I need help with the 2nd TODO comment and the other comments Ill.pdf

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I know that one of the main differences between the rough and smooth endoplasmic reticulum is the presence or absence of ribosomes. The structures also have different shapes, rough ER is flat, and the smooth ER is more round. My question is what specific process causes the two different forms of ER to divide and be so different?At what point does the ER decide to stop making ribosomes and begin to initiate the synthesis of smooth ER? Solution Answer :- The rough ER is specialized with function of protein synthesis and smooth ER with lipid synthesis. When cell is need of lipids and need less proteins the rough ER undergoes loss of ribosomes and gets transformed into smooth ER. ER are actually the membrane extensions of nuclear memebrane..

Education

I need help with the 2nd //TODO comment and the other comments I'll post at the end. I keep
exceeding my memory when I write something for it
public static ArrayList doArrayListRemoveFromEnd(int numItems) {
System.out.print("doArrayListRemoveFromEnd: ");
ArrayList list = new ArrayList<>();
// TODO Write code that adds integers 0 through (numitems - 1)
// to list, inside a loop.
for(int i=0; i<=numItems-1; i++){
list.add(i);
}
long startTime = getTimestamp();
// TODO Write code that removes the last element from list,
// repeatedly until the list is empty.
int last = list.size();
long endTime = getTimestamp();
long totalTime = endTime - startTime;
System.out.println(totalTime);
return list;
}
// TODO Write code that removes the greatest element from set,
// repeatedly until the set is empty.
// TODO Write code that removes the first element from list,
// repeatedly until the list is empty.
// TODO Write code that gets the element at the last index from list.
// TODO Write code that checks if integer (numItems - 1)
// is a member of set.
Solution
public static ArrayList doArrayListRemoveFromEnd(int numItems) {
System.out.print("doArrayListRemoveFromEnd: ");

ArrayList list = new ArrayList<>();
// TODO Write code that adds integers 0 through (numitems - 1)
// to list, inside a loop.
for(int i=0; i<=numItems-1; i++){
list.add(i);
}
long startTime = getTimestamp();
// TODO Write code that removes the last element from list,
// repeatedly until the list is empty.
for(int i=0; i<=numItems-1; i++){
int last=list.size()-1;
list.remove(last);
}
long endTime = getTimestamp();
long totalTime = endTime - startTime;
System.out.println(totalTime);
return list;
}
// TODO Write code that removes the greatest element from set,
// repeatedly until the set is empty.
for(int i=0; i<=numItems-1; i++){
int pos = 0;
int max=list.get(0);
for (int i = 1; i < list.size(); i++) {
if(list.get(i) > max) {
max=list.get(i);
pos=i;
}
}
list.remove(pos);
}
// TODO Write code that removes the first element from list,
// repeatedly until the list is empty.

for(int i=0; i<=numItems-1; i++){
list.remove();
}
// TODO Write code that gets the element at the last index from list.
int element=list.get(list.size()-1);
// TODO Write code that checks if integer (numItems - 1)
// is a member of set.
for(int i=0; i<=numItems-1; i++){
{
if(item==list.get(i))
{
return true;
}
}
return false;

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In C++ Write a recursive function to determine whether or not a Linked List is in sorted order (smallest value to largest value). Add the following function prototypes to your List class under the section for access functions, then implement the functions below the class definition: public: /**Access Functions*/ bool isSorted(); //Wrapper function that calls the isSorted helper function to determine whether //a list is sorted in ascending order. //We will consider that a list is trivially sorted if it is empty. //Therefore, no precondition is needed for this function private: bool isSorted(Nodeptr node); //Helper function for the public isSorted() function. //Recursively determines whether a list is sorted in ascending order. #include //for NULL #include #include using namespace std; template //list stores generic list data, not any specific C++ type class List { private: struct Node { listdata data; Node* next; Node* previous; Node(listdata data): data(data), next(NULL), previous(NULL){} }; typedef struct Node* Nodeptr; Nodeptr first; Nodeptr last; Nodeptr iterator; int size; public: /**Constructors and Destructors*/ List(); //Default constructor; initializes and empty list //Postcondition: numeric values equated to zero, or strings should be empty. List(const List &list); ~List(); //Destructor. Frees memory allocated to the list //Postcondition: position NodePtr at next Node /**Accessors*/ listdata getFirst(); //Returns the first element in the list //Precondition:NodePtr points to the first node on list listdata getLast(); //Returns the last element in the list //Precondition: make new node first on the list listdata getIterator(); bool isEmpty(); //Determines whether a list is empty. int getSize(); //Returns the size of the list /**Manipulation Procedures*/ void startIterator(); void advanceIterator(); void removeLast(); //Removes the value of the last element in the list //Precondition: list is not empty, not the first element. //Postcondition: one remaining node void removeIterator(); void removeFirst(); //Removes the value of the first element in the list //Precondition: list is not empty //Postcondition: no nodes left void insertLast(listdata data); //Inserts a new element at the end of the list //If the list is empty, the new element becomes both first and last //Postcondition: next equal to null void insertFirst(listdata data); //Inserts a new element at the start of the list //If the list is empty, the new element becomes both first and last //Postcondition:point nodePtr next node on list /**Additional List Operations*/ bool offEnd(); void printList(); //Prints to the console the value of each element in the list sequentially //and separated by a blank space //Prints nothing if the list is empty void insertIterator(listdata data); bool operator==(const List &list); }; // constructor definition template List::List(): first(NULL), last(NULL), iterator(NULL), size(0) {} template List::List(const List &list): size(list.size) { if(list..

In C++Write a recursive function to determine whether or not a Lin.pdf

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PROBLEM STATEMENT: In this assignment, you will complete DoubleEndedList.java that implements the ListInterface as well as an interface called DoubleEndedInterface which represents the list's entries by using a chain of nodes that has both a head reference and a tail reference. Be sure to read through the code and understand the implementation. WHAT IS PROVIDED: - A driver class to test your code. You should not modify this file! - A list interface (ListInterface.java) - A double ended interface (DoubleEndedInterface.java) - An incomplete DoubleEndedList class (DoubleEndedList.java) WHAT YOU NEED TO DO: 4. Complete the DoubleEndedList class 4. Run the driver and make sure your output is exactly the same as mine (at the bottom of Driver.java) \} // end else numberofEntries--; else throw new IndexOut0fBoundsException("Illegal position given to remove operation."); return result; // Return removed entry }//endreturnreve public T replace(int givenPosition, T newEntry) \{ T replace(int givenPosition, T newEntry) \{ if ((givenPosition >=1)&& (givenPosition <= numberOfEntries)) \{ // Assertion: The list is not empty Node desiredNode = getNodeAt (givenPosition); ToriginalEntry = desiredNode.getData(); desiredNode.setData(newEntry); return originalEntry; f // end if else throw new IndexOut0fBoundsException("Illegal position given to replace operation."); replace if (( givenPosition >=1)&& (givenPosition <= number0fEntries)) \{ // Assertion: The list is not empty Node desiredNode = getNodeAt ( givenPosition); T originalEntry = desiredNode. getData( ); desiredNode.setData(newEntry); return originalentry; \} // end if throw new Index0ut0fBoundsException("Illegal position given to replace operation."); \} // end replace public T getEntry(int givenPosition) \{ if ((givenPosition >=1) \&\& (givenPosition < = number0fEntries ) ) \{ // Assertion: The list is not empty return getNodeAt (givenPosition). getData(); else // end if throw new IndexOut0fBoundsException("Illegal position given to getEntry operation."); \} // end getEntry public boolean contains ( T anEntry) \{ boolean found = false; Node currentNode = firstNode; while (!found && (currentNode != null)) \{ if (anEntry.equals (currentNode.getData())) else found = true; \} // end while currentNode = currentNode. getNextNode () ; return found; \} // end contains public int getLength() \{ return numberofEntries; \} // end getLength public boolean isEmpty() \{ return number0fEntries ==0; \} // end isEmpty public T[] toArray() \{ // The cast is safe because the new array contains null entries aSuppressWarnings ("unchecked") T[] result =(T[]) new 0bject [numberofEntries]; // Unchecked cast int index =0; Node currentNode = firstNode; while ((index < numberOfEntries) \&\& (currentNode != null)) \& result [ index ]= currentNode. getData () ; currentNode = currentNode.getNextNode ( ); index++; 3 // end while return result; 3 // end toArray // Returns a reference to the node at a given position. // Precondition: L.

PROBLEM STATEMENTIn this assignment, you will complete DoubleEnde.pdf

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I need to fill-in //TODO\'s in .cpp file and in .h file Could someone help me at least with few of them to give me an idea how deal with it. ***SinglyLinkedList.cpp #include #include #include \"SinglyLinkedList.h\" void test_constructor() { SinglyLinkedList lst = {100, 200, 300, 400, 500}; assert(*lst.at(0) == 100); assert(*lst.at(1) == 200); assert(*lst.at(2) == 300); assert(*lst.at(3) == 400); assert(*lst.at(4) == 500); assert(lst.size() == 5); } void test_remove() { SinglyLinkedList lst = {100, 200, 300, 400, 500}; lst.remove(2); assert(*lst.at(0) == 100); assert(*lst.at(1) == 200); assert(*lst.at(2) == 400); assert(*lst.at(3) == 500); assert(lst.size() == 4); } void test_insert() { // TODO } void test_push_back() { // TODO } void test_push_front() { // TODO } void test_append() { // TODO } void test_sum() { // TODO } int main() { test_constructor(); test_remove(); test_insert(); test_push_back(); test_push_front(); test_append(); test_sum(); } ***SinglyLinkedList.h #include #include template class SinglyLinkedList { // Nested class representing the nodes in the list. class SinglyLinkedListNode { public: // The value stored in this node. T value; // The next node in the sequence. SinglyLinkedListNode *next; SinglyLinkedListNode(T value) : value(value), next(nullptr) {} SinglyLinkedListNode(T value, SinglyLinkedListNode *next) : value(value), next(next) {} // Return the size (length) of the linked list. std::size_t size(); }; SinglyLinkedListNode *head; SinglyLinkedListNode *tail; public: // Constructs a new SinglyLinkedList from an initializer_list of type T[]. // This is mostly for convenience, especially when testing. SinglyLinkedList(std::initializer_list items) : head(nullptr), tail(nullptr) { if (items.size() == 0) { return; } // initializer_lists were designed to be used iteratively, // so thats what we do. // Can you think of how to write this recursively? auto it = items.begin(); while (it != items.end()) { this->push_back(*it++); } } // Return a pointer to the value at the given index. // If the index is larger than the size of the list, // return a nullptr. // // ASIDE: We will cover exceptions later. T* at(std::size_t i); // Pushes a new node onto the back of the list. void push_back(T value); // Pushes a new node onto the front of the list. void push_front(T value); // Return the size (length) of the linked list. std::size_t size(); // Remove the specified node from the list. void remove(std::size_t i); // Insert the value at the index. void insert(std::size_t i, T value); // Append the given list to this one. void append(SinglyLinkedList list); }; template T* SinglyLinkedList::at(std::size_t i) { // TODO } template void SinglyLinkedList::push_back(T value) { // TODO // Make sure that this is a O(1) operation! } template void SinglyLinkedList::push_front(T value) { // TODO // Make sure that this is a O(1) operation! } template void SinglyLinkedList::remove(std::size_t i) { // TODO // Don\'t forget to not only unlink the node, b.

I need to fill-in TODOs in .cpp file and in .h file Could some.pdf

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Using the C++ programming language 1. Implement the UnsortedList class to store a list of numbers that are input into the list from data.txt. - create a main.cpp file that gets the numbers from the file - insert the number 7 into the list - insert another number 300 into the list - delete the number 6 from the list - print out the following: --the entire list - the greatest - the least 2. Attach the main.cpp, UnsortedList.cpp, the ItemType.h, and the output file two called outfile2.txt Use the files below: // listDriver.cpp // Test driver #include #include #include #include #include #include \"unsorted.h\" using namespace std; void PrintList(ofstream& outFile, UnsortedType& list); int main() { ifstream inFile; // file containing operations ofstream outFile; // file containing output string inFileName; // input file external name string outFileName; // output file external name string outputLabel; string command; // operation to be executed int number; ItemType item; UnsortedType list; bool found; int numCommands; // Prompt for file names, read file names, and prepare files cout << \"Enter name of input command file; press return.\" << endl; cin >> inFileName; inFile.open(inFileName.c_str()); cout << \"Enter name of output file; press return.\" << endl; cin >> outFileName; outFile.open(outFileName.c_str()); cout << \"Enter name of test run; press return.\" << endl; cin >> outputLabel; outFile << outputLabel << endl; if (!inFile) { cout << \"file not found\" << endl; exit(2) } inFile >> command; numCommands = 0; while (command != \"Quit\") { if (command == \"PutItem\") { inFile >> number; item.Initialize(number); list.PutItem(item); item.Print(outFile); outFile << \" is in list\" << endl; } else if (command == \"DeleteItem\") { inFile >> number; item.Initialize(number); list.DeleteItem(item); item.Print(outFile); outFile << \" is deleted\" << endl; } else if (command == \"GetItem\") { inFile >> number; item.Initialize(number); item = list.GetItem(item, found); item.Print(outFile); if (found) outFile << \" found in list.\" << endl; else outFile << \" not in list.\" << endl; } else if (command == \"GetLength\") outFile << \"Length is \" << list.GetLength() << endl; else if (command == \"IsFull\") if (list.IsFull()) outFile << \"List is full.\" << endl; else outFile << \"List is not full.\" << endl; else if (command == \"MakeEmpty\") list.MakeEmpty(); else if (command == \"PrintList\") PrintList(outFile, list); else cout << command << \" is not a valid command.\" << endl; numCommands++; cout << \" Command number \" << numCommands << \" completed.\" << endl; inFile >> command; }; cout << \"Testing completed.\" << endl; inFile.close(); outFile.close(); return 0; } void PrintList(ofstream& dataFile, UnsortedType& list) // Pre: list has been initialized. // dataFile is open for writing. // Post: Each component in list has been written to dataFile. // dataFile is still open. { int length; ItemType item; list.ResetList(); length = list.GetLength(); for .

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show that the ratio of two consecutive local maximum displacements is a constant Solution Under-damped Case: If the damping is so small that b << n or r /2m<< Kim, then, p = (b2- n2) = - (n2 – b2) = (n2- b2) or p = jw, where w = (n2- b2). Obviously, p is an imaginary quantity, but w is a real quantity. Then from equation (2) X = Since x is a real quantity, therefore (A1 + A2) and j(A1 – A2) must also be real quantities. But A1 and A2 are the complex quantities, therefore (A1 + A2) = a0 sin and j(A1 – A2) = a0 cos in above .equation (3) x = e -bt [ a0 sin cos wt + a0 cos sin wt]x =ao e-bt sin (wt+).

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Researchers from a pharmaceutical company have developed a new drug that they believe will relieve long-term depression. They wish to carry out a clinical trial to compare its efficacy with an existing drug. It is known that of the drugs currently available, some will alleviate symptoms while they are being taken. Which of the following three designs would you recommend? . A group-comparative design . A matched pairs design . A crossover design Briefly justify your choice, commenting on each of the three designs. Solution In a group comparative design, experimental units are randomly assigned to treatment conditions. Randomization provides some control for lurking variables. A cross-over design is not possible because medicines are supposed to relieve long-term depression. In a matched pairs design, experimental units within each pair are assigned to different treatment levels. And here, this design should be used..

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Question: A corp is considering modernizing its production facility by investing in new equipment and selling the old equipment. The following information has been collected on this investment. Old Equipment New Equipment Cost $80,800 Cost $38,560 Accumulated depreciation $40,400 Estimated useful life 8 years Remaining life 8 years Salvage value in 8 years $4,592 Current salvage value $10,600 Annual cash operating costs $29,600 Salvage value in 8 years $0 Annual cash operating costs $36,000 Depreciation is $10,100 per year for the old equipment. The straight-line depreciation method would be used for the new equipment over an eight-year period with salvage value $4,592. a)Determine the cash payback period (Ignore income taxes). (cash payback period in years) (I know i am supposed to divide the cost of capital investment by net annual cash flow, but every time i get the amounts wrong) b)Calculate the anual rate of return c)Calculate the net present value assuming a 18% rate of return (Ignore income taxes) On this answer I was unable to calculate the annual cash flows so that i could get to the part where I use the PV table d)Should the company purchase the new equipment? Solution 4.076 (2 ,00,995.728 / 4.076) 49,312.00YearsPV @ 18%01.00010.84720.71830.60840.51650.43760.37070.31480.2661 to 8 New equipment$$Cost38,5601.00038,560Life8 yearsScrap @ Yr 8Nil0.2660Depreciation (P.A.)4,820Op. cost (P.A.)36,00040,8204.0761,66,382.322,04,942.32Annualized cost(2,04,942.32 / 4.076)50,580.25 4.076Old equipmentPV factor$$Original cost80,800Less : Acc. dep40,400Current cost40,4001.00040,400Life8 yearsScrap @ Yr 8(4,592)0.266-1,221.47Depreciation (P.A.)10,100Op. cost (P.A.)29,60039,7004.0761,61,817.22,00,995.728Annualized cost (2 ,00,995.728 / 4.076) 49,312.00.

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If an ADC gives only 8 different outputs for any input signal within its range/span, what\'s the resolution of the ADC? Explain why and show your calculation. Solution Intended Audience This document is intended for users who operate and interpret the results of a DAQ measurement system. Overview Instrument manufacturers usually supply specifications for their equipment that define its accuracy, precision, resolution and sensitivity. Unfortunately, not all of these specifications are uniform from one to another or expressed in the same terms. Moreover, even when they are given, do you know how they apply to your system and to the variables you are measuring? Some specifications are given as worst-case values, while others take into consideration your actual measurements. Accuracy Accuracy can be defined as the amount of uncertainty in a measurement with respect to an absolute standard. Accuracy specifications usually contain the effect of errors due to gain and offset parameters. Offset errors can be given as a unit of measurement such as volts or ohms and are independent of the magnitude of the input signal being measured. An example might be given as ±1.0 millivolt (mV) offset error, regardless of the range or gain settings. In contrast, gain errors do depend on the magnitude of the input signal and are expressed as a percentage of the reading, such as ±0.1%. Total accuracy is therefore equal to the sum of the two: ±(0.1% of input +1.0 mV). An example of this is illustrated in Table 1. Table 1. Readings as a function of accuracy Conditions: input 0-10 V, Accuracy = ±(0.1% of input + 1 mV) Precision Precision describes the reproducibility of the measurement. For example, measure a steady state signal many times. In this case if the values are close together then it has a high degree of precision or repeatability. The values do not have to be the true values just grouped together. Take the average of the measurements and the difference is between it and the true value is accuracy. Resolution Resolution can be expressed in two ways: It is the ratio between the maximum signal measured to the smallest part that can be resolved - usually with an analog-to-digital (A/D) converter. It is the degree to which a change can be theoretically detected, usually expressed as a number of bits. This relates the number of bits of resolution to the actual voltage measurements. In order to determine the resolution of a system in terms of voltage, we have to make a few calculations. First, assume a measurement system capable of making measurements across a ±10 V range (20 V span) using a 16-bits A/D converter. Next, determine the smallest possible increment we can detect at 16 bits. That is, 216 = 65,536, or 1 part in 65,536, so 20 V÷65536 = 305 microvolt (µV) per A/D count. Therefore, the smallest theoretical change we can detect is 305 µV. Unfortunately, other factors enter the equation to diminish the theoretical number of bits that can be used, such as noise.

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Karla Tanner opens a web consulting business called Linkworks and completes the following transactions in its first month of operations. Prepare journal entries for each transaction and identify the financial statement impact of each entry. The financial statements are automatically generated based on the journal entries recorded. The financial statements report the cumulative impact of all transactions recorded as of the financial statement date. Input the cumulative amount of a) Net Income (Loss), b) Total Assets, c) Total Liabilities, and d) Total Equity that would be reported on the financial statements immediately after each transaction is recorded. (Hint: You can check your work by selecting the date on the trial balance tab.) The first two transactions are completed for you!Apr.1Tanner invested $80,000 cash along with office equipment valued at $26,000 in the company in exchange for common stock.Apr.2The company prepaid $9,000 cash for 12 months\' rent for office space. The company\'s policy is record prepaid expenses in balance sheet accounts.Apr.3The company made credit purchases for $8,000 in office equipment and $3,600 in office supplies. Payment is due within 10 days.Apr.6The company completed services for a client and immediately received $4,000 cash.Apr.9The company completed a $6,000 project for a client, who must pay within 30 days.Apr.13The company paid $11,600 cash to settle the account payable created on April 3.Apr.19The company paid $2,400 cash for the premium on a 12-month insurance policy. The company\'s policy is record prepaid expenses in balance sheet accounts.Apr.22The company received $4,400 cash as partial payment for the work completed on April 9.Apr.25The company completed work for another client for $2,890 on credit.Apr.28The company paid $5,500 cash in dividends.Apr.29The company purchased $600 of additional office supplies on credit.Apr.30The company paid $435 cash for this month’s utility bill. Solution Transaction: Net Income Total Assets Total Liabilities Total Equity Where can you go to find each of your answers? Income statement Balance sheet Balance sheet Balance sheet Apr. 1 Tanner invested $80,000 cash along with office equipment valued at $26,000 in the company in exchange for common stock. $0 $106,000 $0 $106,000 Apr. 2 The company prepaid $9,000 cash for 12 months’ rent for office space. The company\'s policy is record prepaid expenses in balance sheet accounts. 0 106,000 0 106,000 Apr. 3 The company made credit purchases for $8,000 in office equipment and $3,600 in office supplies. Payment is due within 10 days. 0 117,600 11,600 106000 Apr. 6 The company completed services for a client and immediately received $4,000 cash. 4000 121,600 11,600 110000 Apr. 9 The company completed a $6,000 project for a client, who must pay within 30 days. 10000 127,600 11,600 116000 Apr. 13 The company paid $11,600 cash to settle the account payable created on April 3. 10000 116,000 0 116000 Apr. 19 The company paid.

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How many total bits are required for a direct-mapped cache with 2048 entries each holding an 8 word (of 32 bits or 4 bytes) block, assuming a 32-bit address? Discuss with the help of a diagram the logic to check a hit or miss (that is, presence or absence) of a required memory block in the cache described in 1(a), and access a specific word of the block in case of a hit. Label all the fields in a cache word. What is virtual memory? What is the fundamental difference between the memory hierarehy of Q.1 (a) and this one? Explain how briefly how it helps multiple users share the limited memory, and still get the illusion of using a dedicated resource? Solution 2. (a) Virtual Memory: In computing,virtual memoryis a memory management technique that is implemented using both hardware and software. It mapsmemory addresses used by a program, calledvirtual addresses, intophysical addressesin computer memory. Main storage as seen by a process or task appears as a contiguous address space or collection of contiguous segments. The operating system manages virtual address spaces and the assignment of real memory to virtual memory. Address translation hardware in the CPU, often referred to as a memory management unit orMMU, automatically translates virtual addresses to physical addresses. Software within the operating system may extend these capabilities to provide a virtual address space that can exceed the capacity of real memory and thus reference more memory than is physically present in the computer.The primary benefits of virtual memory include freeing applications from having to manage a shared memory space, increased security due to memory isolation, and being able to conceptually use more memory than might be physically available, using the technique of paging. Virtual memory makes application programming easier by hiding fragmentation of physical memory; by delegating to the kernel the burden of managing the memory hierarchy(eliminating the need for the program to handle overlays explicitly); and, when each process is run in its own dedicated address space, by obviating the need to relocate program code or to access memory with relative addressing. Memory virtualization can be considered a generalization of the concept of virtual memory. Virtual memory is an integral part of a modern computer architecture; implementations require hardware support, typically in the form of a memory management unit built into theCPU. While not necessary, emulators and virtual machines can employ hardware support to increase performance of their virtual memory implementations. Consequently, older operating systems, such as those for the mainframes of the 1960s, and those for personal computers of the early to mid-1980s (e.g. DOS), generally have no virtual memory functionality, though notable exceptions for mainframes of the 1960s include: and the operating system for the Apple Lisa is an example of a personal computer operating system of the 1980s that features vir.

How many total bits are required for a direct-mapped cache with 2048 .pdf

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Which of the following statements is (are) trueI The budget is .pdf

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Which of the following is the most appropriate way to determine how .pdf

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Which of the following is a rational number A. pi B. 1.425 C. squ.pdf

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When running reports for Mariette’s Marketing in QuickBooks Online, you notice that the income amount is unusually high but A/R balances look okay. Which of the following errors could be the reason? Solution ANSWER: There is a possbility that it received Payments against invoices, then posted deposits to income accounts. Because the deposited customer payments to an income account thus such payments have not gone towards the customers invoices..

Where is a thyme corpuscle found specifically sinus capsule medulla .pdf

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what are the 3 mechanisms by which RTKs can activate signaling enzym.pdf

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Thermodynamics Answer provided Please answer the questions with solution if you get the correct answers. Thank you Water (p 998 kg/m3 and u 0.001002 Pa s flows through the pipe system shown in the diagram on the next page. At Point 1, the water is already flowing at a velocity of 2.5 m/s through a 0.5 inch inner diameter smooth pipe. The pressure of the water at this point is 120 kPa. A pump with an efficiency of 80% is used to increase the pressure of the water. Note that after the pump there is a sudden change in diameter to 1 inch (smooth pipe). Referring to the diagram, the pressure at Point 2 is 275 kPa. The water then flows through an unknown length of pipe until a sharp exit open to the atmosphere. You may neglect Fioss at the sudden change in diameter from 05 to 1 inch a Determine the pump power (in kW) required to give a pressure of 275 kPa at Point 2 (161 12 W) b) Determine the length of pipe after Point 2 (Lin m). (824 140 m) Solution Volumetric flow rate is calculated first from inlet, by continuity of mas this must ramin const. = 2.5m/s * Area = 3.167 * 10-4m3/s Total head = given height plus friction losses estimate l/d for galve = approx 450, f = .03 for laminar flow head loss due to gate valve is .03*450 * V2/2g = 2.8 m add to head = 12.8 m pressure head = (274-120)*1000/( 998*9.81) =15.83 friction loss at 4 elbows: (l/d) =30*4 =120, head loss approx 1.1 m add to head , 14 m. Total head 30 m approx Power rhoQgH = 998* 3.167* 9.81* H = 116 w approx after factoring 80% efficiency 161 may be type o or some other friction factor used.

This figure displays five small hypothetical proteins. The a-helix se.pdf

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Solve the following initial value problem using Laplace transform.pdf

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