Thursday, November 28, 2019

Black Holes Essay Research Paper Our solar free essay sample

Black Holes Essay, Research Paper Our solar system consists of 10 planets go arounding around the Sun. The Sun serves as a magnet that uses its gravitative pull to keep the solar system together. If the Sun were to vanish, what would keep the planets together? The reply might be a black hole. A black hole is a theorized organic structure whose gravitation is so strong that even light can # 8217 ; t flight from within it ( Shipman 64 ) . If light can # 8217 ; t flight from a black hole, so it must be unseeable # 8211 ; therefore how can we cognize that the black holes be? How do they organize and where can we happen them? This paper will discourse the theory behind the black holes and physical grounds of their being. In order to understand black hole # 8217 ; s belongingss better, lets review basic rules of gravitation. Lets assume that a individual standing on a planet # 8217 ; s surface throws a stone in the air. We will write a custom essay sample on Black Holes Essay Research Paper Our solar or any similar topic specifically for you Do Not WasteYour Time HIRE WRITER Only 13.90 / page The stone will lift up to a point until the gravitation will draw it back, doing the stone autumn. If the individual will throw sway difficult plenty, it will get away planet # 8217 ; s gravitation. The velocity at which the stone will go forth a gravitative pull of a planet is called the # 8220 ; escape speed # 8221 ; . The flight speed differs on the planet # 8217 ; s mass ; the more mass the planet has # 8211 ; the higher flight speed will be. A black hole has so much mass concentrated in a little radius that its flight speed is greater than the speed of visible radiation ( Bunn ) . Since it is impossible for anything to go faster than visible radiation, it means that nil can get away a black hole ( Gribbin and White 75 ) . Black holes may organize after a star is overwhelmed by its gravitative force, that it can # 8217 ; t maintain from fall ining. During their life-time stars remain at a changeless size, because they contain a balance of forces: heat generated by firing atomic fuel expands the star outward while the force of gravitation pulls it in. This is illustrated in Figure 1, taken from the book Black Holes, Quasars, and the Universe. Figure 1. Excess force per unit area in the hot nucleus ( white pointer ) counterbalances the weight of the envelope ( solid pointer ) . The interior constantly loses energy to the envelope and finally to outer infinite because of the flow of radiation from the nucleus, through the envelope to the photosphere, and to infinite as the Sun radiances ( Shipman 26 ) . When a star exhausts its atomic fuel and collapses under its ain weight, it begins to shrivel in size. If the nucleus of the star is monolithic plenty the fall ining star will shrivel to a point where the gravitation will go strong plenty to pin down even light ( Shipman 24 ) . Such strong gravitation disturbs infinite and causes a black hole to hold some certain belongingss like # 8220 ; event skyline # 8221 ; . Event skyline is # 8220 ; a spherical surface that marks boundary of the black hole # 8221 ; . Equally shortly as affair base on ballss through the skyline it cant acquire back out, it will travel closer to black hole # 8217 ; s centre # 8211 ; nearing uniqueness ( Bunn ) . In 1969, American relativist John Wheeler named these massive collapsed stars as # 8220 ; black holes # 8221 ; ( Gribbin and White 74 ) . The thought that a star can shrivel and ensue in a great concentration of mass goes back to the eighteenth century. In the early eighteenth century, Isaac Newton researched and experimented with light. From his experiments he concluded the corpuscular theory of visible radiation, which states that light consists of bantam atoms that move in consecutive lines at great velocities ( Compton # 8217 ; s Multimedia Encyclopedia ) . The Gallic mathematician Pierre Simon de Laplace, in 1796, reasoned that light atoms could non get away from a monolithic organic structure ( Shipman 65 ) . The scientists disregarded Laplace # 8217 ; s theory, until Albert Einstein in 1916 came up with the theory of relativity ( Shipman 65 ) . In theory of relativity Einstein stated that # 8220 ; gravitation is non a force but a curving field in the space-time continuum that is created by the presence of mass # 8221 ; ( Compton # 8217 ; s Multimedia Encyclopedia ) . Not long after Einstein developed the th eory of relativity, the German uranologist Karl Schwarzschild calculated how compressed an object with a given mass ( in this instance a star ) should be in order to organize a black hole ( Shipman 65 ) . His equation became known the Schwarzschild radius, which shows to what critical radius a given mass should be compressed to go a black hole ( Gribbin and White 77 ) . In 1939, the United States physicists J. Robert Oppenheimer, Hartland S. Snyder and Volkoff showed that it is possible for monolithic stars to fall in and organize black holes ( Bunn and Shipman 65 ) . In 1970 # 8217 ; s, the British scientist Steven Hawking developed a theory that black holes are non wholly black ( Bunn ) . Peddling noticed that black holes comply with the 2nd jurisprudence of thermodynamics. The 2nd jurisprudence of thermodynamic says that â€Å"the information of an stray system ever increases, and that when two systems are joined together, the information of the combined system is greater than the amount of the informations of the single systems† ( Ferris 229 ) . It means that â€Å"the country of event skyline increases whenever affair fell into a black hole† . This was researched by pupil at Princeton named Jacob Bekenstein. Such a proposal was logical, but it had a defect in following with the 2nd jurisprudence of thermodynamics. If a organic structure has entropy it besides must hold a temperature, which means that black holes should breathe radiation. But how can black holes emit anything when by the definition nil can get away from their gravitative pull? When Hawking was sing Moscow in 1973 he had a opportunity to discourse black holes with two taking Soviet scientists Yakov Zeldovich and Alexander Starobinsky. They convinced Peddling that â€Å"according to the quantum mechanical uncertainness rule, revolving black holes should make and breathe particles† ( Ferris 230 ) . Peddling decided to cipher how much radiation is emitted from revolving black holes. He found out from his computations that even non – revolving black holes should breathe radiation. However this radiation does non straight comes out of black hole itself. The reply lies in quantum mechanics theory, which tells that â€Å"the atoms do non come from within the black hole, but from the empty â€Å"space† merely outside the black hole’s event horizon† ( Ferris 231 ) . Since black holes are unseeable, uranologists have been seeking to turn up them by detecting their effects. Black holes have enormous gravitative pull, affair and light atoms around them are attracted towards the centre. As affair and light atoms approach a black hole they # 8220 ; organize a swirling accumulation phonograph record, like H2O traveling down the plunghole of a bath, with gas stacking up and acquiring hot as gravitative energy is converted into energy of gesture # 8221 ; ( Gribbin and White 137 ) . When light enters a violent spin under strong gravitative pull, it emits quickly pulsating and noticeable X raies. In 1965 uranologists observed intense X raies coming from the configuration Cygnus ( about 10,000 light old ages off ) . When satellite engineering was born, in 1971, the universe # 8217 ; s first X-ray orbiter pinpointed the exact location of these X raies. It was found to be a monolithic but unseeable object that uranologists have named Cygnus X-1. Since Cyg nus X-1 exhibit all the conjectural belongingss of a black hole, there is a strong belief that Cygnus X-1 might be the first identified black hole ( Gribbin and White 138 ) . At present clip, the Hubble Space Telescope provides us with images that prove the being of black holes. The Figure 2 shows a recent image taken by the Hubble Space Telescope of a nearby galaxy, which has a monolithic black hole. Figure 2. Astronomers have obtained an unprecedented expression at the nearest illustration of galactic cannibalism # 8211 ; a monolithic black hole hidden at the centre of a nearby giant galaxy that is feeding on a smaller galaxy in a dramatic hit. Such pyrotechnics were common in the early existence, as galaxies formed and evolved, but are rare today ( The Space Telescope Science Institute ) . During the 2nd half of the twentieth century, due to scientific attempt and new orbiter engineering, research has contributed to happen more about black holes. As a consequence, scientists have found many of the black holes # 8217 ; belongingss and physical grounds of their being. However since we can # 8217 ; t animate a black hole, the best cogent evidence would be if we were able to travel near a black hole and detect it at close scope. Unfortunately such a mission is impossible right now because suspected nearest black hole is 10,000 light old ages off. For now we can merely detect at great distance the consequence that black holes impose on environing infinite. Bibliography Bunn, Ted. # 8220 ; Black Holes Frequently Asked Questions. # 8221 ; hypertext transfer protocol: //physics7.berkeley.edu/BHfaq.html # q5 ( Sept 1995 ) . Compton # 8217 ; s Multimedia Encyclopedia. Version 2.0P. Compact disc read-only memory. Compton # 8217 ; s Learning Company, 1991. Eisenhamer, Jonathan and Levay, Zolt. # 8220 ; Hubble Provides Multiple Positions of How to Feed a Black Hole. # 8221 ; hypertext transfer protocol: //oposite.stsci.edu /pubinfo/pr/1998/14 ( 14 May 1998 ) . Gribbin, John and White, Michael. Stephen Hawking # 8211 ; A Life in Science. London: Penguin Books, Ltd. , 1992. Ferris, Timothy. Physicss, Astronomy, and Mathematicss. New York: Back Bay Books, 1991. Shipman, Harry. Black Holes, Quasars, and the Universe. Boston: Houghton Mifflin Company, 1976.

Sunday, November 24, 2019

What Hiring Managers Really Think of Tattoos in the Workplace

What Hiring Managers Really Think of Tattoos in the Workplace Approximately 14% of the American population have at least one tattoo.  Tattoo culture is on the rise  and the trend has finally made it’s way into mainstream society. Despite this new found popularity, there still seems to be a stigma regarding tattoos in the workplace. According to Skinfo.com, People with tattoos are more likely to have limited career potential and are less likely to get hired. In fact, there are no laws in place protecting people with tattoos from discrimination. And it’s not just hiring managers either. Skinfo also found that 43% of people feel that visible tattoos in the workplace are inappropriate.Luckily, large companies such as IKEA, Target, Google, and UPS are more tattoo friendly.So if you have tattoos, how do you go about finding and keeping a job? Skinfo has compiled an infographic on everything you need to know about tattoos in the workplace. You can use this to plan for the good, the bad, and the ugly. Good luck!Source:[BusinessInsider ]

Thursday, November 21, 2019

Economic Essay Example | Topics and Well Written Essays - 1000 words - 6

Economic - Essay Example The following table and chart can help us get an idea: â€Å"The main engine of growth is the accumulation of human capital† and â€Å"the main source of differences in living standards among nations is the difference in human capital. Physical capital plays an essential but decidedly subsidiary role† (Bardhan and Udry, 2008, p.266). Human capital has several aspects, including education, training and health. Human capital is considered to be a key determination of economic growth. Investment in human capital benefits individuals and society as a whole. Countries with skilled people grew faster. While develop economics have always emphasized the importance of education in the develop process. Better education generates private gains to the individual and public benefits to society. By the expansion of knowledge it can enable the people to overcome ignorance and superstitions. Without literacy or, to be more precise, perfect education, economic growth in any country is impossible. It can become only lopsided and flawed unless corrective reassures are taken. Jobless growth implies situation where the overall economic growth but does not expand sufficient employment opportunities. ‘Futureless growth’ implies a situation where growth occurs but its effect is not positive on future generation and is adverse. According to Mahatma Gandhi, â€Å"education, which does not teach us to discriminate between good or bad, to assimilate the one and eschew the other, is a misnomer† (Rao, 1991, p.2). The purpose of literacy and education is not only to make our children literate and merely develop individual skills but also to make them appropriate human being with positive and integrated elements of their respective societies. Education means drawing out the best of the body, mind and spirit in man. He believed learning by doing and was of the opinion that the knowledge through education, subjects

Wednesday, November 20, 2019

STRATEGIC REVIEW OF TESCO PLC UK Essay Example | Topics and Well Written Essays - 2500 words

STRATEGIC REVIEW OF TESCO PLC UK - Essay Example By understanding the environment in which you operate (external to your company or department), you can take advantage of the opportunities and minimize the threats. (RAPIDBI) Specifically the PEST or PESTLE analysis is a useful tool for understanding risks associated with market growth or decline, and as such the position, potential and direction for a business or organization. Political Environment: Political forces influence the legislations and government rules and regulations under which the firm operates. Every company faces political constraints in the form of antitrust laws, fair trade decisions, and tax programs, minimum usage legislations, pollution and pricing policies, administrative activities and many other actions, aimed at protecting the consumers and the local environment. In 2001, The Department of Trade and Industry (DTI) introduced a Supermarkets' Code of Practice to regulate trading relationships between the four largest UK supermarkets and their suppliers. Economic Environment: Economic factors clearly indicate the nature and direction of the economy in which a firm operates. Every market is unique and consumption patterns change along with the wealth of the consumers in various market segments. For strategic planning all the economic trends at national and international levels have to be considered. Tesco makes a significant contribution to economic activity in all the places in which it operates. Every week over 400,000 staff serves over 30 million customers in 13 countries. The company has to its credit a track record of providing value for customers, creating jobs and training, providing opportunities for suppliers and regenerating deprived areas. The business relationship of Tesco is with nearly 2,000 own-brand primary suppliers in 98 countries. Social Environment: The social environment is an important factor as changes in the values, beliefs, attitudes, opinions and lifestyles in society create potential opportunities for an organization. For a company to grow, it is necessary to take advantage of societal changes. The cultural, demographic, religious, educational and ethnic conditioning of individuals in society affects the social environment. One of the most important values in which Tesco believes is to treat people how it would like to be treated. The company strives hard to achieve this by being a good employer and by playing its part in local communities. People believe that they the company can use its size and success to be a force for good. This challenge is indeed accepted by Tesco with enthusiasm and commitment. Technological Environment: Technological environment means the trends and developments in the technological field that might: improve production, create new product opportunities, and render the existing ones.

Monday, November 18, 2019

Take Home Exam For The Class Economics of Race & Gender Essay - 1

Take Home Exam For The Class Economics of Race & Gender - Essay Example This class, Economics of Race and Gender has helped me understand how important diversity is to a country and the businesses that operate in that country. Diversity does not only impact the social structure within an economy but it also impacts the educational structure, economic structure and the legal structure. Race and Gender have evolved over the years, previously women and people from ethnic minorities did not interact with the general population, they did minor jobs and the women stayed at home. Times have changed now, women get out of their homes more often and ethnic minorities have more legal rights than the general population. The advantages of group presentations are that they are an easier method to help the students understand what is being taught. It allows the students to conduct research on the topic before hand. These presentations encourage discussions in class and students can question their fellow students more easily then they can question the professor. The disadvantages are that it is usually only that group which needs to present that goes over the topic and conducts research. Other students may become complacent and not take an interest in the class. The three movies presented in class were: Freedom Writers, Rossie the Riveter and Crash. Freedom Writers is a movie about a teacher who inspires her class to write about their daily lives. The students come from poor backgrounds and have experienced much pain in life, they are discouraged and violent, and the teacher asks them to write about their daily lives and things which have happened to them in the past, events which altered their lives. She compiles these diaries into a book and gets them published into a book. These children with the help of their teacher learn some lesson of life such as to be tolerant. Rossie the Riveter revolves around a woman who worked in the factories at a time when women would hardly leave the house. This

Friday, November 15, 2019

Simply Supported And Cantilever Beams

Simply Supported And Cantilever Beams A beam is a structural member which safely carries loads i.e. without failing due to the applied loads. We will be restricted to beams of uniform cross-sectional area. Simply Supported Beam A beam that rests on two supports only along the length of the beam and is allowed to deflect freely when loads are applied. Note see section A of unit. Cantilever Beam A beam that is supported at one end only. The end could be built into a wall, bolted or welded to another structure for means of support. Point or Concentrated Load A load which acts at a particular point along the length of the beam. This load is commonly called a force (F) and is stated in Newtons (N). A mass may be converted into a force by multiplying by gravity whose value is constant at 9.81 m/s2. Uniformly Distributed Load (UDL) A load which is spread evenly over a given length of the beam. This may be the weight of the beam itself. The UDL is quoted as Newtons per metre (N/m). Beam Failure If excessive loads are used and the beam does not have the necessary material properties of strength then failure will occur. Failure may occur in two ways:- Calculating Shear Forces (we must use the shear force rule). When looking right of a section : downward forces are positive and upward forces are negative. When looking left of a section: downward forces are negative and upward forces are positive. Starting at point A and looking left: (note: the negative sign (-) means just to the left of the position and the positive sign (+) means just to the right of the position.) SFA = 0 kN SFA + = 6 kN An alternative method of drawing the shear force diagram is to follow the directions of each force on the line diagram.SFB = 6 kN SFB + = 6 kN SFC = 6 kN SFC + = 6 kN SFD = 6 kN SFD + = 6 12 = -6 kN SFE = 6 12 = -6 kN SFE + = 6 12 = -6 kN SFF = 6 12 = -6 kN SFF + = 6 12 = -6 kN SFG = 6 12 = -6 kN SFG + = 6 12 + 6 = 0 kN Note: the shear force at either end of a simply supported beam must equate to zero. Calculating Bending Moments (we must use the bending moment rule). When looking right of a section : downward forces are negative and upward forces are positive. When looking left of a section: downward forces are negative and upward forces are positive. section F F section F + F + Hogging Beam Sagging Beam Starting at point A and looking left: BMA = 0 kNm BMB = (6 x 1) = 6 kNm BMC = (6 x 2) = 12 kNm BMD = (6 x 3) = 18 kNm BME = (6 x 4) + ( -12 x 1) = 12 kNm BMF = (6 x 5) + ( -12 x 2) = 6 kNm BMG = (6 x 6) + ( -12 x 3) = 0 kNm Note: the bending moment at either end of a simply supported beam must equate to zero. The following page shows the line, shear force and bending moment diagrams for this beam. Simply Supported Beam with Point Load 6 m F E D C G B A 6 kN 6 kN F =12 kN Shear Force Diagram (kN) 0 0 -6 6 0 Line Diagram 12 12 18 6 0 6 Bending Moment Diagram (kNm) Max Tensile Stress SAGGING (+ve bending) Max Compressive Stress F F A maximum bending moment of 18 kNm occurs at position D. Note the shear force is zero at this point. Simply Supported Beam with Distributed Load UDL = 2 kN/m F E D C G B A 6 m RA The force from a UDL is considered to act at the UDL mid-point. e.g. if we take moments about D then the total force from the UDL (looking to the left) would be: (2 x 3) = 6 kN. This force must be multiplied by the distance from point D to the UDL mid point as shown below. e.g. Take moments about D, then the moment would be: (-6 x 1.5) = -9 kNm 1.5m UDL = 2 kN/m D C B A 3 m Taking moments about point D (looking left) We must first calculate the reactions RA and RG. We take moments about one of the reactions to calculate the other, therefore to find RA: Take moments about RG ÃŽ £Clockwise moments (CM) = ÃŽ £Anti-clockwise moments (ACM) RA x 6 = 2 x 6 x 3 RA = 6 kN now, ÃŽ £Upward Forces = ÃŽ £Downward Forces RA + RG = 2 x 6 6 + RG = 12 RG = 6 kN section F + F F F + Calculating Shear Forces (we must use the shear force rule). When looking right of a section : downward forces are positive and upward forces are negative. When looking left of a section: downward forces are negative and upward forces are positive. Starting at point A and looking left: (note: the negative sign (-) means just to the left of the position and the positive sign (+) means just to the right of the position.) SFA = 0 kN SFA + = 6 kN SFB = 6 (21) = 4 kN SFB + = 6 (21) = 4 kN SFC = 6 (22) = 2 kN SFC + = 6 (22) = 2 kN SFD = 6 (23) = 0 kN SFD + = 6 (23) = 0 kN SFE = 6 (24) = -2 kN SFE + = 6 (24) = -2 kN SFF = 6 (25) = -4 kN SFF + = 6 (25) = -4 kN SFG = 6 (26) = -6 kN SFG + = 6 (26) + 6 = 0 kN Note: the shear force at either end of a simply supported beam must equate to zero. Calculating Bending Moments (we must use the bending moment rule). When looking right of a section : downward forces are negative and upward forces are positive. When looking left of a section: downward forces are negative and upward forces are positive. section F F section F + F + Hogging Beam Sagging Beam Starting at point A and looking left: BMA = 0 kNm BMB = (6 x 1) + (-2 x 1 x 0.5) = 5 kNm BMC = (6 x 2) + (-2 x 2 x 1) = 8 kNm BMD = (6 x 3) + (-2 x 3 x 1.5) = 9 kNm BME = (6 x 4) + (-2 x 4 x 2) = 8 kNm BMF = (6 x 5) + + (-2 x 5 x 2.5 = 5 kNm BMG = (6 x 6) + + (-2 x 6 x 3) = 0 kNm Note: the bending moment at either end of a simply supported beam must equate to zero. The following page shows the line, shear force and bending moment diagrams for this beam. Simply Supported Beam with Distributed Load 4 2 0 -2 -4 UDL = 2 kN/m 6 m F E D C G B A Shear Force Diagram (kN) 0 0 -6 6 0 Line Diagram 8 8 9 5 0 Bending Moment Diagram (kNm) 5 6 kN 6 kN Max Tensile Stress SAGGING (+ve bending) Max Compressive Stress F F A maximum bending moment of 9 kNm occurs at position D. Note the shear force is zero at this point. Simply Supported Beam with Point Loads 6 m F E D C G B A RA RG F = 15 kN F = 30 kN We must first calculate the reactions RA and RG. We take moments about one of the reactions to calculate the other, therefore to find RA: Take moments about RG ÃŽ £Clockwise moments (CM) = ÃŽ £Anti-clockwise moments (ACM) RA x 6 = (15 x 4) + (30 x 2) RA = 20 kN now, ÃŽ £Upward Forces = ÃŽ £Downward Forces RA + RG = 15 + 30 20 + RG = 45 RG = 25 kN section F + F F F + Calculating Shear Forces (we must use the shear force rule). When looking right of a section : downward forces are positive and upward forces are negative. When looking left of a section: downward forces are negative and upward forces are positive. Starting at point A and looking left: (note: the negative sign (-) means just to the left of the position and the positive sign (+) means just to the right of the position.) SFA = 0 kN SFA + = 20 kN SFB = 20 kN SFB + = 20 kN SFC = 20 kN SFC + = 20 -15 = 5 kN SFD = 20 -15 = 5 kN SFD + = 20 -15 = 5 kN SFE = 20 -15 = 5 kN SFE + = 20 -15 30 = -25 kN SFF = 20 -15 30 = -25 kN SFF + = 20 -15 30 = -25 kN SFG = 20 -15 30 = -25 kN SFG + = 20 -15 30 + 25 = 0 kN Note: the shear force at either end of a simply supported beam must equate to zero. Calculating Bending Moments (we must use the bending moment rule). When looking right of a section : downward forces are negative and upward forces are positive. When looking left of a section: downward forces are negative and upward forces are positive. section F F section F + F + Hogging Beam Sagging Beam Starting at point A and looking left: BMA = 0 kNm BMB = (20 x 1) = 20 kNm BMC = (20 x 2) = 40 kNm BMD = (20 x 3) + (-15 x 1) = 45 kNm BME = (20 x 4) + (-15 x 2) = 50 kNm BMF = (20 x 5) + (-15 x 3) + (-30 x 1) = 25 kNm BMG = (20 x 6) + (-15 x 4) + (-30 x 2) = 0 kNm Note: the bending moment at either end of a simply supported beam must equate to zero. The following page shows the line, shear force and bending moment diagrams for this beam. 0 20 -25 0 Shear Force Diagram (kN) 5Simply Supported Beam with Point Loads 6 m F E D C G B A 20 kN 25 kN F = 15 kN F = 30 kN Bending Moment Diagram (kNm) 0 0 45 40 20 50 25 Max Tensile Stress SAGGING (+ve bending) Max Compressive Stress F F A maximum bending moment of 50 kNm occurs at position E. Note the shear force is zero at this point. Simply Supported Beam with Point and Distributed Loads (1) 6 m F E D C G B A RA RG 15 kN 30 kN UDL = 10 kN/m We must first calculate the reactions RA and RG. We take moments about one of the reactions to calculate the other, therefore to find RA: Take moments about RG ÃŽ £Clockwise moments (CM) = ÃŽ £Anti-clockwise moments (ACM) RA x 6 = (15 x 4) + (10 x 2 x 3) + (30 x 2) RA = 30 kN now, ÃŽ £Upward Forces = ÃŽ £Downward Forces RA + RG = 15 + (10 x 2) + 30 30 + RG = 65 RG = 35 kN section F + F F F + Calculating Shear Forces (we must use the shear force rule). When looking right of a section : downward forces are positive and upward forces are negative. When looking left of a section: downward forces are negative and upward forces are positive. Starting at point A and looking left: (note: the negative sign (-) means just to the left of the position and the positive sign (+) means just to the right of the position.) SFA = 0 kN SFA + = 30 kN SFB = 30 kN SFB + = 30 kN SFC = 30 kN SFC + = 30 15 = 15 kN SFD = 30 15 (10 x 1) = 5 kN SFD + = 30 15 (10 x 1) = 5 kN SFE = 30 15 (10 x 2) = -5 kN SFE + = 30 15 (10 x 2) 30 = -35 kN SFF = 30 15 (10 x 2) 30 = -35 kN SFF + = 30 15 (10 x 2) 30 = -35 kN SFG = 30 15 (10 x 2) 30 = -35 kN SFG + = 30 15 (10 x 2) 30 + 35 = 35 kN Note: the shear force at either end of a simply supported beam must equate to zero. Calculating Bending Moments (we must use the bending moment rule). When looking right of a section : downward forces are negative and upward forces are positive. When looking left of a section: downward forces are negative and upward forces are positive. section F F section F + F + Hogging Beam Sagging Beam Starting at point A and looking left: BMA = 0 kNm BMB = (30 x 1) = 30 kNm BMC = (30 x 2) = 60 kNm BMD = (30 x 3) + (-15 x 1) + (-10 x 1 x 0.5) = 70 kNm BME = (30 x 4) + (-15 x 2) + (-10 x 2 x 1) = 70 kNm BMF = (30 x 5) + (-15 x 3) + (-10 x 2 x 2) + (-30 x 1) = 35 kNm BMG = (30 x 6) + (-15 x 4) + (-10 x 2 x 3) + (-30 x 2) = 0 kNm Notes: the bending moment at either end of a simply supported beam must equate to zero. The value of the maximum bending moment occurs where the shear force is zero and is therefore still unknown (see Shear Force diagram). The distance from point A to this zero SF point must be determined as follows:- x = 2 15 20 x = 1.5 m Total distance from point A = 2 + 1.5 = 3.5 m therefore, BM max = (30 x 3.5) + (-15 x 1.5) + (-10 X 1.5 x 0.75) = 71.25 kNm The following page shows the line, shear force and bending moment diagrams for this beam. 70 71.25 35 30 60 70 0 0 Simply Supported Beam with Point and Distributed Loads (1) 2 m x 30 -5 Shear Force Diagram (kN) 0 -35 15 0 6 m F E D C G B A 30 kN 35 kN 15 kN 30 kN UDL = 10 kN/m 20 kN Bending Moment Diagram (kNm) Max Tensile Stress SAGGING (+ve bending) Max Compressive Stress F F A maximum bending moment of 71.25 kNm occurs at a distance 3.5 m from position A. Simply Supported Beam with Point and Distributed Loads (2) 1 m RB 12 m E D C F B A 8 kN RE UDL = 6 kN/m UDL = 4 kN/m 12 kN We must first calculate the reactions RB and RE. We take moments about one of the reactions to calculate the other, therefore to find RB. Take moments about RE ÃŽ £Clockwise moments (CM) = ÃŽ £Anti-clockwise moments (ACM) (RBx10)+(6x1x0.5) = (4 x 4 x 9) + (8 x 7) + (12 x 3) + (6 x 3 x 1.5) RB = 26 kN now, ÃŽ £Upward Forces = ÃŽ £Downward Forces RB + RE = (4 x 4) + 8 + 12 + (6 x 4) 26 + RE = 60 RE = 34 kN Calculating Shear Forces Starting at point A and looking left: SFA = 0 kN SFA + = 0 kN SFB = -4 x 1 = -4 kN SFB + = (-4 x 1) + 26 = 22 kN SFC = (-4 x 4) + 26= 10 kN SFC + = (-4 x 4) + 26 8 = 2 kN SFD = (-4 x 4) + 26 8 = 2 kN SFD + = (-4 x 4) + 26 8 12 = -10 kN SFE = (-4 x 4) + 26 8 12 (6 x 3) = -28 kN SFE + = (-4 x 4) + 26 8 12 (6 x 3) + 34 = 6 kN SFF = (-4 x 4) + 26 8 12 (6 x 4) + 34 = 0 kN SFF + = (-4 x 4) + 26 8 12 (6 x 4) + 34 = 0 kN Calculating Bending Moments Starting at point A and looking left: BMA = 0 kNm BMB = (-4 x 1 x 0.5) = -2 kNm BM 2m from A = (-4 x 2 x 1) + (26 x 1) = 18 kNm BM 3m from A = (-4 x 3 x 1.5) + (26 x 2) = 34 kNm BMC = (-4 x 4 x 2) + (26 x 3) = 46 kNm BMD = (-4 x 4 x 6) + (26 x 7) + (-8 x 4) = 54 kNm BM 9m from A = (-4 x 4 x 7) + (26 x 8) + (-8 x 5) + (-12 x 1) + (-6 x 1 x 0.5) = 41 kNm BM 9m from A = (-4 x 4 x 8) + (26 x 9) + (-8 x 6) + (-12 x 2) + (-6 x 2 x 1) = 22 kNm BME = (-4 x 4 x 9) + (26 x 10) + (-8 x 7) + (-12 x 3) + (-6 x 3 x 1.5) = -3 kNm BMF = (-4 x 4 x 10) + (26 x 11) + (-8 x 8) + (-12 x 4) + (-6 x 4 x 2) + (34 x 1) = 0 kNm Point of Contraflexure At any point where the graph on a bending moment diagram passes through the 0-0 datum line (i.e. where the BM changes sign) the curvature of the beam will change from hogging to sagging or vice versa. Such a point is termed a Point of Contraflexure or Inflexion. These points are identified in the following diagram. It should be noted that the point of contraflexure corresponds to zero bending moment. Turning Points The mathematical relationship between shear force and corresponding bending moment is evidenced on their respective graphs where the change of slope on a BM diagram aligns with zero shear on the complementary shear force diagram. Thus, at any point on a BM diagram where the slope changes direction from upwards to downwards or vice versa, all such Turning Points occur at positions of Zero Shear. Turning points are also identified in the following diagram. Simply Supported Beam with Point and Distributed Loads (2) 1 m 26 kN 12 m E D C F B A 8 kN 34 kN UDL = 6 kN/m UDL = 4 kN/m 12 kN 2 6 2 -4 22 -10 Shear Force Diagram (kN) 0 -28 10 0 F F SAGGING (+ve bending) -3 22 41 54 46 34 18 -2 Bending Moment Diagram (kNm) 0 0 F F HOGGING (-ve bending) Points of Contraflexure The maximum bending moment is equal to 54 kNm and occurs at point D where the shear force is zero. Turning points occur at -2 kNm and -3 kNm. Cantilever Beam with Point Load 6 m F E D C G B A RA 12 kN Free End Fixed End In this case there is only one unknown reaction at the fixed end of the cantilever, therefore: ÃŽ £Upward Forces = ÃŽ £Downward Forces RA = 12 kN Calculating Shear Forces Starting at point A and looking left: SFA = 0 kN SFA + = 12 kN SFB = 12 kN SFB + = 12 kN SFC = 12 kN SFC + = 12 kN SFD = 12 kN SFD + = 12 kN SFE = 12 kN SFE + = 12 kN SFF = 12 kN SFF + = 12 kN SFG = 12 kN SFG + = 12 12 = 0 kN Note: the shear force at either end of a cantilever beam must equate to zero. Calculating Bending Moments NB for simplicity at this stage we shall always look towards the free end of the beam. Starting at fixed end, point A, and looking right towards the free end: (the same results may be obtained by starting at point G and looking right) BMA = -12 x 6 = -72 kNm BMB = -12 x 5 = -60 kNm BMC = -12 x 4 = -48 kNm BMD = -12 x 3 = -36 kNm BME = -12 x 2 = -24 kNm BMF = -12 x 1 = -12 kNm BMG = 0 kNm Notes: the maximum bending moment in a cantilever beam occurs at the fixed end. In this case the 12kN force in the beam is trying to bend it downwards, (a clockwise moment). The support at the fixed end must therefore be applying an equal but opposite moment to the beam. This would be 72 kNm in an anti-clockwise direction. See the following diagram. The value of the bending moment at the free end of a cantilever beam will always be zero. -12 -24 -36 -48 -60 -72 Bending Moment Diagram (kNm) 0 0 12 125 Shear Force Diagram (kN) 0 0 72 kNm 72 kNm 6 m F E D C G B A 12 kN 12 kN The following shows the line, shear force and bending moment diagrams for this beam. F F HOGGING (-ve bending) Max Tensile Stress Max Compressive Stress A maximum bending moment of -72 kNm occurs at position A. Cantilever Beam with Distributed Load UDL = 2 kN/m 6 m F E D C G B A RA To calculate the unknown reaction at the fixed end of the cantilever: ÃŽ £Upward Forces = ÃŽ £Downward Forces RA = 2 x 6 RA = 12 kN Calculating Shear Forces Starting at point A and looking left: SFA = 0 kN SFA + = 12 kN SFB = 12 (2 x 1) = 10 kN SFB + = 12 (2 x 1) = 10 kN SFC = 12 (2 x 2) = 8 kN SFC + = 12 (2 x 2) = 8 kN SFD = 12 (2 x 3) = 6 kN SFD + = 12 (2 x 3) = 6 kN SFE = 12 (2 x 4) = 4 kN SFE + = 12 (2 x 4) = 4 kN SFF = 12 (2 x 5) = 2 kN SFF + = 12 (2 x 5) = 2 kN SFG = 12 (2 x 6) = 0 kN SFG + = 12 (2 x 6) = 0 kN Note: the shear force at either end of a cantilever beam must equate to zero. Calculating Bending Moments Starting at fixed end, point A, and looking right towards the free end: (the same results may be obtained by starting at point G and looking right) BMA = -2 x 6 x 3 = -36 kNm BMB = -2 x 5 x 2.5 = -25 kNm BMC = -2 x 4 x 2 = -16 kNm BMD = -2 x 3 x 1.5 = -9 kNm BME = -2 x 2 x 1 = -4 kNm BMF = -2 x 1 x 0.5 = -1 kNm BMG = 0 kNm The following page shows the line, shear force and bending moment diagrams for this beam. Cantilever Beam with Distributed Load8 6 4 2 36 kNm 36 kNm 12 105 Shear Force Diagram (kN) 0 0 -1 -4 -9 -16 -25 -36 Bending Moment Diagram (kNm) 0 0 6 m F E D C G B A 12 kN UDL = 2 kN/m F F HOGGING (-ve bending) Max Tensile Stress Max Compressive Stress A maximum bending moment of -36 kNm occurs at position A. Cantilever Beam with Point and Distributed Loads RG 2 m 10 kN B C D E A F G 4 m UDL = 10 kN/m To calculate the unknown reaction at the fixed end of the cantilever: ÃŽ £Upward Forces = ÃŽ £Downward Forces RG = (10 x 6) + 10 RG = 70 kN Calculating Shear Forces Starting at point A and looking left: SFA = 0 kN SFA + = 0 kN SFB = -10 x 1 = -10 kN SFB + = -10 x 1 = -10 kN SFC = -10 x 2 = -20 kN SFC + = (-10 x 2) + (-10) = -30 kN SFD = (-10 x 3) + (-10) = -40 kN SFD + = (-10 x 3) + (-10) = -40 kN SFE = (-10 x 4) + (-10) = -50 kN SFE + = (-10 x 4) + (-10) = -50 kN SFF = (-10 x 5) + (-10) = -60 kN SFF + = (-10 x 5) + (-10) = -60 kN SFG = (-10 x 6) + (-10) = -70 kN SFG + = (-10 x 6) + (-10) + 70 = 0 kN Note: the shear force at either end of a cantilever beam must equate to zero. Calculating Bending Moments Starting at point A, and looking left from the free end: (the same results may be obtained by starting at point G and looking left) BMA = 0 kNm BMB = -10 x 1 x 0.5 = -5 kNm BMC = -10 x 2 x 1 = -20 kNm BMD = (-10 x 3 x 1.5) + (-10 x 1) = -55 kNm BME = (-10 x 4 x 2) + (-10 x 2) = -100 kNm BMF = (-10 x 5 x 2.5) + (-10 x 3) = -155 kNm BMG = (-10 x 6 x 3) + (-10 x 4) = -220 kNm The following page shows the line, shear force and bending moment diagrams for this beam. 70 kN 2 m 10 kN B C D E A F G 4 m UDL = 10 kN/m 0 0 Shear Force Diagram (kN) -60 -70 -10 -20 -40 -50 220 kNm 220 kNm -30Cantilever Beam with Point and Distributed Loads 0 0 Bending Moment Diagram (kNm) -220 -5 -20 -55 -100 -155 F F HOGGING (-ve bending) Max Tensile Stress Max Compressive Stress A maximum bending moment of -220 kNm occurs at position G.

Wednesday, November 13, 2019

Intra-Organizational Collaboration Essay -- Business, Collaborative Or

Intra-organizational Collaboration Organizations need to recognize the necessity to flatten hierarchical structures. In creating a horizontal structure leaders naturally acquire a greater ability to adapt. Leaders need to have an intensive focus on people. Recognizing the need to tie both compensation and advancement potential with individual’s ability to function in a T-shaped management style. Organizations need to be willing to change the way they hire in order to promote intra-organizational collaboration. Horizontal organizations require the establishment of high performance teams that have legitimate decision-making power. Leaders must be on board and willing to walk the talk as it relates to creating a team centric organization. They must invite honest feedback, designing mechanism whereas constituents can deliver honest feedback without fear of reprisal. In organizations of the future leaders need to be brave admitting mistakes openly and using them as opportunities to learn. Leaders also need to understand that feedback in a team centric organization is not a personal attack. Leaders need to treat negative feedback as a simple business case. Team centric leaders should also encourage others and not be afraid to ask for assistance when necessary. (Guttman, 2009, pg.47-51) People First The most important aspect of collaboration is people. As indicated previously one of the guiding principles for this leaders’ guide, â€Å"The successful ventures are truly creative collaborations of talented people committed to beating the odds.† (Hill, 2000, pp. 46) In order to be truly collaborative constituents need to exist in an environment built on mutual trust. This is the only way leaders can establish a team centric cul... ...petency, in order to collaborate better constituents need to have feedback provided on a continuous basis. â€Å"To get any traction, leaders need to set criteria and metrics for cross-unit contributions. Overarching concepts-â€Å"teamwork†-need to be translated into concrete behaviors that are ranked from the least to the most desirable.† Hansen, 2009, pp. 102) The success of any method is contingent upon its execution but one popular method is the 360-analysis. This type of performance analysis allows for a holistic view of a constituents performance both within his or her unit and their impact on the broader organization. â€Å"Leaders pursuing disciplined collaboration practice consequence management: if people perform well both within and across their units, they are promoted. If they fall short on either dimension, they face the consequences.† (Hansen, 2009, pp. 104)

Sunday, November 10, 2019

Pareting Skills Essay

1)What is positive parenting? Positive parenting is when you encourage good behavior, good communication, and help your children maintain high self-esteem. 2)What is discipline? How does it differ from punishment? Discipline is to teach your child from their mistake and it’s different from punishment because punishment is when your child doesn’t abide by the rules nor does something they were told not to do. 3)What is active listening? Why is it used by parents? Active listening is when the child learns from being given feedback about what they do. Parents use active listening so their child learns from what they did and what they can do to What is guidance? Provide an example of a parent providing guidance to a child? Guidance is guiding your child in the right direction. An example of a parent providing guidance to a child is if your child is yelling in the library instead of yelling back at him telling him to shut up, tell him to talk lower. 4)Where can families and parents find support and resources? Families and parents find support and resources by going to classes for caring for infants, churches, schools, doctors’ offices, and community centers. Do you think positive parenting techniques work? Why or why not? Yes I think positive parenting techniques work because I feel that children respond positively when parents are not negative. 1.Choose one of the three scenarios presented at the beginning of the module. Describe how you would handle the situation. †¢You come home from work and find that your middle school-aged daughter hasn’t finished her homework, even though she was supposed to have the homework done so that you could have a family movie night Since she didn’t do her homework before I got home; while everyone is watching the movie for movie night she will be doing her homework and will not be able to join the family until she is finished 1)Why are consequences an important part of positive parenting? Consequences are an important part of positive parenting because if your child doesn’t have any consequences for their actions they will keep doing the things they do.

Friday, November 8, 2019

Venus Flytrap Facts (Dionaea muscipula)

Venus Flytrap Facts (Dionaea muscipula) The Venus flytrap (Dionaea muscipula) is a rare carnivorous plant that captures and digests its prey with fleshy, hinged jaws. These jaws are actually modified portions of the plants leaves. The plant gets its common name for Venus, the Roman goddess of love. This refers either to the plant traps supposed resemblance to female genitalia or to the sweet nectar it uses to lure its victims. The scientific name comes from Dionaea (daughter of Dione or Aphrodite, the Greek goddess of love) and muscipula (Latin for mousetrap). Fast Facts: Venus Flytrap Scientific Name: Dionaea muscipulaCommon Names: Venus flytrap, tippity twitchetBasic Plant Group: Flowering plant (angiosperm)Size: 5 inchesLifespan: 20-30 yearsDiet: Crawling insectsHabitat: North and South Carolina coastal wetlandsPopulation: 33,000 (2014)Conservation Status: Vulnerable Description The Venus flytrap is a small, compact flowering plant. A mature rosette has between 4 and seven leaves and reaches a size up to 5 inches. Each leaf blade has a petiole capable of photosynthesis and a hinged trap. The trap contains cells that produce the red pigment anthocyanin. Within each trap are trigger hairs that sense touch. The edges of the trap lobes are lined with stiff protrusions which lock together when the trap closes to prevent prey from escaping. Habitat The Venus flytrap lives in damp sandy and peaty soil. It is native only to the coastal bogs of North and South Carolina. The soil is poor in nitrogen and phosphorus, so the plant needs to supplement photosynthesis with nutrients from insects. North and South Carolina get mild winters, so the plant is adapted to cold. Plants that do not undergo winter dormancy eventually weaken and die. Northern Florida and western Washington host successful naturalized populations. Diet and Behavior While the Venus flytrap relies on photosynthesis for most of its food production, it requires supplementation from proteins in prey to meet its nitrogen requirements. Despite its name, the plant primarily catches crawling insects (ants, beetles, spiders) rather than flies. In order for prey to be captured, it must touch the trigger hairs inside the trap more than once. Once triggered, it only takes about a tenth of a second for the trap lobes to snap shut. Initially the fringes of the trap loosely hold the prey. This allows very small prey to escape, as they arent worth the energy expenditure of digestion. If the prey is large enough, the trap fully closes to become a stomach. Digestive hydrolase enzymes are released into the trap, nutrients are absorbed through the leafs interior surface, and 5 to 12 days later the trap opens to release the remaining chitin shell of the insect. Large insects can damage the traps. Otherwise, each trap can only function a few times before the leaf dies and must be replaced. Suitable prey must be small enough to fit within the trap but large enough to supply enough nutrients. de-kay / Getty Images Reproduction Venus flytraps are capable of self-pollination, which occurs when pollen from the plants anthers fertilize a flowers pistil. However, cross-pollination is common. The Venus flytrap does not capture and eat insects that pollinate its flowers, such as sweat bees, checkered beetles, and long-horned beetles. Scientists arent entirely certain how the pollinators avoid being trapped. It could be that the color of the flowers (white) attracts pollinators, while the color of the traps (red and green) attracts prey. Other possibilities include scent differences between the flower and trap, and flower placement above the traps. After pollination, the Venus flytrap produces black seeds. The plant also reproduces by dividing into colonies from rosettes that form beneath mature plants. Conservation Status The IUCN lists the Venus flytraps conservation status as vulnerable. The population of plants in the species natural habitat is decreasing. As of 2014, an estimated 33,000 plants remained, all within a 75 mile radius of Wilmington, NC. Threats include poaching, fire prevention (the plant is fire resistant and relies on periodic burning to control competition), and habitat loss. In 2014, North Carolina Senate Bill 734 made collecting wild Venus flytrap plants a felony. Care and Cultivation The Venus flytrap is a popular houseplant. While its an easy plant to keep, it has certain requirements. It must be planted in acidic soil with good drainage. Usually, it is potted in a mixture of sphagnum peat moss and sand. Its important to water the plant with rainwater or distilled water to provide the proper pH. The plant needs 12 hours of direct sunlight per day. It should not be fertilized and should only be offered an insect if it appears unhealthy. In order to survive, a Venus flytrap requires exposure to a period of cooler temperatures to simulate winter. While the Venus flytrap will grow from seed, it is usually cultivated by dividing the rosettes in the spring or summer. Commercial propagation for nurseries occurs in vitro from plant tissue culture. Many interesting mutations for size and color are available from nurseries. Uses In addition to cultivation as a houseplant, Venus flytrap extract is sold as a patent medicine named Carnivora. The American Cancer Society states that Carnivora is sold as an alternative treatment for skin cancer, HIV, rheumatoid arthritis, herpes, and Crohns disease. However, the health claims have not been supported by scientific evidence. The purified active ingredient in the plant extract, plumbagin, does show antitumor activity. Sources DAmato, Peter (1998). The Savage Garden: Cultivating Carnivorous Plants. Berkeley, California: Ten Speed Press. ISBN 978-0-89815-915-8.Hsu YL, Cho CY, Kuo PL, Huang YT, Lin CC (Aug 2006). Plumbagin (5-Hydroxy-2-methyl-1,4-naphthoquinone) Induces Apoptosis and Cell Cycle Arrest in A549 Cells through p53 Accumulation via c-Jun NH2-Terminal Kinase-Mediated Phosphorylation at Serine 15 in Vitro and in Vivo. J Pharmacol Exp Ther. 318 (2): 484–94. doi:10.1124/jpet.105.098863Jang, Gi-Won; Kim, Kwang-Soo; Park, Ro-Dong (2003). Micropropagation of Venus fly trap by shoot culture. Plant Cell, Tissue and Organ Culture. 72 (1): 95–98. doi:10.1023/A:1021203811457Leege, Lissa (2002) How Does the Venus Flytrap Digest Flies? Scientific American.Schnell, D.; Catling, P.; Folkerts, G.; Frost, C.; Gardner, R.; et al. (2000). Dionaea muscipula. The IUCN Red List of Threatened Species. 2000: e.T39636A10253384. doi:10.2305/IUCN.UK.2000.RLTS.T39636A10253384.en

Wednesday, November 6, 2019

CASE BAGBY COPY COMPANY essays

CASE BAGBY COPY COMPANY essays 1. Discuss the tradeoffs that Bagby faces in choosing between specialized and broad task assignment. Bagby Copy Company manufactures 10 different copiers. The main part of these copiers is a wiring bundle. This device is plugged into various components during the assembly process. They can assign each major task in this process to different employees using a broad task assignment or one individual can be assigned the task of producing the completed bundle using a specialized task assignment. Some of the advantages that Bagby's managers will obtain if they divide the total task of the manufacturing process into specific jobs or tasks are: Exploiting comparative advantage: Specialized task assignments will permit managers at Bagby to match people with jobs based on skills and training so this will permit employees concentrate on their particular specialties. For example, Bagby can hire engineers to design and develop a product and business people to do the marketing. The principle of comparative advantage suggest that this specialization will often produce higher output than using individuals to perform a broad tasks. Lower cost-training expenses: With specialized task assignment, each employee is trained to complete one basis function. With broad task assignment, employees are trained to complete more than one function, this can be very expensive. For instances, suppose at Bagby the designing function requires an engineer, while in the line of production function requires a person with a lower education. Specialized task assignment allows Bagby's managers to hire one engineer and one person without an advanced degree. With broad task assignment, the level of education required is usually the highest level, so it will cost more for Bagby to hire two persons with college degree than one. Broad task assignment is more expensive than specialized task assignment. Some of the costs of specialized task as...

Monday, November 4, 2019

Humans, Technology, Nature, and Spirituality Essay

Humans, Technology, Nature, and Spirituality - Essay Example The main element of religion was compounded in explaining phenomena that confound human to his or her spiritual ability. In most cases, science often attempts to explain natural processes in line with numerous universal laws while applying scientific methods. In most cases, these scientific explanations have since created dire conflicts with numerous religious beliefs. The film BARAKA has also compounded its theme in numerous cultures that are tied to one specific thing, nature. Nature provides humanity to the full understanding and acknowledgement of their mighty creator. Through spiritual belief of different religions, different people connect to their creator by appreciating nature. The same is not true with technology that often tends to provide explanation and pieces evidences on why certain things are usually the way they are. In other words, technology only provides a platform of explaining nature but do not brings connectivity or binding factors between humans and a given ele ment.The film also brings the element of interconnectivity without explanation, that is, human beings can connect to nature without anyone providing convincing explanations to the necessity to such connectivity. This kind of connectivity is only brought forth by spirituality. However, the recording that the same can be appreciated by human being is facilitated by technology. The emergence of technology only tried to expound on the existence of such relationship, but it has never replaced the binding factor, which is the spirituality.

Friday, November 1, 2019

Tau Beta Pi, Honor Society Membership Application Essay

Tau Beta Pi, Honor Society Membership Application - Essay Example This has also lead to growth in my career whereby I have been able to acquire a supervisory role within a period of six months an idea supported by Bureau of Labor Statistics (92). I am also a critical thinker and therefore I have managed to lead my team into making new discoveries and eventually patenting them. Besides many other qualities, I am competent and a hard worker who has never been caught in the web of incomplete projects or undone work. I have always wanted to be a great role model to many aspiring engineers. I believe TBP being the best engineering honor society (TBP p.1), is the best channel to achieve my dream. I would also wish to get a scholarship to further my studies and become a better professional. I will be of great benefit to TBP by contributing in innovation and hands-on activities of the society. I will also be actively involved in coming up with new ideas to keep the society’s quarterly magazine updated. Upon becoming a member of the institution, I will join the MindSET program and participate actively in its activities at various levels of study. I will encourage the students to follow the courageous spirit of the seven astronauts who died in their mission to discovery as documented by (Freiman & Neil,