gains, losses, and just plain changes that occur at each stage of the life cycle. all of the above. Save Question 3 (3 points) Question 3 Saved What is the biggest drawback of the life-span perspective of human development? Question 3 options: It overemphasizes positive development in old age. It is too different from the way that non-scientists think about development It relies too much on universal stages of adult development. It is difficult to figure out what should (or should not) be considered “development”. All of the above. Save Question 4 (3 points) Question 4 Saved There are several big picture meta-theoretical themes used in this class—as presented in the syllabus and textbook, and as explained by the instructor. Find the theme that is NOT part of the assumptions from the meta-theory used in this class. Question 4 options: Development is promoted by person-centered contexts and interactions. People carry the past forward into a dynamic present. Genetics are at the root of behavioral development. Development is influenced by an active changing context and an active developing person. All people face fundamental developmental tasks, although there are many healthy pathways through them. Save Question 5 (3 points) Question 5 Saved All psychological theories of human development deal with certain key developmental issues. These key issues include \"nature vs. nurture,\" and \"human nature as good vs. bad,\" as well as: Question 5 options: Cross-cultural universality versus context specificity; the continuity of development versus the discontinuity of stages. Infant as a blank slate versus the infant coming with stuff; adult as good due to social constraints or evil due to selifsh instincts. The nature of environmental stimulus versus an organism\'s responses; conditioned versus unconditioned responses. Individual existence versus personal stages of growth; collective everyday life versus historic ages of cultural evolution. Save Question 6 (3 points) Question 6 Saved What does it mean to say that \"development is contextually and historically embedded\"? Question 6 options: An individual\'s development takes place within a specific historical time. Development may differ for different generations. Societal context shapes people\'s development. As people live through history, these events change their pathways. All of the above. Save Question 7 (3 points) Question 7 Saved Some people have a mechanistic world view. They .

Across the field of human development, “development” is defined as .pdf

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A linked list is a linear data structure consisting of a set of nodes, where each one except the last one points to the next node in the list. (Appendix A provides more information about linked lists.) Suppose we have the set of 5 nodes shown in the illustration below. These nodes have been scrambled up and placed in a MARIE program as shown below. Write a MARIE program to traverse the list and print the data in order as stored in each node. Solution Address Label (hex) JUMP start 00D Addr, Hex 0006 //Top of list pointer, address of node 1 00E Node2, Hex 0032 //Node data is the character 2. 00F Hex 0008 //address of node 3. 010 Node4, Hex 0034 //character \"4\". 011 Hex 000A //address of node 5 012 Node1, Hex 0031 //Character 1 013 Hex 0002 //address of node2 014 Node3 Hex 0033 //Character 3 015 Hex 0004 // addresss of node 4 016 Node5 Hex 0035 //Character 5 017 Hex 0000 //end of the list Next Hex 0000 //pointer to next element ONE Hex 0001 //one Start LOAD Addr //LOAD LINK HEAD STORE Next // LOAD Next Loop SKIPCOND 400 //if AC==0 - end of list reached JUMP NextElem JUMP End NextElem CLEAR //Next is not null ADDI Next // Load data value from current pointter to AC OUTPUT // Display it LOAD ONE // AC <-- 1 ADD Next // AC <--- Next+1 (pointer field address) STORE Next // AC ---> Next CLEAR // AC <---- 0 ADDI Next // AC <--[Next] (load next address from pointer field) STORE Next // AC--> Next (Next loop) JUMP Loop End HALT.

A linked list is a linear data structure consisting of a set of nodes.pdf

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12. The white “Spirit” black bear, Ursys anerucabys kermodei, differs from the ordinary black bear by a single amino acid change in the melanocortin 1 receptor gene. In this population, the gene has two forms (or alleles): the “white: allele b and the “black” allele B. The trait is recessive: white bears have two copies of the while allele of this gene (bb), whereas a bear is black if it has one ore two copies of the black allele (Bb or BB). Both colors morphs and all three genotypes are found together in the bear population of the northwest coast of British Columbia. If processing the white allele has no effect on growth, survival, reproductive success, or mating patterns of individual bears, then the frequency of individuals with 0, 1, or 2 copies of the white allele (b) in the population will follow a binomial distribution. To investigate, Hendrick and Ritland sampled and genotyped 87 bears from the northwest coast: 42 were BB, 24 were Bb, and 21 were bb. Assume that this is a random sampled of bears. a. Calculate the fraction of b alleles in the population (remember, each bear has two copies of the gene). proportion from the data b. With your estimate of the fraction of b alleles, and assuming a binomial distribution, calculate the expected frequency of bears with 0, 1, and 2 copies. c. Compare the observed and expected frequencies in a graph. Describe how they differ. Solution.

12. The white “Spirit” black bear, Ursys anerucabys kermodei, differ.pdf

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With respect to the study of law What is the purpose of discovery? Who can be deposed? Who must answer interrogatories? Solution discovery is the civil process where both parties to a case to be interogated and find collect the facts or relevant information to the case. it is a civil process where the authorities will personally interact with both the parties individually and collects the relevant information relates to the issue. both the parties will rise their voices and they explained their views and perceptions towards the issue. the purpose is to konw and identify the truths with pleas and perceptions of the individual related to the case. here just hearings will happens. the police parties or official parties will participate in the interogation and questions the victims and accused people in the case. they will express their views and the reality to the officials..

With respect to the study of lawWhat is the purpose of discovery.pdf

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Why do you need proteins to allow ions to cross the cell membrane and enter the muscle cell? Solution The transfer of ions are required for the conductance of muscle impulse. The proteins are attached to, or associated with the membrane of a cell or an organelle. Facilitated diffusion is the spontaneous passage of molecules or ions across a biological membrane passing through specific transmembrane integral proteins. The use of proteins along with different proteins used in the transport of ions are listed below:.

Why do you need proteins to allow ions to cross the cell membrane an.pdf

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When you digest a sample of your protein with Chymotrypsin you get a different set of fragments: MNY LMRNF CDEKICY LVSKMQL IGRASTF HPPRGATW a) How does the Edman Degradation method work? b) What is the correct sequence of your protein? c) What are alternate ways you could have determined the sequence of your protein? Solution Edman degradation is a chemical method used for the sequencing of amino acids in a peptide. This method is developed by Pehr Edman in 1960. In this method, N-terminal is labeled or cleaved form the peptide and number of residues are identified The Edman degradation is a very important reaction for protein sequencing, because it allows the ordered amino acid composition of a protein to be discovered. Automated Edman sequencers are now in widespread use, and are able to sequence peptides up to approximately 50 amino acids long. A reaction scheme for sequencing a protein by the Edman degradation follows: 1.Break anydisulfide bridgesin the protein with areducing agentlike2-mercaptoethanol. Aprotecting groupsuch asiodoacetic acidmay be necessary to prevent the bonds from re-forming. 2.Separate and purify the individual chains of the protein complex, if there are more than one. 3.Determine the amino acid composition of each chain. 4.Determine the terminal amino acids of each chain. 5.Break each chain into fragments under 50 amino acids long. 6.Separate and purify the fragments. 7.Determine the sequence of each fragment. 8.Repeat with a different pattern of cleavage. 9.Construct the sequence of the overall protein. Step – 1: The phenyl isothiocyanate reagent attacks and removes the N-terminal amino acid unit in the form of a substituted thiohydantoin heterocycle. Characteristic hydantoin derivatives of all the amino acids have been made and catalogued, so identification of the terminal unit is accomplished easily by comparison. Step – 2: The terminal amino group adds to the reactive isothiocyanate function to give a substituted thiourea. This moiety is conformationally mobile, and the sulfur atom may approach the carboxyl carbonyl of the N-terminal amino acid. Step – 3: Anhydrous HCl protonates the amide oxygen atom, activating it to nucleophilic attack by sulfur. Next, a thiazolinone heterocycle incorporating the N-terminal unit is cleaved, leaving behind a shorter peptide chain. This heterocyclic ring is unstable and undergoes acid-catalyzed rearrangement to the isomeric thiohydantoin (in the green shaded box). Digestion into peptide fragments Peptides longer than about 50-70 amino acids long cannot be sequenced reliably by the Edman degradation. Because of this, long protein chains need to be broken up into small fragments that can then be sequenced individually. Digestion is done either by endopeptidases such as trypsin or pepsin or by chemical reagents such as cyanogen bromide. Different enzymes give different cleavage patterns, and the overlap between fragments can be used to construct an overall sequence. Reaction The pe.

When you digest a sample of your protein with Chymotrypsin you get a.pdf

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What type of rock can be dated with radiometric methods? Solution the metamorphic rock and Igneous rock & melt, all are vey slowly cooling and we can measure radioactivity decay unless or until it cool down below it’s the closure temperature.and more frequently used radiometric dating technique are depending on breakdown of the Potassium [40K] to -argon [40Ar]. sharp measurement of 40K to-- 40Ar amount in the igneous rock can tell you the real amount of time has been passed since this rock crystallized.for the sedimentary rocks we used Carbon-14 method for young rock( >50,00 years old rock) . This method is depend upon uptake of the naturally occured radioactive isotope like carbon-14,when living organism dead than stop accumulating carbon-14 atomes in body,over a time dead living things incorprated into the sedimentary rock, and geo-scientist calculate radiometric dating by using this carbon-14 atoms.this method used to measure the age of meteorites, Earth , ages of fossils, Moon, also early man, ages of mineral deposits, timing of glaciations, recurrence the rates of volcanic, earthquakes & eruptions Geologists scientist use this radiometric dating method to estimate or determine that how long ago back rocks were formed, & give information of ages of the fossils that contained within rocks. Naturally occurring various radioactive elements in the universe, and we know that Radioactive atoms are generally inherently unstable ,in radioactive \"large atoms\" decay into the stable small daughter atoms.in general When the molten rock cool & forming rock are know as igneous rocks a lot of radioactive atom were trapped inside the rock. than they starte decaying at the perticular expected rate. By determine the number of unstable atoms are left in rock & compare to the number of stable molecule of daughter atoms in a rock, geologist can be estimated the amount of the time is over since the perticularly rock formed.but the Fossils are frequently found in sedimentary rock,& not in the igneous rock. Sedimentary rocks can be use to dated a radioactively,carbon, but the problem is carbon decays vey quickly, this method can only work for the younger rocks age about 50 thousand years. So that in order to measure or date rock & older fossils, geologist scientists used layer of the volcanic ash & igneous rock below and above the fossil.geo- Scientists date the igneous rock by using elements which slow to decay, like uranium & potassium..

What type of rock can be dated with radiometric methodsSolution.pdf

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What isSuperfortressand the Flying Fortress their roles in recent history? What isSuperfortressand the Flying Fortress their roles in recent history? What isSuperfortressand the Flying Fortress their roles in recent history? Solution SUPER FORTRESS: The Boeing B-29 Superfortress is a four-engine propeller-driven heavy bomber designed by Boeing which was flown primarily by the United States during World War II and the Korean War. It was one of the largest aircraft operational during World War II and featured state of the art technology. It was the single most expensive weapons project undertaken by the United States in World War II, exceeding the cost of the Manhattan Project by between 1 and 1.7 billion dollars. It featured apressurized cabin, all dual wheeled, tricycle landing gears, and a remote, computer-controlled fire-control system that directed four machine gun turrets that could be operated by a single gunner and a fire-control officer. A manned tail gun installation was semi- remote. The name “Superfortress” continued the pattern Boeing started with its well-known predecessor, the B-17 Flying Fortress. Designed for the high-altitude strategic bomber role, the B-29 also excelled in low-altitude nighttimeincendiary bombing missions. One of the B-29’s final roles during World War II was carrying out the atomic bomb attacks on Hiroshima and Nagasaki. Due to the B-29’s highly advanced design, unlike many other World War II-era bombers, the Superfortress remained in service long after the war ended, with a few even being employed as flying television transmitters for the Stratovisioncompany. The B-29 served in various roles throughout the 1950s. The Royal Air Force flew the B-29 as the Washington until phasing out the type in 1954. The Soviet Union produced an unlicensed reverse-engineered copy as the Tupolev Tu-4. The B-29 was the progenitor of a series of Boeing-built bombers, transports, tankers, reconnaissance aircraft and trainers including the B-50 Superfortress (the first aircraft to fly around the world non-stop) which was essentially a re-engined B-29. The type was finally retired in the early 1960s. The B-29 production total was 3,970 aircraft. Dozens of B-29s remain as static displays but only two examples, Fifi and DOC, have been restored to flying status; with Doc flying again for the first time from McConnell AFB on July 17, 2016. FLYING FORTRESS: The Boeing B-17 Flying Fortress is a four-engine heavy bomber developed in the 1930s for the United States Army Air Corps (USAAC). Competing against Douglas and Martin for a contract to build 200 bombers, the Boeing entry outperformed both competitors and exceeded the air corps\' performance specifications. Although Boeing lost the contract because the prototype crashed, the air corps ordered 13 more B-17s for further evaluation. From its introduction in 1938, the B-17 Flying Fortress evolved through numerous design advances.[5][6] The B-17 was primarily employed by the United State.

What isSuperfortressand the Flying Fortress their roles in recen.pdf

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Volvox was formerly classified as a complex colony, but now scientists classify it as a multicellular organism. How is Volvox different from a colonial species like Scientism\'s What are the structural differences and similarities between Elodea (plant) and human cheek cells (animal)? Compare the check cell and the amoeba. How are they similar? What do amoebas have that cheek cells don\'t? Cheek cells come from a highly complex multicellular organism; why are they so simple compared to the amoeba? Solution Answer :-Q7:-volvox is a colonial alga of fresh water. Here the number, position, shaped is definite and is called coenobium. Volvox is motile. The other colonial alaga like scenedesmus are non motile and here the number, position and shape of cells need not be fixed..

Volvox was formerly classified as a complex colony, but now scientist.pdf

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\"Using netbeans java\" The purpose of this exercise is to effectively demonstrate, that if an external code receives the address of an instance of a class, it can change the state of the object of that class, even it is against the intentions of the programmer who implemented the class. In order to prevent such an intrusion, accessors of reference type fields should not return the field itself, but rather a copy of it. 1. The ArrayPractice class has two private data fields, both are arrays. Declare the fields as -- an array to store integer numbers of type int named numbers -- an array to store Rectangle references (type is Rectangle) named boxes \"Do not instantiate the fields at the declaration!\" 2. Each field has an associated accessor method as usual 3. There is a second accessor method associated to boxes named as getCopyOfBoxes, see the definition later. 4. Define a method named loadArrays( ). The method is void and takes no parameter. The method assigns the entries of the integer array and those of the Rectangle array such that for each array it --runs a for loop to the length of the array --entries of the int array assigned a random integer between 1 and 100 --each entry of the Rectangle array is assigned a new Rectangle object with random integer length and width values. Select the integers again between 1 and 100. Note that the individual Rectangle objects do not have names. 5. The class has a default constructor and an initializer constructor such that it takes one parameter, an int type value named size. The constructor -instantiates the fields numbers and boxes such that the array length is size for both -calls the loadArray() method 6. Define a method named copyNumbers( ). The method takes no parameter and returns an array integers, a copy of the numbers field. In the method declare and instantiate an array (a local variable) named say, temp, with the same type and to the same length as those of the field. The method runs a for loop to load the temp array such that it assigns each temp entry the corresponding numbers entry index-by-index. The method returns temp. 7. Define a method named copyBoxes( ). This method is entirely analogous to the previous method copyNumbers except in the for loop of we do NOT assign boxes[k] to temp[k], but rather a new Rectangle which is a copy of the boxes[k]. This copy must be instantiated by the copy constructor you added to the Rectangle class in Exercise 1 above. That is, the value assigned to temp[k] will be new Rectangle(boxes[k]) 8. Complete the definition of the second accessor method getCopyOfBoxes() mentioned above so as to return copyBoxes( ) rather than boxes itself. \"Using netbeans java\" The purpose of this exercise is to effectively demonstrate, that if an external code receives the address of an instance of a class, it can change the state of the object of that class, even it is against the intentions of the programmer who implemented the class. In order to prevent such an intrus.

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This is the last homework problem i\'m having trouble with. Can anyone help me and explain what to do. i cannot seem to get it right. Thank you so much for the help!! (The 17\'s are not correct, i just put them is as placeholders!!!)*** Solution 2pr = x where r is the radius of the circle. r = x/2p Area of the circle = p (x/2p)2 = x2/4p. 4l = 17-x where l is the length of the side of the square. l = (17-x)/4 Area of the square = l2 Net area = (17-x)2/16 total area = x2/4p + (17-x)2/16 A(x) = x2/4p + (17-x)2/16 A\'(x) = x/2p + [-(17-x)/8] A\'(x) = 0, x/2p + [-(17-x)/8] = 0 8x - 2p(17-x) = 0 8x + 2px - 34p = 0 x = 34p/(8+2p) = 17p/(4+p). The wire should be cut so that 7.476 feet are used for the circle and 9.524 feet are used for the square. The radius of the circle = 1.190 feet Each side of the square = 2.381 feet.

This is the last homework problem im having trouble with. Can anyo.pdf

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Theodore Roosevelts mediation in the Russo-J(APanese War reflected .pdf

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The problem is called \"Name Search\" and it has to be in C# 6. Name Search In the Chap07 folder of the Student Sample Programs, you will find the following files: • GirlNames.txt—This file contains a list of the 200 most popular names given to girls born in the United States from 2000 through 2009. • BoyNames.txt—This file contains a list of the 200 most popular names given to boys born in the United States from 2000 through 2009. Create an application that reads the contents of the two files into two separate arrays or Lists. The user should be able to enter a boy’s name, a girl’s name, or both, and the application should display Solution Form1.cs using System; using System.Collections.Generic; using System.ComponentModel; using System.Data; using System.Drawing; using System.Linq; using System.Text; using System.Threading.Tasks; using System.Windows.Forms; using System.IO; namespace NameSearch { public partial class Form1 : Form { public Form1() { InitializeComponent(); //Intitialize the form. } private void btnSearch_Click(object sender, EventArgs e) {//On clicking the search button.. string fileName;//fileName variable string fileName2;//second filename variable string searchName;//search name variable from user input string searchName2;//second search name variable from user input string nameType;//type of name(boy/girl) try//try to.. { if (txtBoysInput.Text != \"\" || txtGirlsInput.Text != \"\") //If either of the textboxes are not empty.. { if (txtBoysInput.Text != null && txtGirlsInput.Text == \"\") { //if the boys text box is not empty fileName = \"BoyNames.txt\"; nameType = \"boy\'s\"; //set the file name, and the name type to reflect the masculine gender searchName = txtBoysInput.Text; //set the search name variable to the boy\'s contents. parseSearchSingleFile(fileName, searchName, nameType); //Parse the file into a list, and search it. } else if (txtGirlsInput.Text != null && txtBoysInput.Text == \"\") {//if the girls textbox is not empty and the boys is empty.. fileName = \"GirlNames.txt\"; nameType = \"girl\'s\"; //set the file name, and the name type to reflect the feminine gender searchName = txtGirlsInput.Text; //set the search name to the user\'s input parseSearchSingleFile(fileName, searchName, nameType); //parse the file into a list, and search the list. } else if (txtGirlsInput.Text != null && txtBoysInput.Text != null) { fileName = \"GirlNames.txt\"; searchName = txtBoysInput.Text; fileName2 = \"BoyNames.txt\"; searchName2 = txtGirlsInput.Text; //set the file names, and collect the user input for the search names. parseSearchBothFiles(fileName, fileName2, searchName, searchName2); //parse the files, and search both of them. } } else { MessageBox.Show(\"Enter a name, please.\"); //Show the messagebox for empty textboxes. AKA prompt for user entry. } } catch (Exception ex) { MessageBox.Show(\"Error Code: \" + ex); //if something goes wrong, show the error code, please. } } private void parseSearchSingleFile(string fileNa.

The problem is called Name Search and it has to be in C# 6. .pdf

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The displacement d of an object at time t seconds is given by the equation d = 5 cos (pi/2 t) Describe motion of the object What is the maximum displacement from its resting position What is the time required for one oscillation What is the frequency Solution d = 5cos(pi*t/2) a) Motion is sinusoidal with time period and amplitude b) Max. displacement = +5 c) Period = 2pi/ (pi/2) = 4 sec d) frequency = 1/period = 1/4 per sec.

The displacement d of an object at time t seconds is given by the equ.pdf

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