Inventor classwhat is the purpose of using assemply constraints.pdf

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Inventor class what is the purpose of using assemply constraints? Solution You use assembly constraints and the purpose is to create parametric relationships between parts in the assembly. Just as you use 2D constraints to control 2D geometry, you use 3D constraints in an assembly to position parts in relation to other parts.

Molecules like glucose usually need a special transporter protein to provide a tunnel through cell membranes. Explain why using the chemical properties of the cell membrane. Solution Hi, The cell envelope is made up of lipid bilayer. This bilayer is hydrophobic in nature ( non polar) and does not allow all molecules to pass freely across the membrane. Only small uncharged molecules ( O2, CO2) can pass through it. Even the small uncharged polar molecules such as H2O can pass through it it. But the larger uncharged polar molecules cannot pass through it freely, such as Glucose. The charged molecules such as Na+, H+ are polar in nature and hence cannot pass through the non-polar lipid bilayer regardless of size. They all need specialized transporters to facilitate their transport without interaction with hydrophobic bilayer..

Missy Crane opened a public relations firm called Goth on August 1, 2.pdf

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Missy Crane opened a public relations firm called Goth on August 1, 2018 The During Septembe following amounts summarize her business on August 31, 2018 2018, the business completed the following transactions 1 (Click the icon to view the transactions.) EE (click the icon to view the amounts) Analyze the effects of the transactions on the accounting equation of Goth. Analyze the events chronologically, one transaction at a time. Beginning with transaction a, calculate the balance in each account after analyzing the effect of the transaction on the accounting equation. (Complete onily the necessary answer boxes for your transaction lines [Do not enter any zeros for your transaction lines ] Carry down all balances to the \"Bal.\" line, including zero balance accounts, entening a \"O\" for any zero balances Enter a decrease in an account with a minus sign or parentheses. Abbreviations used AP Accounts Payable /R - Accounts Receivable: Adv Advertising, Cap -Crane, Capital; Exp Expense, Liab Liabilities Rev. Revenue, Sup = Supplies withdr = Crane, withdrawals ) ASSETS LIAB.+ EQUITY Cash + A/R + Office + Land = A/P + Cap Service Rent Adv Sup RevEXp 3,400 Exp Bal 2300 3.400 12.000 = 8,0006 300 0 Bal Solution Assets = Liabilities + Equity cash + A/r + Office supplies + land = A/p + cap - withdrawls + Service revenue - Rent expense - Adve expense bal 2300 3400 0 12000 8000 6300 0 3400 0 0 a 16000 16000 bal 18300 3400 0 12000 8000 22300 0 3400 0 0 b 1600 1600 bal 19900 3400 0 12000 8000 22300 0 5000 0 0 c (8000) (8000) bal 11900 3400 0 12000 0 22300 0 5000 0 0 d 600 600 bal 11900 3400 600 12000 600 22300 0 5000 0 0 e 2100 (2100) bal 14000 1300 0 12000 0 22300 0 5000 0 0 f (1100) 1100 bal 12900 1300 0 12000 0 22300 1100 5000 0 0 g 5000 5000 bal 12900 6300 0 12000 0 22300 1100 10000 0 0 h.1 (1100) 1100 h.2 (450) 450 bal 11350 6300 0 12000 0 22300 1100 10000 1100 450 Journal entries Capital Cr. 16000 Service revenue Cr. 1600 Cash Cr. 8000 Accounts payable Cr. 600 A/R Cr. 2100 Cash Cr. 1100 Service revenue Cr. 5000 Cash Cr. 1100 2. Advertising expense Dr. 450 Cash Cr. 450 Assets = Liabilities + Equity cash + A/r + Office supplies + land = A/p + cap - withdrawls + Service revenue - Rent expense - Adve expense bal 2300 3400 0 12000 8000 6300 0 3400 0 0 a 16000 16000 bal 18300 3400 0 12000 8000 22300 0 3400 0 0 b 1600 1600 bal 19900 3400 0 12000 8000 22300 0 5000 0 0 c (8000) (8000) bal 11900 3400 0 12000 0 22300 0 5000 0 0 d 600 600 bal 11900 3400 600 12000 600 22300 0 5000 0 0 e 2100 (2100) bal 14000 1300 0 12000 0 22300 0 5000 0 0 f (1100) 1100 bal 12900 1300 0 12000 0 22300 1100 5000 0 0 g 5000 5000 bal 12900 6300 0 12000 0 22300 1100 10000 0 0 h.1 (1100) 1100 h.2 (450) 450 bal 11350 6300 0 12000 0 22300 1100 10000 1100 450.

Justify that the following circuit represents the functionality of a .pdf

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Justify that the following circuit represents the functionality of a Full-Adder circuit. Solution 1) Sender, receiver, medium, message, and protocol. 2) Internet draft A working Internet document (a work in progress) with no official status and a six-month lifetime. Proposed Standard: It was a standard developed by IETF as a less mature and well defined standard. 3) a. If one connection fails, the other connections will still be working. b. The other devices will still be able to send data through the hub; there will be no access to the device which has the failed connection to the hub. c. All transmission stops. d. The failed connection may disable the whole network unless it is a dual ring. 4) Application Layer: Packet Network Layer: IP Datagram DataLink Layer: Frame 5) Application Layer: sagarsetti4u@gmail.com Network Layer:Logical Addresses Data Link Layer: Link Layer Addresses 6) No, multiplexing/ de-multiplexing at transport layer does notmean combining several upper- layer packets (from the sameor different applications) into one transport-layer packet. Itonly means that each of the transport-layer protocols such asTCP can carry a packet from any application-layer protocolthat needs its service. However, a transport-layer packet cancarry one, and only one, packet from an application layerprotocol. For instance, UDP can carry a message from FTPin one user datagram and a message from HTTP in anotheruser datagram 7) a) Print your Boarding Pass and check in your Luggage. b) Head to Security c) Find your gate/terminal d) Wait for Your Plane e) Wait for Your Announcement f) Get the Boarding pass checked g) Enter the aircraft h) Get Settled Down We can Prove that it is a full adder by constructing Truth Table. Initially Full adder Truth Table is given by x y z Sum Carry 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 1 1 0 1 1 0 0 1 0 1 0 1 0 1 1 1 0 0 1 1 1 1 1 1 Now Let us See For (xy) Cin and For (xy) Cin+xy x y Cin (xy) (xy) Cin (xy) Cin xy (xy) Cin+xy 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 1 0 1 1 0 0 0 0 1 1 1 0 1 0 1 1 0 0 1 1 0 0 0 1 0 1 1 0 1 0 1 1 1 0 0 0 0 1 1 1 1 1 0 1 0 1 1 Hence we can see that (xy) Cin and Sum are identical. (xy) Cin+xy and Carry are Identical. Hence it a Full Adder x y z Sum Carry 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 1 1 0 1 1 0 0 1 0 1 0 1 0 1 1 1 0 0 1 1 1 1 1 1.

IncompressibilityI cant understand some parts in a book.The .pdf

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Incompressibility I can\'t understand some parts in a book. \"The pointwise incompressibility condition i.s. the condition of vanishing dilatational strain, provides certain constraints to the assumed displacement finite element formulation. As an example the displacements for a four node rectangular plane strain element can be assumed as These three conditions of constraint will limit the element motion to only two possible deformation modes which, of course, cannot represent more general state of strain.\" My questions: I understand that these constraints limit the element motion, but why just two possible deformation modes are available? How can I know which of them? Solution.

If you were to illuminate an Elodea leaf with light at 550 nm, would.pdf

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If you were to illuminate an Elodea leaf with light at 550 nm, would you expect CO2 consumption by the leaf? Why? Solution Answer: CO2 consumption by the Elodea leaf depends on the rate of photosynthesis. The light at 550nm corresponds to green light which is not favourable for photosynthesis. The chlorophyll pigment reflects green light and absorbs other wavelengths of light between 400-700nm for photosynthesis. Se we expect less consumption of CO2 at 550nm.

Identify the three major types of controls that organizations can us.pdf

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Identify the three major types of controls that organizations can use to protect their information resources ? Solution 1. At secure organizations, information security is supported by senior management. Support includes making resources and budget available for information security, as well as clear statements by senior management that information security is a priority for the organization. Since senior managers establish priorities and set the tone for an organization, it is difficult to be a secure organization without their clear and consistent support. As a result of the recent spate of high-profile security breaches, most senior managers now understand the importance of information security and will support information security efforts. 2. Secure organizations regularly identify and document how sensitive data --customer and/or proprietary -- flows in, through and out of the organization. This enables an organization to focus its time, effort and money on protecting its sensitive data. Conversely, it\'s difficult for an organization to protect what it doesn\'t know about, and organizations struggle to protect their data if they don\'t perform this exercise 3. Secure organizations create and maintain a formal, documented inventory of all systems that process, transmit or store sensitive data -- including the operating system, if it\'s physical or virtualized, and what major applications have been installed. Without such an inventory, an organization can\'t fully understand what systems it must protect. Having such an inventory allows an organization to quickly determine whether a particular security vulnerability is relevant to the organization\'s systems. 4. Secure organizations segment sensitive systems from non-sensitive systems through jump servers, firewall rules, router ACLs or switch VLANs. This minimizes the attack surface for an organization\'s sensitive systems and allows access to the systems to be tightly controlled and logged. 5. Secure organizations have a strong change-control process that is rigorously enforced. Changes, including emergency changes, are fully documented then formally reviewed and approved. Unapproved changes can lead to security vulnerabilities that nobody knows about until there\'s a breach. 6. Secure organizations have a strong configuration management process. Sensitive systems are hardened and built only with necessary functionality via an automated build process or a managed configuration software tool such as Puppet or Chef. After the initial build, configuration software tools, which regularly check the configuration of systems, are used to ensure systems stay hardened or strong change control is used to maintain system configuration and prevent server creep..

I have C++ question that I do not know how to do, Can you teach me t.pdf

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I have C++ question that I do not know how to do, Can you teach me this question: Draw a CLASS diagram for Binary Search Tree (BTS) : can you explain me each step, thank you Solution //program to convert Binary Tree to Binary Search Tree #include #include // A binary tree node structure struct node { int data; struct node *left; struct node *right; }; void storeInorder (struct node* node, int inorder[], int *index_ptr) { // Base Case if (node == NULL) return; // first store the left subtree storeInorder (node->left, inorder, index_ptr); // Copy the root\'s data inorder[*index_ptr] = node->data; (*index_ptr)++; // increase index for next entry // finally store the right subtree storeInorder (node->right, inorder, index_ptr); } /* A helper function to count nodes in a Binary Tree */ int countNodes (struct node* root) { if (root == NULL) return 0; return countNodes (root->left) + countNodes (root->right) + 1; } // Following function is needed for library function qsort() int compare (const void * a, const void * b) { return ( *(int*)a - *(int*)b ); } /* A helper function that copies contents of arr[] to Binary Tree. This functon basically does Inorder traversal of Binary Tree and one by one copy arr[] elements to Binary Tree nodes */ void arrayToBST (int *arr, struct node* root, int *index_ptr) { // Base Case if (root == NULL) return; /* first update the left subtree */ arrayToBST (arr, root->left, index_ptr); /* Now update root\'s data and increment index */ root->data = arr[*index_ptr]; (*index_ptr)++; /* finally update the right subtree */ arrayToBST (arr, root->right, index_ptr); } // This function converts a given Binary Tree to BST void binaryTreeToBST (struct node *root) { // base case: tree is empty if(root == NULL) return; /* Count the number of nodes in Binary Tree so that we know the size of temporary array to be created */ int n = countNodes (root); // Create a temp array arr[] and store inorder traversal of tree in arr[] int *arr = new int[n]; int i = 0; storeInorder (root, arr, &i); // Sort the array using library function for quick sort qsort (arr, n, sizeof(arr[0]), compare); // Copy array elements back to Binary Tree i = 0; arrayToBST (arr, root, &i); // delete dynamically allocated memory to avoid meory leak delete [] arr; } /* Utility function to create a new Binary Tree node */ struct node* newNode (int data) { struct node *temp = new struct node; temp->data = data; temp->left = NULL; temp->right = NULL; return temp; } /* Utility function to print inorder traversal of Binary Tree */ void printInorder (struct node* node) { if (node == NULL) return; /* first recur on left child */ printInorder (node->left); /* then print the data of node */ printf(\"%d \", node->data); /* now recur on right child */ printInorder (node->right); } /* Driver function to test above functions */ int main() { struct node *root = NULL; /* Constructing tree given in the above figure 10 / \\ 30 15 / \\ 20 5 */ root = newNode(10); root->left = newNode(30.

I need proper and details explanation for this case study Financial statement part (i, ii & iii Area of concern) Analyse the financial statements that have been prepared by Home Range Ltd’s financial director. In particular, comment on the following aspects of the company: i. Areas of concern in financial performance, focusing mainly on information from the income statement ii. Areas of concern in financial health, focusing on ratios dependent on the income statement and the balance sheet iii. Areas of concern in cash flow management, focusing on information available from the cash flow statement. Assess the value of Clare Lombardo’s parent’s recent concern that ‘the company has not spent its significant cash resources wisely in the past year’ (the year to 31 December 2017). What practical steps should Clare Lombardo now undertake in order to help to maximise the value of the business for resale in about 3 years? Case study Home Range Ltd Clare Lombardo comes from a family that has bought, renovated and sold or rented houses as a side-line to their main forms of employment for a number of years. This family interest has proved to be a profitable small business. Her father is an electrician by trade but has developed skills in other areas such as basic plumbing and carpentry. This has greatly reduced the cost of ‘doing up’ old properties. Clare’s mother works as a part-time bookkeeper so she has enough spare time to keep proper financial records for the business. Clare’s favourite subject at school was Art. While still at school Clare discovered a knack for helping her parents to tastefully furnish renovated properties in preparation for their sale or rent. After school Clare studied interior design while helping run her parent’s business. About 20 years ago Clare was left money by a grandparent. She used this to open her first Home Range shop in an exclusive part of London. It sells high quality home furnishings and accessories to mainly wealthy customers who have limited time. Clare’s luck, good taste and business acumen has proved very successful. She formed the private company Home Range Ltd 18 years ago, and it has grown to consist of a chain of 25 stores in wealthy parts of London and the south east of England. The retail outlets have been complimented by an effective and profitable website. Clare is the majority shareholder while her parents own the remaining shares in the business. Clare has recently decided she wants to sell the business in about three years. She is conscious that her parents are growing old and she also want to spend more time with her young family. Clare’s parents have decided to retire from formal employment in the next few months. They will still run their property business. While they are essentially passive shareholders in Home Range Ltd, Clare still relies on them for common sense advice she can trust. Home Range Ltd has been particularly profitable since the global financial crash beginning in 2007. It appears.

How is Vietas contribution to mathematics distinctly modern in spi.pdf

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How is Vieta\'s contribution to mathematics distinctly modern in spirit? Solution At the end of 16th century, mathematics was placed under the dual aegis of the Greeks, from whom they borrowed the tools of geometry, and the Arabs, who provided procedures for the resolution. At the time of Vieta, algebra therefore oscillated between arithmetic, which gave the appearance of a list of rules, and geometry which seemed more rigorous. Meanwhile, Italian mathematicians Luca Pacioli, Scipione del Ferro, Niccolò Fontana Tartaglia, Ludovico Ferrari, and especially Raphael Bombelli (1560) all developed techniques for solving equations of the third degree, which heralded a new era. On the other hand, the German school of the Coss, the Welsh mathematician Robert Recorde (1550) and the Dutchman Simon Stevin (1581) brought an early algebraic notation, the use of decimals and exponents. However, complex numbers remained at best a philosophical way of thinking and Descartes, almost a century after their invention, used them as imaginary numbers. Only positive solutions were considered and using geometrical proof was common. The task of the mathematicians was in fact twofold. It was necessary to produce algebra in a more geometrical way, i.e., to give it a rigorous foundation; and on the other hand, it was necessary to give geometry a more algebraic sense, allowing the analytical calculation in the plane. Vieta and Descartes solved this dual task in a double revolution. Firstly, Vieta gave algebra a foundation as strong as in geometry. He then ended the algebra of procedures (al-Jabr and Muqabala), creating the first symbolic algebra. In doing so, he did not hesitate to say that with this new algebra, all problems could be solved (nullum non problema solvere). In his dedication of the Isagoge to Catherine de Parthenay, Vieta wrote, \"These things which are new are wont in the beginning to be set forth rudely and formlessly and must then be polished and perfected in succeeding centuries. Behold, the art which I present is new, but in truth so old, so spoiled and defiled by the barbarians, that I considered it necessary, in order to introduce an entirely new form into it, to think out and publish a new vocabulary, having gotten rid of all its pseudo-technical terms...\" Vieta did not know \"multiplied\" notation (given by William Oughtred in 1631) or the symbol of equality, =, an absence which is more striking because Robert Recorde had used the present symbol for this purpose since 1557 and Guilielmus Xylander had used parallel vertical lines since 1575. Vieta had neither much time, nor students able to brilliantly illustrate his method. He took years in publishing his work, (he was very meticulous) and most importantly, he made a very specific choice to separate the unknown variables, using consonants for parameters and vowels for unknowns. In this notation he perhaps followed some older contemporaries, such as Petrus Ramus, who designated the points in geomet.

Fungal associations with plants Examine a type of fungal association .pdf

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Fungal associations with plants Examine a type of fungal association by completing each sentence. Then, arrange the sentences in a logical paragraph order. Solution 1. NUTRIENT The plant provides nutrients to the fungi as fungi doesnot has its green pigment called Chlorophyll,which helps the plant in photosynthesis.As it doesnot has the chlorophyll,the plant helps it by providing the nutrients basically the carbon amount so that it will be live and help the plant in providing lots of nutrients and minerals. 2. MINERAL AND NUTRIENTS The fungi has different roles in the plant system. The fungi makes a symbiotic relationship with the plant.By the help of Mycorrhizae ,the plant obtain phosphate and other minerals like Zinc and Copper . Hyphae are the long extensions which can grow into small soil pores and can allow the Phosphorous nutrients to the plant body. 3. MYCORRHIZAE, ROOTS The soil consists of lot of nutrients and minerals. The plant itself cannot intake all the nutrients,so it takes help of certain symbiotic relationship. By this relationship the plant obtains Phosphate and other minerals by the help of fungus. While the fungus obtains the sugars from the roots of the plant. 4. HYPHAE Hyphae is a lonf,branching, filamentous structure of a fungus that is main mode of vegetative growth. The hyphae helps the plant in distributiing and absorbing the nutrients and minerals from the soil..

For each problem in this homework, your assignment is to determine .pdf

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For each problem in this homework, your assignment is to determine modes of inheritance for a set of phenotypes each of which is determined by genotypes at two independently assorting loci. The data come from an alien species of fly whose genetics is identical to that of earth flies. As part of your assignment, you must confirm your hypothesized modes of inheritance by applying a chi-square analysis. NOTE ABOUT DEGREES OF FREEDOM: In situations where the observed counts and the expected counts for a category are both 0, that category does not figure into the chi-square calculation, so that the degrees of freedom is reduced by one in such situations. The possible modes of inheritance are: (i) Autosomal recessive; (ii) Homozygous lethal (dominant); (iii) Sex-linked recessive. The Sex-linked recessive mode can only occur in females parents. Counts of phenotype pairs for the F1 generation give you clues as to the mode of inheritance of each phenotype. Good luck! ONE PHENOTYPE PER PARENT, TWO PHENOTYPES TOTAL For the Tratagorian Kumbus Fly, that lives on the planet Zeon, parental matings occur only among flies in which each parent has a disease phenotype at one locus and is wild-type at the other locus. Problem 4. A female with Flonasean eye color mates with a male with Muppetrus eye shape, producing the following counts: F1 Generation Males WT-WT Disease-WT WT-Disease Disease-Disease 0 242 0 247 Females WT- WT Disease-WT WT-Disease Disease-Disease 239 0 253 0 Solution It is sex linked recessive mode of inheritance due to which males are not affected and females are affected..

For each story problem, identofy the opertaion and the interpretatio.pdf

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For each story problem, identofy the opertaion and the interpretation of the operationg. Then, consider the context of each story problem and identify what, if anything, might be problematic about the context. Pick three of the story problems that you hae idenitfied as having a problematic context and rewrite each of the story problems. -Jake, Jenn, and Juile was triplets. For their birthday, their parents bought them 7 presents each. How many presents did their parents buy all together? -Terrence scored 27 points in a one-on-one basketball game. Devon scored 34. How many points did Devon win by? -Janice gets $40 a week for allowance If she save up for her money for the next 12 weeks, how much more will she have in all? -Ramon and Julio got detention after school for 90 minutes every day for a week. How much time will they spend in detention all together? Solution -Jake, Jenn, and Juile was triplets. For their birthday, their parents bought them 7 presents each parents buy all together 7+7+7 =21 presents -Terrence scored 27 points in a one-on-one basketball game. Devon scored 34. Devon win by 34-27 =7 points -Janice gets $40 a week for allowance If she save up for her money for the next 12 weeks, she have in all =40*12 =480$ -Ramon and Julio got detention after school for 90 minutes every day for a week. week has 7 days time they will spend in detention all together=90*7=630 minutes.

Explain a. casting b. overloading c. sentinel d. echoSolution.pdf

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Explain: a. casting b. overloading c. sentinel d. echo Solution Casting- Changing the datatype of 1 variable to another datatype is called casting. Each program has its own rule. eg: int can be converted to double. overloading- Multiple function when used with same name but have different signature is called overloading in programs. sentinel- It is the flag value that is used in loop like for loop or while which is used for the termination of loops or recursive algorithm. eg flag=1 while(flag<5) { ..... } echo- It is the keyword that is used to print the message or variable in the PHP language echo \"Hi\"; It will print Hi on screen..

Find all the square roots of the complex number 14i. Write the squar.pdf

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Find all the square roots of the complex number 14i. Write the square roots in polar (re^i) form, with the smaller angle first. Give angles in degress Solution this is the polar form r(cos()+isin()) we compare it with 1-4i rcos =1 rsin=-4 squaring and adding them r^2(cos^2 +sin^2) = 1 +(-4)^2 r^2 = 1+16 r= sqrt(17) =4.1 tan = -4/1 = arc tan(-4) = -76 degree square root of r(cos()+isin()) is ±r(cos(/2)+isin(/2)). square root form is ±(4.1) (cos(-76/2) + isin(-76/2) ±4.1(cos38 -isin38) in polar form 4.1 e^(-38i).

Describe the flow of oxygenated blood from the placenta to the fe.pdf

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Describe the flow of oxygenated blood from the placenta to the fetus\' systemic circulation and deoxygenated blood from the fetus to the placenta. Be sure to include all five structures that allow for fetal circulation to take place effectively. Solution Placenta receives blood from two systems: a uteroplacental system that delivers maternal blood to the placenta; and the fetoplacental system that receives from the fetus. The umbilical vein carries blood from the placenta to the fetus. A majority of it enters liver and 1/3rd is diverted to ductus venosus and carried to inferior vena cava. In liver proper, the blood enters from its inferior border. A branch of the umbilical vein joins the portal vein. The blood is then supplied to the right atrium of the heart. The heart has an opening connecting left and right atrium called foramen ovale. Due to this opening, the blood bypasses the pulmonary circulation and is directly circulated to the body of the fetus, via the aorta. Some of the blood moves to the placenta through iliac and then umbilical arteries into the mother\'s blood circulation for the exchange of gasses and removal of waste products. A part of the blood that enters the atrium of the fetus enters left ventricle and is pumped to the pulmonary artery The pulmonary artery of the fetus is connected to the aorta via ductus atriosus and thus the blood enters aorta and escapes from entering the lungs. The blood in aorta takes the same route as described earlier..

describe two different forms of bindingSolutionTwo forms Co.pdf

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describe two different forms of binding Solution Two forms : Compile time Binding : thus is done at compile time where the symbols are resolved i.e the type is assigned during compile time.also called early or static binding. Runtime binding where the functions, variables methods and properties are assiged type during run time. also called as late or dynamic binding. Examples: Compile time - method overloading Run time - method overriding.

Connect onnect mheducation.co.pdf

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Connect onnect mheducation.com/flow/connect.html?isReg true&returnuri; https%3A%2F%2Fconnect.mheducation com%2Fpa C a Search uiz (Lecture 4) Taxonomy is the branch of systemic biology that studies the evolutionary relatedness among groups of organisms. Answer True False Help Solution 2. False Taxonomy is the science of naming, describing and classifying living organisms 3. Presence of a jaw 4. True.

Consider the following interrupting system. The active-edge inputs o.pdf

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Consider the following interrupting system. The active-edge inputs on PJ7 and PP7 can occur at any time, including at the same time. The object is to count the number of each type of interrupt. How would you best describe the usage of Count in this system? a) This is a perfectly appropriate usage of Count, because there are two permanently allocated variables with private scope, such that each variable counts the number of interrupts for each ISR. b) There is a critical section bug because of the read/modify/write access to a shared global. c) Because both ISRs share the same Count, the system can not distinguish between a PJ7 and a PP7 interrupt. d) Count is initialized each interrupt, so its value is not the total number. e) The acknowledge statements in the two ISRs are not friendly because they affect all 8 bits of the flag register. f) None of the above is true. void interrupt 24 KeyHanJ (void) ( unsigned short static Count-0; PIFJ = 0x80; // acknowledge Countt count the number of PU7 edges void interrupt 56 KeyHanP (void) ( unsigned short static count=0; PIFP = 0x80; // acknowledge Count++count the number of PP7 edges Solution Looking around, we find ourselves to be surrounded by various types of embedded systems. Be it a digital camera or a mobile phone or a washing machine, all of them has some kind of processor functioning inside it. Associated with each processor is the embedded software. If hardware forms the body of an embedded system, embedded processor acts as the brain, and embedded software forms its soul. It is the embedded software which primarily governs the functioning of embedded systems. During infancy years of microprocessor based systems, programs were developed using assemblers and fused into the EPROMs. There used to be no mechanism to find what the program was doing. LEDs, switches, etc. were used to check correct execution of the program. Some ‘very fortunate’ developers had In-circuit Simulators (ICEs), but they were too costly and were not quite reliable as well. As time progressed, use of microprocessor-specific assembly-only as the programming language reduced and embedded systems moved onto C as the embedded programming language of choice. C is the most widely used programming language for embedded processors/controllers. Assembly is also used but mainly to implement those portions of the code where very high timing accuracy, code size efficiency, etc. are prime requirements. Initially C was developed by Kernighan and Ritchie to fit into the space of 8K and to write (portable) operating systems. Originally it was implemented on UNIX operating systems. As it was intended for operating systems development, it can manipulate memory addresses. Also, it allowed programmers to write very compact codes. This has given it the reputation as the language of choice for hackers too. As assembly language programs are specific to a processor, assembly language didn’t offer portability across systems. To overcome this disadv.

Case History- An 11-month old Boy whos blue and short of breathC.pdf

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Case History- An 11-month old Boy who\'s blue and short of breath Chief Complaint: An 11-month old boy brought to the pediatric clinic with difficulty breathing. History: Austin Green, an 11-month old Caucasian boy was brought by his foster parents to the Pediatric clinic with comnplaint of dyspnea (i.e difficulty breathing) and fatigue. These episodes last for several minutes, occur daily, and happen most copmmonly during exertion (i.e exercise). Austin spent the first 9 months of his life with hisbiological parents, but recent social problems have necessitated that he receive foster care.His foster parents have had him for the past month, and they have note that these \"breathing spells\" have become more frequent and severe over period of time. They also note that Austinb\'s shortness of breath diminishes somewhat if he lies down. Interestingly, he frequently assumes a squatting position during these episodes, and this seems to reduce hsi symptoms as well. His foster parents report that Austin has had no nausea, vomiting, change in appetie, or change in bowel or bladder habits. He is currently on no medications and no known allergies. There is no medical information available about any othefr member of Austin\'s biological family. Lab. test results: Hematocrit 43% normal (37-47%) Hemoglobin 14mg/dl (normal 12-16ml/dl) PH =7.21 (normal range=7.35 to 7.45) [HCO^-3]=18meq/L (normal range=22 to 26 me=Eq/L) 1. The cardiologist ordered X-ray that showed decreased blood flow to the lungs. A follow-up echocardiogram revealed the following abnormalities in Austin\'s heart. (A) a narrowing of pulmonic valve, (B) a thickening of right ventriclewall, (C) a ventricular septal hole allowing direct exchange of blood between the right and the left ventricles, (D) an aorta that was shifted in position so that it was over-riding the right venbtricl, allowing it to recieve blood from both the left and the right ventricle. What is the name of Austin\'s condition? 2. What is specifically causing the heart murmur? 3. Why did the chest X-ray show freduced pulmonary blood flow? 4. Austin has a thickening right ventriclewall. what type of hypertropy is this? Why does Austin have this particular type of ventricular hypertrophy? BE SPECIFIC 5. Why di Austin\'s EGG show a right-shufted ventricular depolarization axis? 6. Which way will blood typically flow through Austin\'s ventricular septal defect(i.e from the left ventricle to the right ventricle, or vice versa? explain your answer in detail 7. Why is Austin cyanotic? BE SPECIFIC 8. Why does supplemental oxygen therapy did not correct Austin\'s cyanosis? BE SPECIFIC 9. How might the squatting position help relieve Auatin\'s dyspnea? 10. Why is Austin\'s... (a) arterial pCO2 higher than normal? (b). arterial [HCO^-3] lower than normal? (c). arterial PH lower than normal? 11. Describe treatments for Austin\'s condition? Solution 1. Cyanosis 2. Murmurs casues: problem with heart valve close and open to let blood flow.

Complete this table. At the completion of this table, students will.pdf

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Complete this table. At the completion of this table, students will be able to: compare the nucleotides of DNA and RNA compare the bases found in DNA and RNA determine which compounds are single or double stranded THIS ASSIGNMENT IS DUE BY FEBRUARY 26 AT 11:59 PM. nucleic acid table.doc Text Submission Attach File Solution (i) Compare the nucleotide of DNA and RNA. DNA Nucleotide RNA Nucleotide PURINES -Deoxyadenylate,Deoxyguanylate Adenylate ,Guanylate . PYRIMIDINES-Deoxycytidylate,Deoxythymidylate Cytidylate,Uridylate -DNA contain deoxyribose sugar RNA contain ribose sugar -At 2nd position of ribose sugar hydroxyl ion is absent -At 2nd position of ribose sugar hydroxyl ion is present (ii) Compare the bases found in DNA and RNA , DNA -cytosine , Adenosine ,Guanine ,Thymine RNA- at the place of thymine urecil is present . (iii) DNA is double standered while RNA is a single standered..

Chest pain upon physical exertion, when due to coronary artery disea.pdf

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Chest pain upon physical exertion, when due to coronary artery disease, is termed: Select one: a. Silent angina b. Myocardial infarction c. Stable angina d. Variant (Prinzmetal\'s) angina Solution Stable angina Note - stable angina occur with physical exertion while unstable angina occur withour physical exertion and is more dangerous, Stable angina can be improved by rest and is due to presence of plaque in the vessel supplying blood oxygen to heart whereas unstable angina is kind of heart attack.

Can molds be multicellular If so, in what conditions and under what.pdf

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Can molds be multicellular? If so, in what conditions and under what classigications can they be multicellular? Solution Answer: Molds fall under kingdom Fungi and grows in the form of multicellular hyphae. Molds in general do not fall under specific taxonomic or phylogenetic groups. They can be found under the divisions Zygomycota and Ascomycota. Previously, most molds were classified under Deuteromycota..

C++ program help! Cant find whats wrong with my program.The sa.pdf

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C++ program help! Can\'t find what\'s wrong with my program. The sample output is below and the program I wrote is below that. The program is good but if you enter in a number of minutes like \"120\" it doesn\'t know what to do and crashes. Anything helps. Thank you. -In this project, the “current time” should be stored in 24-hour format. Only when the user actually interacts with the program will the output and input be sometimes in 12-hour format. Implement a main menu that will allow the user to (1) enter the current time, (2) add a forward delta to the “current time”, (3) add a backwards delta to the “current time”, (4) display the current time, (5) toggle the interface mode between 24-hour and 12-hour (should impact both getTime and printTime). The program should identify invalid input for the main menu. When entering the current time, only allow hours between 0 and 23 and minutes between 0 and 59. Accept only ‘a’ and ‘p’ when entering the time in 12-hour mode. When entering time deltas, only allow positive values or 0 (this is true for both forward and backwards time deltas). Sample Output: (user input in italics) 1 - Enter a time 2 - Add delta to time 3 - Subtract delta from time 4 - Display current time 5 - Toggle 24-hour mode 6 - Exit > 1 Enter the time with whitespace separating the hours and minutes: 5 50 1 - Enter a time 2 - Add delta to time 3 - Subtract delta from time 4 - Display current time 5 - Toggle 24-hour mode 6 - Exit > 4 The current time is 05:50 1 - Enter a time 2 - Add delta to time 3 - Subtract delta from time 4 - Display current time 5 - Toggle 24-hour mode 6 - Exit > 2 Enter an increment of hours and minutes (separated by a space): 0 120 The new time would be 07:50 1 - Enter a time 2 - Add delta to time 3 - Subtract delta from time 4 - Display current time 5 - Toggle 24-hour mode 6 - Exit > 2 Enter an increment of hours and minutes (separated by a space): 2 20 The new time would be 08:10 1 - Enter a time 2 - Add delta to time 3 - Subtract delta from time 4 - Display current time 5 - Toggle 24-hour mode 6 - Exit > 5 12-hour mode turned on 1 - Enter a time 2 - Add delta to time 3 - Subtract delta from time 4 - Display current time 5 - Toggle 24-hour mode 6 - Exit > 1 Enter the time with whitespace separating the hours, minutes, and either \'a\' for am or \'p\' for pm: 0 15 a 1 - Enter a time 2 - Add delta to time 3 - Subtract delta from time 4 - Display current time 5 - Toggle 24-hour mode 6 - Exit > 4 The current time is 12:15 am 1 - Enter a time 2 - Add delta to time 3 - Subtract delta from time 4 - Display current time 5 - Toggle 24-hour mode 6 - Exit > 3 Enter a decrement of hours and minutes (separated by a space): 1 15 The new time would be 11:00 pm 1 - Enter a time 2 - Add delta to time 3 - Subtract delta from time 4 - Display current time 5 - Toggle 24-hour mode 6 - Exit > 3 Enter a decrement of hours and minutes (separated by a space): 0 90 The new time would be 10:45 pm *****This is the program I wrote: #in.

Assignment DIn Problem D1 we will use a file to contain the dat.pdf

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Assignment D In Problem D1 we will use a file to contain the data which we will read into the program. In Problem D2 we will build a StringOfCar class which will contain a sequence of Car objects. Also, an operator= member function will be added to the Car class. Problem D1 1.Car constructors and destructor within the Car class definition a.default constructor b.copy constructor c.other constructors d.destructor 2.main 3.Car member functions declared within the Car class but defined later a.setup b.output 4.global functions a.operator== with Car parameters b.input In this problem, we will read cars from a file, rather than typing them in from the keyboard. Do the steps in order to build the solution to this problem. 1.Create a file to use for input. This is good because we will be using the input many times.Create a file containing the following three lines of data (Omit the heading line). 2.Modify the input function. Remove the & from the parameters in the function header, so they are all values rather than references. Move the parameters from the function header and put them within the input function, so they are now all local variables. Remove the parameters from the prototype for the input function, so it matches the function header. In the input function declare a string named carType. In the current file all the carType values are \"Car\". Later we will see different values. Open the file. If the open fails, send a message to stderr and exit the program. Use a loop to process each line from the file. Hint: you can do this with the following while statement, which loops as long as there is data in the file to be read: The peek function will return EOF if there is no more data, so we will leave the loop then.Within this loop, each line in the file will provide the data for a Car. In all the reads within the input function, There are two differences from the read activity in assignment C: Remove the user promp. We do not want it because we are reading from a file. read using the inputFile object, rather than using the stdin object. Read the carType field first. It just indicates that we are building an object of type Car. Read each of the fields: ARR, number, kind, loaded Always read the destination field even if the Car is not loaded. The destination will be NONE in the input file when that is the case. Hint: We need to use getline when reading the destination. using >> skips leading white space before reading the data. getline does not skip this leading whitespace. So, before using getline use the following code: peek looks at the next character you are about to read. If it is a space, get is used to read the space character, to get it out of the way. If the carType is \"Car\", declare a Car object named temp using the constructor that takes 5 parameters. Call the output function for the Car named temp. If the carType is not \"Car\", send an error message and do nothing. At the bottom of the input function, close the file. 3.Call the inp.

Answer the following before beginning the case study. What are the f.pdf

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Answer the following before beginning the case study. What are the four human blood types? _____, ______, ______, _______ What are the three alleles that determine the blood types? ____, _____, _______ What is the genotype of a type O Individual? _________ What are the possible genotypes for a type A individual? ______ and _______ What is the allele interaction between the type O allele and type A allele? _____ What is the allele interaction between the type O allele and Type B allele?_______ What is the allele interaction between the type A allele and type B allele? ______ Could a person who has type A blood, be a parent of a type O child? Explain. Could a person who has type AB blood be a parent of type O child? Explain. Could a person with type B blood be a parent to a type A child? Explain. A second of human blood type is the Rh 0 factor. The presence of Rh D (Rh +) is dominant to the absence of the RHD (Rh-) What are the potential genotypes of an individual who is Rh+______and_______ A person with a negative Rh D factor a has a genotype of_______ Could a person that is Rh+ be a parent to a child that is Rh-? Explain. What do you think is going on with these two families? The hospital would have simple things like Diana\'s blood type on file. How could you use blood type to figure out what is going on? Solution A, B, AB and O IA and IB and the i allele i allele IAIA or IAi IA is dominant over i IB is dominant over i IA and IB alleles are co-dominant Yes, A blood group person with heterozygous genotype when married to A, B or O person can get a baby with O blood group. Example IAi x IAi = IAIA (A type), IAi (A type) , ii (O type) No, Because AB person gives IA and IB alleles which results in a baby which is either A, B or AB type.

A singles tournament had six players. Each player played every other.pdf

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A singles tournament had six players. Each player played every other player only once, with no ties. If Helen won 4 games, Ines won 3 games, Janet won 2 games, Kendra won 2 games and Lara won 2 games, how many games did Monica win? Solution Since there are 6 players, a total of 5+4+3+2+1=15 games are played. So far, 4+3+2+2+2=13 games finished (one person won from each game), so Monica needs to win 15-13 = 2.

Write the overall reaction for the combustion of butane (C4H10) with.pdf

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Write the overall reaction for the combustion of butane (C4H10) with 120% theoretical air. Solution Percentage of Carbon in C4H10 = 48*100/(48+10) % = 82.76% Percentage of Hydrogen = 100 - 82.76 = 17.24 % Theretical Air required = 4.31(2.67C + 8H) = 15.468 kg of air /kg of fuel Actual Air Supplied = 1.2* 15.468 = 18.562 kg/kg of fuel Now since excess air is supplied there is no formation of carbon monoxide C4H10 + 7.15O2 -----> 4CO2 + 5H2O + 0.65O2.

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