Sunday, May 17, 2020
Magnetars Neutron Stars with a Kick
Neutron stars are weird, enigmatic objects out there in the galaxy. Theyve been studied for decades as astronomers get better instruments capable of observing them. Think of a quivering, solid ball of neutrons squished together tightly into a space the size of a city.à One class of neutron stars in particular is very intriguing; theyre called magnetars. The name comes from what they are: objectsà with extremely powerful magnetic fields. While normal neutron stars themselves have incredibly strongà magnetic fields (on the order of 1012 Gauss, for those of you who like to keep track of these things), magnetars are many times more powerful. The most powerful ones can be upwards of a TRILLION Gauss! à By comparison, the magnetic field strength of the Sun is about 1 Gauss; the average field strength on Earth is half a Gauss. (A Gauss is the unit of measurement scientists use to describe the strength of a magnetic field.) Creation of Magnetars So, how do magnetars form? It starts with a neutron star. These are created when a massive star runs out of hydrogen fuel to burn in its core. Eventually, the star loses its outer envelope and collapses. The result is a tremendous explosion called a supernova. During the supernova, the core of a supermassive star gets crammed down into a ball only about 40 kilometers (about 25 miles) across. During the final catastrophic explosion, the core collapses even more, making an incredibly dense ball about 20 km or 12 miles in diameter. That incredible pressure causes hydrogen nuclei to absorb electrons and release neutrinos. Whats left after the core is through collapsing is a mass of neutrons (which are components of an atomic nucleus) with incredibly high gravity and a very strong magnetic field.à To get a magnetar, you need slightly different conditions during the stellar core collapse, which create theà final core that rotates very slowly, but also has a much stronger magnetic field.à Where Do We Findà Magnetars? A couple of dozen known magnetars have been observed, and other possible ones are still being studied. Among the closest is one discovered in a star cluster about 16,000 light-years away from us. The cluster is calledà Westerlund 1, and it contains some of the most massive main-sequence stars in the universe. Some of these giants are so big their atmospheres would reach to Saturns orbit, and many are as luminous as a million Suns. The stars in this cluster are quite extraordinary. With all of them being 30 to 40 times the mass of the Sun, it also makes the cluster quite young. (More massive stars age more quickly.) But this also implies that stars that have already left the main sequence contained at least 35 solar masses. This in of itself is not a startling discovery, however the ensuing detection of a magnetar in the midst of Westerlund 1 sent tremors through the world of astronomy. Conventionally, neutron stars (and therefore magnetars) form when a 10 - 25 solar mass star leaves the main sequence and dies in a massive supernova. However, with all the stars in Westerlund 1 having formed at nearly the same time (and considering mass is the key factor in the aging rate) the original starà must have been greater thanà 40 solar masses. It is not clear why this star did not collapse into a black hole. One possibility is that perhaps magnetars form in a completely different manner from normal neutron stars. Maybe there was a companion star interacting with the evolving star, which made it spend much of its energy prematurely. Much of the mass of the object might have escaped, leaving too little behind to fully evolve into a black hole. However, there is no companion detected. Of course, the companion star could have been destroyed during the energetic interactions with the magnetars progenitor. Clearly astronomers need to study these objects to understand more about them and how they form. Magnetic Field Strength However a magnetar is born, its incredibly powerful magnetic field is its most defining characteristic. Even at distances of 600 miles from a magnetar, the field strength would be so great as to literally rip human tissue apart. If the magnetar floated halfway between the Earth and the Moon, its magnetic field would be strong enough to lift metal objects such as pens or paperclips from your pockets, and completely demagnetize all of the credit cards on Earth. Thats not all. The radiation environment around them would be incredibly hazardous. These magnetic fields are so powerful that acceleration of particles easily produce x-ray emissions and gamma-ray photons, the highest energy light in the universe. Edited and updated by Carolyn Collins Petersen.
Wednesday, May 6, 2020
Examples Of Project Management - 1027 Words
The first step for me as a Project Manager will be to create an agenda with project phases with added details. It all begins with project planning, initiation, execution, monitor and control and, last step which is project closure. As a project manager my list of roles and additions at any time are as follows; â⬠¢ Ownership for project process, project deliverables and communication frameworks. â⬠¢ Manages client requests for technical work. â⬠¢ Creates and maintains project plans schedules. â⬠¢ Leads, manages and reports project progress to stakeholders and senior management. â⬠¢ Single point of contact for client during project. â⬠¢ Gathers project requirements, creates project charter, scope and structure. â⬠¢ Tracks, manages, negotiates andâ⬠¦show more contentâ⬠¦Ã¢â¬ ¢ Categorizing and approving modification requests. â⬠¢ Centralizing the management and tracking of modification activity. â⬠¢ Insuring that software inventory information is maintained. Task Mode Task Name Duration Start Auto Scheduled IS PROJECTS 30 days Tue 7/4/17 Manually Scheduled Intake from Service Manager 5 days Tue 7/4/17 Auto Scheduled Planned Work 30 days Mon 7/31/17 Manually Scheduled IS Supoorted Applications 20 days Wed 8/16/17 Auto Scheduled Intake from Development Team 1 day Tue 7/4/17 Manually Scheduled intake from IO and RD Daily 125 days Tue 7/4/17 Auto Scheduled WEEKLY Management Control Process/ MEETINGS 126 days? Fri 7/7/17 Having the management control process will help reduce downtime, increased customer satisfaction, Identification of impact and interdependencies between owners, resources and schedules. The following is the combined matrix of all stakeholders who hold a key role in the Grantham University Project. As we move along further in this proposal process the scheduling and Work Breakdown Structure (WBS) matrix will be finalized prior to commencement of any work scheduled start dates. The Project Manager also needs to ensure that the correct numbers of databases were ordered from Sales on the roll call. This is often missed on the roll call so the Technical Consultant should investigate to ensure it has been ordered as well. There is a recurring cost that is assessed the project by Hosting when the database isShow MoreRelatedIt Project Management Essay examples1277 Words à |à 6 PagesStrayer University IT Project Management Assessment CIS 517 1. Summarizes how the project manager or team exhibited exceptional and ethical project management practices. Often, the project manager (PM) is faced with an issue that is not easily resolved by theory or the knowledge acquired from formal training. These types of problems are usually not of a technical nature and more often tend to be ethical or human resource issues (Stare, 2011). The satisfactory answer is oftenRead MoreProject Management Essay example2581 Words à |à 11 Pagesmay impact on project implementation are as follows: a. The cabinet approval process. b. Office of Parliamentary counsel. c. Consultation wiyh Local Government and National Competition Policy. d. Consultation with other key Stakeholders. e. Tabling in Parliament. f. Timing of Commencement of the Act. Q.2 - Project stakeholders are individuals and organizations that are actively involved in the project, or whose interests may be affected as a result of project execution or project completion. TheyRead MoreProject Management Essay examples2230 Words à |à 9 PagesProject Management In todays business it is in the best interest of companies to have project managers. Common sense isnt always easily accomplished. Anyone whos ever worked on a project in a technical setting knows this. 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Slacksââ¬â¢ new launch ââ¬â Slack Enterprise Gridâ⬠¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦..â⬠¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦..3 5. Conclusionâ⬠¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦..â⬠¦3 6. Bibliographyâ⬠¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦Ã¢â¬ ¦..4 What is project and project management? AsRead MoreEssay on project management1748 Words à |à 7 Pagesï » ¿ Project Charter Colorado Spring Welcome Home Parade QI CHEN Project Management January 24, 2015 Abstract This project charter is planned to help Oââ¬â¢Donnell Oââ¬â¢Donnell LLP who will lead the project management team take place the parade smoothly. This parade for welcoming home troops will be organized in Colorado Spring which has a long history of military. This project charterââ¬â¢s goal is making sure project management team and sponsorsRead MoreImplementing Project Management Techniques Essay1275 Words à |à 6 Pages There are many methods and techniques for a project manager (PM) can use to run a successful project. Some of these include: identifying the stakeholderââ¬â¢s roles and responsibilities, tracking measureable business outcomes, apply project controls and view monitoring the dynamics of a working project. 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Computer Assignment Many Bits Are Needed For The Opcode
Questions: 1. A digital computer has a memory unit with 24 bits per word. The instruction set consists of 150 different operations. All instructions have an operation code part (opcode) and an address part (allowing for only one address). Each instruction is stored in one word of memory. Answer the following: a. How many bits are needed for the opcode? b. How many bits are left for the address part of the instruction? c. What is the largest unsigned binary number that can be accommodated in one word of memory? 2. Write a program to evaluate the arithmetic statement x = (a( b + c ) d - e) using a stack organized computer with zero-address instructions (so only pop and push can access memory). 3. What is the difference between using direct and indirect addressing? Give an example. 4. Suppose a byte-addressable computer using set associative cache has 221 byes of main memory and a cache of 64 blocks, where each cache block contains 4 bytes. a. If this cache is 2-way set associative, what is the f ormat of a memory address as seen by the cache, that is, what are the sizes of the tag, set, and offset fields? b. If this cache is 4-way set associative, what is the format of a memory address as seen by the cache? 5. Discuss the advantages and disadvantages of dynamic linking. ?6. A computer has 32-bit instructions and 12-bit addresses. Suppose there are 250 two-address instructions. What is the maximum number of one-address instructions that can be added to the instruction set?7. Suppose a disk drive has the following characteristics: 6 surfaces 16,383 tracks per surface 63 sectors per track 512 bytes/sector Tract-to-track seek time of 8.5 milliseconds Rotational speed of 7,200 RPM. a. What is the capacity of the drive? b. What is the access time? 8. Suppose we have the instruction LDA 800. Given the memory as follows and the base register contains the value 100. Memory address value 800 900 900 1000 1000 500 1100 600 1200 800 1300 250 What would be loaded into the AC if the addressing mode for the operand is: a. immediate b. direct c. indirect d. indexed Rationale This assessment task covers from topic 6 to topic 9, and has been designed to ensure that you are engaging with the subject content on a regular basis. More specifically it seeks to assess your ability to: be able to define and appropriately use computer systems terminologies; be able to describe the essential elements of computer organisation and discuss how the elements function; be able to describe and design the basic instruction set architecture of a simple computer; be able to discuss various programming tools available and their relationship to the computer architecture; Answers: As per given information, The word size of the memory is 24 bits long. There are 150 operations supported by the instruction set. Each instruction has an opcode and an address part. All instructions are stored in memory. As there are 150 instructions. So there will be, 8 bits for representing one operation as, Thus, 8 bits are needed for opcode. An instruction is stored in a word with 24 bits. So, there will be (24-8) = 16 bits for an address part in an instruction. The largest unsigned binary number that can fit into one word of the memory is, (111111111111111111111111)2. The postfix expression is, abc+xdxe- The program is, PUSH a PUSH b PUSH c ADD MUL PUSH d MUL PUSH e SUB POP 3. The difference between direct and indirect addressing is, in case of direct addressing the address part of an instruction holds the address of the operand. On the other hand, in case of indirect addressing, the address part contains the address of the word in memory than contains the address of the operand. For example, consider the following information, Memory Address Value 800 900 900 200 An instruction is, LOAD 800 In direct addressing scheme, the address part of instruction LOAD is 800. It will be treated as the address of the operand. Thus 900 will be loaded from memory as an operand. On the other hand, in case of indirect addressing, 800 will be considered as the address of the location that contains the address of the operand. So, control will go to 800, it will fetch 900 as the address of the operand to be loaded. Then it will go to location 900 and will fetch the value 200 as operand. The cache has 64 blocks, the set associative cache has 221 bytes of main memory. The memory is byte addressable. Each block has 4 bytes. The cache is 2 way set associative and The format of the memory address will be, Tag Set Offset 5 1 2 1. The cache is 4 way set associative. The format of the memory address will be, Tag Set Offset 4 2 2 The advantages of dynamic linking are, It reduces overhead during loading. The linking is performed during runtime. Code sharing is easier in case of dynamic linking. The disadvantages of dynamic linking are, There may be aliasing problem. There may be code cracking problems. The computer has 32 bit instructions and 12 bits address. There are 250 two address instructions. So, each of the two addresses instructions will have (32 (2x12)) = 8 bits for opcode. If there are 8 bits for opcode then there will be so, each of the two addresses instructions will have (32 (2x12)) = 8 bits for opcode. If there are 8 bits for opcode then there will be possible instructions. But only 250 are used. So there are 6 leftover combinations of 8 bit opcode. These 6 will be used by 1 address instructions. On the other hand each of the one address instruction will have 12 bits gained from address fields. Thus there will be maximum of number of one address instructions. The disk drive has, Capacity of () bits = (61638363512) bytes = 3023.815 MB The tract to tract seek time is 8.5 ms and rotational speed is 7200 RPM. The access time will be, = = = 8.5 + 4.167 ms = 12.667 ms. The instruction is LDA 800. The value of the base register has value 100. Addressing Modes Calculation Immediate In immediate addressing, 800 will be considered as the operand value. So AC will have 800. Direct In direct addressing, 800 will be considered as the address of the operand. So control will fetch value from memory address 800. AC will have value 900. Indirect In indirect addressing, 800 will hold the address of the memory location that holds the address of the operand. So, control will go to 800, then it will get 900 as the location for operand. It will then fetch value from 900. AC will have value 1000. Indexed In case of indexed scheme, the address of the operand will be (800+100) = 900. AC will have value from 900. So, AC will have value 1000. References Hennessy, J. L., Patterson, D. A. (2012). Computer Architecture, A Quantitative Approach (5th ed.). Elsevier. Mano, M. (1993). Computer System Architecture. Pearson Education. Mano, M. M. (2010). Digital Logic and Computer Design. Pearson. Null, L., Lobur, J. (2014). The Essentials of Computer Organization and Architecture (4th ed.). Jones Bartlett Publishers.
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