Pressure distribution on an ellipto-zhukovsky aerofoil Essay Example

Weight dissemination on an ellipto-zhukovsky aerofoil Paper The weight dissemination around an Ellipto Zhukovsky aerofoil with a harmony of 254 mm at a scope of approaches (- 4? , 7? what's more, 15? ) was resolved and constrain commitments to lift were assessed in a T3 air stream at City University. This was completed at a harmony Reynolds number of 3. 9 x 105. Diagrams for lift and pitching second coefficients were plotted against approaches. A chart for Cm and Cl was likewise plotted from which the streamlined focus was resolved to be 23. 7%. The estimation of lift bend slant was resolved to be 4. 4759. Henceforth the estimation of k (the proportion of the real lift bend incline to the hypothetical one) for this aerofoil was resolved to be 0. 917. The estimation of Cmo was additionally seen as 0. 0172. Example estimations for 15 degrees approach can be found in the informative supplement segment. Rundown OF SYMBOLS Cp Pressure Coefficient Cpu Pressure Coefficient of upper surface Cpl Pressure Coefficient of lower surface Cl Lift Coefficient Cm Moment Coefficient x/c Position of weight tapping on aerofoil isolated by harmony length Px Pressure at tapping x (Pa) Patm Atmospheric Pressure (Pa) ? Thickness of air (kg/m3) I Dynamic consistency ? Kinematics thickness (m/s2) We will compose a custom article test on Pressure dissemination on an ellipto-zhukovsky aerofoil explicitly for you for just $16.38 $13.9/page Request now We will compose a custom exposition test on Pressure dispersion on an ellipto-zhukovsky aerofoil explicitly for you FOR ONLY $16.38 $13.9/page Recruit Writer We will compose a custom exposition test on Pressure dispersion on an ellipto-zhukovsky aerofoil explicitly for you FOR ONLY $16.38 $13.9/page Recruit Writer h Digital manometer perusing ? point of which manometer is slanted D or t Diameter of chamber (mm) h burrow stature (mm) V Velocity of wind stream (m/s) R Molar gas consistent (J/kg. K) T Temperature (K) Re Reynolds Number INTRODUCTION An airfoil is any piece of a plane that is intended to create lift. Those pieces of the plane explicitly intended to create lift incorporate the wing and the tail surface. In present day airplane, the fashioners for the most part give an airfoil shape to even the fuselage. A fuselage may not create a lot of lift, and this lift may not be delivered until the airplane is flying moderately quick, however all of lift makes a difference. The primary effective aerofoil hypothesis was created by Zhukov sky and depended on changing a circle onto an aerofoil-formed shape. This change gave a cusped trailing edge, thus the change was adjusted to get a slim semi-overshadow trailing edge, which offered ascend to the name Ellipto Zhukovsky. At the point when a flood of wind currents past an aerofoil, there are nearby changes in speed around the aerofoil, and therefore changes in static weight as per Bernoullis hypothesis. The conveyance of weight decides the lift, pitching second, structure drag, and focal point of weight of the aerofoil. In our trial we are worried about the impact of weight circulation on lift, pitching second coefficient (Cm), and focus of weight. The focal point of weight can be characterized as the point on the aerofoil where Cm is zero, and along these lines the streamlined impacts by then might be spoken to by the lift and drag alone. A positive weight coefficient suggests a weight more prominent than the free stream esteem, and a negative weight coefficient infers a weight not exactly the free stream esteem (and is regularly alluded to as attractions). Additionally, at the stagnation point, Cp has its most extreme estimation of 1 (which can be seen by plotting Cp against x/c). Zhucovsky guaranteed that the aerofoil produces adequate flow to discourage the back stagnation point from its situation, without dissemination, down to the (sharp) trailing edge. There is adequate proof of a physical sort to legitimize this theory and the accompanying brief portrayal of the Experiment on an aerofoil may serve supportive. The analysis centers around the weight conveyance around the Zhucovsky airfoil at a low speed and the qualities related with an airfoil:â coefficient of lift,â coefficient of pitching momentâ and focal point of weight. The airfoil is made sure about to the two sides of the air stream with pressure tappings made as little as conceivable not to influence the flow,(appendix-photograph 1 . The weight distinction around the airfoil is estimated with twenty-five manometer readings which are recorded for each approach. The manometer liquid is liquor and has a particular gravity of 0. 83 and slanted at an edge of 30 degrees. Cylinder 1 is left open to climatic weight, while tubes 2-13 are the lower surface of the airfoil and cylinders 14-24 are the upper surface of the airfoil. The weight recordings are situated on the airfoil a ways off x/c, noted in the outcomes table and cylinder 35 is the static weight of the air stream. The dynamic weight is given by an advanced manometer. The advanced readout results were utilized for all estimations since they are increasingly exact. Results Raw information and determined qualities for x/c, Cp and Cp(x/c) can be found in the reference section. Diagrams of Cp against x/c for approaches - 4, 7, and 15 degrees can be likewise be found in the reference section. These charts decide the lift coefficient. Tallying the squares strategy was utilized to decide the estimations of Cl. Charts of Cp*(x/c) against x/c for approaches - 4, 7, and 15 degrees can be additionally be found in the supplement. These diagrams decide the pitch second coefficient. Checking the squares strategy was utilized to decide the estimations of Cm. Diagrams of Cl against approach ,Cm against approach, and Cm against Cl can be found in the addendum. Additionally underneath is a rundown of the outcomes: Angle of Attack (degrees) Cl Cm - 4 - 0. 513 0. 153 7 0. 740 - 0. 166 15 0. 946 - 0. 183 Discussion The investigation was directed in a low speed, shut air stream, working at around half of its speed. The aerofoil was mounted in the air stream and its weight recordings associated with a manometer slanted at 30 degrees to the flat. The tallness of the fluid in every manometer tube spoke to the weight following up on every one of the aerofoil recordings. The weight in the working area, and the weight at the venturi delta were considered, and a subsequent air stream speed was shown on an advanced manometer. The Reynolds number was determined (see reference section. Estimations of Cl and Cm for different approaches were gotten from different gatherings leading the investigation, and were utilized to get increasingly precise diagrams. It was additionally discovered that the slant of the Cl against approach chart was 4. 4759, which was not generally near the hypothetical estimation of 7. 105. The streamlined focus was determined at 23. 7% of the harmony length (from the slant of the Cm against Cl chart). It was discovered that the lift expanded with approach, to a limited extent where the aerofoil encounters slow down, and a sensational loss of lift happens. As there was little change in the lower surface weight conveyance, the lift was mostly created because of the upper surface attractions. As the approach builds, the tallness of the upper surface attractions pinnacle should increment, and push ahead, showing that the focal point of weight is pushing ahead. Notwithstanding, tentatively this was not unmistakable, and can be credited to a potential unsettling influence in the weight dissemination around the aerofoil. At zero degrees approach, for a balanced aerofoil, lift and Cm should rise to zero. The explanation that they were not zero implies that the aerofoil more likely than not had a little approach. The inconsistency between the hypothetical and trial estimation of lift bend slant is because of limit layer impacts, and the impact of the thickness of the aerofoil, and along these lines the hypothetical worth should be duplicated by the k esteem (=0. 917) to get the exploratory outcome. End The point of the test was accomplished with a generally decent degree of test precision. The weight conveyance over an aerofoil contributes towards the lift and pitching second coefficient, where the expansion in pull on the upper surface (because of an expanded approach) builds the lift, and pitching second coefficient. The variety of weight circulation additionally influences the area of the focal point of weight. The variables which influenced the weight circulation, were essentially the thickness and the Reynolds number. In any case, with regards to contrasting the outcomes and their hypothetical qualities it is obvious to see that there have been noteworthy mistakes have happened in the examination. These are recorded underneath. Human blunders in perusing of the manometer tubes. Where a few people were included and this prompted various strategies being utilized it would have been best for everybody to take their own arrangement of readings and the normal worth determined utilizing all the information. The most widely recognized mistake without subterranean insect question was parallax and this could have been maintained a strategic distance from by utilizing computerized estimating gadgets. Figuring mistakes I. e. adjusting, transformation mistake and blunder happening when the territory under the diagrams was determined for the coefficient of lift.â Experimental mistakes a portion of the tapping may have been faulty and insufficient tapping were given. Likewise to get a superior lift bend slant there ought to have more approaches. Additionally any deterrents before the air stream, for example, individuals would make pointless choppiness inside the air stream. Informative supplement Specimen Calculations.

Are the Royal Family a Luxury we can no Longer Afford?

Cheers ascend from the happy group as the delegated second we’ve all been hanging tight for at last shows up. Thousands fill the Mall, extending their necks as Queen Elizabeth II ventures out onto her overhang, offering a grin to her revering subjects. Clad in her amazed cap, she reviews the scene: a huge ocean of British banners, unglued in their help of sixty years on the seat, waving so as to a boisterous interpretation of God Save the Queen. It presumably never entered anyone’s heads that the financing was taken from their duties as they wore silly red, white and blue manifestations for the sake of patriotism.No one asked the amount it was costing. Or on the other hand who was paying for everything! I wasn’t disturbed at the time either, nor the school children and laborers, glad for a three day weekend and a reason to observe television. The individuals who went to see it in person couldn’t even observe the Queen appropriately, except if they were squ eezed against the doors of Buckingham Palace. On the off chance that they might, they be able to weren’t ready to see each Swarovski precious stone they had paid for on her outfit. While she was having a great time, waving to her subjects, I figure she was attempting to push one annoying idea to the rear of her mind.That by letting the nation take a vacation day, old fashioned Liz had interfered with us ?1. 2 billion. I’m sure a downturn is the point at which a nation has no cash to spend, which clarifies the activity misfortunes and significant cuts being made. So where is this cash coming from? For reasons unknown, the citizens have just paid for organizing the Jubilee show, the group control and the colossal TV screens so everybody could watch the situation develop from the Mall. They never inquired as to whether we needed to waste our assessments on such a unimportant occasion, rather than something that would profit us, as, maybe, uncovering us from underneath rec ession.The day may have helped open soul, yet soul won’t take care of the tabs, particularly in this season of assumed somberness. The contention from those in control was that the travel industry produced would support the economy. They offered the rights to the inclusion to more than one hundred and forty nations around the globe, trusting it would take care of a significant part of the expense, if not make a benefit, which would go to the Diamond Jubilee Trust. The Trust is intended to profit nations in the Commonwealth, and ‘deliver notable undertakings †¦ that are a fitting and suffering tribute to Her Majesty the Queen’, as per their website.These remember speculations for territories, for example, sport for young people, helping the impaired, and securing our country’s legacy. Regardless of whether the cash was raised is by all accounts an undisclosed issue, maybe an unmistakable sign that the benefit they initially foreseen wasn’t came to. The way that the Diamond Jubilee is being praised overall maybe shows what a noticeable figure the Queen is the entire world over. Perhaps that has gone to her head, as it is no uncertainty her British supporters paying for the driving and security for her.Why would it be a good idea for us to hack up so she can swan off to Timbuktu or something like that? Maybe she’s got onto the way that her position isn’t very what it used to be the point at which the government could arrange a criminal to be decapitated as effectively as calling their house cleaner for some tea. Maybe she’s upset since she has scarcely any forces, other than passing out honors to saints and famous people. She even had a job in the James Bond sketch for the Olympics to support her appeal.It’s dismal to perceive what tricks the old dear has been decreased to simply to increase open gratefulness, however of course, perhaps it’s time she focused and understood that her position doesn’t have a similar status it used to. At the point when you tot up all dear Queenie’s accounts however, it turns out she’s just got a small ?310 million for herself, which must be a terrible hardship for the multi year old. How she adapts to just being 262nd on the Sunday Times Rich List I’ll never know. On the off chance that you can’t be the most extravagant individual in your own nation, at that point what’s the point?All the energetic hours she gives to us, the difficult work she places in she’s still not number one. Appears to be uncalled for, however wouldn’t you fill in as hard if the prize was a personal jet to ship you around the globe, and a group of security to ensure you. I would gladly invest more exertion on the off chance that I was given such sumptuousness consequently. However, I feel that, in spite of her conspicuous cash inconveniences (?310 million of them! ), it’s great she makes good on gatheri ng charge. Paying her way in her own nation is a respectable activity. She doesn’t need to, yet the fact is, she picks to.I’m sure she’d be happy to pay everything as well, if the unique rate orchestrated her wasn’t so engaging. ?1,375 for a royal residence, what a deal! That’s a similar sum my folks pay for their three-room house in Scotland. Something doesn’t very include. Focal London area, a greater number of rooms than I want to tally, and as yet paying not exactly the vast majority in Britain. Most likely not exactly a few people hit hardest by neediness. However, they don’t mind, in any event she pays it, isn't that so? Figures from the Queen’s accounts show that we each paid 52p towards the upkeep of the Royal Family last year.The cash is from charges, which I am certain should go towards running essential open administrations that will profit us. Do the Queen and family fit that bill? I guess they offer an assistance o f sorts: making a special effort to help good cause and show up at significant occasions, demonstrating the country what awesome authority we have. That’s around ?35 million every year we pay to keep Her Maj however, which could pay for any number of things, from a superior equity framework to better clinical research to help spare the lives of her people.Surely she can see (with or without her glasses) that our assessments could be put to all the more likely use. In spite of the fact that, it’ll before long be more than 52p we’re spending, with the great declaration of Kate and Wills anticipating their first youngster. Another person to sprinkle out on. Two private experts at the esteemed King Edward VII clinic inspected Kate subsequent to being conceded with ‘hyperemesis gravidarum’ or intense morning infection. It more likely than not been extremely intense to warrant two specialists, each on a pay of around ?125,000 a year.Surely, as experts, it would have just taken one to affirm she was pregnant and it was causing her evil wellbeing. Such consideration and she’s not blood-identified with the Royals; she’s just wedded to the second-in-line. Maybe it’s on the grounds that this kid, be it a kid or a young lady, will one day be administering our nation. On the other hand, if their spending continues spiraling, there won’t be a lot of a nation to run. Possibly we can no longer bear the cost of the Royal family. Or maybe like Gran’s best china, they’re just brought out on extraordinary events to include some glamour.The rest of the time they sit on an extremely stupendous rack gathering dust until they are required once more. At that point they should be cleaned at extraordinary cost by us to ensure they are looking glorious for their inconsistent appearances. Are the expenses of capacity and upkeep worth keeping up a generally nostalgic Britain on the uncommon case that they are requi red to show up at an occasion? Cutting the financial plan for the Monarchy may should be something that is considered soon, or probably the Royals will wind up out of the extravagance they are so acquainted with.

How to write a programming assignment with Perfection

How to write a programming assignment with Perfection This Programming help blog post for those students who are seeking a degree in software engineering and its applications. Here we are going to mention how to write a programming assignment. Before you start writing your assignment, you need to have some basic knowledge about C programming, C++, Java, and Matlab. But some students dont even have the basic idea about the subjects referenced above, at that point, it will hard for them to finish their Assignments inside the given due date. With the assistance of this programming assignment help blog, programming aspirants can get tips to complete their programming assignments before the deadline. What is Programming? Summary What is Programming?How to write a programming assignment?UnderstandExpected behaviorSupporting Info MappingAssemble TestingTypes of Programming languages and their Uses: PythonJavaRubyHTMLJavaScriptCC++C#Objective CPHPSQLSwiftFuture of Programming Programmers use programming languages to speak with PCs. Various Programming languages exist, and every programming language has its very own intended purpose, however, they all offer different kinds of applications few are fast few are easy to learn. Some languages use to make projects to tackle issues or interpret the information. With a solid requirement for remarkable and various programming, it is basically difficult to make a single all-inclusive programming language that addresses all issues. Programming is regularly improving and even join with different languages after some time, advancing to meet our changing innovative requirements. To tackle the programming assignments our experts give you the best programming assignment help. How to write a programming assignment? Follow the below steps to complete your assignment. In Case you are not able to complete your assignment before the deadline you must hire call tutors for your Programming Assignment Help. Understand Understand the problem statement. What is the problem asking you to do? What are your expected inputs and outputs? Expected behavior Check steps your program need to take to get from the given input to the best output? What is your algorithm for solving the problem? Supporting Info What additional information and requirements does the problem statement give you? Does it suggest certain methods? What information do you already know that can help? What questions do you have? As you learn the answers, add to your info. Mapping Match your supporting data to the means in your supporting data. For each progression, what bit of supporting data will enable you to achieve the progression? You should need to make a table to keep arrange your means and information. Assemble Utilize your data from the past strides to really compose the program. Your calculation and mappings should make this progression generally straightforward. Testing Use the test cases you identified in Step to test your solution. If you encounter errors, recheck your steps and information, look for error messages, and analyze the actual results versus expected results. Types of Programming languages and their Uses: You can check here the Programming languages in detail. also, we are providing the service in all programming languages given below. you can get the best Programming assignment help from call tutors experts with a quality solution Python Python is an open-source programming language use by programming engineers and back-end Web designers. it is used for logical processing, and it is generally easy to learn. Calltutors also provides the python Programming assignment help online. Java It is powerful in Web-based advancement, and it was made in 1995. Numerous organizations in the wellbeing sciences, instruction, and fund businesses use Java. Java enables the downloading of applets from sites, which empower programs to play out extra capacities. Java is little complex so assignment for Java Programming takes more time. Call Tutors experts will help you in understanding the code better also provide you assistance for all Programming assignment help in a better way. Ruby Ruby is an open-source scripting language that coders can utilize freely or related to Ruby on Rails. NASA utilizes Ruby in its work with reproductions. HTML It helps widely in Web development. HTML is the code that fills in as the establishment of Web pages, enabling individuals to make and structure electronic archives for the survey on the web. JavaScript It is used by Web designers and programming specialists to control page components to make them all the more captivating. JavaScript improves HTML, and it is fixed in most Internet programs. C C is a general-purpose, powerful computer programming language use by programming designers and frameworks investigators. Software engineers use C to make applications that join with working frameworks. C++ C++ created in 1983, is another center dimension programming language and fills in as an expansion of C. Software engineers use C++ to make amusements, designs, and office applications. C# C# is a programming language utilized by programming engineers who make applications designed to work with Windows working frameworks. Objective C Objective C is an object-oriented and general-purpose programming language used by mobile developers and software engineers. This language is mostly using by Apple PHP PHP was released to the public in 1995. It stands for Hypertext Preprocessor. We use PHP for creating open-source dynamic web pages. Php is a server-side scripting language, so you can collect data with this and create dynamic web pages. SQL SQL empowers software engineers to make, read, update, and erase data in a database. Organizations use SQL to assemble information. Swift Apple uses the Swift programming language to create and maintain iOS and OS X applications. Apparently, Objective-C and Swift both are using by apple but swift is faster than objective-c. We know Swift Programming is little tough so Assignment to make for Swift programming language take too much time so its better to hire Programming Assignment help Experts to do your Assignment. If you think you have less time to complete your assignment So, you can take programming help from Call tutors experts Future of Programming Moving innovation guarantees that programming languages will keep on developing. Be that as it may, anticipating the eventual outcome of programming can be testing.As more devices and gadgets which work with a PC chip, programming should be used routinely to stay up with the latest and working effectively.Software engineers constantly face the difficulties of shielding gadgets from infections and creating applications that enable clients to utilize their gadgets firmly.The freshest programming languages will be quicker and increasingly instinctive with fewer mistakes and issues. For instance, R is a standout amongst the latest programming dialects and plan by analysts for information investigation. How Call Tutors Experts will provide you Programming Assignment help? Writing computer programs is an Interesting subject and unpredictable too. On the off chance that you need to compose an Assignment identifying with the programming language, at that point you need time and persistence in the event that you dont have these two capacities, at that point. 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The Right to Die in Assisted Suicide - 2517 Words

Initial Thoughts on Physician Assisted Suicide (February 28th, 2013) The promotion of physician assisted suicide has sparked a debate throughout the world. From my point of view, assisted suicide is doctors assist patients who could not endure the pain of diseases and are voluntarily given lethal amount of substances resulting in death. However, physician assisted suicide might be considered to be deviant in many countries currently due to the religions, laws and the negative image. Also, the physicians who assist their patients to suicide might be labelled as killers. For instance, Jack Kevorkian, who was known for successfully assisting more than 130 patients to end their lives, was charged with second degree murder and was†¦show more content†¦The patient who request to assisted in suicide must be diagnosed with incurable illnesses and constantly suffering from severe pain. 2. The patient are mentally healthy and that they understand the alternatives are provided (e.g. continue receiving treatments) yet still want to commit suicide by doctors’ assistance. In addition, further observation should be applied if the patient is diagnosed with depression. 3. Before proceeding, the doctor must not encourage the patient to be assisted suicide unless the patient â€Å"clearly and repeatedly† express his/her own free will to end his/her life (Dixon). 4. The physician must fully understand the patient’s struggles and should have established â€Å"a meaningful relationship† with the patient (Dixon). Also, alternative treatments should be provided by the doctor anytime during assisted suicide if the patient changes his/her mind. 5. Another consulting physician must be involved with the final confirmation from the patient and his/her family. 6. Related documentation should be signed by the patient and family with consent. The documentation should be clarified that the physicians would not be prosecuted resulting in receiving criminal sanctions by assisting the patient in suicide. After all the criteria of carrying out assisted suicide have been met, the method applied in assisted suicide is that patients are injected a fatal dose of medication by assistors. And ideally, the patient’s physicianShow MoreRelatedAssisted Suicide And The Right Of Die Essay1792 Words   |  8 PagesAssisted Suicide and the Right to Die Assisted suicide and the right to die has been a controversial topic in both society and the medical field. Many people are against assisted suicide due to religious or personal beliefs. This topic has many viewpoints and different reasons behind its position. These viewpoints vary from person to person, but only some main points can be covered. 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Freedom much like love is essential to life, a right...

Freedom: much like love is essential to life, a right given to all creatures as soon as they take their first breath, and most importantly; freedom, much like love, will always prevail. Dr. Martin Luther King Jr. thoroughly understood the symbiotic relationship that freedom and love had on equality. Therefore, if Dr. King was still alive he would not advocate the revision Section 4 of the Civil Rights Act of 1965 to include a new formula that would coerce a stronger watch of the federal government to states that were once offenders of equal opportunity as it pertained to voting. The sole reason why Dr. King would not advocate a revision of Section 4 is simple: he dedicated his entire life to forgiveness, peace, and using love to forget the†¦show more content†¦It is an unfair burden and embarrassment to states covered with this formula which mandates an unusual amount of federal oversight. Dr. King would surely concur. Dr. King’s greatest inspiration for his life†™s work was Ghandi: who was a strong believer in â€Å"turning the other cheek† so to speak. â€Å"The weak can never forgive. Forgiveness is the attribute of the strong†, said Ghandi. Dr. King closely followed this rationale as seen through his peaceful protests without any retaliation, or talks of violence. Why then, would a man so devout to love, forgiveness, and equality want to continue to force retributions to past oppressors? The coverage formula used in Section 4 carried out its duties but it has also run its course; it is time for a modern America that can rely on love to create equality. â€Å"He who is devoid of the power to forgive is devoid of the power to love.† (Martin Luther King Jr.) A new America cannot exist without the removal of outdated stipulations in law that would set America free to show its love to the world. The 5-4 vote in the Supreme Court which turned down Section 4 was not an example of second generation barriers as Justice Rut h Ginsburg would suggest; rather, it is an indication of how for America has come from its regrettably dark past to the bright future. (NYtimes) Justice Ginsburg calls the court’s decision a disservice to Dr. King, but that is not the case, it is in fact,Show MoreRelatedWhat Is an Essay?1440 Words   |  6 PagesBuscemi Essay #3 Rough Draft An essay is a creative written piece in which the author uses different styles such as diction, tone, pathos, ethos or logos to communicate a message to the reader using either a personal experience, filled with morals and parables, or a informative text filled with educational terms. Educational terms could mean the usage of complicated and elevated words or simply information you would get in schools. 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Nss Phy Book 2 Answer Free Essays

1 1 2 3 C Motion I 7 (a) From 1 January 2009 to 10 January 2009, the watch runs slower than the actual time by 9 minutes. Therefore, when the actual time is 2:00 pm on 10 January 2009, the time shown on the watch should be 1:51 pm on 10 January 2009. Practice 1. We will write a custom essay sample on Nss Phy Book 2 Answer or any similar topic only for you Order Now 1 (p. 6) D (a) Possible percentage error 10 ? 6 = ? 100% 24 ? 3600 = 1. 16 ? 10 % 1 (b) = 1 000 000 days 10 ? 6 –9 It would take 1 000 000 days to be in error by 1 s. (b) Percentage error 9 = ? 100% 9 ? 24 ? 60 = 6. 94 ? 10–2% 4 (a) One day = 24 ? 60 ? 60 = 86 400 s Practice 1. 2 (p. 15) 1 2 3 4 5 C B D D (b) One year = 365 ? 86 400 = 31 500 000 s 5 Let t be the period of time recorded by a stop-watch. Percentage error = 0. 4 ? 100% ? 1% t t ? 40 s (a) Total distance she travels 2 ? ? 10 2 ? ? 20 2 ? ? 15 + + = 2 2 2 = 141 m (b) Magnitude of total displacement = 10 ? 2 + 20 ? 2 + 15 ? 2 = 90 m Direction: east Her total displacement is 90 m east. The minimum period of time is 40 s. 6 (a) Percentage error error due to reaction time = ? 100% time measured 0. 3 = ? 100% 10 = 3% 6 7 His total displacement is 0. With the notation in the figure below. (b) From (a), the percentage error of a short time interval (e. g. 10 s) measured by a stop-watch is very large. Since the time intervals of 110-m hurdles are very short in the Olympic Games, stop-watches are not used to avoid large percentage errors. Since ZX = ZY = 1 m, ? = ? = 60 °. Therefore, XY = ZX = ZY = 1 m The magnitude of the displacement of the ball is 1 m.  © 8 (a) The distance travelled by the ball will be longer if it takes a curved path. 7 (a) Length of the path = 0. 8 ? 120 = 96 m (b) No matter which path the ball takes, its displacement remains the same. (b) Length of AB along the dotted line 96 = 30. 6 m = (c) Magnitude of Jack’s average velocity 30. 6 ? 2 = = 0. 51 m s–1 120 Practice 1. 3 (p. 23) 1 B Total time 5000 5000 = + = 9821 s 1. 4 0. 8 5000 + 5000 = 1. 02 m s–1 Average speed = 9821 Practice 1. 4 (p. 31) 1 2 C B Final speed = 1. 5 ? 1 – 0. 2 ? 1 = 1. 3 m s–1 2 C Total time = 9821 + 10 ? 60 =10 421 s 5000 + 5000 Average speed = = 0. 96 m s–1 10 421 3 A By a = 3 D When the spacecraft had just finished 1 revolution, the spacecraft returned to its starting point. Therefore, its displacement was zero and its average velocity was also zero. v ? u , t v = u + at 36 = + ( ? 1. 5) ? 2 3. 6 = 7 m s–1 = 7 ? 3. 6 km h–1 = 25. 2 km h–1 Its speed after 2 s is 25. 2 km h–1. 4 5 D (a) Average speed 100 = = 10. m s–1 9. 69 (b) Yes. This is because the magnitude of the displacement is equal to the distance in this case. 4 B Take the direction of the original path as positive. Average acceleration of the ball ? 10 ? 17 = 0. 8 = –33. 8 m s–2 The magnitude of the average acceleration of the ball is 33. 8 m s– 2. v ? u By a = , t 100 ? 0 v ? u 3. 6 t= = = 4. 27 s a 6. 5 6 (a) Two cars move with the same speed, e. g. 50 km h–1, but in opposite directions. (b) A man runs around a 400-m playground. When we calculate his average speed, we can take 400 m as the distance and his average speed is non-zero. But since his displacement is zero (he returns to his starting point), his average velocity is zero. 5 The shortest time it takes is 4. 27 s.  © 6 Time / s –1 4 0 2 4 6 17 8 22 D Average speed 80 + 60 = 5 = 28 km h–1 Average velocity = Speed / m s 2 7 12 v ? u 22 ? 2 a= = 2. 5 m s–2 = t 8 The acceleration of the car is 2. 5 m s–2. 7 (a) I will choose ‘towards the left’ as the positive direction. 80 2 + 60 2 5 (b) 5 = 20 km h–1 C Total time 10 10 = + 2 3 = 8. 33 s v ? u , t u = v ? at = 9 ? (? 2) ? 3 = 15 m s–1 –1 (c) By a = Average speed 20 = 8. 33 = 2. 4 m s–1 Her average speed for the whole trip is 2. m s–1. The initial velocity of the skater is 15 m s . 8 (a) The object initially moves towards the left and accelerates towards the left. It will speed up. 6 7 8 9 10 C C C B A Magnitude of displacement = 2000 2 + 6000 2 = 6324. 6 m Magnitude of average velocity 6324. 6 = 4 ? 3600 = 0. 439 m s–1 6000 tan ? = 2000 ? = 71. 6 ° His average velocity is 0. 439 m s–1 (S 71. 6 ° E). (b) The object initially moves towards the right and accelerates towards the left. It will slow down. Its velocity will be zero and then increases in the negative direction (moves towards the left). Revision exercise 1 Multiple-choice (p. 5) 1 2 3 C D B  © 11 C Total time = 13 min = 780 s 840 ? 2 = 2. 15 m s ? 1 Average speed = 780 (b) Displacement from Sheung Shui to Lok Ma Chau 1000 = ? 6. 3 1 = 6300 m Magnitude of average velocity 6300 = 359 = 17. 5 m s–1 (1M) (1A) (1M) (1A) 12 13 D (HKCEE 2003 Paper II Q3) Conventional (p. 37) 1 Total time left for the two players = 4 ? 60 + 9 + 5 ? 60 + 16 = 565 s Total time they have been playing = 2 ? 60 ? 60 ? 565 = 6635 s (= 110 min 35 s = 1 h 50 min 35 s) (1A) 5 (a) Total distance = 1500 + 40 ? 1000 + 10 ? 1000 = 51 500 m Total time = 2 ? 3600 + 3 ? 60 + 8 = 7388 s Average speed 51 500 = 7388 = 6. 7 m s–1 (1M) (1A) 2 (a) 50 m (1A) (b) Ma gnitude of average velocity of Kitty 50 = (1M) 1? 60 + 15 = 0. 667 m s ? 1 (1A) (1M) (1A) (c) Average speed of the coach 5 + 50 + 5 = 1? 60 + 15 = 0. 8 m s ? 1 (b) Swimming: Average speed 1500 = 21 ? 60 + 28 = 1. 16 m s–1 Cycling: Average speed 40 000 = 1 ? 3600 + 1 ? 60 + 53 = 10. 8 m s–1 Running: Average speed 10 000 = 39 ? 60 + 47 = 4. 19 m s–1 (1M) His average speed was the highest in cycling. (1A) 3 (a) Since she measures the time interval based on 1 cycle of the pendulum, the error (0. 3 s) in measuring the cycle of the pendulum accumulates. is from 8 to 14 s. 1A) (1A) The range of the time interval (10 cycles) (b) When finding the time for one pendulum cycle, Jenny should time more pendulum cycles (e. g. 20) with the stop-watch and divide the time by the number of cycles. (1A) 4 (a) Time required 7. 4 ? 1000 = 20. 6 = 359 s (5 min 59 s) (1M) (1A)  © (c) Yes. Since the time interval of this competition is quite long, (1A) using stop-watch will not result in large percentage error as the reaction time for an average person is only 0. 2 s. (1A) (1M) (c) Total time = 5 min 45 s ? 1 min 58 s = 3 min 47 s = 3 ? 60 + 47 = 227 s v? u a= (1M) t 431 ? 0 = 3. = 0. 527 m s–2 (1A) 227 The average acceleration of the train is 0. 527 m s–2. 6 (a) v = u + at =0+6? 4 = 24 m s–1 = 86. 4 km h 86. 4 km h . –1 –1 (1A) The maximum speed of the car is 8 (1M) (a) Total distance = 8000 + 4000 + 5000 = 17 000 m Total time = 1 ? 3600 + 30 ? 60 + 45 ? 60 (b) v = u + at = 24 + (–4) ? 2 = 16 m s –1 –1 = 57. 6 km h (1A) –1 = 8100 s Average speed 17 000 = 8100 = 2. 10 m s–1 (1M) (1A) (c) The final speed of the car is 57. 6 km h . v? u a= (1M) t 16 ? 0 = 6 = 2. 67 m s–2 2. 67 m s–2. (1A) The average acceleration of the car is (b) 7 (a) Average speed 30 000 = 8 ? 60 = 62. m s–1 The average speed of the train is 62. 5 m s–1. (1M) (1A) (b) Maximum speed 430 = = 119. 4 m s? 1 average speed 3. 6 (1A) The average speed must be smaller than the maximum speed because the train needs to speed up from start and slows down to stop during the trip. (1A) Magnitude of displacement = 3000 2 + 4000 2 = 5000 m Magnitude of average velocity 5000 = = 0. 617 m s–1 8100 4000 tan ? = 3000 (1A) ? = 53. 1 ° His average velocity is 0. 617 m s (N 53. 1 ° E).  © –1 (1A) 9 (a) Distance travelled = 10. 5 ? 3 ? 60 = 1890 m (1M) (1A) 10 (a) Total distance = (120 + 50) ? 1000 = 170 000 m (1M) (1A) b) Circumference of the track =2 r = 2 (400) = 2513 m The distance travelled by Marilyn is 3 1890 m which is about of the 4 circumference. (1A) (b) N ?XYZ is a right-angled triangle. Z ? 50 km 30 ° Y 60 ° X ? ? 120 km Magnitude of displacement (from town X to town Z) = 120 000 2 + 50 000 2 = 130 000 m 120 tan ? = 50 ? = 67. 4 ° Magnitude of displacement AB = 400 2 + 400 2 (1A) (1A) ? = 90 ° ? 67. 4 ° = 22. 6 ° ? = 60 ° ? 22. 6 ° = 37. 4 ° The t otal displacement of the car is 130 000 m (N 37. 4 ° E). = 566 m Magnitude of average velocity 566 = 3 ? 60 = 3. 14 m s 400 tan ? = 400 ? = 45 ° (S 45 ° E). –1 (c) (1A) Total time 170 000 = = 10 200 s 60 3. 6 Magnitude of average velocity 130 000 = 10 200 = 12. 7 m s–1 Its average velocity is 12. 7 m s (N 37. 4 ° E). –1 (1A) (1A) (1M) (1A) Her average velocity is 3. 14 m s–1  © 11 (a) AC = 60 2 + 80 2 = 100 m 80 tan ? = ? = 53. 1 ° 60 (1M) The total displacement of the athlete is 100 m (S53. 1 °W). (1A) 13 (Correct label of velocity with correct direction (towards the left). ) (Correct label of acceleration with correct direction (towards the right). ) (1A) (1A) (a) The coin moves in the following sequence: B A C C A Therefore, it is at A finally. Displacement of the coin = 15 cm (1A) (1M) (1A) (1M) b) Distance travelled by the coin = 15 + 30 + 30 = 75 cm (b) Time / s v / m s–1 0 –6 1 –4 2 –2 3 0 4 +2 5 +4 6 +6 (1A) (1A) (c) (i) Total time = 2 s ? 4 = 8 s Average velocity 15 ? 10 ? 2 = 8 = 0. 0188 m s? 1 (0. 5A ? 6) (1M) (1A) (c) The car will slow down and its speed will drop to zero. After th at the car will move towards the right with increasing speed (uniform acceleration). (1A) (1M) (1A) (1M) (1A) (1M) (1A) A (ii) Average speed 75 ? 10 ? 2 = 8 = 0. 0938 m s? 1 (1M) (1A) 12 (a) Total distance travelled = 60 + 80 + 80 + 60 = 280 m (d) (i) The coin moves in the following sequence: B A C C A B B b) Magnitude of total displacement = 80 + 80 = 160 m 160 m (west). The total displacement of the athlete is Therefore, it is at B finally. zero. the coin is also zero. (1A) (1M) (1A) (1M) (1A) (1M) (1A) (ii) The displacement of the coin is Therefore the average velocity of (c) Total distance travelled = 280 + 60 + 80 = 420 m 14 (a) Total distance = ? r = 5? ? 60 m C = 15. 7 m Total displacement =5+5 = 10 m 80 m  © The total displacement travelled by her is 10 m. (b) Jane’s statement is incorrect. (1A) Since both girls start at X and meet at Y, they have the same displacement. (1A) Betty’s statement is incorrect. 1A) Since both girls return to their starting point, their displacements are zero. (1A) Physics in articles (p. 40) (a) From 19 January 2006 to 28 February 2007, (1A) It takes New Horizons spacecraft a total of 406 days to travel from the Earth to Jupiter. (1A) (b) (i) Average speed total distance travelled = total time of travel (1M) = 8 ? 108 406 ? 24 (1A) (1M) = 8. 21 ? 104 km h? 1 (ii) Average acceleration change in velocity = total time of travel = (8. 23 ? 5. 79)? 10 4 406 ? 24 = 2. 50 ? 104 km h? 2 (1A) (1A) (c) July 2015  © 2 1 2 3 4 5 Motion II 10 (a) The object moves with a constant elocity. Practice 2. 1 (p. 61) D B D D B 30 ? 10 = 10 m s–1 v= 2 (b) The object moves with a uniform acceleration from rest. (c) The object moves with a uniform deceleration, starting with a certain initial velocity. Its velocity becomes zero finally. The velocity of the car at t = 2 s is 10 m s–1. 6 7 C (d) The object first moves with a uniform acceleration from rest, then at a constant velocity, and finally moves with a smaller uniform acceleration again. (a) Total displacement = 4 ? 5 + (? 5) ? (7 ? 5) = 10 m The total displacement from the staircase to her classroom is 10 m. (e) The object moves at a constant velocity and then suddenly moves at constant velocity of same magnitude in the opposite direction. (b) Classroom C 8 (f) The object moves with uniform deceleration from an initial velocity to rest, and continue to move with the uniform acceleration of the same magnitude in opposite direction. 9 (a) The object accelerates. (b) The object first moves with a constant velocity. Then it becomes stationary and finally moves with a higher constant velocity again. 11 (a) The object moves with zero acceleration (with constant velocity of 50 m s–1). (b) The object moves with a uniform cceleration of 5 m s–2. (c) 12 The object moves with uniform deceleration of 5 m s–2. (c) The object decelerates to rest, and then accelerates in opposite direction to return to its starting point. (a) It moves away from the sensor. (d) The object moves with uniform velocity towards the origin (the zero displacement position), passes the origin, and continues to move away from the origin with the same uniform velocity.  © (b) (c) The greatest rate of change in speed 0 ? 3. 5 = 2 = –1. 75 m s–2 (d) Total distance travelled = area under the graph 3. 5 ? 2 2 ? 6 = + 2 2 = 9. 5 m Practice 2. 2 (p. 71) 1 C By v2 = u2 + 2as, 290 3. 6 2 13 (a) =0+2? 1? s s = 3240 m = 3. 24 km 3. 5 km The minimum length of the runway is 3. 5 km. 2 B Cyclist X is moving at constant speed. Time for cyclist X to reach finish line displacement 150 = = = 30 s time 5 For cyclist Y: u = 5 m s–1, s = 250 m, (b) Total distance travelled = area under the graph (12 + 6) ? 3 = 2 = 27 m a = 2 m s–2 By s = ut + 1 2 at , 2 1 250 = 5 ? t + ? 2 ? t2 2 (c) Average speed total distance travelled = time taken 27 = 3 t = 13. 5 s or t = ? 18. 5 s (rejected) Y needs 13. 5 s to reach finish line. Therefore, cyclist Y will win the race. 3 B Since the bullet start decelerates after fired into the wall, we could just consider the displacement of the bullet in the wall. To prevent the bullet from penetrating the wall, the bullet must stop in the wall. = 9 m s–1 14 (a) She moves towards the motion sensor. (b) The highest speed of the girl in the journey is 3. 5 m s–1.  © By v2 = u2 + 2as, 0 = 500 + 2 ? (? 800 000) ? s 2 8 By v = u + at, 14 = u + 2 ? 5 u = 4 m s–1 s = 0. 156 m = 15. 6 cm 15. 8 cm The minimum thickness of the wall is 15. 8 m. By v2 = u2 + 2as, 142 = 42 + 2 ? 2 ? s s = 45 m 4 C When the dog catches the thief at t = 5 s, its total displacement is 30 m. The dog is sitting initially, so u = 0. 1 By s = ut + at2, 2 1 30 = 0 + a(5)2 2 The displacement of the girl is 45 m. 9 (a) v = u + at = 0 + 20 ? 0. 3 = 6 m s? 1 The horizontal speed of the ball travelling towards the goalkeeper is 6 m s? 1. a = 2. 4 m s–2 Its acceleration is 2. 4 m s–2. (b) By v2 = u2 + 2as, 02 ? 62 a= = –22. 5 m s? 2 2 ? 0. 8 The deceleration of the football should be 22. 5 m s? 2. 5 6 D 90 36 ? v? u = 3. 6 3. 6 = 1. 5 m s–2 a= t 10 By v = u + 2as, 2 2 10 (a) The reaction time of the cyclist is 0. 5 s. s= v ? u = 2a 2 2 90 3. 6 36 3. 6 2 ? 1. 5 ? 2 2 = 175 m (b) Braking distance (2. ? 0. 5)? 15 = 11. 25 m = 2 Thinking distance = 15 ? 0. 5 = 7. 5 m Stopping distance = 11. 25 + 7. 5 = 18. 75 m child. 20 m The distance travelled by the motorcycle is 175 m and its acceleration is 1. 5 m s . –2 7 (a) Thinking distance = speed ? reaction time 108 = ? 0. 8 = 24 m 3. 6 Therefore, the bicycle would not hit the (b) Since the car decelerates uniformly, braking distance v+u = ? t 2 108 +0 = 3. 6 ? (3 ? 0. 8) 2 = 33 m 11 By v = u2 + 2as, 0 = 32 + 2 ? (–0. 5) ? s s=9m 8m Therefore, the golf ball can reach the hole. 2 12 (a) (i) By v = u + at, 0 = u + (–4)(4. 75) u = 19 m s–1 The initial velocity of the car is 19 m s–1. (c) Stopping distance = thinking distance + braking distance = 24 + 33 = 57 m  © (ii) By v2 = u2 + 2as, 0 = 19 + 2 ? (–4) ? s s = 45. 1 m 2 3 C For option A, apply equation v2 = u2 – 2gs and take s = 0 (the ball returns to the second floor), v = –u = –10 m s–1 (vertically downwards) The displacement of the car before it stops in front of the traffic light is 45. 1 m. This is the same velocity as the initial velocity of option B. Therefore, in both ways the ball has the same vertical speed when it reaches the ground. (b) By v = u + 2as, 17 = 0 + 2 ? 3 ? s s = 48. 2 m 2 2 2 The displacement of the car between starting from rest and moving at 17 m s is 48. 2 m. –1 4 B Take the upward direction as positive. 1 By s = ut + at2, 2 1 0 = u ? 30 + ? (? 10) ? 302 2 u = 150 m s–1 13 (a) By v2 = u2 + 2as, v2 = 0 + 2 ? 0. 1 ? 500 v = 10 m s–1 His speed is 10 m s . –1 (b) Consider the first section. By v = u + at, v? u t= a 10 ? 0 = 0. 1 = 100 s Consider the second section. 1 By s = ut + at2, 2 1 800 = 10t + ? 0. 5t2 2 t = 40 s or t = –80 s (rejected) The speed of the bullet is 150 m s–1 when it is fired. 5 Speed of stone Equation used t=1s t=2s t=3s t=4s v = u + at Distance travelled by the stone 1 s = ut + at 2 2 m 20 m 45 m 80 m 10 m s–1 20 m s 30 m s –1 –1 40 m s–1 Total time taken = 100 + 40 = 140 s It takes 140 s for Jason to travel downhill. 6 1 By s = ut + at2, 2 1 10 = 0 + (10) t2 2 t = 1. 41 s v = u + at Practice 2. 3 (p. 83) 1 2 D D = 0 + 10(1. 41) = 14. 1 m s–1 It takes 1. 41 s for a diver to drop from a 10-m platform. His speed is 14. 1 m s–1 when he enters the water.  © 7 Take the upward direction as positive. By v = u + 2as, 4 = 0 + (2)(–10)s s = 0. 8 m 2 2 2 Besides, since Y spends a shorter time to reach its highest point, it should be fired after X. 10 (a) By s = ut + The highest position reached by the puppy is 0. m above the ground. 8 (a) Consider the boy’s downward journey. Take the downward direction as positive. 1 By s = ut + at2, 2 1 0. 5 = 0 + (10) t2 2 t = 0. 316 s 1 2 at , 2 1 120 = 8t + ? 10 ? t2 2 t = 4. 16 s or t = ? 5. 76 s (rejected) It takes 4. 16 s to reach the ground. (b) v = u + at = 8 + 10 ? 4. 16 = 49. 6 m s–1 Its speed on hitting the ground is 49. 6 m s–1. 11 (a) Distance between the ceiling and her hands = 6 – 2 – 1. 2 = 2. 8 m Hang-time of the boy = 0. 316 ? 2 = 0. 632 s (b) Let s be her vertical displacement when she jumps. As the maximum jumping speed is 8 m s–1, i. e . u = 8 m s–1. By v2 = u2 + 2as, v2 ? 2 s= 2a 2 0 ? 82 = (upwards is positive) 2 ? (? 10) s = 3. 2 m 2. 8 m Therefore, the indoor playground is not safe for playing trampoline. 1 (a) By s = ut + at2, 2 1 132 = 0 ? t + ? 10 ? t2 2 t = 5. 14 s The vehicle can experience a free fall in the Zero-G facility for 5. 14 s. (b) Take the upward direction as positive. By v = u + 2as, 0 = u + 2 ? (–10) ? 0. 5 u = 3. 16 m s–1 2 2 2 The jumping speed of the boy is 3. 16 m s–1. 9 Take the upward direction as positive. (a) By v2 = u2 + 2as, 0 = u2 + 2(–10)(200) u = 63. 2 m s–1 The velocity of the firework X is 63. 2 m s–1 when it is fired. 12 (b) By v = u + at, = 63. 2 + (–10)t t = 6. 32 s It takes 6. 32 s for the firework X to reach that height. (c) From (a) and (b), for firework Y to explode at 130 m above the ground, the speed of Y should be smaller than that of X. Therefore, Y should be fired at a (b) By v2 = u2 + 2as, v2 = 02 + 2 ? 10 ? 132 v = 51. 4 m s? 1 The speed of the vehicle before it comes to a stop is 51. 4 m s? 1.  © lower speed. (c) Take the upward direction as positive. By v = u + at, –v = v – gt 2v = gt If the stone is projected with a speed of 2v, let the new time of travel be t?. (–2v) = (2v) – gt? v t? = 4 ( ) g = 2t Its new time of travel is 2t. 6 B Take the upward direction as positive. 1 s = ut + at2 2 1 = (10)(4) + (–10)(4)2 2 = –40 m The distance between the sandbag and the ground is 40 m when it leaves the balloon. Revision exercise 2 Multiple-choice (p. 87) 1 D By v2 = u2 + 2as, 0 = 102 + 2a(25 – 10 ? 0. 2) a = –2. 17 m s–2 His minimum deceleration is 2. 17 m s–2. 2 3 D B Consider the rock released from the 2nd floor. By v2 = u2 + 2as, v2 = 2as floor. Note that s2 = 3. 5s. (v2)2 = 2as2 = 3. 5(2as) = 3. 5v2 v2 = 1. 87v (as u = 0) Then consider the rock released from the 7th 7 8 D C Take the downward direction as positive. u = 200 m s–1, v = 5 m s–1, a = ? 0 m s–2 By v = u + at, 5 = 200 + (? 20)t t = 9. 75 s The rockets should be fired for at least 9. 75 s. Both C and D satisfy this requirement. But for D, after firing for 10. 2 s, v = u + at = 200 + (–20)(10. 2) = –4 m s–1 i. e. it flies away from the Moon with 4 m s–1 upwards. It c annot land on the Moon. Therefore, the correct answer is C. 4 5 A C The stone returns to the ground with the same speed (but in opposite direction). 9 10 D D  © 11 12 13 (HKCEE 2006 Paper II Q1) (HKCEE 2007 Paper II Q2) (HKCEE 2007 Paper II Q33) (b) (i) Conventional (p. 89) 1 (a) The reaction time of the driver is 0. 6 s. (b) v a= t = 0 ? 12 3. 6 ? . 6 (1A) (Correct axes with label) from t = 1. 20 s to 1. 25 s) from t = 1. 45 s to 1. 50 s) (1A) (1A) (1A) (A straight line with slope = 0. 35 m s–1 (A straight line with slope = –0. 35 m s–1 (1A) (1M) = –4 m s–2 The acceleration of the car is –4 m s–2. (c) The stopping distance of the car is the area under graph. Stopping distance 12 ? (3. 6 ? 0. 6) =12 ? 0. 6 + 2 = 25. 2 m The stopping distance of the car is shorter than 27 m. The driver will not be charged with driving past a red light. (1A) (1A) (1M) (ii) 2 (a) The object moves away from the motion sensor with uniform velocity at 0. 35 m s–1 from t = 1. 20 s to 1. 25 s. 1A) From t = 1. 25 s to 1. 45 s, the object moves with negative acceleration. (1A) Then, from t = 1. 45 s to 1. 50 s, the object changes its moving direction and moves towards the motion sensor again with a uniform velocity of –0. 35 m s–1. (1A) (Correct axes with labels) (1A) (Correct graph with the acceleration of ? 0. 35 ? 0. 35 about 1. 40 ? 1. 30 = –7 m s–2 at t = 1. 30 s to 1. 40 s) (1A) !  © 3 (a) (b) Total displacement of the car = area bound by the v? t graph and the time axis 1 1 = (5 ? 5) ? (20 ? 3) 2 2 = ? 17. 5 m (1M) (1A) (c) Yes, the car moves 12. 5 m forwards from t = 0 to t = 5 s. Therefore, it hits the roadblock. 1A) 5 Take the upward direction as positive. (a) From point A to the highest point: (Correct axes with labels) (Correct shape of minibus’ graph) (Correct shape of sports car’s graph) (Correct values) (1A) (1A) (1A) (1A) By v2 = u2 + 2as, 0 = 42 + 2 (–10) s s = 0 . 8 m By v = u + at, 0 = 4 + (–10)t t = 0. 4 s (1M) From the highest point to the trampoline: 1 s = ut + at2 (1M) 2 1 = 0 + (–10)(1. 2 – 0. 4)2 2 = –3. 2 m (1A) 3. 2 m above the trampoline. (1A) The maximum height reached by him is (1M) (b) From the graph in (a), the two vehicles have the same velocity at t ? 2. 3 s after passing the traffic light. (1A) (1M) (c) The area under graph is the displacement of the cars. Consider their displacements at t = 3 s, For the sports car: 1 s = ? 15 ? 3 = 22. 5 m 2 For the minibus: 1 s = ? (7 + 13) ? 3 = 30 m 2 The minibus will take the lead 3 s after passing the traffic light. (1A) (b) Height of point A above the trampoline (1A) = 3. 2 – 0. 8 = 2. 4 m (1M) (1A) 6 (a) Initial velocity v = 90 km h–1 90 = m s–1 3. 6 = 25 m s–1 Thinking distance =v? t = 25 ? 0. 2 =5m The thinking distance is 5 m. (1A) (1M) 4 (a) The car moves forward with uniform acceleration at ? 1 m s? 2 from t = 0 s to t = 5 s. (1A) (1A) Then the car changes its moving direction. From t = 5 s to t = 8 s, it moves backwards with a uniform acceleration of ? 6. 67 m s . ?2 Its instantaneous velocity is 0 at t = 5 s. (1A) †  © (b) By v2 = u2 + 2as, v2 ? u2 a= 2s 2 0 ? 25 2 = 2 ? (80 ? 5) = ? 4. 17 m s–2 4. 17 m s–2. (1M) (c) The slope of the graph is the magnitude of the acceleration of the apple. speed / m s? 1 7. 75 (1A) (1A) Hence, the deceleration of the car is (c) By v2 = u2 + 2as, s= v ? u 2a 0 2 ? 25 2 = 2 ? ( ? 4. 17 ? 2) 2 2 (1M) 0 0. 775 time / s (Correct labelled axes) (2A) (1A) (Straight line with a slope of 10 m s? 2) = 37. 5 m Braking distance = 37. 5 m Stopping distance = 37. 5 + 5 = 42. m (1M) (d) The two graphs have no difference. (1A) (1A) 8 (a) Take the downward direction as positive. By v2 = u2 + 2gs, v = u + 2 gs 2 The driver could not stop before the traffic light. Therefore, his claim is incorrect. (1A) (1M) 7 (a) Take the downward direction as positive. 1 By s = ut + gt2, 2 1 3 = 0 ? t + ? 10 ? t2 2 3? 2 t= = 0. 775 s 10 (1M) = 0 2 + 2 ? 10 ? (40 ? 3) = 27. 2 m s–1 cushion is 27. 2 m s? 1. 1 (b) (i) By s = ut + gt2, 2 1 40 – 3 = 0 + ? 10 ? t2 2 t = 2. 72 s (1A) The speed of the residents landing on the (1M) (1A) The apple travels in air for 0. 775 s. (1A) (b) By v2 = u2 + 2as, v = 2 ? 10 ? 3 (1M) 1A) –1 = 7. 75 m s? 1 The speed of the apple is 7. 75 m s when the apple just reaches the ground. The time of travel in air is 2. 72 s. u+v (ii) By s = t, (1M) 2 2s t= u+v 2? 3 = t 27. 2 + 0 = 0. 221 s (1A) The time of contact is 0. 221 s.  © (c) (b) Slope of the graph from t = 0 to t = 0. 28 s 2. 3 ? 0 = 0. 28 ? 0 = 8. 21 m s–2 The acceleration of the ball due to gravity is 8. 21 m s–2. (1M) (1A) (c) (Correct labeled axes) (Correct shape) (Correct values) (1A) (1A) (1A) (i) 9 (a) t = 2 s: Displacement of the trolley = 0. 7 ? 0. 15 = 0. 55 m t = 3. 4 s: (1A) Displacement of the trolley = 1. 175 ? 0. 15 = 1. 025 m t = 4. 9 s: 1A) Displacement of the trolley = 0. 6 ? 0 . 15 = 0. 45 m (1A) (b) It moves away from the motion sensor with a changing speed from t = 2 s to t = 3. 4 s. (Correct sign) (Correct shape) (1A) (1A) (1A) (1A) (1A) (ii) The method does not work Then it rests momentarily at t = 3. 4 s. After that, it moves towards the motion since ultrasound will be reflected by the transparent plastic plate. (1A) (c) sensor with a changing speed. 1 By s = ut + at2, 2 1 ? 0. 1 = 0. 7 ? 2. 9 + ? a ? (2. 9)2 2 a = ? 0. 507 m s? 2 (1A) (1M) 11 (a) (i) The ball is held 0. 15 m from sensor before being released. The ball hits the ground which is 1. m from the sensor. (1A) (1A) Therefore, the ball drops a height of 0. 95 m. which are 0. 45 m, 0. 65 m and 0. 775 m from the sensor in its first 3 rebounds. (1A) The acceleration of the trolley is ? 0. 507 m s? 2. (ii) The ball rebounds to the positions 10 (a) The motion sensor is protruded outside the table to avoid the reflection of ultrasonic signal from table. (1A)  © At the 1st rebound, the ball rises up (1. 1 ? 0. 45) = 0. 65 m. nd The average acceleration is 66. 6 m s–2. (1A) (1A) (1A) (c) v / m s? 1 6. 32 At the 2 rebound, the ball rises up (1. 1 ? 0. 65) = 0. 45 m. rd At the 3 rebound, the ball rises up (1. 1 ? 0. 75) = 0. 325 m. (b) (i) The ball hits the ground with velocities of 3. 9 m s , 3. 25 m s and 2. 75 m s–1 in its first 3 rebounds. (3A) 3. 9 (1M) 0. 95 ? 0. 55 (1A) –1 –1 t3 t1 t2 t4 t5 t/s (ii) Acceleration = slope of graph = = 9. 75 m s–2 ?6. 32 (3 straight lines) (Correct slopes) (1A) (1A) 12 Take the downward direction as positive. 1 (a) By s = ut + gt2, (1M) 2 1 2 = 0 ? t + ? 10 ? t2 2 2? 2 t= = 0. 632 s (1A) 10 It takes 0. 632 s from t1 to t2. (Correct labels of time and velocity)(1A) 13 (a) Speed v = 70 km h–1 70 = m s–1 3. 6 = 19. 4 m s–1 d Reaction time = v 6 = 19. 4 = 0. 309 s The reaction time of the man was 0. 09 s. (1M) (b) At t2, v = u + at (1A) = 0 + 10 ? 0. 632 = 6. 32 m s –1 –1 (1 M) Shirley’s speed is 6. 32 m s when she lands on the trampoline at t2. At t4, she leaves the trampoline at the same speed. Therefore, from t3 to t4, by v2 = u2 + 2as, a= v2 ? u2 2s (? 6. 32) 2 ? 0 2 = 2 ? 0. 3 (b) By v2 = u2 + 2as, v2 ? u2 a= 2s 2 0 ? 19. 4 2 = 2 ? 48 = –3. 92 m s–2 3. 92 m s–2. (1M) (1M) (1A) The average deceleration of the car was (c) (1A) Speed v = 80 km h–1 80 = m s–1 3. 6 = 22. 2 m s–1 = 66. 6 m s–2  © Thinking distance = vt = 22. 2 ? 0. 309 = 6. 86 m By v = u + 2as, braking distance s v2 ? u2 = 2a 2 0 ? 22. 2 2 = 2 ? ? 3. 92) 2 2 (1A) Take the upward direction as positive. 1 s = ut + at2 (1M) 2 1 = 7 ? 1. 75 + ? (–10) ? 1. 752 2 = –3. 06 m (negative means the water is below the spring board) The spring board is 3. 06 m above the water. Alternative method: (1A) = 62. 9 m Therefore, the stopping distance = 6. 86 + 62. 9 = 69. 8 m (1A) Consider the upward motion and downward motion separatel y. For the upward motion, she takes 0. 7 s to reach the highest point from the spring board. Take the upward direction as positive. 1 By s = ut + at2, (1M) 2 1 s1 = 7 ? 0. 7 + ? (–10) ? 0. 72 2 = 2. 45 m For the downward motion, she takes 1. 5 s from the highest point to enter water. Take the downward direction as positive. By s = ut + 1 2 gt , 2 1 s2 = 0 + ? 10 ? 1. 052 = 5. 51 m 2 (1A) This stopping distance is greater than the initial distance between the car and the boy. (1A) Therefore, the car would have knocked down the boy if the car had travelled at 80 km h? 1 or faster. (d) A drunk has a longer reaction time. (1A) This means that the thinking distance, and thus the stopping distance (sum of thinking distance and braking distance), increases. (1A) (1M) (1A) 14 (a) Take the upward direction as positive. By v = u + at, u = 0 ? (? 10) ? 0. 7 = 7 m s–1 board is 7 m s . 1 Therefore the height of the spring board above the water = s2 – s1 = 5. 51 – 2. 4 5 = 3. 06 m (1A) (1M) (1A) The speed of Belinda leaving the spring (b) Total time taken from the spring board to the water = 0. 7 + 1. 05 = 1. 75 s (c) v = u + at = 0 + (? 10) ? 1. 05 = ? 10. 5 m s–1 is 10. 5 m s–1.  © The speed of the diver entering the water (d) Deceleration of car Y = slope of the graph during 0. 5 s? 8. 5 s = 0 ? 19. 4 = –2. 43 m s–2 8. 5 ? 0. 5 (1A) The deceleration of car Y is 2. 43 m s–2. (c) Thinking distance = area under the graph during 0? 0. 5 s = 19. 4 ? 0. 5 = 9. 7 m (1A) (Correct shape) (Correct times) (Correct velocities) 1A) (1A) (1A) Braking distance = area under the graph during 0. 5 s? 8. 5 s 1 = ? 19. 4 ? (8. 5 – 0. 5) 2 = 77. 6 m distance are 9. 7 m and 77. 6 m respectively. (1A) The thinking distance and the braking (e) (See the figure in (d). ) (Correct slope – parallel to that in (d). ) (1A) (Correct position – above that in (d). ) (1A) 15 (a) Speed 70 km h–1 70 = m s–1 3 . 6 = 19. 4 m s –1 (d) The coloured area is equal to the difference in the stopping distances travelled by cars X and Y. (1A) (e) (1M) Stopping distance of car X = area under the graph during 0? 5 s 1 = ? 19. 4 ? 5 = 48. 5 m 2 Coloured area = 9. 7 + 77. 6 – 48. = 38. 8 m 50 m Since the difference in stopping distances of the cars is smaller than the initial separation of the cars, the two cars do not collide with each other before they stop. (1A) (1M) (1M) Distance travelled by car Y in 2 s = vt = 19. 4 ? 2 = 38. 8 m 50 m Since the distance between the cars is greater than the distance that car Y can travel in 2 s, the driver of car Y obeys the rule. corresponding v–t graph. Deceleration of car X = slope of the graph during 0? 5 s (1A) (1M) (b) Deceleration of a car is the slope of their 0 ? 19. 4 = 5? 0 = –3. 88 m s–2 The deceleration of car X is 3. 88 m s–2. (1A) 16 a) From t = 0 s to t = 5 s, the car moves with a uniform acceleration of 17 ? 0 = 3. 4 m s–2. 5 (1A)  © From t = 5 s to t = 20 s, the car moves with a constant velocity of 17 m s–1. (1A) From t = 20 s to t = 28 s, the car moves with a uniform acceleration of 0 ? 17 = ? 2. 125 m s–2. 28 ? 20 at rest. (1A) (b) s = ut + 1 2 at 2 1 = 0 + ? 17. 5 ? (8 ? 60)2 2 = 2 016 000 m (2016 km) (1M) (1A) The Shuttle travels 2 016 000 m (2016 km) in the first 8 minutes. From t = 28 s to t = 30 s, the car remains (1A) 19 (a) (i) The cyclist is using first gear when the acceleration is greatest before braking. shortest time. (1A) (1A) (1M) (1M) (1A) b) (ii) The cyclist uses second gear for the (b) Distance travelled = area under straight line PQ (8 + 6) ? 2 = 2 = 14 m The cyclist travels 14 m in second gear. (c) The acceleration during t = 18 s? 20 s 0? 9 = (1M) 20 ? 18 = ? 4. 5 m s–2 The deceleration is 4. 5 m s . –2 (1A) (Correct shape) (Correct time instants) (Correct accelerations) (1A) (1A) (1A) (1A) (1A) 20 21 (c) Yes. (HKCEE 200 5 Paper I Q1) 1 (a) s = ut + at2 2 1 = 0 + ? 10 ? (500 ? 10? 3)2 2 = 1. 25 m Therefore the minimum height the (1M) The car changes direction at t = 30 s. Its velocity changes from positive to negative, showing a change in its travelling direction. 1A) (1M) (1A) (1A) laptop must fall for it to be ‘saved’ is 1. 25 m. (b) v = u + at = 0 + 10 ? (500 ? 10 ) = 5 m s? 1 the ground is 5 m s–1. ?3 (1M) (1A) 17 18 (HKCEE 2002 Paper I Q8) (a) v = u + at = 0 + 17. 5 ? 8 ? 60 = 8400 m s–1 minutes is 8400 m s–1. The speed of the computer when it hits The speed of the Shuttle after the first 8  © (c) Most falls are likely to be from below this height, effect. (1A) (1A) (1A) so the protection will not have taken Physics in articles (p. 96) (a) 2. 45 m (b) (i) By v2 = u2 + 2as, u = v ? 2as u2 = 0 ? 2(? 10)(2. 45 + 0. 07 ? 1. 09) u = 5. 35 m s? 1 2 2 (1A) (1M) Take the upward direction as positive. 22 (a) Any one from: Rate of change of displacement Displacement per unit time (1A) (b) The velocity of a braking car is decreasing (with time) (1A) so the car has negative acceleration. (1A) Its displacement is (still) increasing with time, so its velocity is (still) positive In this case, the acceleration and velocity are in opposite directions. (1A) (1A) (1A) The vertical speed of Javier Sotomayor is 5. 35 m s? 1 when he leaves the ground. (ii) Take the upward direction as positive. Consider the upward journey. By v = u + at, v ? u 0 ? 5. 35 t= = = 0. 54 s a ? 10 (1M) (c) i) Consider the downward journey. 1 By s = ut + at2, (1M) 2 1 ? (2. 45 + 0. 07 ? 0. 71) = 0 + (? 10) t2 2 t = 0. 60 s The time that he stays in the air = (0. 54 + 0. 60) = 1. 14 s Alternative method: (1A) (Correct graph) (1A) Take the upward direction as positive. 1 By s = ut + at2, (1M) 2 (0. 71 ? 1. 09) = 5. 35t + 1 (? 10)t 2 (1M) 2 t = 1. 14 s or t = ? 0. 07 s (rejected) (ii) Vertical distance travelled = area under the graph from 4. 0 s to 10. 0 s (70 + 130)? 6 = 2 (1M) (1A) The time that he stays in the air is 1. 14 s. = 600 m (1A) The vertical distance travelled by the rocket between t = 4. 0 s and t = 10. s is 600 m.  © 3 1 2 3 4 C C Force and Motion 6 (a) The MTR train is accelerating in the forward direction. The man tends to move at his original speed (smaller speed), so he would move backwards relative to the MTR train. (b) The MTR train is slowing down. The man tends to move at his original speed (greater speed), so he would move forwards relative to the MTR train. (c) The MTR train is moving forwards at constant velocity. The man moves forwards with the same constant velocity, so he would remain at rest relative to the MTR train. (d) The MTR train is turning a corner. The Practice 3. 1 (p. 104) (b), (e), (f) 5 a) Stretching a rubber band (b) Standing on the floor (c) Walking time (e) (f) A compass A rubbed plastic ruler attracts small bi ts of paper (d) Exists in every object on the earth at any 7 man tends to move at his original direction, so he would move outwards relative to the MTR train. In space, the gravitational force acts on the spaceship is negligible. When the rockets are shut down, they do not exert a force on the spaceship. Therefore, no net force acts on the spaceship. By Newton’s first law, the spaceship is in uniform motion and can travel far out in space. 8 Joan moves on the ice surface with a constant velocity. Practice 3. 2 (p. 111) 1 2 3 4 5 C C D C (a) No. Athletes would hit the wall of the stadium if it is too close to the finishing line. (b) The mat is used to protect the athletes if they hit the wall after passing the finishing line. Practice 3. 3 (p. 122) 1 2 3 4 5 D A B A D  © 6 (a) 7 (a) Horizontal component = 40 + 30 cos 30 ° = 66. 0 N Vertical component = 30 sin 30 ° = 15 N Resultant = 66 2 + 15 2 = 67. 7 N Let ? be the angle between the resultant Resultant’s magnitude is 67 N and the angle between the resultant and the horizontal is 13 °. (b) and the horizontal. 15 tan = ? = 12. 8 ° 66 Resultant’s magnitude is 67. N and the angle between the resultant and the horizontal is 12. 8 °. (b) Horizontal component = 40 + 30 cos 45 ° = 61. 2 N Vertical component = 30 sin 45 ° = 21. 2 N Resultant’s magnitude is 65 N and the angle between the resultant and the horizontal is 19 °. (c) Resultant = 61. 2 2 + 21. 2 2 = 64. 8 N Let ? be the angle between t he resultant and the horizontal. 21. 2 tan = ? = 19. 1 ° 61. 2 Resultant’s magnitude is 64. 8 N and the angle between the resultant and the horizontal is 19. 1 °. (c) Resultant’s magnitude is 60 N and the angle between the resultant and the horizontal is 25 °. (d) Horizontal component = 40 + 30 cos 60 ° = 55 N Vertical component = 30 sin 60 ° = 26. 0 N Resultant = 55 2 + 26. 0 2 = 60. 8 N Let ? be the angle between the resultant and the horizontal. 26. 0 ? = 25. 3 ° tan = 55 Resultant’s magnitude is 60. 8 N and the angle between the resultant and the Resultant’s magnitude is 50 N and the angle between the resultant and the horizontal is 37 °. horizontal is 25. 3 °.  © (d) Resultant = 40 2 + 30 2 = 50 N Let ? be the angle between the resultant and the horizontal. 30 tan = ? = 36. 9 ° 40 Resultant’s magnitude is 50 N and the angle between the resultant and the horizontal is 36. 9 °. Hence, the angle between the two 5-N forces is 120 °. Alternative method: By tip-to-tail method, the two 5-N forces and the resultant 5-N force form an equilateral triangle. It is known that each angle of an equilateral triangle is 60 °. Therefore, the angle between the two 5-N forces is 120 °. 8 (a) 10 (b) Resultant force = 2 ? 400 = 800 N The resultant force provided by the cable is 800 N. 11 For the 2-kg mass: (c) 9 R = weight ? cos ? = 20 cos 30 ° = 17. 3 N Suppose the two forces act in the direction as shown. T = 20 N Therefore we have: Vertical component Fx = 5 sin ? Horizontal component Fy = 5 ? 5 cos ? = 5 ? 1 ? cos ? ) (magnitude of the resultant)2 = Fx2 + Fy 2 52 = (5 sin ? )2 + [5 ? (1 ? cos ? )]2 1 = sin ? + 1 ? 2 cos ? + cos ? 2 2 2T cos 45 ° = W 2 ? 20 ? cos 45 ° = W cos ? = 0. 5 W = 28. 3 N ? = 60 °  © 12 (a) 2T sin 10 ° = 500 T = 1440 N The tension of the string is 1440 N. 3 4 5 6 B C A Net force = ma = 40 ? 0. 5 = 20 N C By v2 – u2 = 2as, 0 à ¢â‚¬â€œ u2 = 2a(20) ? u2 = 40a u2 a=? 40 Resistance = ma = 12 ? ? u2 = –0. 03u2 40 (b) Component of force = T cos 10 ° = 1440 ? cos 10 ° = 1420 N The component of the force that pulls the car is 1420 N. 13 (a) 7 8 ‘A bag of sugar weighs 10 N. ’ or ‘A bag of sugar has a mass of 1 kg. By F = ma, F 800 000 a= = = 2 m s–2 m 4 ? 10 5 (b) As the mass is stationary, the net force acting on it is zero. When it flies horizontally, its acceleration is 2 m s–2. 100 ( )? 0 v? u (a) a = = 3. 6 = 4. 63 m s–2 t 6 The acceleration of the car is 4. 63 m s–2. (c) (i) y-component of F1 = weight of mass = 10 N 9 y-component of F1 = F1 sin 30 ° F1 sin 30 ° = 10 N F1 = 20 N x-component of F1 = F1 cos 30 ° = 20 cos 30 ° = 17. 3 N (b) F = ma = 1500 ? 4. 63 = 6945 N The force provided by the car engine is 6945 N. 10 (a) (ii) y-component of F2 = 0 x-component of F2 = x-component of F1 = 17. 3 N How to cite Nss Phy Book 2 Answer, Papers

Social Work Social Policy And Social Welfare Social Work Essay Essay Example

Social Work Social Policy And Social Welfare Social Work Essay Paper With mention to alterations in Government policy and political orientations of public assistance, debate the significance of the displacement from Victorian Pauper to 21st century service user and its impact on societal work pattern and values. This assignment has used a historical timeline of Governmental alterations to policies and Torahs as a background to debate the displacement from Victorian Pauper to 21st century service user. The divide between the rich and hapless has ever been an issue that different authoritiess have faced, covering with it otherwise, for illustration Clement Attlee s labour Government in 1945 introduced the public assistance province to give every British citizen screen, irrespective of income or deficiency of it. Those who lacked occupations and places would be helped. We will write a custom essay sample on Social Work Social Policy And Social Welfare Social Work Essay specifically for you for only $16.38 $13.9/page Order now We will write a custom essay sample on Social Work Social Policy And Social Welfare Social Work Essay specifically for you FOR ONLY $16.38 $13.9/page Hire Writer We will write a custom essay sample on Social Work Social Policy And Social Welfare Social Work Essay specifically for you FOR ONLY $16.38 $13.9/page Hire Writer The definition of a Pauper harmonizing to the Collins lexicon is person who is comparatively hapless, in comparing to the general population or historically eligible for public charity. The definition of a service user is person who at some point uses or receives wellness or societal attention services. ( General Social Care Council ) The term service user is criticised, as critics Adams et Al ( 2009 ) believe it focuses on one component of the person, connoting dependence, without taking into history other facets and argue that the term places the service user in a disempowered place in their relationship with a professional, with power shacking with this professional. The National Network of Service Users: Determining our lives nevertheless sees the term service user as positive, it s an person who uses the services, they confer power making a stronger voice and holding a greater ability to determine services. ( Levin 2004 ) Modern British societal policy has its foundation in the Poor Laws, dating from 1598 to 1948. The Poor Law ( 1601 ) provided a mandatory hapless rate and helped put the hapless to work. However as the Parish was the basic country of disposal, and Torahs were enforced otherwise from parish to parish with no set criterions of attention doing incompatibilities between countries. The Poor Law amendment act ( 1834 ) modified the bing system. Poor Law Unions were introduced, parishes were grouped together, and those Unions would be the duty of a Board of Guardians. The Guardians were responsible for the disposal of hapless alleviation for their vicinity, instead than go forthing the duty of disposal in the custodies of single parishes and townships. Workhouses were introduced and encouraged, one workhouse in each brotherhood to give hapless alleviation. This Act stated that no able bodied individual was to have any other aid other than in the workhouse. The intent of the workhouse was to allow persons come in and go forth as they liked and they would have free nutrient and adjustme nt, nevertheless as clip passed concern grew with respect to the seeming overuse of the workhouse. As a consequence the eligibility standards for entry to the workhouse was so altered. Life in the workhouse was to be made every bit barbarous as it was outdoors. The deplorable being offered, and the stigma attached to being an inmate , ensured that merely the genuinely deprived used them. A jail manner system of segregation for work forces and adult females meant that even households had to be separated, wholly different from twenty-first Century societal work values which espouse a stronger committedness to maintaining households together, with kid protection services and kid public assistance bureaus supplying support to guarantee household saving. ( Payne, 2005 ) It was in 1869 that The Charity Organisation Society ( COS ) was formed to unite the many smaller beginnings of alleviation and do proviso more efficient and effectual. The COS perceived that charitable aid was needed and believed that their purpose was to make all households, but were diffident how the money had antecedently been spent. COS set out a strategy of fiscal aid, presenting local commissions, who so raised financess and distributed these to households in demand. Similar to many charities today, there are still many households who do nt inquire for aid because of faith, linguistic communication, pride or because they are non cognizant that aid may be available. ( Family Action ) The purpose of COS was to carry charities to meet their resources, which might so be distributed more consistently. However the persons helped had to be deemed capable of going ego back uping. Worthiness was considered before any charitable aid was given. Those who were nt considered worthy were le ft to destitution, the Society efficaciously make up ones minding that as they deemed there to be no hope for their salvation, that to assist them would be a waste of limited resources which would be better spent elsewhere with persons or households who could later raise themselves out of poorness and dependance. ( Campling, 1996 ) The COS theoretical account was open uping in taking into history the effects if they helped every person, as they thought this would take to dependency and exaggeration in order to have money. COS was besides formed with the aim of accomplishing a lessening in the charitable outgo as a consequence of greater efficiency and the economic system of graduated table, and in this sense the COS reflected the wider political orientation of the industrial revolution. Current Social Work objectives likewise seek to accomplish value for money , with the Audit committee specifying value for money as the best possible balance of economic system, efficiency and effectivity. Whilst the twenty-first century societal worker enterprises to cover reasonably with the demands of everyone, however, the distribution of demands is uneven and alterations invariably. To guarantee quality is consistent attention program reappraisals are monitored and service users may be involved in quality circles, prosecuting them in make up ones minding what s best. Direct payment strategies are besides offered to a minority group of service users to allow them make up ones mind on and custom-make their ain soci etal attention. The chief job Social Services face is the possible impact of the dynamic and variable economic and political environment in which they must work. Whilst seeking to supply individualized attention bundles they have to guarantee that it s effectual in regard to cost. Besides there is turning acknowledgment that a figure of minority groups may be excluded from accessing services such as Social Work services and, besides those services which have antecedently non been provided in culturally appropriate ways. ( Making ends meet, 2010 ) Appropriate stairss will hence necessitate to be taken to implement anti prejudiced pattern and anti oppressive pattern, when sing how to supply aid in accessing services for minority groups. The COS rapidly found that more than fiscal assistance was needed to assist paupers. Emotional and other practical aid was besides required, for illustration aid with finding employment. COS voluntaries were trained to offer such extra aid and, hence, the formal preparation offered by COS can be seen as the precursor of modern societal work preparation and makings. They adopted an attack which attempted to analyze the job. Working with the person and household to assist them accomplish a permanent solution so all could be kept in their familiar environments. This attack was really clip consuming, and the attack the worker took to look into the person was really obnoxious. This attack is the footing of the current Social Work casework attack which is now extremely criticised. Holman ( 1993 ) suggests that the casework attack merely masks societal and political duties in person s lives, hence assisting to keep their state of affairs. There are now other attacks that focus on cut downing inequality. Which investigate the societal and political grounds every bit good as the person s as to why they are in poorness. A new pronunciamento for Social Work now highlights the demand to utilize a aggregation of attacks as the demand to battle poorness and favoritism is greater than of all time. Many who tried to utilize the COS principles found it hard to ignore persons who needed aid. Other attacks were recently introduced to assist more persons. The Settlement House Movement ( 1884 ) was one of these. Its rules focused on university voluntaries working with the hapless in their trim clip, offering instruction. Its purpose was to accomplish common regard between the categories. This attack focused on authorising the hapless, assisting them to assist themselves. Society besides benefitted from this theoretical account. It focused on a more structured analysis of poorness and its impact on human behavior by rehearsing intercessions at a community degree. This is needed now to assist little communities and the persons within it. The nature of societal work pattern so changed and focused on persons. A important component was hearing client s voices and the uncomparable cognition of the professionals working with them to assist whichever manner they could. ( Adams et al, 2009 ) Using these theoretical accounts the Government laid the footing of the hereafter societal services. The major concern being that all countries should be given the same services. These new services were provided off from The Poor Law to hedge the association. Current Social Work still has its incompatibilities, nevertheless the White Paper Tackling Health Inequalities Programme of Action ( Department of Health, 2003 ) , focuses on a figure of ways to equalize entree to healthcare, for illustration working with people who face overlapping wellness jobs for case older people who have ill wellness and are in poorness. Social Workers are concentrating on secondary bar, as this type of bar can impact more persons. A major study produced sing the public assistance of persons was the Beveridge Report ( Department of Health, 1942 ) . This study focused on how Britain could be rebuilt after the war. In 1945 labor was elected and promised to present a public assistance province. The public assistance province involved presenting new services. These included the National Health Services and Housing Acts. The public assistance province was produced to promote the proviso of services for the populace. ( Laybourn, 1995 ) Glasby ( 2005 ) looked at old reforms and how the hereafter would be in big societal attention. It evaluated all of import studies to see how societal work could be improved. One study that wedged policy and pattern during the 1960 s was the Seebohm study ( 1968 ) . This study highlighted the jobs of poorness and was tasked to reexamine the administration and duties of the Local Authority Social Services in England every bit good as to see what alterations were desirable to procure an effectual household service. ( Seebohm, 1968, pg11. ) Prior to this study Social Work was spread across assorted Local Authorities and different Government subdivisions. This caused insufficiencies in the quality of proviso. Access was really hard. For illustration, scope and quality of proviso of services were inconsistent besides the Seebohm Report highlighted a hapless coordination of information between these services. The study recommended a new Local Authority section supplying a community based and household orientated service, which would be available for all . When this recommendation was brought into action new Social Services Departments were formed. The Seebohm Report did highlight potency jobs. It stated that holding separate sections for kids and grownups might later do it hard to handle the household s demands as a whole. The Barclay Report ( 1982 ) looked into the function of a societal worker. In its gap line it stated that excessively much was expected of societal workers. It found that it was a profession that was confused about its function and because of intense media examination was fighting with work burden. It found that there was an on-going demand for societal workers to carry through many maps including advancing community webs, working with other services and moving as an advocator for clients. The study did knock societal work sections for taking a reactive stance towards societal jobs, covering with those demands which are forced upon their attending but neglecting to develop overall programs which link the voluntary, statutory and private services in an country into a coherent program which is still a job today. ( Department of Health, 1982, pg.38 ) Social Services Departments find it hard to assist every demand as they do nt hold illimitable resources. They need to utilize other ser vices and work with them closely, the aid of Interprofessional instruction will guarantee that other professionals have an apprehension of societal workers functions. The Barclay Report produced really similar recommendations to that of the Seebohm Report ( 1968 ) . Although it highlighted that the community attack may hold more success now, as there is a greater capacity for persons to be more independent and do their ain determinations. The community attack focuses on the local community and societal workers would detect persons in the context of their community. This attack uses local Centres and pools resources, making less impact on the Social Work services so their resources can be spread further. Reports such as these have highlighted how of import good societal work is, and how much it is needed. There are many jobs involved in the profession. There is still stigma attached to the term service user merely as there was to the term pauper. Whilst researching the different Acts of the Apostless and public assistance political orientations that have been introduced throughout the timeline I have used ( see paragraph one, page one ) I have found that persons still have jobs accessing aid. There are households who still may be disinclined to inquire for aid because of the stigma of making so. New attacks have introduced service user engagement by inquiring them what help they want and specifying the quality of aid they receive. A recent study by Beresford, Shamash, Forrest and Turner ( 2007 ) researched service users future vision for grownup services. They found that the procedure of accessing societal attention was often negative for service users furthermore the appraisals wer e really dependent on the quality of the staff transporting it out. All societal workers should work to one high criterion. It should nt be a lottery of if you get a good one or non. A cosmopolitan tool could be implemented so that all service users were asked the same inquiries and could foreground their specific jobs and demands whilst utilizing the tool. Service users besides highlighted the fact that entree to their societal worker was low and many of the service users questioned had spreads in their services doing them experience insecure. Reports researched for this assignment have all found that service users know what they want and can easy foreground the jobs they face or have had antecedently. One study found that while public assistance bureaucratism has been condemned by authoritiess for a long piece, service users still place jobs ( Determining Our Lifes, 2007 ) . There is still societal exclusion. Social Services Departments may lend to it as they help persons merely plenty, happening the quickest manner to assist them non needfully the best manner in the long tally because of finite resources. Using different attacks, for illustration the community attack would assist at different degrees so less accent is on Social Services Departments resources. Huge betterments are still needed. For illustration kid poorness is acquiring worst. The Report Monitoring poorness and societal exclusion ( 2009 ) found that kids who live in low-income families, where at least one grownup plants, is at the highest it has of all ti me been. This addition has affected the Governments kid poorness marks. The recession affected making the marks greatly. It is critical now to retrieve from the recession but besides to retrieve from underlying jobs that were at that place antecedently before the economic downswing began. Reports like Determining Our Lifes ( 2007 ) found that service users feel more responsible and confident about the aid they are having when they have been more involved in the determination processes. A study by Beresford et Al, ( 2007 ) found that service users would wish a watchdog with a board of service users and professionals so they could be involved in judging the quality of attention they receive. The Race Equality Act ( 2006 ) sets the context for anti- discriminatory pattern within which societal workers operate. However, whilst it could be critiqued that some advancement has been made as a consequence with regard to those of different civilization and faith, go oning inequalities would propose much more advancement remains to be made. To reason societal work has changed significantly and come on to assist all persons demands to go on. However as a profession it needs a larger voice to speak about the jobs they face hence acquiring excess aid to guarantee that service users and paupers have even fewer similarities. There has been a great displacement from pauper to service user. Service users have much more freedom and rights now. Albeit there are still similarities which need to be focused on to better the services available. Social Service Departments besides need to decline to allow policies be imposed when they do nt better on what is already implemented. Rights are now profiting service users but we need to guarantee this continues.