School Vouchers essays

School Vouchers essays Education School Vouchers There has been a lot of debate recently over the use of school vouchers. Voucher programs offer students attending both public and private schools tuition vouchers. It gives taxpayers the freedom to pick where their tax dollars go. In theory, good schools will thrive with money and bad schools will lose students and close its doors. Most people feel that taking taxpayer money from public schools and using this money as vouchers for private schools is a violation of the constitution. Most private schools in America right now are run by religious organizations. There has been a lot of controversy over this issue mainly because of the importance of an education in a modern society. School choice initiatives are based on the premise that allowing parents to choose what schools their children attend is not only the right thing to do, but is also an important way for improving education. Instead of a one-size-fits-all model, School choice programs offer parents various options from which to pick the educational settings they believe will work best for their child. However, there is Supporters of school vouchers claim that it levels the educational playing field for lower income families who would have the option to send their kids away from an ineffective poorly funded public schools. Some lower class families feel that their kids would have a better chance with a tuition voucher to go to a private school where more money is spent on education. Many feel that vouchers would undermine public schools, by taking away public money for smaller class sizes, teacher training and innovative curriculum. Also, many feel that vouchers would erode the support for public education. In Milwaukee, voucher schools say they do not give special services to students with disabilities. Most of the voucher schools refused to sign a letter that they will ...

Lines of Longitude in Geography

Lines of Longitude in Geography Longitude is the angular distance of any point on Earth measured east or west of a point on Earths surface. Where Is Zero Degrees Longitude? Unlike latitude, there is no easy point of reference such as the equator to be designated as zero degrees in the longitude system. To avoid confusion, the worlds nations have agreed that the Prime Meridian, which passes through the Royal Observatory in Greenwich, England, will serve as that reference point and be designated as zero degrees. Because of this designation, longitude is measured in degrees west or east of the Prime Meridian. For example, 30Â °E, the line passing through eastern Africa, is an angular distance of 30Â ° east of the Prime Meridian. 30Â °W, which is in the middle of the Atlantic Ocean, is an angular distance of 30Â ° west of the Prime Meridian. There are 180 degrees east of the Prime Meridian and coordinates are sometimes given without the designation of E or east. When this is used, a positive value represents coordinates east of the Prime Meridian. There are also 180 degrees west of the Prime Meridian and when W or west is omitted in a coordinate a negative value such as -30Â ° represents coordinates west of the Prime Meridian. The 180Â ° line is neither east nor west and approximates the International Date Line. On a map (diagram), lines of longitude are the vertical lines running from the North Pole to the South Pole and are perpendicular to lines of latitude. Every line of longitude also crosses the equator. Because longitude lines are not parallel, they are known as meridians. Like parallels, meridians name the specific line and indicate the distance east or west of a 0Â ° line. Meridians converge at the poles and are farthest apart at the equator (about 69 miles (111 km) apart). Development and History of Longitude For centuries, mariners and explorers worked to determine their longitude in an effort to make navigation easier. Latitude was determined easily by observing the inclination of the sun or the position of known stars in the sky and calculating the angular distance from the horizon to them. Longitude could not be determined in this way because Earths rotation constantly changes the position of stars and the sun. The first person to offer a method for measuring longitude was the explorer Amerigo Vespucci. In the late 1400s, he began measuring and comparing the positions of the moon and Mars with their predicted positions over several nights at the same time (diagram). In his measurements, Vespucci calculated the angle between his location, the moon, and Mars. By doing this, Vespucci got a rough estimate of longitude. This method did not become widely used however because it relied on a specific astronomical event. Observers also needed to know the specific time and measure the moon and Mars positions on a stable viewing platform- both of which were difficult to do at sea. In the early 1600s, a new idea to measure longitude was developed when Galileo determined that it could be measured with two clocks. He said that any point on Earth took 24 hours to travel the full 360Â ° rotation of Earth. He found that if you divide 360Â ° by 24 hours, you find that a point on Earth travels 15Â ° of longitude every hour. Therefore, with an accurate clock at sea, a comparison of two clocks would determine longitude. One clock would be at the home port and the other on the ship. The clock on the ship would need to be reset to local noon each day. The time difference would then indicate the longitudinal difference traveled as one hour represented a 15Â ° change in longitude. Shortly thereafter, there were several attempts to make a clock that could accurately tell time on the unstable deck of a ship. In 1728, clockmaker John Harrison began working on the problem and in 1760, he produced the first marine chronometer called Number 4. In 1761, the chronometer was tested and determined to be accurate, officially making it possible to measure longitude on land and at sea. Measuring Longitude Today Today, longitude is more accurately measured with atomic clocks and satellites. The Earth is still divided equally into 360Â ° of longitude with 180Â ° being east of the Prime Meridian and 180Â ° west. Longitudinal coordinates are divided into degrees, minutes and seconds with 60 minutes making up a degree and 60 seconds comprising a minute. For example, Beijing, Chinas longitude is 116Â °2330E. The 116Â ° indicates that it lies near the 116th meridian while the minutes and seconds indicate just how close it is to that line. The E indicates that it is that distance east of the Prime Meridian. Although less common, longitude can also be written in decimal degrees. Beijings location in this format is 116.391Â °. In addition to the Prime Meridian, which is the 0Â ° mark in todays longitudinal system, the International Date Line is also an important marker. It is the 180Â ° meridian on the opposite side of the Earth and is where the eastern and western hemispheres meet. It also marks the place where each day officially begins. At the International Date Line, the west side of the line is always one day ahead of the east side, no matter what time of day it is when the line is crossed. This is because the Earth rotates east on its axis. Longitude and Latitude Lines of longitude or meridians are the vertical lines running from the South Pole to the North Pole. Lines of latitude or parallels are the horizontal lines running from the west to the east. The two cross each other at perpendicular angles and when combined as a set of coordinates they are extremely accurate in locating places on the globe. They are so accurate that they can locate cities and even buildings to within inches. For example, the Taj Mahal, located in Agra, India, has a coordinate set of 27Â °1029N, 78Â °232E. To view the longitude and latitude of other places, visit the collection of Locate Places Worldwide resources on this site.

Rolls royce anual report Coursework Example | Topics and Well Written Essays - 1000 words

Rolls royce anual report - Coursework Example The external auditor of the company has given an opinion according to which the group financial statements of the company present a true and fair view for the financial year ending December 31, 2012. The auditor reports states that the group financial statements are adequately prepared in accordance with the International Financial Reporting Standard (IFRS) and Companies Act 2006. It is also mentioned in the audit report that the financial statements of the company have been prepared in accordance with the Generally Accepted Accounting Principle (GAAP) as adopted by European Union. The audit report forms an integral part of the financial statements of a company, especially a public listed company. The primary purpose of the audit report is to bring into the knowledge of the users of the financial statements whether these financial statements are prepared in accordance with the applicable financial reporting framework. [Readyratios.com (2011)] Users of the financial statement usually do not have the time to thoroughly analyze the authenticity of the financial statements. They take reliance of from the audit report that whether the financial statements are giving a true and fair view or not. ... 69% 3% Current ratio 1.33 1.2 1.70 Inventory turnover period 109 days 112 days 50 days Payables’ turnover period 247 days 262 days 20 days Gearing ratio 1.967 2.634 4% P/E ratio 7.1 16.43 9.0 x Answer to Question No. 4 2012 2011 Variation % variation Change in million ? Sales 12,161 11,124 1,037 9.32% Rise Operating Profit 1,373 1,186 187 15.77% Rise Share Price (pence) 874 755 118 15.66% Rise Answer to Question No. 5 During the current financial year, the return on equity (ROE) has considerably increased during the current financial year. The return on equity is calculated by dividing the net profit attributable to shareholders by the shareholders equity. [Investopedia, 2012] The ratio is quite essential from the investor’s point of view as it represents how well a company is earning on its shareholder equity, which mainly comprises of issued capital and retained earnings. The ROE of the company has increased during the financial year 2011 which is due to the fact that the net profit of Rolls Royce has increase by a staggering 170% as it increased from 848 million to 2,295 million. The main reason behind this increase in the net profit is one unusual item of 669 million which is due to the disposal of a segment of the business during the current financial year. It would also be worth mentioning that the sales and operating profit of the company has increased by 9.32% and 15.77% during the current financial year. These escalations in figures have further accentuated the return on equity during the current year. The ratio is better than the industry average which is a sign for positive financial outlook. The gross profit has increased marginally during the current financial year. The reason behind the marginal increase in the gross profit margin is the fact that

Customer relationship Case Study Example | Topics and Well Written Essays - 2500 words

Customer relationship - Case Study Example Abrupt usage of such principles had ultimately made customer the king in the global economy of today. As almost all of the market follows the perfect competition, so the decision precisely depends upon the buyer to choose the services of which company they want to purchase. As the customer has been the most important aspect that has to be cared by the companies, the concept of customer relationship becomes immensely important. The present day companies try and develop stable and trust worthy relationship with the esteemed customers so that the relationship is maintained and retained. The modern theory suggests that it is very difficult to have a new customer in favour of a company due to increased competition. Therefore, the onus lies on the companies to develop a mutually benefitting relationship so that the customers as well as the organisations benefit and the existing customers are retained apart from attracting newer customers. Established in 2001, the Lebara Group had objective to offer exciting and innovating telecom solutions to the families and the friends of the customer. The group falls under the greater preview of Lebara Foundation, which is a charitable trust with the aim to assist the deprived children all across the globe. The vision and the mission statement of the organisation says that the charitable trust aims to provide good life to the deprived children in terms of basic necessities like that of food, shelter, treatment (for both mental as well as physical illness), education and various skill sets required to get settled in life. Lebara Mobile was first launched in 2004 in the land of Netherlands as the low cost international mobile service. The group offers prepaid mobile SIM cards in at least eight European nations which include the advanced nations like Netherlands, Australia, Denmark, Norway, Spain, Switzerland, United Kingdom and Sweden. The various unique selling propositions of the products of the company include: Very low rates Instant connections High quality networks Multilingual customer services Reliable service No access codes No hidden charges No lost minutes (Lebara, n.d.) The company claims to have 24 X 7 reliability monitoring. Also it has a dedicated routing and switching teams to retain the high quality. But the most important factor of the Lebara Telecom Group is that of its low cost. And the factor is guaranteed by its team of global carrier. The international calls of the mobile group are routed by the London Network Operations Centre. Customer Relationship of Lebara The customer relationship of Lebara Group is quite commendable. As the group has presence in more than 8 countries, also the policies adopted by the group are aimed to retain the existing customers along with attracting new clients. The group has declared objective of providing the customers with value for money along with ensuring all of its connections with the best possible quality. As the objective of the group is to connect the customers with their friends and relatives all across the globe, so it takes special measure to ensure that the customers get the convenience of direct-dial international mob

Mother Teresa Essay Example for Free

Mother Teresa Essay Mother Teresa was determined to help as many poor and homeless people as possible. She was always thinking of the many people in need. Everyone said about her â€Å"either move out of her way or help the poor.† They said that because she was always helping the poor and she wouldn’t let anyone or anything get in her way. She never cared about herself getting sick, all she cared about was the people in need. The reason why I want to be like Mother Teresa is because I want to make a difference in people’s lives just like she did. Also, I want to be like her because she was always thinking of others. Mother Teresa was very ambitious because she never rested until all of her patients were comforted because she saw the face of Jesus in all of them. God talked to Mother Teresa a lot. Once when she was on a train she heard God say to her, â€Å"get off the train and help the people in need.† So she did. She would always obey God’s orders. Whenever a person came up to her to ask for help she would help them with anything like if they needed food, care, or help with anything she was always there. That is another reason why I want to be like Mother Teresa. Mother Teresa is very worthy of imitation. Every day of Mother Teresa’s life since she became a nun, she was working with people. Mother Teresa served the poorest of the poor for over 45 years. Thanks to Mother Teresa, now there are many places for the homeless, poor, the unwanted, and lots more. Pope John Paul II said â€Å" Mother Teresa marked the history of our century with courage. She served all human beings by promoting their dignity and respect, and made those who had been defeated by life feel the tenderness of God.†

The Anchoress of England: Julian of Norwichs Portrait of Christ as Mot

The Anchoress of England: Julian of Norwich's Portrait of Christ as Mother      Ã‚  Ã‚  Ã‚  Ã‚   When speaking of medieval literature, Chaucer, Gower and Langland are quite often the most noted. However, recent studies have provided modern scholars with a wide variety of medieval women writers from all over Europe and a few in England. The most widely anthologized English female writer is Julian of Norwich. Julian was an anchoress, and as Marcelle Thiebaux notes, "The anchorite movement was widespread in England from the eleventh to the fifteenth centuries. Both men and women chose this extreme form of asceticism, which was favored and encouraged by the crown, the church, and the laity. Anchorholds were small, narrow cells attached to churches or friaries" (442). 1[1] The process of becoming an anchoress was difficult and complicated, but suffice it to say that after the process was completed "the anchoress was sealed up, never to re-emerge into the world. Penance, meditation, reading, and in some cases writing were the anchorite's sole activities" (Thiebaux 442).   This was the case for Julian of Norwich. She was "well read in Scripture, dwelling especially on the Psalms, the gospels, and the epistles of Paul and John, ...and was the first English woman to write a book" (Thiebaux 443-44). Her Book of Showings to the Anchoress Julian of Norwich 2[2]   possesses literary and religious value, and the work lends itself quite naturally to a feminist reading. In her clear, lucid, prose style, combined with the images of the medieval mystic, Julian establishes herself as an independent, female religious authority and she gives a staunch affirmation of the divinity of God with this unique view point: the motherhood of God.      Ã‚  Ã‚  Ã‚   In her fi... ...7. All biblical references come from the Geneva Bible (which is based on the Jerome Bible) but were checked and crossed referenced with the Jerome Bible with help of Professor Behunin as the Jerome Bible is in Latin. 7[7] It is interesting to note that there might be a biblical correlation to the hazelnut. The name Hazel appears in the Bible, and in Jewish the name Hazel is a feminine name and means "one who sees God." ( Harrison, R.K. Biblical Hebrew England: Hodder & Stoughton, 1986.) 8[8] Sir Gawain and the Green Knight. Trans. Marie Borroff. Norton Anthology of British Literature Vol. 1, New York: WW Norton, 1993. 9[9] The "five fives" as they are known in medieval literature and religion can be found in Sir Gawain and the Green Knight. Trans. Marie Borroff. Norton Anthology of British Literature Vol. 1, New York: WW Norton, 1993, lines 640-654.   

An Analysis of Robert Francis’ Poem The Hound Essay

In Robert Francis’ poem â€Å"The Hound,† the writer creates a sense of fear by comparing how mysterious and unpredictable dogs are to life using an extended metaphor to show life as uncertain and one has to wait to see what it brings at them. In lines 1-5, the writer states: â€Å"Life the hound/Equivocal/Comes at a bound/Either to rend me/Or to befriend me.† The speaker compares the habits of a hound to life to show even though they are not alike, they are similar in the way of not knowing the next thing that can happen. They are both questionable and suspicious because no one knows what will happen next and there is a possibility of it tearing one apart. Additionally, it can be a friend to one and everything will go well. This enforces how uncertain life is towards human beings. Meanwhile, the speaker continues in lines 6-11 by saying: â€Å"I cannot tell/ The hound’s intent/ Till he has sprung/ At my bare hand/ With teeth or tongue.† There is a sense of tension that is created in these lines by the use of imagery in order to show how no one is sure of what will follow. There is a possibility that life won’t be very friendly and will come at one â€Å"with teeth or tongue.† This can seem like a negative thing because it could come and bite you or that phrase could mean it is coming with happiness and joy thus promising good things to come. Finally, the speaker says in the last two lines â€Å"Meanwhile I stand/And wait the event.† This suggests the speaker is passive and waiting for something to happen. This is because one doesn’t know what will happen thus will not actively participate. In life, one might not always participate because life is uncertain and it can either be a friend or destroy the person.

College Physics 9e

1 Introduction ANSWERS TO MULTIPLE CHOICE QUESTIONS 1. Using a calculator to multiply the length by the width gives a raw answer of 6783 m 2 , but this answer must be rounded to contain the same number of signi? cant ? gures as the least accurate factor in the product. The least accurate factor is the length, which contains either 2 or 3 signi? cant ? gures, depending on whether the trailing zero is signi? cant or is being used only to locate the decimal point. Assuming the length contains 3 signi? cant ? gures, answer (c) correctly expresses the area as 6. 78 ? 10 3 m 2 .However, if the length contains only 2 signi? cant ? gures, answer (d) gives the correct result as 6. 8 ? 10 3 m 2 . Both answers (d) and (e) could be physically meaningful. Answers (a), (b), and (c) must be meaningless since quantities can be added or subtracted only if they have the same dimensions. According to Newton’s second law, Force = mass ? acceleration . Thus, the units of Force must be the product of the units of mass (kg) and the units of acceleration ( m s 2 ). This yields kg ? m s 2, which is answer (a). The calculator gives an answer of 57. 573 for the sum of the 4 given numbers.However, this sum must be rounded to 58 as given in answer (d) so the number of decimal places in the result is the same (zero) as the number of decimal places in the integer 15 (the term in the sum containing the smallest number of decimal places). The required conversion is given by: ? 1 000 mm ? ? 1. 00 cubitus ? h = ( 2. 00 m ) ? ? = 4. 49 cubiti ? 1. 00 m ? ? 445 mm ? This result corresponds to answer (c). 6. The given area (1 420 ft 2 ) contains 3 signi? cant ? gures, assuming that the trailing zero is used only to locate the decimal point. The conversion of this value to square meters is given by: 1. 00 m ? 2 2 2 A = (1. 2 ? 10 3 ft 2 ) ? ? ? = 1. 32 ? 10 m = 132 m ? 3. 281 ft ? Note that the result contains 3 signi? cant ? gures, the same as the number of signi? cant ? gures in the least accurate factor used in the calculation. This result matches answer (b). 7. You cannot add, subtract, or equate a number apples and a number of days. Thus, the answer is yes for (a), (c), and (e). However, you can multiply or divide a number of apples and a number of days. For example, you might divide the number of apples by a number of days to ? nd the number of apples you could eat per day. In summary, the answers are (a) yes, (b) no, (c) yes, (d) no, and (e) es. 2 2. 3. 4. 5. 1 http://helpyoustudy. info 2 Chapter 1 8. The given Cartesian coordinates are x = ? 5. 00, and y = 12. 00 , with the least accurate containing 3 signi? cant ? gures. Note that the speci? ed point (with x < 0 and y > 0 ) is in the second quadrant. The conversion to polar coordinates is then given by: r = x 2 + y 2 = ( ? 5. 00 ) + (12. 00 ) = 13. 0 2 2 tan ? = y 12. 00 = = ? 2. 40 x ? 5. 00 and ? = tan ? 1 ( ? 2. 40 ) = ? 67. 3 ° + 180 ° = 113 ° Note that 180 ° was added in the last step to yield a s econd quadrant angle. The correct answer is therefore (b) (13. 0, 113 °). 9. Doing dimensional analysis on the ? st 4 given choices yields: (a) [ v] ?t ? ? ? 2 = LT L = 3 T2 T (b) [ v] ?x2 ? ? ? = LT = L? 1T ? 1 L2 (c) ? v 2 ? ( L T )2 L2 T 2 L2 ? ?= = = 3 T T T [t ] (d) ? v 2 ? ( L T )2 L2 T 2 L ? ?= = = 2 L L T [ x] Since acceleration has units of length divided by time squared, it is seen that the relation given in answer (d) is consistent with an expression yielding a value for acceleration. 10. The number of gallons of gasoline she can purchase is # gallons = total expenditure 33 Euros ? cost per gallon ? Euros ? ? 1 L ? ? ? 1. 5 L ? ? 1 quart ? ? ? ? ? 5 gal ? 4 quarts ? ? 1 gal ? ? ? ? ? so the correct answer is (b). 1. The situation described is shown in the drawing at the right. h From this, observe that tan 26 ° = , or 45 m h = ( 45 m ) tan 26 ° = 22 m 26 h Thus, the correct answer is (a). 12. 45 m Note that we may write 1. 365 248 0 ? 10 7 as 136. 524 80 ? 10 5. Th us, the raw answer, including the uncertainty, is x = (136. 524 80  ± 2) ? 10 5. Since the ? nal answer should contain all the digits we are sure of and one estimated digit, this result should be rounded and displayed as 137 ? 10 5 = 1. 37 ? 10 7 (we are sure of the 1 and the 3, but have uncertainty about the 7). We see that this answer has three signi? cant ? ures and choice (d) is correct. ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 2. Atomic clocks are based on the electromagnetic waves that atoms emit. Also, pulsars are highly regular astronomical clocks. http://helpyoustudy. info Introduction 3 4. (a) (b) (c) ~ 0. 5 lb ? 0. 25 kg or ~10 ? 1 kg ~ 4 lb ? 2 kg or ~10 0 kg ~ 4000 lb ? 2000 kg or ~10 3 kg 6. Let us assume the atoms are solid spheres of diameter 10? 10 m. Then, the volume of each atom is of the order of 10? 30 m3. (More precisely, volume = 4? r 3 3 = ? d 3 6 . ) Therefore, since 1 cm 3 = 10 ? 6 m 3, the number of atoms in the 1 cm3 solid is on the order of 10 ? 1 0 ? 30 = 10 24 atoms. A more precise calculation would require knowledge of the density of the solid and the mass of each atom. However, our estimate agrees with the more precise calculation to within a factor of 10. Realistically, the only lengths you might be able to verify are the length of a football ? eld and the length of a house? y. The only time intervals subject to veri? cation would be the length of a day and the time between normal heartbeats. In the metric system, units differ by powers of ten, so it’s very easy and accurate to convert from one unit to another. 8. 10. ANSWERS TO EVEN NUMBERED PROBLEMS . 4. 6. 8. 10. 12. 14. 16. 18. (a) L T2 (b) L All three equations are dimensionally incorrect. (a) (a) (a) (a) (a) kg ? m s 22. 6 3. 00 ? 108 m s 346 m 2  ± 13 m 2 797 (b) (b) (b) (b) (b) Ft = p 22. 7 2 . 997 9 ? 108 m s 66. 0 m  ± 1. 3 m 1. 1 (c) 17. 66 (c) (c) 22. 6 is more reliable 2. 997 925 ? 108 m s 3. 09 cm s (a) (b) (c) (d) 5. 60 ? 10 2 km = 5. 60 ? 10 5 m = 5. 60 ? 10 7 cm 0. 491 2 km = 491. 2 m = 4. 912 ? 10 4 cm 6. 192 km = 6. 192 ? 10 3 m = 6. 192 ? 10 5 cm 2. 499 km = 2. 499 ? 10 3 m = 2. 499 ? 10 5 cm 20. 22. 24. 26. 10. 6 km L 9. 2 nm s 2 . 9 ? 10 2 m 3 = 2 . 9 ? 108 cm 3 2 . 57 ? 10 6 m 3 ttp://helpyoustudy. info 4 Chapter 1 28. 30. 32. 34. ? 108 steps ~108 people with colds on any given day (a) (a) 4. 2 ? 10 ? 18 m 3 ? 10 29 prokaryotes (b) (b) ~10 ? 1 m 3 ~1014 kg (c) ~1016 cells (c) The very large mass of prokaryotes implies they are important to the biosphere. They are responsible for ? xing carbon, producing oxygen, and breaking up pollutants, among many other biological roles. Humans depend on them! 36. 38. 40. 42. 44. 46. 48. 2. 2 m 8. 1 cm ? s = r12 + r22 ? 2r1r2 cos (? 1 ? ?2 ) 2. 33 m (a) 1. 50 m (b) 2. 60 m 8. 60 m (a) and (b) (c) 50. 52. 54. y= (a) y x = tan 12. 0 °, y ( x ? . 00 km ) = tan 14. 0 ° d ? tan ? ? tan ? tan ? ? tan ? 1. 609 km h (b) 88 km h (d) 1. 44 ? 10 3 m (c) 16 km h Assumes population of 300 million, average of 1 can week per person, and 0. 5 oz per can. (a) ? 1010 cans yr 7. 14 ? 10 ? 2 gal s A2 A1 = 4 ? 10 2 yr (b) (b) (b) (b) ? 10 5 tons yr 2. 70 ? 10 ? 4 m 3 s V2 V1 = 8 ? 10 4 times (c) 1. 03 h 56. 58. 60. 62. (a) (a) (a) ? 10 4 balls yr. Assumes 1 lost ball per hitter, 10 hitters per inning, 9 innings per game, and 81 games per year. http://helpyoustudy. info Introduction 5 PROBLEM SOLUTIONS 1. 1 Substituting dimensions into the given equation T = 2? ionless constant, we have g , and recognizing that 2? is a dimen- [T ] = [ ] [ g] or T= L = L T2 T2 = T Thus, the dimensions are consistent . 1. 2 (a) From x = Bt2, we ? nd that B = [ B] = [ x] L = 2 [t 2 ] T x . Thus, B has units of t2 (b) If x = A sin ( 2? ft ), then [ A] = [ x ] [sin ( 2? ft )] But the sine of an angle is a dimensionless ratio. Therefore, [ A] = [ x ] = L 1. 3 (a) The units of volume, area, and height are: [V ] = L3, [ A] = L2 , and [h] = L We then observe that L3 = L2 L or [V ] = [ A][h] Thus, the equation V = Ah is dimensionally correct . (b) Vcylinder = ? R 2 h = (? R 2 ) h = Ah , where A = ?R 2 Vrectangular box = wh = ( w ) h = Ah, where A = w = length ? width 1. 4 (a) L ML2 2 2 m v 2 = 1 m v0 + mgh, [ m v 2 ] = [ m v0 ] = M ? ? = 2 ? ? 2 T ? T? 1 2 L ? M L while ? mgh ? = M ? 2 ? L = . Thus, the equation is dimensionally incorrect . ? ? ? T ? T ? In the equation 1 2 2 (b) L L but [at 2 ] = [a][t 2 ] = ? 2 ? ( T 2 ) = L. Hence, this equation ? ? T ? T ? is dimensionally incorrect . In v = v0 + at 2, [ v] = [ v0 ] = L In the equation ma = v 2, we see that [ ma] = [ m][a] = M ? 2 ? ?T Therefore, this equation is also dimensionally incorrect . 2 ? = ML , while [ v 2 ] = ? L ? = L . ? ? ? 2 T2 ? T ? T? 2 (c) . 5 From the universal gravitation law, the constant G is G = Fr 2 Mm. Its units are then [G ] = [ F ] ? r 2 ? ( kg ? m s2 ) ( m 2 ) m3 ? ?= = kg ? kg kg ? s 2 [ M ][ m ] http://helpyoustudy. info 6 Chapter 1 1. 6 (a) Solving KE = p2 for the momentum, p, gives p = 2 m ( K E ) where the numeral 2 is a 2m dimensionless constant. Dimensional analysis gives the units of momentum as: [ p] = [ m ][ KE ] = M ( M ? L2 T 2 ) = M 2 ? L2 T 2 = M ( L T ) Therefore, in the SI system, the units of momentum are kg ? ( m s ) . (b) Note that the units of force are kg ? m s 2 or [ F ] = M ? L T 2 . Then, observe that [ F ][ t ] = ( M ?L T 2 ) ? T = M ( L T ) = [ p ] From this, it follows that force multiplied by time is proportional to momentum: Ft = p . (See the impulse–momentum theorem in Chapter 6, F ? ?t = ? p , which says that a constant force F multiplied by a duration of time ? t equals the change in momentum, ? p. ) 1. 7 1. 8 Area = ( length ) ? ( width ) = ( 9. 72 m )( 5. 3 m ) = 52 m 2 (a) Computing ( 8) 3 without rounding the intermediate result yields ( 8) (b) 3 = 22. 6 to three signi? cant ? gures. Rounding the intermediate result to three signi? cant ? gures yields 8 = 2. 8284 > 2. 83 Then, we obtain ( 8) 3 = ( 2. 83) = 22. 7 to three signi? ant ? gures. 3 (c) 1. 9 (a) (b) (c) (d) The answer 22. 6 is more reliable because rounding in part (b) was carried out too soon. 78. 9  ± 0. 2 has 3 significant figures with the uncertainty in the tenths position. 3. 788 ? 10 9 has 4 significant figures 2. 46 ? 10 ? 6 has 3 significant figures 0. 003 2 = 3. 2 ? 10 ? 3 has 2 significant figures . The two zeros were originally included only to position the decimal. 1. 10 c = 2 . 997 924 58 ? 108 m s (a) (b) (c) Rounded to 3 signi? cant ? gures: c = 3. 00 ? 108 m s Rounded to 5 signi? cant ? gures: c = 2 . 997 9 ? 108 m s Rounded to 7 signi? cant ? gures: c = 2 . 997 925 ? 08 m s 1. 11 Observe that the length = 5. 62 cm, the width w = 6. 35 cm, and the height h = 2. 78 cm all contain 3 signi? cant ? gures. Thus, any product of these quantities should contain 3 signi? cant ? gures. (a) (b) w = ( 5. 62 cm )( 6. 35 cm ) = 35. 7 cm 2 V = ( w ) h = ( 35. 7 cm 2 ) ( 2. 78 cm ) = 99. 2 cm 3 continued on next page http://helpyoustudy. info Introd uction 7 (c) wh = ( 6. 35 cm )( 2. 78 cm ) = 17. 7 cm 2 V = ( wh ) = (17. 7 cm 2 ) ( 5. 62 cm ) = 99. 5 cm 3 (d) In the rounding process, small amounts are either added to or subtracted from an answer to satisfy the rules of signi? cant ? gures.For a given rounding, different small adjustments are made, introducing a certain amount of randomness in the last signi? cant digit of the ? nal answer. 2 2 2 A = ? r 2 = ? (10. 5 m  ± 0. 2 m ) = ? ?(10. 5 m )  ± 2 (10. 5 m )( 0. 2 m ) + ( 0. 2 m ) ? ? ? 1. 12 (a) Recognize that the last term in the brackets is insigni? cant in comparison to the other two. Thus, we have A = ? ?110 m 2  ± 4. 2 m 2 ? = 346 m 2  ± 13 m 2 ? ? (b) 1. 13 C = 2? r = 2? (10. 5 m  ± 0. 2 m ) = 66. 0 m  ± 1. 3 m The least accurate dimension of the box has two signi? cant ? gures. Thus, the volume (product of the three dimensions) will contain only two signi? cant ? ures. V = ? w ? h = ( 29 cm )(17. 8 cm )(11. 4 cm ) = 5. 9 ? 10 3 cm 3 1. 14 (a) The sum is rounded to 797 because 756 in the terms to be added has no positions beyond the decimal. 0. 003 2 ? 356. 3 = ( 3. 2 ? 10 ? 3 ) ? 356. 3 = 1. 14016 must be rounded to 1. 1 because 3. 2 ? 10 ? 3 has only two signi? cant ? gures. 5. 620 ? ? must be rounded to 17. 66 because 5. 620 has only four signi? cant ? gures. (b) (c) 1. 15 5 280 ft ? ? 1 fathom ? 8 d = ( 250 000 mi ) ? ? ? = 2 ? 10 fathoms ? 1. 000 mi ? ? 6 ft ? The answer is limited to one signi? cant ? gure because of the accuracy to which the conversion from fathoms to feet is given. . 16 v= t = 186 furlongs 1 fortnight ? 1 fortnight ? ? 14 days ? ? ? 1 day ? ? 220 yds ? ? 8. 64 ? 10 4 s ? ? 1 furlong ? ? 3 ft ? ? 1 yd ? ? 100 cm ? ? ? 3. 281 ft ? ? ? giving v = 3. 09 cm s ? ? 3. 786 L ? ? 1 gal ? ? 10 3 cm 3 ? ? 1 m 3 ? = 0. 204 m 3 ? ? 1 L ? ? 10 6 cm 3 ? ? ? 1. 17 ? 9 gal 6. 00 firkins = 6. 00 firkins ? ? 1 firkin ? (a) 1. 18 1. 609 km ? 2 5 7 = ( 348 mi ) ? 6 ? ? = 5. 60 ? 10 km = 5. 60 ? 10 m = 5. 60 ? 10 cm ? 1. 000 mi ? ? 1. 609 km ? 4 h = (1 612 ft ) ? 2 ? = 0. 491 2 km = 491. 2 m = 4. 912 ? 10 cm 5 280 ft ? ? ? 1. 609 km ? 3 5 h = ( 20 320 ft ) ? = 6. 192 km = 6. 192 ? 10 m = 6. 192 ? 10 cm 5 280 ft ? ? (b) (c) continued on next page http://helpyoustudy. info 8 Chapter 1 (d) ? 1. 609 km ? 3 5 d = (8 200 ft ) ? ? = 2 . 499 km = 2 . 499 ? 10 m = 2 . 499 ? 10 cm ? 5 280 ft ? In (a), the answer is limited to three signi? cant ? gures because of the accuracy of the original data value, 348 miles. In (b), (c), and (d), the answers are limited to four signi? cant ? gures because of the accuracy to which the kilometers-to-feet conversion factor is given. 1. 19 v = 38. 0 m ? 1 km ? ? 1 mi ? ? 3 600 s ? ? = 85. 0 mi h ? s ? 10 3 m ? ? 1. 609 km ? 1 h ? Yes, the driver is exceeding the speed limit by 10. 0 mi h . mi ? 1 km ? ? 1 gal ? ? = 10. 6 km L ? gal ? 0. 621 mi ? ? 3. 786 L ? ? ? 1. 20 efficiency = 25. 0 r= 1. 21 (a) (b) (c) diameter 5. 36 in ? 2. 54 cm ? = ? ? = 6. 81 cm 2 2 ? 1 in ? 2 A = 4? r 2 = 4? ( 6. 81 cm ) = 5. 83 ? 10 2 cm 2 V= 4 3 4 3 ? r = ? ( 6. 81 cm ) = 1. 32 ? 10 3 cm 3 3 3 ? ? 1 h ? ? 2. 54 cm ? ? 10 9 nm ? ? 3 600 s ? ? 1. 00 in ? ? 10 2 cm ? = 9. 2 nm s ? 1. 22 ? 1 in ? ? 1 day rate = ? ? 32 day ? ? 24 h ? This means that the proteins are assembled at a rate of many layers of atoms each second! 1. 3 ? m ? ? 3 600 s ? ? 1 km ? ? 1 mi ? 8 c = ? 3. 00 ? 10 8 ? = 6. 71 ? 10 mi h s ? ? 1 h ? ? 10 3 m ? ? 1. 609 km ? ? ? 2 . 832 ? 10 ? 2 m3 ? Volume of house = ( 50. 0 ft )( 26 ft )(8. 0 ft ) ? ? 1 ft 3 ? ? ? 100 cm ? = 2 . 9 ? 10 2 m3 = ( 2 . 9 ? 10 2 m3 ) ? = 2 . 9 ? 10 8 cm3 ? 1m ? ? 1. 25 1. 26 2 2 ? 1 m ? 43 560 ft ? ? 1 m ? ? ? = 3. 08 ? 10 4 m3 Volume = 25. 0 acre ft ? ? ? ? ? 3. 281 ft ? 1 acre ? ? 3. 281 ft ? ? ? ? ? 1 Volume of pyramid = ( area of base )( height ) 3 3 1. 24 ( ) = 1 ? (13. 0 acres )( 43 560 ft 2 acre ) ? ( 481 ft ) = 9. 08 ? 10 7 f ? 3? ? 2 . 832 ? 10 ? 2 m3 ? 3 = ( 9. 08 ? 10 7 ft 3 ) ? 5 ? = 2 . 57 ? 10 m 1 ft3 ? ? 1. 27 Volume of cube = L 3 = 1 quart (Where L = length of one side of the cube. ) ? 1 gallon ? ? 3. 786 liter ? ? 1000 cm3 ? i = 947 cm3 Thus, L 3 = 1 quart ? ? 4 quarts ? ? 1 gallon ? ? 1 liter ? ? ? ? ( ) and L = 3 947 cm3 = 9. 82 cm http://helpyoustudy. info Introduction 9 1. 28 We estimate that the length of a step for an average person is about 18 inches, or roughly 0. 5 m. Then, an estimate for the number of steps required to travel a distance equal to the circumference of the Earth would be N= or 3 2? ( 6. 38 ? 10 6 m ) Circumference 2?RE = ? ? 8 ? 10 7 steps 0. 5 m step Step Length Step Length N ? 108 steps 1. 29. We assume an average respiration rate of about 10 breaths/minute and a typical life span of 70 years. Then, an estimate of the number of breaths an average person would take in a lifetime is ? ? breaths ? 10 7 ? min n = ? 10 ( 70 yr ) ? 3. 156 ? yr s ? ? 160 s ? = 4 ? 108 breaths ? ? ? min ? 1 ? ? ? or n ? 108 breaths 1. 30 We assume that the average person catches a cold twice a year and is sick an average of 7 days (or 1 week) each time. Thus, on average, each person is sick for 2 weeks out of each year (52 weeks).The probability that a particular person will be sick at any given time equals the percentage of time that person is sick, or probability of sickness = 2 weeks 1 = 52 weeks 26 The population of the Earth is approximately 7 billion. The number of people expected to have a cold on any given day is then 1 Number sick = ( population )( probability of sickness ) = ( 7 ? 10 9 ) ? ? = 3 ? 108 or ? 108 ( ? ? ? 26 ? 1. 31 (a) Assume that a typical intestinal tract has a length of about 7 m and average diameter of 4 cm. The estimated total intestinal volume is then ? ?d 2 ? ? ( 0. 04 m ) Vtotal = A = ? ( 7 m ) = 0. 009 m 3 ? 4 ? 4 ? 2 The approximate volume occupied by a single bacterium is Vbacteria ? ( typical length scale ) = (10 ? 6 m ) = 10 ? 18 m 3 3 3 If it is assumed that bacteria occupy one hundredth of the total intestinal volume, the estimate of the number of microorganisms in the human intestinal tract is n= (b) 3 Vtotal 100 ( 0. 009 m ) 100 = = 9 ? 1013 or n ? 1014 10 ? 18 m 3 Vbacteria The large value of the number of bacteria estimated to exist in the intestinal tract means that they are probably not dangerous. Intestinal bacteria help digest food and provide important nutrients. Humans and bacteria enjoy a mutually bene? ial symbiotic relationship. Vcell = 3 4 3 4 ? r = ? (1. 0 ? 10 ? 6 m ) = 4. 2 ? 10 ? 18 m 3 3 3 1. 32 (a) (b) Consider your body to be a cylinder having a radius of about 6 inches (or 0. 15 m) and a height of about 1. 5 meters. Then, its volume is Vbody = Ah = (? r 2 ) h = ? ( 0. 15 m ) (1. 5 m ) = 0. 11 m 3 or ? 10 ? 1 m 3 2 continued on next page http://helpyoustudy. info 10 Chapter 1 (c) The estimate of the number of cells in the body is then n= Vbody Vcell = 0. 11 m 3 = 2. 6 ? 1016 or ? 1016 ? 18 3 4. 2 ? 10 m 1. 33 A reasonable guess for the diameter of a tire might be 3 f t, with a circumference (C = 2? r = ?D = distance travels per revolution) of about 9 ft. Thus, the total number of revolutions the tire might make is n= total distance traveled ( 50 000 mi )( 5 280 ft mi ) = 3 ? 10 7 rev, or ~ 10 7 rev = distance per revolution 9 ft rev 1. 34 Answers to this problem will vary, dependent on the assumptions one makes. This solution assumes that bacteria and other prokaryotes occupy approximately one ten-millionth (10? 7) of the Earth’s volume, and that the density of a prokaryote, like the density of the human body, is approximately equal to that of water (103 kg/m3). (a) estimated number = n = Vtotal Vsingle prokaryote 10 )V ? ?7 Earth Vsingle prokaryote (10 )(10 m ) ? ? (length scale) (10 m ) ?7 3 Earth ? 7 6 3 ? 6 3 (10 ) R 3 ? 10 29 (b) (c) 3 kg ? ? ? ? mtotal = ( density )( total volume) ? ?water ? nVsingle ? = ? 10 3 3 ? (10 29 )(10 ? 6 m ) ? 1014 kg ? ? prokaryote ? ? m The very large mass of prokaryotes implies they are important to the biosphere. They are responsible for ? xing carbon, producing oxygen, and breaking up pollutants, among many other biological roles. Humans depend on them! x = r cos? = 2 . 5 m cos 35 ° = 2. 0 m 1. 35 The x coordinate is found as and the y coordinate ) y = r sin? = ( 2 . 5 m ) sin 35 ° = 1. m ( 2 1. 36 The x distance out to the ? y is 2. 0 m and the y distance up to the ? y is 1. 0 m. Thus, we can use the Pythagorean theorem to ? nd the distance from the origin to the ? y as d = x 2 + y2 = ( 2. 0 m ) + (1. 0 m ) 2 = 2. 2 m 1. 37 The distance from the origin to the ? y is r in polar coordinates, and this was found to be 2. 2 m in Problem 36. The angle ? is the angle between r and the horizontal reference line (the x axis in this case). Thus, the angle can be found as tan ? = y 1. 0 m = = 0. 50 x 2. 0 m and ? = tan ? 1 ( 0. 50 ) = 27 ° The polar coordinates are r = 2. 2 m and ? = 27  ° 1. 8 The x distance between the two points is ? x = x2 ? x1 = ? 3. 0 cm ? 5. 0 cm = 8. 0 cm and the y distance between them is ? y = y2 ? y1 = 3. 0 cm ? 4. 0 cm = 1. 0 cm. The distance between them is found from the Pythagorean theorem: d= 1. 39 ? x + ? y = (8. 0 cm ) + (1. 0 cm ) = 2 2 2 2 65 cm 2 = 8. 1 cm Refer to the Figure given in Problem 1. 40 below. The Cartesian coordinates for the two given points are: x1 = r1 cos ? 1 = ( 2. 00 m ) cos 50. 0 ° = 1. 29 m y1 = r1 sin ? 1 = ( 2. 00 m ) sin 50. 0 ° = 1. 53 m x2 = r2 cos ? 2 = ( 5. 00 m ) cos ( ? 50. 0 °) = 3. 21 m y2 = r2 sin ? 2 = ( 5. 00 m ) sin ( ? 50. 0 °) = ? 3. 3 m continued on next page http://helpyoustudy. info Introduction 11 The distance between the two points is then: ? s = ( ? x ) + ( ? y ) = (1. 29 m ? 3. 21 m ) + (1. 53 m + 3. 83 m ) = 5. 69 m 2 2 2 2 1. 40 Consider the Figure shown at the right. The Cartesian coordinates for the two points are: x1 = r1 cos ? 1 y1 = r1 sin ? 1 x2 = r2 cos ? 2 y2 = r2 sin ? 2 y (x1, y1) r1 ?s ?y y1 y2 The distance between the two points is the length of the hypot enuse of the shaded triangle and is given by ? s = ( ? x ) + ( ? y ) = 2 2 q1 ( x1 ? x2 ) + ( y1 ? y2 ) 2 2 (x2, y2) r2 ? x q2 x1 x2 x or ? s = (r 2 1 cos 2 ? 1 + r22 cos 2 ? ? 2r1r2 cos ? 1 cos ? 2 ) + ( r12 sin 2 ? 1 + r22 sin 2 ? 2 ? 2r1r2 sin ? 1 sin ? 2 ) = r12 ( cos 2 ? 1 + sin 2 ? 1 ) + r22 ( cos 2 ? 2 + sin 2 ? 2 ) ? 2r1r2 ( cos ? 1 cos ? 2 + sin ? 1 sin ? 2 ) i Applying the identities cos 2 ? + sin 2 ? = 1 and cos ? 1 cos ? 2 + sin ? 1 sin ? 2 = cos (? 1 ? ?2 ) , this reduces to ? s = r12 + r22 ? 2r1r2 ( cos ? 1 cos ? 2 + sin ? 1 sin ? 2 ) = 1. 41 (a) r12 + r22 ? 2r1r2 cos (? 1 ? ?2 ) With a = 6. 00 m and b being two sides of this right triangle having hypotenuse c = 9. 00 m, the Pythagorean theorem gives the unknown side as b = c2 ? a2 = ( 9. 00 m )2 ? ( 6. 00 m )2 = 6. 1 m (c) sin ? = b 6. 71 m = = 0. 746 c 9. 00 m (b) tan ? = a 6. 00 m = = 0. 894 b 6. 71 m 1. 42 From the diagram, cos ( 75. 0 °) = d L Thus, d = L cos ( 75. 0 °) = ( 9. 00 m ) cos ( 75. 0 °) = 2. 33 m L 9 . 00 m 75. 0 d http://helpyoustudy. info 12 Chapter 1 1. 43 The circumference of the fountain is C = 2? r , so the radius is C 15. 0 m = = 2. 39 m 2? 2? h h Thus, tan ( 55. 0 °) = = which gives r 2. 39 m r= h = ( 2. 39 m ) tan ( 55. 0 °) = 3. 41 m 1. 44 (a) (b) sin ? = cos ? = opposite side so, opposite side = ( 3. 00 m ) sin ( 30. 0 ° ) = 1. 50 m hypotenuse adjacent side so, adjacent side = ( 3. 00 m ) cos ( 30.  ° ) = 2 . 60 m hypotenuse (b) (d) The side adjacent to ? = 3. 00 sin ? = 4. 00 = 0. 800 5. 00 1. 45 (a) (c) (e) The side opposite ? = 3. 00 cos ? = tan ? = 4. 00 = 0. 800 5. 00 4. 00 = 1. 33 3. 00 1. 46 Using the diagram at the right, the Pythagorean theorem yields c = ( 5. 00 m ) + ( 7. 00 m ) = 8. 60 m 2 2 5. 00 m c q 7. 00 m 1. 47 From the diagram given in Problem 1. 46 above, it is seen that tan ? = 5. 00 = 0. 714 7. 00 and ? = tan ? 1 ( 0. 714 ) = 35. 5 ° 1. 48 (a) and (b) (c) See the Figure given at the right. Applying the de? nition of the tangent fun ction to the large right triangle containing the 12.  ° angle gives: y x = tan 12. 0 ° [1] Also, applying the de? nition of the tangent function to the smaller right triangle containing the 14. 0 ° angle gives: y = tan 14. 0 ° x ? 1. 00 km (d) From Equation [1] above, observe that x = y tan 12. 0 ° [2] Substituting this result into Equation [2] gives y ? tan 12. 0 ° = tan 14. 0 ° y ? (1. 00 km ) tan 12. 0 ° continued on next page http://helpyoustudy. info Introduction 13 Then, solving for the height of the mountain, y, yields y= 1. 49 (1. 00 km ) tan 12. 0 ° tan 14. 0 ° tan 14. 0 ° ? tan 12. 0 ° = 1. 44 km = 1. 44 ? 10 3 m Using the sketch at the right: w = tan 35.  ° , or 100 m w = (100 m ) tan 35. 0 ° = 70. 0 m w 1. 50 The ? gure at the right shows the situation described in the problem statement. Applying the de? nition of the tangent function to the large right triangle containing the angle ? in the Figure, one obtains y x = tan ? Also, applying the d e? nition of the tangent function to the small right triangle containing the angle ? gives y = tan ? x? d Solving Equation [1] for x and substituting the result into Equation [2] yields y = tan ? y tan ? ? d The last result simpli? es to or y ? tan ? = tan ? y ? d ? tan ? y ? tan ? = y ? tan ? ? d ? tan ? ? tan ? or [2] [1]Solving for y: y ( tan ? ? tan ? ) = ? d ? tan ? ? tan ? y=? 1. 51 (a) d ? tan ? ? tan ? d ? tan ? ? tan ? = tan ? ? tan ? tan ? ? tan ? Given that a ? F m , we have F ? ma . Therefore, the units of force are those of ma, [ F ] = [ ma] = [ m][a] = M ( L T 2 ) = M L T-2 (b) L M? L [F ] = M ? 2 ? = 2 ? ? T ? T ? 1 so newton = kg ? m s2 1. 52 (a) mi ? mi ? ? 1. 609 km ? km = ? 1 ? = 1. 609 h ? h ? ? 1 mi ? h mi ? mi ? ? 1. 609 km h ? km = ? 55 ? = 88 h ? h ? ? 1 mi h ? h mi mi ? mi ? ? 1. 609 km h ? km ? 55 = ? 10 ? = 16 h h ? h ? ? 1 mi h ? h (b) vmax = 55 (c) ?vmax = 65 http://helpyoustudy. info 14 Chapter 1 1. 3 (a) Since 1 m = 10 2 cm , then 1 m 3 = (1 m ) = ( 10 2 cm ) = (10 2 ) cm 3 = 10 6 cm 3, giving 3 3 3 ? 1. 0 ? 10 ? 3 kg ? 3 mass = density volume = ? ? 1. 0 m 3 ? 1. 0 cm ? ( )( ) ( ) ? 10 6 cm3 ? ? kg ? 3 = ? 1. 0 ? 10 ? 3 3 ? 1. 0 m 3 ? ? = 1. 0 ? 10 kg 3 ? cm ? ? 1m ? ( ) As a rough calculation, treat each of the following objects as if they were 100% water. (b) (c) (d) 3 kg 4 cell: mass = density ? volume = ? 10 3 3 ? ? ( 0. 50 ? 10 ? 6 m ) = 5. 2 ? 10 ? 16 kg ? ? m ? 3 ? 3 4 kg 4 kidney: mass = density ? volume = ? ? ? r 3 ? = ? 10 3 3 ? ? ( 4. 0 ? 10 ? 2 m ) = 0. 27 kg ? ? ? ? m ? 3 ? 3 ? ? ?y: mass = density ? olume = ( density ) (? r 2 h ) 2 kg = ? 10 3 3 ? ? (1. 0 ? 10 ? 3 m ) ( 4. 0 ? 10 ? 3 m ) = 1. 3 ? 10 ? 5 kg ? ? m ? ? 1. 54 Assume an average of 1 can per person each week and a population of 300 million. (a) number cans person ? number cans year = ? ? ? ( population )( weeks year ) week ? ? ? ?1 ? ? can person ? 8 ? ( 3 ? 10 people ) ( 52 weeks yr ) week ? ? 2 ? 1010 cans yr , or ~10 10 cans yr (b) number of tons = ( weight can )( number cans year ) ? oz ? ? 1 lb ? ? 1 ton ? 10 can ? ? 0. 5 ? ? 2 ? 10 ? can ? ? 16 oz ? ? 2 000 lb ? yr ? ? 3 ? 10 5 ton yr , or ~10 5 ton yr Assumes an average weight of 0. oz of aluminum per can. 1. 55 The term s has dimensions of L, a has dimensions of LT? 2, and t has dimensions of T. Therefore, the equation, s = k a m t n with k being dimensionless, has dimensions of L = ( LT ? 2 ) ( T ) m n or L1T 0 = L m T n? 2 m The powers of L and T must be the same on each side of the equation. Therefore, L1 = Lm and m =1 Likewise, equating powers of T, we see that n ? 2 m = 0, or n = 2 m = 2 Dimensional analysis cannot determine the value of k , a dimensionless constant. 1. 56 (a) The rate of ? lling in gallons per second is rate = 30. 0 gal ? 1 min ? ?2 ? ? = 7. 14 ? 10 gal s 7. 0 min ? 60 s ? continued on next page http://helpyoustudy. info Introduction 15 (b) 3 1L Note that 1 m 3 = (10 2 cm ) = (10 6 cm 3 ) ? 3 ? 3 ? 10 cm ? = 10 3 L. Thus, ? ? rate = 7. 14 ? 10 ? 2 (c) t= gal ? 3. 786 L ? ? 1 m 3 ? ?4 3 ? ? = 2. 70 ? 10 m s s ? 1 gal ? ? 10 3 L ? ? 1h ? Vfilled 1. 00 m 3 = = 3. 70 ? 10 3 s ? ? = 1. 03 h ? 4 3 rate 2. 70 ? 10 m s ? 3 600 s ? 1. 57 The volume of paint used is given by V = Ah, where A is the area covered and h is the thickness of the layer. Thus, h= V 3. 79 ? 10 ? 3 m 3 = = 1. 52 ? 10 ? 4 m = 152 ? 10 ? 6 m = 152 ? m 25. 0 m 2 A 1. 58 (a) For a sphere, A = 4? R 2 .In this case, the radius of the second sphere is twice that of the ? rst, or R2 = 2 R1. Hence, A2 4? R 2 R 2 ( 2 R1 ) 2 = = 2 = = 4 2 2 A1 4? R 1 R 1 R12 2 (b) For a sphere, the volume is Thus, V= 4 3 ? R 3 3 V2 ( 4 3) ? R 3 R 3 ( 2 R1 ) 2 = = 2 = = 8 3 3 3 V1 ( 4 3) ? R 1 R 1 R1 1. 59 The estimate of the total distance cars are driven each year is d = ( cars in use ) ( distance traveled per car ) = (100 ? 10 6 cars )(10 4 mi car ) = 1 ? 1012 mi At a rate of 20 mi/gal, the fuel used per year would be V1 = d 1 ? 1012 mi = = 5 ? 1010 gal rate1 20 mi gal If the rate increased to 25 mi gal, the annual fuel consumption would be V2 = d 1 ? 012 mi = = 4 ? 1010 gal rate2 25 mi gal and the fuel savings each year would be savings = V1 ? V2 = 5 ? 1010 gal ? 4 ? 1010 gal = 1 ? 1010 gal 1. 60 (a) The amount paid per year would be dollars ? ? 8. 64 ? 10 4 s ? ? 365. 25 days ? 10 dollars annual amount = ? 1 000 ? ? = 3. 16 ? 10 s ? ? 1. 00 day ? ? yr yr ? ? Therefore, it would take (b) 10 ? 10 12 dollars = 3 ? 10 2 yr, 3. 16 ? 10 10 dollars yr or ~10 2 yr The circumference of the Earth at the equator is C = 2? r = 2? 6. 378 ? 10 6 m = 4. 007 ? 10 7 m ( ) continued on next page http://helpyoustudy. info 16 Chapter 1 The length of one dollar bill is 0. 55 m, so the length of ten trillion bills is m ? 12 12 = ? 0. 155 ? ? (10 ? 10 dollars ) = 1? 10 m. Thus, the ten trillion dollars would dollar ? ? encircle the Earth 1 ? 1012 m n= = = 2 ? 10 4 , or ~10 4 times C 4. 007 ? 10 7 m 1. 61 (a) (b) ? 365. 2 days ? ? 8. 64 ? 10 4 s ? 1 yr = (1 yr ) ? = 3. 16 ? 10 7 s ? ? ? 1 day ? 1 yr ? ? ? Consider a segment of the surface of the Moon which has an area of 1 m2 and a depth of 1 m. When ? lled with meteorites, each having a diameter 10? 6 m, the number of meteorites along each edge of this box is n= length of an edge 1m = = 10 6 meteorite diameter 10 ? 6 m The total number of meteorites in the ? led box is then N = n 3 = 10 6 3 = 10 18 At the rate of 1 meteorite per second, the time to ? ll the box is 1y ? = 3 ? 10 10 yr, or t = 1018 s = (1018 s ) ? ? ? 7 ? 3. 16 ? 10 s ? 1. 62 ~1010 yr ( ) We will assume that, on average, 1 ball will be lost per hitter, that there will be about 10 hitters per inning, a game has 9 innings, and the team plays 81 home games per season. Our estimate of the number of game balls needed per season is then number of balls needed = ( number lost per hitter ) ( number hitters/game )( home games/year ) games ? hitters ? ? innings ? = (1 ball per hitter ) 10 ? 81 ? ? ? year ? inning ? ? game ? = 7300 balls year o r ~10 4 balls year 1. 63 The volume of the Milky Way galaxy is roughly ? ?d2 ? ? VG = At = ? t ? 10 21 m 4 ? 4 ? ? ( ) (10 m ) 2 19 or VG ? 10 61 m3 r If, within the Milky Way galaxy, there is typically one neutron star in a spherical volume of radius r = 3 ? 1018 m, then the galactic volume per neutron star is V1 = 3 4 3 4 ? r = ? ( 3 ? 1018 m ) = 1 ? 10 56 m 3 3 3 or V1 ? 10 56 m 3 The order of magnitude of the number of neutron stars in the Milky Way is then n= VG 10 61 m 3 ? V1 10 56 m 3 or n ? 10 5 neutron stars http://helpyoustudy. info 2 Motion in One DimensionQUICK QUIZZES 1. 2. (a) (a) 200 yd (b) 0 (c) 0 False. The car may be slowing down, so that the direction of its acceleration is opposite the direction of its velocity. True. If the velocity is in the direction chosen as negative, a positive acceleration causes a decrease in speed. True. For an accelerating particle to stop at all, the velocity and acceleration must have opposite signs, so that the speed is decreasing. I f this is the case, the particle will eventually come to rest. If the acceleration remains constant, however, the particle must begin to move again, opposite to the direction of its original velocity.If the particle comes to rest and then stays at rest, the acceleration has become zero at the moment the motion stops. This is the case for a braking car—the acceleration is negative and goes to zero as the car comes to rest. (b) (c) 3. The velocity-vs. -time graph (a) has a constant slope, indicating a constant acceleration, which is represented by the acceleration-vs. -time graph (e). Graph (b) represents an object whose speed always increases, and does so at an ever increasing rate. Thus, the acceleration must be increasing, and the acceleration-vs. -time graph that best indicates this behavior is (d).Graph (c) depicts an object which ? rst has a velocity that increases at a constant rate, which means that the object’s acceleration is constant. The motion then changes t o one at constant speed, indicating that the acceleration of the object becomes zero. Thus, the best match to this situation is graph (f). 4. Choice (b). According to graph b, there are some instants in time when the object is simultaneously at two different x-coordinates. This is physically impossible. (a) The blue graph of Figure 2. 14b best shows the puck’s position as a function of time. As seen in Figure 2. 4a, the distance the puck has traveled grows at an increasing rate for approximately three time intervals, grows at a steady rate for about four time intervals, and then grows at a diminishing rate for the last two intervals. The red graph of Figure 2. 14c best illustrates the speed (distance traveled per time interval) of the puck as a function of time. It shows the puck gaining speed for approximately three time intervals, moving at constant speed for about four time intervals, then slowing to rest during the last two intervals. 5. (b) 17 http://helpyoustudy. info 1 8 Chapter 2 (c) The green graph of Figure 2. 4d best shows the puck’s acceleration as a function of time. The puck gains velocity (positive acceleration) for approximately three time intervals, moves at constant velocity (zero acceleration) for about four time intervals, and then loses velocity (negative acceleration) for roughly the last two time intervals. 6. Choice (e). The acceleration of the ball remains constant while it is in the air. The magnitude of its acceleration is the free-fall acceleration, g = 9. 80 m/s2. Choice (c). As it travels upward, its speed decreases by 9. 80 m/s during each second of its motion. When it reaches the peak of its motion, its speed becomes zero.As the ball moves downward, its speed increases by 9. 80 m/s each second. Choices (a) and (f). The ? rst jumper will always be moving with a higher velocity than the second. Thus, in a given time interval, the ? rst jumper covers more distance than the second, and the separation distance between th em increases. At any given instant of time, the velocities of the jumpers are de? nitely different, because one had a head start. In a time interval after this instant, however, each jumper increases his or her velocity by the same amount, because they have the same acceleration. Thus, the difference in velocities stays the same. . 8. ANSWERS TO MULTIPLE CHOICE QUESTIONS 1. Once the arrow has left the bow, it has a constant downward acceleration equal to the freefall acceleration, g. Taking upward as the positive direction, the elapsed time required for the velocity to change from an initial value of 15. 0 m s upward ( v0 = +15. 0 m s ) to a value of 8. 00 m s downward ( v f = ? 8. 00 m s ) is given by ? t = ? v v f ? v0 ? 8. 00 m s ? ( +15. 0 m s ) = = = 2. 35 s a ? g ? 9. 80 m s 2 Thus, the correct choice is (d). 2. In Figure MCQ2. 2, there are ? ve spaces separating adjacent oil drops, and these spaces span a distance of ? x = 600 meters.Since the drops occur every 5. 0 s, the ti me span of each space is 5. 0 s and the total time interval shown in the ? gure is ? t = 5 ( 5. 0 s ) = 25 s. The average speed of the car is then v= ? x 600 m = = 24 m s ? t 25 s making (b) the correct choice. 3. The derivation of the equations of kinematics for an object moving in one dimension (Equations 2. 6 through 2. 10 in the textbook) was based on the assumption that the object had a constant acceleration. Thus, (b) is the correct answer. An object having constant acceleration would have constant velocity only if that acceleration had a value of zero, so (a) is not a necessary condition.The speed (magnitude of the velocity) will increase in time only in cases when the velocity is in the same direction as the constant acceleration, so (c) is not a correct response. An object projected straight upward into the air has a constant acceleration. Yet its position (altitude) does not always increase in time (it eventually starts to fall back downward) nor is its velocity always dir ected downward (the direction of the constant acceleration). Thus, neither (d) nor (e) can be correct. http://helpyoustudy. info Motion in One Dimension 19 4. The bowling pin has a constant downward acceleration ( a = ? g = ? 9. 80 m s 2 ) while in ? ght. The velocity of the pin is directed upward on the upward part of its ? ight and is directed downward as it falls back toward the juggler’s hand. Thus, only (d) is a true statement. The initial velocity of the car is v0 = 0 and the velocity at time t is v. The constant acceleration is therefore given by a = ? v ? t = ( v ? v0 ) t = ( v ? 0 ) t = v t and the average velocity of the car is v = ( v + v0 ) 2 = ( v + 0 ) 2 = v 2. The distance traveled in time t is ? x = vt = vt 2. In the special case where a = 0 ( and hence v = v0 = 0 ) , we see that statements (a), (b), (c), and (d) are all correct. However, in the general case ( a ? , and hence v ? 0 ), only statements (b) and (c) are true. Statement (e) is not true in either ca se. The motion of the boat is very similar to that of a object thrown straight upward into the air. In both cases, the object has a constant acceleration which is directed opposite to the direction of the initial velocity. Just as the object thrown upward slows down and stops momentarily before it starts speeding up as it falls back downward, the boat will continue to move northward for some time, slowing uniformly until it comes to a momentary stop. It will then start to move in the southward direction, gaining speed as it goes.The correct answer is (c). In a position versus time graph, the velocity of the object at any point in time is the slope of the line tangent to the graph at that instant in time. The speed of the particle at this point in time is simply the magnitude (or absolute value) of the velocity at this instant in time. The displacement occurring during a time interval is equal to the difference in x-coordinates at the ? nal and initial times of the interval ? x = x t f ? x ti . 5. 6. 7. ( ) The average velocity during a time interval is the slope of the straight line connecting the points on the curve corresponding to the initial and ? al times of the interval ? v = ? x ? t = ( x f ? xi ) ( t f ? ti ) ? . Thus, we see how the quantities in choices (a), (e), (c), and (d) ? ? can all be obtained from the graph. Only the acceleration, choice (b), cannot be obtained from the position versus time graph. 8. From ? x = v0 t + 1 at 2, the distance traveled in time t, starting from rest ( v0 = 0 ) with constant 2 acceleration a, is ? x = 1 at 2 . Thus, the ratio of the distances traveled in two individual trials, one 2 of duration t1 = 6 s and the second of duration t 2 = 2 s, is 2 2 ? x2 1 at 2 ? t 2 ? ? 2 s ? 1 2 = 1 2 =? ? =? ? = ? x1 2 at1 ? 1 ? ? 6 s ? 9 and the correct answer is (c). 2 9. The distance an object moving at a uniform speed of v = 8. 5 m s will travel during a time interval of ? t = 1 1 000 s = 1. 0 ? 10 ? 3 s is given by ? x = v ( ? t ) = (8. 5 m s ) (1. 0 ? 10 ? 3 s ) = 8. 5 ? 10 ? 3 m = 8. 5 mm so the only correct answer to this question is choice (d). 10. Once either ball has left the student’s hand, it is a freely falling body with a constant acceleration a = ? g (taking upward as positive). Therefore, choice (e) cannot be true. The initial velocities of the red and blue balls are given by viR = + v0 and viB = ? 0 , respectively. The velocity of either ball when it has a displacement from the launch point of ? y = ? h (where h is the height of the building) is found from v 2 = vi2 + 2a ( ? y ) as follows: 2 vR = ? viR + 2a ( ? y ) R = ? ( + v0 ) 2 + 2 ( ? g ) ( ? h ) = ? 2 v0 + 2 gh http://helpyoustudy. info 20 Chapter 2 and 2 vB = ? viB + 2a ( ? y ) B = ? ( ? v0 ) 2 + 2 ( ? g ) ( ? h ) = ? 2 v0 + 2 gh Note that the negative sign was chosen for the radical in both cases since each ball is moving in the downward direction immediately before it reaches the ground.From this, we see that choice (c) is tr ue. Also, the speeds of the two balls just before hitting the ground are 2 2 2 2 vR = ? v0 + 2 gh = v0 + 2 gh > v0 and vB = ? v0 + 2 gh = v0 + 2 gh > v0 Therefore, vR = vB , so both choices (a) and (b) are false. However, we see that both ? nal speeds exceed the initial speed and choice (d) is true. The correct answer to this question is then (c) and (d). 11. At ground level, the displacement of the rock from its launch point is ? y = ? h , where h is the 2 height of the tower and upward has been chosen as the positive direction.From v 2 = vo + 2a ( ? y ) , the speed of the rock just before hitting the ground is found to be 2 2 v =  ± v0 + 2a ( ? y ) = v0 + 2 ( ? g ) ( ? h ) = (12 m s )2 + 2 ( 9. 8 m s2 ) ( 40. 0 m ) = 30 m s Choice (b) is therefore the correct response to this question. 12. Once the ball has left the thrower’s hand, it is a freely falling body with a constant, non-zero, acceleration of a = ? g . Since the acceleration of the ball is not zero at any point o n its trajectory, choices (a) through (d) are all false and the correct response is (e). ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS . Yes. The particle may stop at some instant, but still have an acceleration, as when a ball thrown straight up reaches its maximum height. (a) (b) 6. (a) No. They can be used only when the acceleration is constant. Yes. Zero is a constant. In Figure (c), the images are farther apart for each successive time interval. The object is moving toward the right and speeding up. This means that the acceleration is positive in Figure (c). In Figure (a), the ? rst four images show an increasing distance traveled each time interval and therefore a positive acceleration.However, after the fourth image, the spacing is decreasing, showing that the object is now slowing down (or has negative acceleration). In Figure (b), the images are equally spaced, showing that the object moved the same distance in each time interval. Hence, the velocity is constant in Figure ( b). At the maximum height, the ball is momentarily at rest (i. e. , has zero velocity). The acceleration remains constant, with magnitude equal to the free-fall acceleration g and directed downward. Thus, even though the velocity is momentarily zero, it continues to change, and the ball will begin to gain speed in the downward direction.The acceleration of the ball remains constant in magnitude and direction throughout the ball’s free ? ight, from the instant it leaves the hand until the instant just before it strikes the 4. (b) (c) 8. (a) (b) http://helpyoustudy. info Motion in One Dimension 21 ground. The acceleration is directed downward and has a magnitude equal to the freefall acceleration g. 10. (a) Successive images on the ? lm will be separated by a constant distance if the ball has constant velocity. Starting at the right-most image, the images will be getting closer together as one moves toward the left.Starting at the right-most image, the images will be getting fa rther apart as one moves toward the left. As one moves from left to right, the balls will ? rst get farther apart in each successive image, then closer together when the ball begins to slow down. (b) (c) (d) ANSWERS TO EVEN NUMBERED PROBLEMS 2. 4. 6. (a) (a) (a) (d) 8. (a) (d) 10. 12. (a) (a) (d) 14. 16. (a) 2 ? 10 4 mi 10. 04 m s 5. 00 m s ? 3. 33 m s +4. 0 m s 0 2. 3 min L t1 2 L ( t1 + t 2 ) 1. 3 ? 10 2 s (b) 13 m (b) (b) 64 mi ? L t 2 (c) 0 (b) (b) (b) (e) (b) ? x 2 RE = 2. 4 7. 042 m s 1. 25 m s 0 ? 0. 50 m s (c) ? 1. 0 m s (c) ? 2. 50 m s a) The trailing runner’s speed must be greater than that of the leader, and the leader’s distance from the ? nish line must be great enough to give the trailing runner time to make up the de? cient distance. (b) t = d ( v1 ? v2 ) (c) d2 = v2 d ( v1 ? v2 ) 18. (a) Some data points that can be used to plot the graph are as given below: x (m) t (s) (b) (c) 5. 75 1. 00 16. 0 2. 00 35. 3 3. 00 68. 0 4. 00 119 5. 00 192 6. 00 41. 0 m s , 41. 0 m s , 41. 0 m s 17. 0 m s , much smaller than the instantaneous velocity at t = 4. 00 s l http://helpyoustudy. info 22 Chapter 2 20. 22. 24. (a) 20. 0 m s , 5. 00 m s (b) 263 m 0. 91 s (i) (a) (ii) (a) 0 0 (b) (b) 1. 6 m s 2 1. 6 m s 2 500 x (m) (c) (c) 0. 80 m s 2 0 26. The curves intersect at t = 16. 9 s. car police officer 250 0 0 4. 00 8. 00 12. 0 16. 0 20. 0 t (s) 28. 30. a = 2. 74 ? 10 5 m s 2 = ( 2. 79 ? 10 4 ) g (a) (b) (e) 32. (a) (d) 34. 36. 38. 40. (a) (a) (a) (a) (c) 42. 44. 46. 48. 95 m 29. 1 s 1. 79 s v 2 = vi2 + 2a ( ? x ) f 8. 00 s 13. 5 m 22. 5 m 20. 0 s 5. 51 km 107 m v = a1t1 (c) a = ( v 2 ? vi2 ) 2 ( ? x ) f (d) 1. 25 m s 2 (b) 13. 5 m (c) 13. 5 m (b) (b) (b) (b) No, it cannot land safely on the 0. 800 km runway. 20. 8 m s, 41. 6 m s, 20. 8 m s, 38. 7 m s 1. 49 m s 2 ? = 1 a1t12 2 2 ? xtotal = 1 a1t12 + a1t1t 2 + 1 a2 t 2 2 2 (a) Yes. (b) vtop = 3. 69 m s (c) ?v downward = 2. 39 m s (d) No, ? v upward = 3. 71 m s. The two rocks have the same acceleratio n, but the rock thrown downward has a higher average speed between the two levels, and is accelerated over a smaller time interval. http://helpyoustudy. info Motion in One Dimension 23 50. 52. (a) (a) (c) 21. 1 m s v = ? v0 ? gt = v0 + gt v = v0 ? gt , d = 1 gt 2 2 29. 4 m s ? 202 m s 2 4. 53 s vi = h t + gt 2 (b) (b) 19. 6 m d = 1 gt 2 2 (c) 18. 1 m s, 19. 6 m 54. 56. 58. 60. 62. 64. (a) (a) (a) (a) (b) (b) (b) (b) 44. 1 m 198 m 14. m s v = h t ? gt 2 See Solutions Section for Motion Diagrams. Yes. The minimum acceleration needed to complete the 1 mile distance in the allotted time is amin = 0. 032 m s 2 , considerably less than what she is capable of producing. (a) (c) y1 = h ? v0 t ? 1 gt 2 , y2 = h + v0 t ? 1 gt 2 2 2 2 v1 f = v2 f = ? v0 + 2 gh (d) 66. (b) t 2 ? t1 = 2 v0 g y2 ? y1 = 2 v0 t as long as both balls are still in the air. 68. 70. 3. 10 m s (a) (c) 3. 00 s (b) v0 ,2 = ? 15. 2 m s v1 = ? 31. 4 m s, v2 = ? 34. 8 m s 2. 2 s only if acceleration = 0 (b) (b) ? 21 m s Yes, for all initial velocities and accelerations. 72. 74. (a) (a)PROBLEM SOLUTIONS 2. 1 We assume that you are approximately 2 m tall and that the nerve impulse travels at uniform speed. The elapsed time is then ? t = 2. 2 (a) 2m ? x = = 2 ? 10 ? 2 s = 0. 02 s v 100 m s At constant speed, c = 3 ? 108 m s, the distance light travels in 0. 1 s is ? x = c ( ? t ) = ( 3 ? 108 m s ) ( 0. 1 s ) ? 1 mi ? ? 1 km ? 4 = ( 3 ? 10 7 m ) ? ? = 2 ? 10 mi 3 ? 1. 609 km ? ? 10 m ? (b) Comparing the result of part (a) to the diameter of the Earth, DE, we ? nd 3. 0 ? 10 7 m ? x ? x = = ? 2. 4 DE 2 RE 2 ( 6. 38 ? 10 6 m ) ( with RE = Earth’s radius ) http://helpyoustudy. info 24 Chapter 2 2. 3Distances traveled between pairs of cities are ? x1 = v1 ( ? t1 ) = (80. 0 km h ) ( 0. 500 h ) = 40. 0 km ? x2 = v2 ( ? t 2 ) = (100 km h ) ( 0. 200 h ) = 20. 0 km ? x3 = v3 ( ? t3 ) = ( 40. 0 km h ) ( 0. 750 h ) = 30. 0 km Thus, the total distance traveled is ? x = ( 40. 0 + 20. 0 + 30. 0 ) km = 90. 0 km, a nd the elapsed time is ? t = 0. 500 h + 0. 200 h + 0. 750 h + 0. 250 h = 1. 70 h. (a) (b) v= ? x 90. 0 km = = 52. 9 km h ? t 1. 70 h ?x = 90. 0 km (see above) v= v= ? x 2. 000 ? 10 2 m = = 10. 04 m s ? t 19. 92 s 2. 4 (a) (b) 2. 5 (a) ?x 1. 000 mi ? 1. 609 km ? ? 10 3 m ? = ? ? = 7. 042 m s ? t 228. 5 s ? 1 mi ? 1 km ? Boat A requires 1. 0 h to cross the lake and 1. 0 h to return, total time 2. 0 h. Boat B requires 2. 0 h to cross the lake at which time the race is over. Boat A wins, being 60 km ahead of B when the race ends. Average velocity is the net displacement of the boat divided by the total elapsed time. The winning boat is back where it started, its displacement thus being zero, yielding an average velocity of zero . (b) 2. 6 The average velocity over any time interval is ? x x f ? xi = ? t t f ? ti ? x 10. 0 m ? 0 v= = = 5. 00 m s ? t 2. 00 s ? 0 v= (a) (b) (c) (d) (e) v= v= v= v= ? x 5. 00 m ? 0 = = 1. 25 m s ? 4. 00 s ? 0 ? x 5. 00 m ? 10. 0 m = = ? 2. 50 m s ? t 4. 00 s ? 2. 00 s ? x ? 5. 00 m ? 5. 00 m = = ? 3. 33 m s ? t 7. 00 s ? 4. 00 s 0? 0 ? x x2 ? x1 = = = 0 ? t t 2 ? t1 8. 00 s ? 0 2. 7 (a) (b) 1h ? Displacement = ? x = (85. 0 km h ) ( 35. 0 min ) ? ? ? + 130 km = 180 km ? 60. 0 min ? 1h ? The total elapsed time is ? t = ( 35. 0 min + 15. 0 min ) ? ? ? + 2. 00 h = 2. 83 h ? 60. 0 min ? so, v= ? x 180 km = = 63. 6 km h ? t 2. 84 h http://helpyoustudy. info Motion in One Dimension 25 2. 8 The average velocity over any time interval is ? x x f ? xi = ? t t f ? ti ? x 4. 0 m ? 0 v= = = + 4. 0 m s ? t 1. 0 s ? 0 ? ? 2 . 0 m ? 0 v= = = ? 0. 50 m s ? t 4. 0 s ? 0 v= (a) (b) (c) (d) v= v= ? x 0 ? 4. 0 m = = ? 1. 0 m s ? t 5. 0 s ? 1. 0 s ? x 0? 0 = = 0 ? t 5. 0 s ? 0 2. 9 The plane starts from rest ( v0 = 0 ) and maintains a constant acceleration of a = +1. 3 m s 2 . Thus, we ? nd the distance it will travel before reaching the required takeoff speed ( v = 75 m s ) , from 2 v 2 = v0 + 2a ( ? x ) , as ? x = 2 v 2 ? v0 ( 75 m s ) ? 0 = = 2. 2 ? 10 3 m = 2. 2 km 2 2a 2 (1. 3 m s ) 2 Since this distance is less than the length of the runway, the plane takes off safely. 2. 10 (a) The time for a car to make the trip is t = cars to omplete the same 10 mile trip is ? t = t1 ? t 2 = (b) ? x ? x ? 10 mi 10 mi ? ? 60 min ? ? =? ? ? = 2. 3 min v1 v2 ? 55 mi h 70 mi h ? ? 1 h ? ?x . Thus, the difference in the times for the two v When the faster car has a 15. 0 min lead, it is ahead by a distance equal to that traveled by the slower car in a time of 15. 0 min. This distance is given by ? x1 = v1 ( ? t ) = ( 55 mi h ) (15 min ). The faster car pulls ahead of the slower car at a rate of vrelative = 70 mi h ? 55 mi h = 15 mi h Thus, the time required for it to get distance ? x1 ahead is ? t = ? x1 = vrelative ( 55 mi h ) (15 min ) 15. 0 mi h = 55 minFinally, the distance the faster car has traveled during this time is ? x2 = v2 ( ? t ) = 2. 11 (a) ( 70 mi h ) ( 55 min ) ? ? 1h ? ? = 64 mi ? 60 min ? From v 2 = vi2 + 2a ( ? x ) , with vi = 0 , v f = 72 km h , and ? x = 45 m, the acceleration of the f cheetah is found to be km ? ? 10 3 m ? ? 1 h 72 ? 0 h ? ? 1 km ? ? 3 600 s v 2 ? vi2 f a= = = 4. 4 m s 2 2 ( ? x ) 2 ( 45 m ) continued on next page 2 http://helpyoustudy. info 26 Chapter 2 (b) The cheetah’s displacement 3. 5 s after starting from rest is 1 1 2 ? x = vi t + at 2 = 0 + ( 4. 4 m s 2 ) ( 3. 5 s ) = 27 m 2 2 2. 12 (a) (b) (c) (d) 1 = v2 = ( ? x )1 + L = = + L t1 ( ? t )1 t1 ( ? x )2 ? L = = ? L t2 ( ? t )2 t 2 ( ? x ) total ( ? x )1 + ( ? x )2 + L ? L 0 = = = 0 = t1 + t 2 t1 + t 2 t1 + t 2 ( ? t ) total +L + ? L total distance traveled ( ? x )1 + ( ? x )2 2L = = = ( ave. speed )trip = t1 + t 2 t1 + t 2 t1 + t 2 ( ? t ) total vtotal = The total time for the trip is t total = t1 + 22 . 0 min = t1 + 0. 367 h , where t1 is the time spent traveling at v1 = 89. 5 km h. Thus, the distance traveled is ? x = v1 t1 = vt total, which gives 2. 13 (a) (89. 5 km h ) t1 = ( 77. 8 km h ) ( t1 + 0. 367 h ) = ( 77. 8 km h ) t1 + 28. 5 km or, (89. 5 km h ? 77. km h ) t1 = 28. 5 km From which, t1 = 2 . 44 h for a total time of t total = t1 + 0. 367 h = 2. 81 h (b) The distance traveled during the trip is ? x = v1 t1 = vt total, giving ? x = v ttotal = ( 77. 8 km h ) ( 2. 81 h ) = 219 km 2. 14 (a) At the end of the race, the tortoise has been moving for time t and the hare for a time t ? 2 . 0 min = t ? 120 s. The speed of the tortoise is vt = 0. 100 m s, and the speed of the hare is vh = 20 vt = 2 . 0 m s. The tortoise travels distance xt, which is 0. 20 m larger than the distance xh traveled by the hare. Hence, xt = xh + 0. 20 m which becomes or vt t = vh ( t ? 120 s ) + 0. 0 m ( 0. 100 m s ) t = ( 2 . 0 m s ) ( t ? 120 s ) + 0. 20 m t = 1. 3 ? 10 2 s This gives the time of the race as (b) 2. 15 xt = vt t = ( 0. 100 m s ) (1. 3 ? 10 2 s ) = 13 m The maximum allowed time to complete the trip is t total = total distance 1600 m ? 1 km h ? = ? ? = 23. 0 s required average speed 250 km h ? 0. 2 78 m s ? The time spent in the ? rst half of the trip is t1 = half distance 800 m ? 1 km h ? = ? ? = 12 . 5 s v1 230 km h ? 0. 278 m s ? continued on next page http://helpyoustudy. info Motion in One Dimension 27 Thus, the maximum time that can be spent on the second half of the trip is t 2 = t total ? 1 = 23. 0 s ? 12 . 5 s = 10. 5 s and the required average speed on the second half is v2 = 2. 16 (a) ? 1 km h ? half distance 800 m = = 76. 2 m s ? ? = 274 km h t2 10. 5 s ? 0. 278 m s ? In order for the trailing athlete to be able to catch the leader, his speed (v1) must be greater than that of the leading athlete (v2), and the distance between the leading athlete and the ? nish line must be great enough to give the trailing athlete suf? cient time to make up the de? cient distance, d. During a time t the leading athlete will travel a distance d2 = v2 t and the trailing athlete will travel a distance d1 = v1t .Only when d1 = d2 + d (where d is the initial distance the trailing athlet e was behind the leader) will the trailing athlete have caught the leader. Requiring that this condition be satis? ed gives the elapsed time required for the second athlete to overtake the ? rst: d1 = d2 + d giving or v1t = v2 t + d or t = d ( v1 ? v2 ) (b) v1t ? v2 t = d (c) In order for the trailing athlete to be able to at least tie for ? rst place, the initial distance D between the leader and the ? nish line must be greater than or equal to the distance the leader can travel in the time t calculated above (i. e. , the time required to overtake the leader).That is, we must require that D ? d2 = v2 t = v2 ? d ( v1 ? v2 ) ? ? ? or D? v2 d v1 ? v2 2. 17 The instantaneous velocity at any time is the slope of the x vs. t graph at that time. We compute this slope by using two points on a straight segment of the curve, one point on each side of the point of interest. (a) (b) (c) (d) vt=1. 00 s = vt=3. 00 s = 10. 0 m ? 0 = 5. 00 m s 2 . 00 s ? 0 ( 5. 00 ? 10. 0 ) m = ? 2 . 50 m s ( 4. 0 0 ? 2 . 00 ) s ( 5. 00 ? 5. 00 ) m vt=4. 50 s = = 0 ( 5. 00 ? 4. 00 ) s 0 ? ( ? 5. 00 m ) vt=7. 50 s = = 5. 00 m s (8. 00 ? 7. 00 ) s http://helpyoustudy. info 28 Chapter 2 2. 18

Physician Assisted Suicides essays

Physician Assisted Suicides essays Euthanasia. . ., Specifically, Physician Assisted Suicides. . . Imagine yourself suffering a debilitating disease, such as cancer, of which there is no cure. Imagine the disease spreading through your body and producing pain, so severe, that there is no drug to stop it. The medical doctors have all told you there is nothing more they can do. All that is left is for you to lie dying in a hospital or home and rithing in pain. Given these circumstances, would it comfort you to know you have a choice in how much pain you want to suffer and when YOU want to end it all? Im talking about Euthanasis. . . Euthanasia is the act of taking ones life, either passively (by quitting eating or drug/alcohol abuse) or actively (. (DEFINE) The simple . . . . . . .. Specifically, physicians assisted suicides. Suicide is legal in all 50 states. Meaning, one cannot be prosecuted for taking their own life. Yet, Physician assisted suicides are illegal in all but one state, Oregon. Public opinion for Physician Assisted Suicide is overwhelming in favor allowing the patient the right to have a doctor end their life. A Gallop poll, conducted in April, 1996 and again in June 1999, found that 75% and 76% respectively of adults in this country favor allowing doctors to end the lives of the terminally ill, upon request from the patient. However, in 1997, the Supreme Court struck down the constitutional right of all Americans to assisted suicides. Physician assisted suicides in the United States should be a legal right for all Americans and protected under the constitution. Id like to walk you through what the pro-life proponents believe and then identify, what I believe, would ensure a physician assisted suicide program should entail to ensure som e type of integrity. I think with specific guidelines, patients should have the right to choose and medical doctors should be allowed to assist the terminally ill to end their life. I be...

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There's a guide called"College Search Step-by-Step" for students who are unsure of how to go about the search process.It will ask specific questions about what you are looking for in a school and give you a "snapshot" of your preferences to help you decide where to apply. On the main college search page,you can find college matches based on different criteria that you’re interested in, or you can search for schools directly.You can also compare schools, but the amount of data is less staggering than some other sites, so it may be easier to digest. College InSight If you’re a more scientifically minded person, this site will probably appeal to you.You have the option to start your search witha specific college, a topic that interests you, or with building your own table of different variablesto compare schoolsacross any metric you choose. This site has a huge amount of data for each school that includes variables such as economic and racial diversity of the student body and extensive statistics on debt and financial aid.This is a helpful resource for comparing colleges side by side based on hard data.You can choose a â€Å"focus† school and a â€Å"comparison† school, and the site will list the data for each school in table format.It's a good way to get an objective viewpoint on what each school is like. If you’re not as interested in looking at a bunch of data you may find this site to be overwhelming, but if you’re trying to findan easy way to compare school stats without having to create a profilethis is the place to go. This is just a screenshot of the top of the comparison charts - there are many more statistics that show up below this. College Navigator This site (put out by the National Center for Educational Statistics) doesn’t have the prettiest interface, but it’s useful for finding reliable data about any college in the country.If you already know what school you want, you can search for it directly.You can also search by state or even area code.There are filters for degree level and type of college as well, so you can search for just public or private colleges or just 4-year colleges. Each college has a page that lists relevant data including information about cost and financial aid, admissions, programs offered, graduation rates, athletics, and other general statistics. The downside of this site is that the information can be overwhelming and difficult to digest because there is so much data.If you prefer something less dry and scientific you might want to look elsewhere, but if you’re just looking for the facts with no frills this is a good resource. The Inside Scoop: Sites With the Best Student Feedback and College Matchmaking This next section highlights sites that are best for when you already have an idea of which schools you're thinking about and are interested in comparing them and learning more about whether they're really a good fit. Cappex This is a great site for comparing schools, calculating admissions chances, calculating financial aid, and just figuring out if a school is a good match for you in general.You will be asked to create a profile, which then allows the site to figure out which schools might work best based on what matters most to you.There are also student reviews, so in addition to comparing hard data you will also be able to compare how people feel about the school.There’s even a feature that helps you plan campus visits! This site has one of the best balances between comprehensive data and a user-friendly experience that tells you what you really want to know about colleges. It will even take you as far as your application! Each school that you put on your list of favorites has a link that takes you directly to the school's website where you can begin the application process. Chegg On Chegg's college matchmaking site, you can create a profile and see which schools might be right for you.This site can be very helpful in your college search because it lets you keep a running list of schools that you’re interested in once you create a profile. Design-wise, it’s nice to look at, andthe statistics are presented in a very user-friendly, simple way that’s easy to understand.You’ll also find grades based on student reviews for everything from campus dining to academics to the party scene, so you can get an inside look at what people really think about the school. Based on these grades, the site has lists of schools that have the â€Å"best quality of life†, â€Å"best merit aid†, and many other categories that may help you narrow down your search.There’s also a tool on each college page for you to enter your GPA and SAT/ACT scores and see your chances of admission.Oneissue with student ratings is that they are very subjective and may not always reflect the truth about the school.Before you take these as fact, make sure you also check out the school’s main website and see what they have to say about themselves. This site contains less hard data, but is relatively easy to use and makes the college search a little more fun (they even give grades for the attractiveness of the student body...very important in choosing your dream school). Niche This site gives you access totons of statistics as well as student reviews and letter grades for different aspects of student life(this is actually where the college grades on Chegg, a site I recommend later, come from).You’ll also see lists of the best schools in different categories based on reviews. If you’re looking for a comprehensive overview of what’s offered by colleges, this is a great resource.There are a lot of statistics to wade through, but you can also create a free profile and get matched up with colleges.This site is nice becauseit can get really technical and specific,but it also gives you the tools to search for schools without feeling too overwhelmed.You can keep a running list of schools that you’re interested in, and the site will help you notify representatives for these schools of your interest if you think any of them are especially good matches. One aspect of Niche that I'm not so crazy about is the amount ofdistracting promotions and ads for other related sitesthat are involved. It makes things a bit more cluttered and confusing to navigate and kind of turned me off from using this site despite the fact that it presents a lot of great information. College Confidential You might be familiar with College Confidential for its (sometimes less than trustworthy) discussion forums, but it also has a great college search feature.You can fill out your preferences in over 20 different categories including location, majors, Greek life, party scene, and special services.Based on your answers and how important these factors are to you, you’ll get matched with schools that fit you best (out of a database of over 4,500 schools that includes schools outside of the US). You can also create a list of schools that you like by â€Å"pinning† your favorites.To narrow down your search even further, you can compare your top choices in different areas to see how they measure up against each other.You can save and share your lists if you want as well! I’d recommend this as a starting point if you're not sure which schools you’re interested in - it will help you figure out what’s most important to you and give you some initial ideas. It's super easy to pin and compare schools, although the information pages are a little confusing and not as nice to look at as the school matching interface. If you click on a school, you are linked out to another site called "College View" for the statistics, which makes things a bit harder to navigate compared to other sites where all information is internal. Unigo Get it? Go to Uni? It sounds like the Brits are responsible for this one, which made me suspicious, but this site is pretty cool.It has a modern design and includes tons of reviews in written and video format from students at the colleges.Based on student feedback, schools are rated on a scale of 1-10 for a bunch of different factors including campus safety, political activity, arts culture, Greek life, intellectual life, and more. Each college page has financial aid and admissions statistics, a built in map of the surrounding area, and statistics on student life such as the number of campus organizations and Greek houses.You can also make a list of schools and compare them across different metrics like tuition and selectivity. It does seem like they’re still working on adding more statistics about academics, so this site is a bit light on data in some areas.Still, it’s fun to use and will give you a different perspective than other sites with its multimedia format; there are lots of pictures of schools from students in addition to the video reviews. Taking Action: The Best Sites to Learn About Paying for and Applying to College Once you have a pretty solid list of which colleges you want to apply to, you can use these sites to help you navigate college applications and paying for school. Fastweb This site is oriented towards helping students search for scholarships, but it will also help you find colleges that may interest you. You can set up an account for free that will allow you to locate colleges and scholarships that fit your specific needs.There are helpful blog articles on the site as well that have advice on scholarships and colleges. You can also search for student loans on Fastweb - you just have to answer a few questions and you’ll get a list of all your loan choices.This site is a nice user-friendly means of finding scholarships and loans that will help you to avoid getting too overwhelmed with your options. It’s also a good site to keep in mind going forward, since it has resources for career planning and finding internship opportunities as well. If you're practically-minded and are serious about finding scholarships, you should definitely try this site out. You may want to use another platform for the college search process because Fastweb doesn't have as much information or tools for finding colleges as it does for finding scholarships and loans. Peterson’s This site is good for learning about application timelines and getting yourself organized for college as well as searching for schools that meet your criteria.There are articles with helpful tips about choosing colleges, applying to college, and getting financial aid and scholarships. You'll get fast stats about colleges and suggestions about other schools that you may like based on your searches. This isn’t my favorite site in terms of its presentation of college statistics. However, it does provide advice in other areas that is more specific than other sites and may help answer questions you have about the application process and how to actually follow through on your college decision. How to Start Your College Search Your mind is probably reeling from all of these options, so to make it easier, below is how we recommend you get started. The three sites described in this section are the best ones for beginning your college search, and I'll explain exactly how to use them. Step 1: Develop a Preliminarly List of Schools With Cappex As we mentioned above, Cappex is an excellent site for helping you figure out which types of schools you're interested in and which fit you well. Below are the steps to get the most out of Cappex. 1. Create a profile - you'll answer a lot of questions about your college preferences including location, size, selectivity, and religious affiliation as well as questions about your GPA and scores. This gives the site an idea of what colleges may fit with your needs. You also have the opportunity to enter in schools that interest you if you already have options in mind. 2. Play around with your Dashboard: this is where you'll see all the colleges that might fit your preferences. You'll be asked to select a region and a major so that the results are more tailored to you (don't worry - you can choose undecided if you aren't sure what you want to major in yet). You'll see a slider of different schools that looks like this: If you hover over any of the schools, you'll get an at-a-glance stats panel that shows you whether the school is public or private, its tuition rates, and the number of students. The first row of schools is for colleges that you may be interested in, and the second row is for colleges that are interested in you based on your level of high school achievement. 3. Click on the link inside the stats panel for any school that catches your interest. Once you get to the next page, you'll have a ton of information at your fingertips. One helpful feature that you probably want to check first is "Your Fit", which is the fourth button down on the left side panel. This gives you an easy way to tell if the college matches up well with the preferences you indicated in your profile. 4. Explore the college a little more. Think about other things that are important to you and whether they fit well with what the college offers. You'll notice that in each college profile you can comb through a huge database of information by clicking on the options on the left. Categories include everything from campus life to student reviews to application deadlines. Before you get too confused from all the information, try making a list of what you think will make you happy at college and target your browsing to the areas that matter most to you. 5. If you're feeling really good about a school, you can add it to the running list of colleges that is a part of your Cappex profile. Just click the "yes" button at the top of the screen where it asks if you're interested in the school. This will also notify the school that you are interested. If you fill out your profile completely, Cappex will also give you your chances of admission to a college if you click on "yes calculate my chances". Here's what your college list will look like: Notice that you can rank colleges by how much you like them, and if you update your profile you'll see your chances of admission as well. The list also makes it easy to compare colleges. If you click on the "compare" tab you can compare colleges side by side in admissions, tuition, diversity, and campus life (this could be a good way to give your parents the hard facts about why you prefer one school over another that they think is better). 6. After you're satisfied with your list, you can apply to schools by clicking on the "apply to college" link beneath every list entry. This will send you directly to the admissions site for the school and make it easy for you to get your application started. Step 2: Refine Your College List Using Chegg Once you've made a preliminary list of colleges with Cappex, we recommend next using Chegg to refine your list and see if there are any schools you missed. Below are the steps for getting the most out of Chegg. 1. Login through Facebook or sign up manually for a profile. Make sure you fill out as much information as you can in your profile so that your college matches will be as accurate as possible. 2. Click on the "Colleges" tab at the top of the screen, and you'll get to a page where you can search for schools by name or sort schools by your preferences. You can also look at lists of schools that are provided for you on the site based on student ratings for different things like "best food" or "most beautiful campus". You should also look at your "Matches", which are colleges selected for you based on the preferences you indicated in your profile. Just switch to the matches tab by clicking on it at the top of the screen: 3. Once you see a college that looks promising, go to the page for the school and start exploring the statistics and ratings. You can even look at your chances of admission based on your GPA and scores. Again, I would recommend writing down some factors that are important to you for a college before you dive into the stats so that you have a more focused approach. I think a good thing about this site is that it doesn't go too heavy on the nitty gritty statistics. It boils it down to the basic things you probably want to know and also gives you perspective from real students. 4. Add any college you think sounds good to your list of schools. Once you add a college to your list, you will have the option to fill out information that will allow colleges you are interested in to contact you. You can then compare up to five schools side by side based on scores, admission requirements, tuition, and other basic facts. This should help you narrow your search down significantly. 5. Take a look at other features of the site - you can also use Chegg to find scholarships and internships! Step 3: Search for Scholarships Using Fastweb Fastweb isthe best way to find scholarships, which is a super important part of the college application process. You can also find colleges on Fastweb, but it's not as streamlined as the other options. We recommend using Fastweb after you've used Cappex and Chegg to get a list of colleges you're interested in applying to. 1. Fill out a free profile, providing as many details as possible about your interests and strengths as possible. This will help you get matched to scholarships that are appropriate for your specific talents. 2. Click on "see my matches" in your profile to take a look at all the scholarships that are available to you: You can sort scholarships by deadline, amount, and provider. This is an awesome resource - instead of traversing the internet, you can see all of your potential scholarships right here. 3. If you click on the link for a scholarship and think you're interested, you can use the dropdown menu on the right to mark it as one that you "might apply" or "will apply" to;this will add it to your list under one of those tabs. This will make it easy for you to keep track of which scholarships interest you and which ones you promised yourself you would apply for (because we all know how easy it is to say you're going to apply and how hard it is to actually follow through). 4. You can also use this site to search for colleges (switch to the "colleges" tab at the top of the screen) and create a list of schools you're interested in. The list will look similar to the scholarship matches page: As I mentioned, the college search features on this site aren't quite as comprehensive as the others because there's no way to compare schools and there aren't as many student reviews or user-friendly statistics. You CAN use it to make a list, but I'd say Fastweb is mostly for scholarships. 5. If you want, start looking into your student loan options! This is a really daunting process that is made easier by the Student Lending Center feature under the Student Loans section of Fastweb. You can also get help with navigating the financial aid process under the Financial Aid tab and finding internships under the Career Planning tab. Basically, for anything involving the practical concerns associated with college, Fastweb is the best place to go. Conclusion The college search is a challenging, sometimes confusing process, especially when you're not sure where to start. I've given you the top ten sites I would recommend for finding colleges that fit your needs and a more detailed guide to the three sites that I think will be the most useful. Here are my overall rankings for the best sites to use in your college search: 1. Cappex2. Chegg3. Fastweb4. Niche5. College Confidential6. Peterson's7. College Board's Big Future8. College InSight9. Unigo10. College Navigator All of these sites will help you out in one way or another to narrow down your search, but I believe the top three have the best combination of user friendly format, solid information, and fun interactive features that are also very helpful to students. Remember to focus solely on what is most important to you in your search. You may have never heard of a school, but that doesn't mean it won't be a perfect fit. Be honest with yourself and you'll end up making a great choice! What's Next? Aiming high in your college search? Read this article on how to get into the top schools in the nation. If you're worried about your chances of getting into college period, check out this list of the easiest schools to get into. Building a killer college application is hard work. Find out the how to create a versatile application and what not to do in your college essay. The Common Application makes it easy to apply to a bunch of different schools but only fill out one application! Find out which schools accept the Common App. Want to improve your SAT score by 160 points or your ACT score by 4 points?We've written a guide for each test about the top 5 strategies you must be using to have a shot at improving your score. Download it for free now:

Marketing Across Cultures Case Study Example | Topics and Well Written Essays - 250 words

Marketing Across Cultures - Case Study Example In fact, besides gender, culture is probably the biggest determinant of global marketing techniques. For example, Asians are known to be very conservative in fashion and indulgence. They do not like â€Å"loud† clothes, muscle cars, and they are cuisine is minimalist in nature. A global fashion giant may design and market summer clothes in North America, Europe and Latin America but launch entirely different designs of the same clothes in Japan. This is because it is aware of what sells in Japan, and it is dictated by culture. Film studio companies launch movies in some parts of Asia and Eastern Europe with some parts edited to avoid antagonising those populations and record low sales. Culture is therefore a major factor in global marketing; it is central to most global marketing strategies (Migliore, 2011:40). Introduction of a laptop in China and Thailand would probably require the sale of operating systems in languages that are common in those languages. For example, English is not the dominant language in the two countries, so the operating system would have to be in a culturally conducive language. In Nigeria, on the other hand, English is the main form of official communication; therefore the operating system would be in English. In China and Thailand, it is possible that certain colours are associated with negative connotations like bad omen (Migliore, 2011:52). Consequently, these colours would be avoided when introducing laptops lest sales be negatively affected. In Nigeria, on the other hand, laptops can be introduced in all possible colours because it means nothing to most