Zara IT for Fast Fashion Essay Example | Topics and Well Written Essays - 2000 words

Zara IT for Fast Fashion - Essay Example The case study "Zara IT for Fast Fashion" indicates that the company relies on an outdated operating system for the store terminals and lacks readily available network across all the stores. The dilemma that the management is grappling with is determining whether they should upgrade the present system and loose the reliability that characterises the current system or proceed with the current DOS-based system that is devoid of compatibility to the future improvements. The case study focuses on Zara, the largest retail chain for Inditex, Spanish retail giant, and the manner in which it has been successful due to its flexibility, excellent fashions and application of a simple business model (McAfee, Dessain and Sjoman, 2007). The head of the IT department in the company, Salgado, is grappling with a dilemma on whether to upgrade the IT capabilities and infrastructure for the retailers or retain the current system. The case study indicates that the company relies on an outdated operating system (OS) for the store terminals and lacks readily available network across all the stores. The Microsoft Disk Operating System (MS-DOS) seems to be working well as no major challenges have been witnessed in the course of its long time use. One of the major concerns expressed by Salgado in the long run is that Zara Point of Sale (POS) is running on MS-DOS that is not supported by Microsoft Inc. According to Capell (2008), DOS has been replaced by Windows since 2001, the same time that Microsoft Inc stopped supporting DOS.

Discrimination Against Lesbians in Britain

Discrimination Against Lesbians in Britain Lesbianism and the problems of identification in contemporary Britain In Britain, lesbian women may not have had to campaign to have their sexual activities decriminalised, as homosexual had needed to do; yet their lives were not free of problems. However, men even homosexual men had more social, economic, and political power and status than British women did as a whole have. Men had a greater control of and over their own bodies than either heterosexual or lesbian women had in the immediate post-war period, and indeed before then as well. British lesbian women would have to campaign for greater rights as both women do, and as lesbians to challenge the discrimination and expected gender roles from a patriarchal and heterosexual dominated society. In other words, British lesbians had to counter sexual discrimination as well as orientation discrimination before they were able to feel fully secure in publicising their sexual orientation and identification. Changing public perceptions of their different gender roles and sexual orientation was, as the main lesbian rights groups realised was not going to be easy, as their experiences during the 1970s proved.[8] In Britain the 1960s was a decade that brought about some wide-ranging social changes and promised further changes for the future. Lesbian women could certainly regard the availability of the contraceptive pill and the legalisation of abortion as being an improvement for the choices that all women were able to make in relation to their own bodies. During the 1960s expectations about the gender roles of women began to change, as feminism meant that fewer women were prepared to become wives and mothers without having a career first. Lesbian women also realised they did not have to put up with marriage and children just because it was expected of them to do so. The decriminalisation of male homosexual acts at the end of the 1960s could also be viewed as a demonstration that British society was slowly becoming less illiberal in its attitudes towards people that embraced alternative lifestyles. For the more radical lesbians the social changes in the 1960s were the start of the process of fighting discrimination, rather than the end of the process. British lesbian women and gay men were encouraged to start gay pride movements by the apparent success such groups in the United States were having in altering social attitudes, especially in cities like New York and San Francisco.[9] British lesbian and gay pride organisations began in earnest during the early 1970s with the objectives of enabling their members to have pride in their orientation identification, as well as aiming to reduce the level of social discrimination which their own members had to endure. The hope was that reduced social discrimination and reduced fear of such prejudices would allow all lesbian women to readily admit their sexuality with pride. These lesbian pride organisations as a whole found that changing social attitudes towards them and their sexual orientation was a slow process, after all generations of social prejudices could not be expected to vanish overnight.[10] The effectiveness or otherwise of lesbian and gay pride movements since the 1970s has been an area of much debate. Depending on which criteria are used to judge the gay pride movements the achievements of these organisations will alter.[11] If judging the achievements of the lesbian and gay pride movements solely in terms of their ability to have anti-discrimination legislation passed these movements were undoubtedly a failure before the election of the New Labour government in 1997.[12] The only lesbian women to gain from anti-discrimination legislation between the early 1970s and 1997 did so because they also came under gender equality, race relations and disability legislation. Governments from the early 1970s did not believe that legislation was needed to prevent homophobic discrimination, leaving lesbian and gay pride organisations on their own to lower such prejudice aimed at their members. New Labour has taken more steps than any previous governments to pass legislation to red uce homophobic prejudices and promote the rights of all lesbians and gays. New Labour legislation has included passing legislation to end discrimination upon the grounds of sexual orientation to enhance the opportunities for all lesbians and gays to be openly identified as such without fear of homophobic discrimination. Besides specific legislation to protect lesbian and gay rights, lesbians can also use the Human Rights Act of 1998 to make sure that their orientation being publicly identifiable is not detrimental to their rights or their safety.[13] Under New Labour, lesbian women have finally received legal equality with heterosexual men and women. For instance, lesbian women and gay men are no longer dismissed from Britain’s armed forces if they openly admit their sexuality. Lesbian women and gay men now also have the right to ‘marry’ each other in civil partnerships. The introduction of civil partnerships means that lesbian women in long term partnerships hav e the same rights as married and co-habitant heterosexual couples in terms of property, taxation and inheritance rights, as well as the same benefit and pension entitlements. Lesbian women now have the same rights as heterosexual women when it comes to the custody or adoption of children.[14] In respect of the lesbian pride movements they probably had a more immediate impact on changing social attitudes towards them than they did in influencing governments to pass legislation which tackled discrimination against them. Lesbian and gay pride organisations did not use the same tactics to publicise their sexual orientation. The more radical lesbian and gay pride organisations were happy to shock heterosexuals in British society with the tone and the methods in which they demonstrated their identifications of sexual orientations. More radical groups were even prepared to ‘out’ famous people to make headlines and raise public awareness of lesbian and gay issues.[15] Other more moderate lesbian and gay pride groups were less keen on showing their sexual orientation and identification in such an ostentatious manner. The more moderate lesbian pride groups would have preferred to be open about their sexual orientation and identification without using over the top publi city stunts. Moderate lesbian groups would have fitted in with insider pressure groups, which attempt to achieve their objectives behind closed doors, rather than in public. Radical lesbian and gay pride groups are examples of outsider groups that have little influence with governments and rely on publicity to bring their objectives to public attention.[16] The campaigns of lesbian pride groups were not as successful in changing public perceptions of lesbian women as a majority of those groups would have hoped from the early 1970s. This was especially the case during the period of Conservative governments between 1979 to 1997. The Conservative party more than the Labour and Liberal/Liberal Democrats parties like to represent traditional family values instead of promoting the rights and the lifestyles of lesbian and gay pride groups. When in office the Conservatives prevented lesbian and gay groups’ form educating school pupils about their sexual orientations and identification through the infamous section 28. Lesbian and gay groups lobbied New Labour to have the section removed as soon as possible. The media did not always help lesbian and gay groups convey the messages to the public that they wanted to be sent out, especially right wing newspapers did not want to change public perceptions of lesbian women. The right wing newspap ers often portrayed the objectives of lesbian and gay groups in the most negative perspective possible.[17] Despite the efforts of some politicians, religious groups, and parts of the media the number of people who happen to be publicly prepared to be known as being lesbian and gay in Britain has increased noticeably since the 1970s. It is harder to argue out right that such a situation is entirely due to the attempts of lesbian and gay pride groups to alter the attitudes of British society. After all changed reactions towards their orientation and gender identification could have resulted from the campaigning of such groups, or alternatively could just have been a continuation of increasing indifference to how other people chose to live their lives.[18] Thus in conclusion, lesbian women in Britain have not always felt comfortable or able to openly display their sexual orientation, and their gender identification due to the nature of British society. The immediate post-war society in Britain has remained predominantly patriarchal and heterosexual in terms of social and gender relationships as well as expected behaviours, roles, and identifications. That situation meant lesbianism did exist in Britain, it was just well hidden. The 1960s were a decade that kick started the process of altering British society, and provided the stimulus for lesbian and gay pride groups to emerge during the early 1970s. The legacies, influence, and success of such groups are highly debatable. Although these groups certainly raised the profile of lesbians and gays in Britain they actually had little affect on the mainstream political agenda prior to New Labour gaining office in 1997. Lesbian and gay pride groups may have attempted to increase acceptance of their lifestyles and their alternative identifications, yet changing public perceptions of lesbian women has not been universal or overwhelmingly favourable. The way in socialisation operates in such a haphazard way means that the predominantly heterosexual and patriarchal nature of British society will continue for a considerable amount of time, even if the acceptance of lesbian women should continue to improve, and more fluid gender roles should develop further. Bibliography Abercrombie N, Hill S Turner B S (2000) Penguin Dictionary of Sociology 4th edition, Penguin, London Eatwell R Wright A (2003) Contemporary Political Ideologies 2nd Edition, Continuum, London Hobsbawm, E (1994) Age of Extremes, the Short Twentieth Century 1914-1991, Michael Joseph, London Whitaker’s, (2007) Whitaker’s Almanack – today’s world in a single volume, A C Black, London Young H, (2003) supping with the Devils – Political writing from Thatcher to Blair, Guardian Books, London 1 Footnotes [1] Abercrombie, Hill, Turner, 2000 p. [2] Judt, 2007 p.485 [3] Martin, 2003 p. 233 [4] Abercrombie, Hill, Turner, 2000 p. 314 [5] Abercrombie, Hill, Turner, 2000 p. 314 [6] Abercrombie, Hill, Turner, 2000 p. 314 [7] Sandbrook, 2005 p. 599 [8] Eatwell Wright, 2003 p. 214 [9] Hobsbawm, 1994 p. 428 [10] Abercrombie, Hill, Turner, 2000 p. 314 [11] Judt, 2007 p. 785 [12] Abercrombie, Hill Turner, 2000, p. 314 [13] Young, 2003 p. 216 [14] Whitaker’s, 2007 p. 604 [15] Judt, 2007 p. 785 [16] Abercrombie, Hill, Turner, 2000 p. 336 [17] Abercrombie, Hill, Turner, 2000 p. 314 [18] Judt, 2007 p. 785

Galahad as a Christ Figure Essay -- God Religion Galahad Essays Papers

Galahad as a Christ Figure Very few people can call themselves a Christ figure. There are so many elements that go into being a Christ figure. Galahad had all those elements. He was enraptured, he saw 'the wonders of the Holy Grail', and he had disciples. Also, he made life better for human beings by sacrificing himself for others, he preformed miracles, and he heard holy voices and saw holy visions. Because of all that he did Galahad was a proven Christ figure of his time. Galahad would hear holy voices and see visions throughout his journeys. Galahad saw a vision of angels praying about a silver table, meaning that he was close to the end of his search for the Holy Grail. Many times voices would help him and warn him. At one time he was about to kill one of the knights he had been fighting when a voice told him, 'If those two knights had known you as well as I do, they would let you alone.' Hearing this, Galahad rode off without killing either of the knights. The voice saved him from killing his own father. 'For once caution was the better part of valor!' Performing miracles to make life better for human beings is a characteristic that Christ figures possess. When the silver table appeared on Galahad?s boat, he needed help taking it off. He called to an old crippled man to help him carry it. The old man told Galahad that he had been crippled for far too long and such a task would be impossible for him. Finally, the cripple obeyed Galahad and when he stood up he ...

When Kids Get Life

In the Frontline video â€Å"When Kids Get Life† we were introduced to 5 cases in the state of Colorado where teenage boys had been sentenced to life imprisonment without parole. After watching the video I found myself struggling to have an objective opinion on the issue presented, mostly because of personal experiences being a victim of childhood abuse and also having a family member (my brother) murdered. I felt the video to be very one sided but I do find myself agreeing with the point the producers were trying to make.I feel that teenagers should have more opportunities at rehabilitation from crimes committed before the age of 21 then those criminals that are convicted after the age of 21. I also feel strongly that when it comes to teenagers and violent crimes that great emphasis needs to be placed on the motive for the crime, for example if there was long term abuse or neglect as well as any substance abuse involved, and what kind of support if any the child has ever had in their lives.In my opinion the age at which a person should be given life imprisonment is 21. I developed this opinion for three reasons. The first being my personal experience, there were two men involved in my brother’s murder one man was 26 at the time and the other was just barely 18, neither man was sentenced to any long term prison time but of the two the 18 year old has shown greater signs of rehabilitation.I have also had a lot of exposure to the darker side of society and I have seen more improvement come from the younger ‘criminals’ then I have from the older ones. My second reason is the overwhelming amount of scientific evidence regarding brain development and mental processes. According to the Time magazine article ‘What Makes Teens Tick’ Dr. Jay Giedd states that â€Å"The very last part of the brain to be pruned or shaped to its adult dimensions is the prefrontal cortex, home of the executive functions. This area of the brain is the part that allows adults to weigh the consequences of their actions.A teenager may understand the principles of right and wrong but lack the ability to realize the ramifications of any wrong they might do. In another article by Lee Bowman of the Scripps Howard News Service Deborah Yurgelun-Todd of Harvard Medical School and McClean Hospital says that â€Å"[When] shown a set of people’s faces contorted in fear, adults named the right emotion, but teens seldom did, often saying the person was angry. Yurgelun-Todd and her team performed this test using the fMRI and discovered an amazing difference in the parts of the brain being used. The adults used both the prefrontal cortex and the amygdala to process what they saw and younger teens relied entirely on the amygdala while older teens (oldest being 17) showed a progressive shift toward using the frontal cortex My third reason is the hormone factor according to an article published by the American Bar Association.One of the hor mones which has the most dramatic effect on the body in adolescence is testosterone. Testosterone is closely associated with aggression; it increases tenfold in adolescent boy. Considering all of this information I feel that 21 would be a better age to consider legal culpability of a person. I do not feel that teenage offenders of violent crime should go unpunished but life in prison seems to be an excessive punishment for a teenager incapable of comprehending the consequences of his actions.My last issue to address is the circumstances by which an offender should receive life imprisonment. I feel that when it comes to cases of long term abuse of any kind when a teenager is feeling pushed into a corner and the only way out is to ‘kill or be killed’ there will undoubtedly be a negative outcome. There is a long list of possible effects and none of them are positive anything from drug and alcohol abuse, to self harm, to suicide, to homicide. One researcher Phyllis L.Crocke r of Cleveland-Marshall College of Law wrote that â€Å"the nexus between poverty, childhood abuse and neglect, social and emotional dysfunction, alcohol, and drug abuse and crime is so tight in the lives of many capital defendants as to form a kind of social historical profile†. According to Dr. Chris Mallett, Public Policy Director at Bellefaire Jewish Children’s Bureau in Ohio more that 30% of death row juvenile offenders had experienced six or more distinct areas of childhood trauma with an overall average of four such experiences per offender.Mallett also found that such mitigating evidence was presented to juries in fewer than half of the offenders’ trials . That fact I find astonishing I feel passionately that the motivation behind a teenager’s violent act should play a very hefty role in the prosecution of any said act. The cases highlighted in the video ‘When Kids Get Life† were very disturbing to me because several of the cases invol ved long term sexual and psychological abuse and it appeared that no one took that into consideration at the time of trial.In conclusion I feel that no violent crime should be excused however life imprisonment should be reserved for those over the age of 21, or for the truly psychotic individuals out there in society. Any teenager that is convicted of a violent crime should be given prison time but then after a determined amount of time re-evaluate the person psychologically and determine level of rehabilitation. I believe that people (even criminals) can change in both directions good and bad, and teenagers have an even greater capacity to change for the better if guided in the right direction.

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

Word Choice Dessert vs. Desert - Proofread My Paper

Word Choice Dessert vs. Desert - Proofread My Paper Word Choice: Dessert vs. Desert We’ve all been there. You’re in a restaurant with friends. You finish your main course and order a â€Å"desert.† Moments later, the waiter returns and buries your table beneath a truckload of sand, ruining the evening for everyone. Mmmm sand. [Photo: Simon A. Eugster]OK, this probably hasn’t happened in real life. But that’s because â€Å"desert† and â€Å"dessert† are pronounced differently enough that you wouldn’t normally mix them up. Written down it’s different, and we see these terms confused on a regular basis. Sometimes this is a just a typo (so remember to proofread!) but it can also be an issue with spelling, so make sure you know the difference between these words. Dessert (Sweet and Delicious) This is the easy one, since â€Å"dessert† only has one meaning. Specifically, it’s used to describe sweet food eaten after the main part of a meal: For dessert, I ate my own body weight in ice cream. As such, you should only use â€Å"dessert† in reference to food. Getting hungrier now You sometimes see people write â€Å"just desserts,† but this is actually a mistake; the correct term is â€Å"just deserts,† meaning â€Å"that which is deserved.† By comparison, â€Å"just desserts† would mean something like â€Å"only puddings,† which is only useful if you run a diner for people with a sugar addiction. Desert (Dry and Sandy) The most common use of â€Å"desert† is to identify a place with little to no rainfall, typically somewhere hot and sandy: The Mojave is the hottest desert in the United States. It really is very, very hot there. [Photo: Theschmallfella]We also use this sense of â€Å"desert† as an adjective, such as when describing something associated with the desert: From his campaign in North Africa during WWII, Rommel became known as the â€Å"Desert Fox.† Meanwhile, the verb â€Å"desert† is pronounced a little differently despite having the same spelling. This sense of â€Å"desert† means â€Å"to abandon†: The prisoner got away after the guard deserted his post. Dessert or Desert? It’s easy to avoid mistakes with â€Å"dessert† and â€Å"desert† as long as you remember that â€Å"dessert† only has one meaning (it becomes even easier if you imagine the double-â€Å"s† in â€Å"dessert† stands for â€Å"something sweet†). Once you’ve ruled that out, whether you’re using â€Å"desert† as a noun or a verb, the spelling is the same. Just keep in mind that: Dessert (noun) = The final course in a meal (â€Å"something sweet†) Desert (noun) = A dry, sandy area of land Desert (verb) = To leave or abandon

Benefits of Uniforms Within Schools essays

Benefits of Uniforms Within Schools essays Uniforms are commonplace in many important areas within society and are a source of affiliation and pride for those who wear them. Sport teams from house league to the professional level have uniforms. Medical professionals wear uniforms, as do public safety workers. Every school is a learning institution which imprints upon its students strong morals and values. They strive to create an environment in which all of this can be achieved. The philosophy of St. Marys Catholic Secondary School is to develop in all its members, a genuine self-love, love of thy neighbour morale. It is believed that to achieve this dynamic, the individuality of each student must be respected by allowing for the development of the whole person to his or her fullest potential. Effective schools are characterized by a sense of pride on the part of the students. The presence of uniform clothing within an educational institution is undoubtedly beneficial to the school and its students. Uniforms are a concrete wa y to successfully accomplish this. Not only do they significantly increase the safety of the institution, but they also positively influence classroom behaviour, and curtail the effects of peer pressure. A safe and disciplined learning environment is paramount in educating students. The quality of education today is being hampered as a result of the rising presence of gang violence within schools. Uniform dress, however, has combated this problem. As a result of the implementation of uniforms, teachers no longer have to be concerned with student apparel and can devote more time and effort to instructing students. A uniform policy prohibits clothing that bears gang colour or insignia. This protocol significantly reduces the risk of gang related violence within educational institutions. Not only do uniforms virtually eliminate precarious gang confrontations within schools, they also make it possible for staff and students to easily identify intruder...

Witness for the Prosecution

Witness for the Prosecution There has been a murder in 1950s England. Miss Emily French, a woman approaching age 60, was found dead in her house on Friday October the 14th. Her housekeeper was away that evening and Miss Emilys   only other friend, Leonard Vole, was the last person to see her alive. The murder occurred at approximately 9:30 at night. Leonard Vole insists he was at his own home at that time, however the housekeeper, Janet Mackenzie, says she heard him speaking with Miss Emily French at 9:25 when Janet briefly returned home to pick up a sewing pattern. Leonard Vole has retained the services of a solicitor, Mr. Mayhew, and barrister, Sir Wilfred Robarts, QC. Leonard Vole is an extremely likeable man with a story that could either be 1.) the most believable tale of a nice man down on his luck who made friends with an older woman or 2.) the perfect set-up for the chance to inherit close to a million pounds. When Miss Emily French’s last will and testament names Leonard as the sole beneficiary of her estate, it seems Leonard will be found guilty. Only Leonard’s wife, Romaine, has a chance of persuading the jury of Leonard’s innocence. But Romaine has a few secrets and a hidden agenda of her own and she isn’t sharing the details with anyone. Production Details Setting: Sir Wilfred Robart’s offices, English Courtroom Time: 1950s Cast Size: This play can accommodate 13 actors with numerous non-speaking small roles as the jury and courtroom attendants. Male Characters: 8 Female Characters: 5 Characters that could be played by either males or females: 0 Content Issues:  Stabbing Roles Carter is Sir Wilfred’s clerk. He is an older gentleman who prides himself on keeping good time and good order of his boss’s offices. Greta is Sir Wilfred’s typist. She is described as â€Å"adenoidal† and flighty. She is easily distracted by the people who come into the office, especially if she has read about them in the newspaper. Sir Wilfred Robarts, QC is the well-respected barrister on Leonard Vole’s case. He prides himself on reading people and their intentions perfectly the first time he meets them. He is knowledgeable and puts genuine effort into each case he tries. Mr. Mayhew is the solicitor on Leonard Vole’s case. He assists Sir Wilfred in office work and provides another pair of eyes and ears to examine the evidence and consider strategies. His knowledge and opinions are invaluable assets for the case. Leonard Vole appears to be the all-around good-natured sort of man one would enjoy befriending. He has dreams and aspirations that will not come to fruition in his current financial situation, but he is not a complainer. He has the ability to endear himself to anyone, especially to women. Romaine is Leonard’s wife. Their marriage is not technically legal, as she is still married (on paper) to a man from her native Germany. Although Leonard insists that Romaine loves him and is devoted to him, she is a difficult woman to read. She has her own agenda and is skeptical that anyone will be able to help her. Mr. Myers, QC is the prosecuting barrister. He and Sir Wilfred, who often find themselves opposite one other in court, have a contentious relationship and. Both manage to keep civil tongues and behave when they appear in front of the judge, but their mutual animosity is evident. Mr. Justice Wainwright is the judge in Leonard Vole’s case. He is fair and handles the barristers and witnesses with a firm hand. He is not above inserting his opinion or telling a story if need be. Janet Mackenzie was Miss Emily French’s housekeeper and companion for twenty years. She has an unyielding personality. She is not charmed by Leonard Vole and has a very dim opinion of him as a person. Other Smaller Roles and Non-speaking Roles Inspector Hearne Plain Clothes Detective Third Juror Second Juror Foreman of the Jury Court Usher Clerk of the Court Alderman Judge’s Clerk Court Stenographer Warder Barristers (6) Policeman Dr. Wyatt Mr. Clegg The Other Woman Production Notes Set. The two must-have sets for Witness for the Prosecution are Sir Wilfred’s office and the courtroom. For this show – no minimalistic approaches. The sets ought to be built and dressed according to resemble a formal barrister’s office and courtroom of the time period. Costumes must be period specific and of note are the traditional wigs and robes worn in British courtrooms by the barristers, judges, and solicitors. Because the time span of the play is six weeks, some actors will need several costume changes. The playwright provides a specific note on doubling up the roles actors may play in order for smaller casts to still achieve the â€Å"spectacle† of the courtroom. She offers a template for the roles that may be reduced or be cast by the using the same actor. This template is available in the script offered from Samuel French. However, Christie stresses that the same actress that plays Greta should not play the role of â€Å"The Other Woman.† Even though the two characters never appear onstage at the same time, Christie does not want the audience to think that it is part of the plot and that Greta is in fact The Other Woman. Christie goes on to offer suggestions that â€Å"local amateurs† be used to fill out the courtroom scene or even that the audience be invited to sit on the stage. Playwright Agatha Christie (1890 – 1976) is beloved and renowned mystery writer from England. She is best known for her novels and such characters as Miss Marple, Hercule Pirot, and Tommy and Tuppence. Her stories focus on mysteries and murder; where the truth is found in the details and the characters are never who they first appear to be. Her play Mousetrap claims the title of longest running play with a production history that spans over 60 years. Agatha Christie is so prolific and popular that only Shakespeare and the Bible have only outsold her works. Samuel French holds the production rights for Witness for the Prosecution.

Population And Region of South and East Asia Essay

Population And Region of South and East Asia - Essay Example Nearly every mainland is made up of river valleys, highlands, coastal plains and mountainous regions. The increasing human population in the region has encroached on the tropical forests due to increasing demand for settlement and cultivation land. Although large forest lands still exist, much of the frontier has been taken up by human settlement. The settlement patterns in the region as originally influenced by natural water sources such as freshwater lakes and rivers because they were a source of food as well as a medium of transportation. Environmental factors The people of the region have greatly affected the environmental conditions in the region. Majority of the region is located within the humid tropics with the climate being determined as monsoonal. The region has attracted a number of animals such as the orangutan, the Sumatran rhino, Asian elephant, and Malayan tapir. The region has three major species of tiger namely the Malayan tiger, Sumatran tiger, and the Indochinese t iger. Some unique species of animals are found in this region such as the Komodo dragon which has been regarded as the largest living species of lizards (United Arab Emirates News Agency). Due to the surging population in the region, some of the animals in the region have been described as the endangered species due to their diminishing chances of survival. The people have destroyed the natural habitats of the animals such as the tiger and elephants besides hunting them for their precious skin and other body parts.

Manhattan New School Essay Example | Topics and Well Written Essays - 1750 words

Manhattan New School - Essay Example School's professional teachers are hired from Bank Street College of Education, Columbia University Teachers College, New York University and Fordham University. Manhattan New School resides in the heart of Manhattan in the former PS190, which was built in 1903 (About School, Manhattan New School). According to Karen Ruzzo, school's principal, "We pride ourselves at being a highly literate community, and we also understand our responsibility to prepare students to cope with the multiple demands of an ever-changing society. As a result, children learn within real-world contexts. Along with reading and writing, instructions in mathematics, science, social studies, technology, music and art engage young learners in meaningful explorations that develop critical thinking skills" (Karen Ruzzo, Mission Statement). School's vision is to grow in the near future as a role model organization for the greater national cause, while providing the students with the opportunity to investigate a range of big ideas, to ask and answer important questions, and to develop the self-management strategies that enable them to negotiate their daily life. This shows that the school's vision is broadly based and its process creates a commitment to lifelong learning. The Manhattan New School's goals include, committing to ensure that all students benefit from a shared educational experience, and continuing to develop strategies to ensure school-wide collaboration, continuity and accountability. Although all goals direct school towards its future vision, the later set certainly is very vital for its long-term vision's success. The school's objectives for its strategies to attain its long-term vision are to establish continuity of instruction both on and across grade levels, and aligning best instructional practices in all curriculum areas with positive performance outcomes for all students. SWOT Analysis: Strengths Recognition of programs and vision Empirical and interdisciplinary education, and opportunities for student leadership development Recognized educational activities with active community outreach Significant and ongoing faculty and alumni involvement at intermediate level Small classes Extraordinary emphasis on personal attention Friendly, supportive, and comfortable in-house environment Integration of technology into the curriculum Classes primarily taught by professionals incorporation of ecological sustainability due to diversified culture Early adoption Attractive campus building and facilities while having state of the art architecture Weaknesses Lower than optimal enrollment of out-of-state and international students Insufficient diversity among students and teachers Low ratio of spending per student Average class size increasing everyday Limited school capacity Not enough programs for extra curricular activities Too many language courses Opportunities Government's importance for schools playing a larger role in community development Higher market share due to increasing population and awareness Capacity to respond for future growth Capacity to help improve intermediate education Increased demand for professional and diversified education Increasing flow of funding Greater expertise in the use of technology in teaching Access to all parts of New York City Threats Lack of sufficient funding Replica of academic programs by

Paraphrase Essay Example | Topics and Well Written Essays - 2000 words - 1

Paraphrase - Essay Example The book contains case studies explaining experiences of people who executed strategies from previous editions. Jacobson (64) illustrates development of AdWords mechanisms, human-machine relationships along with verified strategies regarding ad group interface, split testing and opt in landing page technologies. Latest edition overviews modern technologies that might be combined to guarantee AdWords success. Like, modified third-party tools and thorough analysis of free tool for examining and augmenting web links from Google. The readers acknowledges the significance of Website Optimizer and AdWords prize on card valued $25. This is an interesting technique to advertise a product/ brand and earn after the struggle. It provides essential knowledge for online marketing and boosts the abilities of shareholders for steering this digital marketing tool. Users can create business ads adjacent to search results. It curtails advertising expenses as products information is effortlessly retrieved. As per Jacobson, Optimizer tool teaches parties to run a landing page linked with technology, order forms, assisting them to boost revenues although they will be hidden from other websites. Jacobson illustrates beneficial cost-effective economic plans through conversion tracking to curtail costs. Google made online marketing atmosphere user-friendly to boost efficiency. It has executed a policy to stop showing URL; this step of limited interface was taken after instances when users were deceived and became victims of AdWords advertisers, which makes the online ad tool complex. Conversely, this user-friendly, technological and cost cutting type of advertising has some advantages and disadvantages. Drawbacks are imperative as they lead to constraints which equivocates damage claims from firms who think their trademarks are trespassed. The anticipated costs associated with such claims against Google might

Entrepreneurship and Small Business Development Essay - 3

Entrepreneurship and Small Business Development - Essay Example The focus is on Leadership.This section is purely theoretical identification and review with proposition(s) proposed (approximate 700 words). Characteristics/attributes associated with entrepreneurship. The emphasis is on that of the business entrepreneur and/or factors that are likely to impact on the entrepreneur and/or the creation of a new venture or venture activities. Typical topics include: motivation, leadership, invention, innovation, and approaches to change, but these are not limited. A critical review of the existing literature in the knowledge field with an in-depth insight, focusing on the selected topical area. Various theories/conceptual perspectives centered on the selected topical area are identified, reviewed, discussed and synthesised. A proposition/propositions being proposed based on literature reviewed. 3. Practical Evidence (15%): Collect information about the entrepreneur focused on the theme you selected in the literature review and review relevant material related to the selected entrepreneur (approximate 850 words). Describe and discuss the process you undertook to collect the secondary data about the chosen business entrepreneur, e.g., mention library, Internet, and any other sources that you consulted. Some appreciation of the limitation of this approach. A detailed description of the business entrepreneur and/or the venture created. This should be based entirely on material (via the secondary research) related to the conceptual material presented in literature review. The description of the identified examples/evidence in practice is focused, reflecting and coherent with appropriate and sufficient evidence to support. 4. Analysis (20%): A synthesis of the material on the selected entrepreneur and the review of the literature. Discussions are to generate insights into the entrepreneur from the theory (approximate 850 words). A focused analysis and synthesis of the material on the selected entrepreneur

Tax Law Research Worksheet Paper Example | Topics and Well Written Essays - 500 words

Tax Law Worksheet - Research Paper Example Rulings in agreement with administrative rules together with court rulings permit for the formation of public agencies that regulate and govern statutes and judicial decisions. The purpose of these administrative agencies aid in administering rulings that are being geared towards implementation of the duties and powers of the laws stipulated. Additionally, it prevents against impunity regarding the violation of laws and regulations enforced by these administrative agencies. Administrative rulings and court cases have helped in the design of a system of checks and balances that aims to minimize the risks of bureaucracy (Tax Law, 2003). In many occasions, there has been inconsistencies and conflict about court rulings. If I had a choice to choose the judgment to go with, then I would rely on the hierarchical rule between the two court rulings. The hierarchical rule will dictate the case ruling that will dominate over the other and that is the position I will take. It is because a court case ruling based on the internal revenue code will always prevail over a case ruling basing on other tax law. It is because the code is supreme over the other tax laws (Tax Law, 2003). It is possible to use timing strategies to have an impact (AGI) Adjusted Gross Income deductions, tax credits, itemized deductions and dependency exceptions. It is possible by strategically timing when the year ends. It means that there will be less uncertainty to the short-term viewpoint for tax law as opposed to the beginning of the next year. With this brief tax reprieve comes the opportunity to form a tax saving plan that is sound. It is essential to time strategically the moment to implement the plan. For example, one can reap benefits on tax retirement savings by making a maximum amount allowable at the end of every year. On the other hand, income shifting strategies impacts or alters these deductions by reducing the central income tax

Managing diversity and equal opportunities Essay - 2

Managing diversity and equal opportunities - Essay Example mmission (2011a) The Equality Act 2010 requires companies to give equal payment for work of same proportion and value to both male and female workers. This act came as a replacement to older acts like Equal Pay Act 1970 and Sex Discrimination Act 1975. Elimination of any form of inequality in pay is extremely important as far as achieving gender impartiality and respecting the dignity of women is concerned. Judges and committees will always interpret the law that promotes equality in pay, purposively because this stipulation is based on the foundation of the law enacted by the European Union (TSO, 2014). The domestic law in the UK has the obligation to adhere to the law enacted by the European Union that imposes rightful obligation in terms of equality in payment which will have an unswerving impact. Therefore, while considering the claims for equal pay clause under the Equality Act 2010, the UK based courts are obligated to consider the pertinent provisions enacted by the European Union. Therefore, if a claimant does not get full assistance from the domestic laws the individual has the option of relying on the European Union law in the British Court (Equality and Human Rights Commission, 2011a). While the British amendments are more generalized in nature requiring companies to eliminate any form of gender discrimination associated with unequal pay, the European amendment is more detailed in nature which eliminates any loopholes associated with gender pay inequality. The European amendment is evident in article 141 stipulated within the European Community Treaty which requires every Member State to make sure that the law of equal pay for male and female workers for work of same value and proportion is pursued. The law states that the ordinary minimum wage is kept same for male and female worker who does work of similar nature, value or quantity (Trueel, 2010). The European amendment urges organizations to supervise the cases where pay is determined on the basis

Tax Law Research Worksheet Paper Example | Topics and Well Written Essays - 500 words

Tax Law Worksheet - Research Paper Example Rulings in agreement with administrative rules together with court rulings permit for the formation of public agencies that regulate and govern statutes and judicial decisions. The purpose of these administrative agencies aid in administering rulings that are being geared towards implementation of the duties and powers of the laws stipulated. Additionally, it prevents against impunity regarding the violation of laws and regulations enforced by these administrative agencies. Administrative rulings and court cases have helped in the design of a system of checks and balances that aims to minimize the risks of bureaucracy (Tax Law, 2003). In many occasions, there has been inconsistencies and conflict about court rulings. If I had a choice to choose the judgment to go with, then I would rely on the hierarchical rule between the two court rulings. The hierarchical rule will dictate the case ruling that will dominate over the other and that is the position I will take. It is because a court case ruling based on the internal revenue code will always prevail over a case ruling basing on other tax law. It is because the code is supreme over the other tax laws (Tax Law, 2003). It is possible to use timing strategies to have an impact (AGI) Adjusted Gross Income deductions, tax credits, itemized deductions and dependency exceptions. It is possible by strategically timing when the year ends. It means that there will be less uncertainty to the short-term viewpoint for tax law as opposed to the beginning of the next year. With this brief tax reprieve comes the opportunity to form a tax saving plan that is sound. It is essential to time strategically the moment to implement the plan. For example, one can reap benefits on tax retirement savings by making a maximum amount allowable at the end of every year. On the other hand, income shifting strategies impacts or alters these deductions by reducing the central income tax