Essay On Non Technical Manager - 896 Words

Security for Non-Technical Managers Information security is one of the weakest links on the information system management. Thus, non-technical managers lack knowledge on the subject. This increases the risk of threats by driving them to comply with the organization security policy. To protect, the organization information security assets, non- technical managers should be exposed to the three main areas of accountability such as Confidentiality, Integrity, and Availability (CIA). By doing so, it helps to create a pro-active environment to preserve the confidentiality of the information, maintain its integrity, and also ensure its availability. Since the value of information is so high, companies are thriving all the time to develop†¦show more content†¦To prevent employees from doing that, they should not be allowed the login on any social media using company equipment on company time. By establishing and enforcing such as policies, the company can reach their main goal of keeping Information secure. Next, preserving Integrity of information refers to protecting information from being modified by unauthorized parties. For example, in today’s world news, it is very difficult to assess the integrity of the information from each group that has its own version and interpretation of their data. So, if the information is correct then the value of it will be considered high. On the other hand, the value will be nothing and worse of that, it could affect the credibility and the loss of your business. A generic definition of risk is: â€Å"a probability or threat of damage, injury, liability, loss, or any other negative occurrence that is caused by external or internal vulnerabilities of an asset and thereby cause harm to the organization†. Therefore, it is almost impossible to quantify the damage of security breaches can cause to any businesses if management neglects the execution of security policies. In addition, there are many techniques to preserve the integrity of the business data among the public key encryption, symmetric encryption, digital signature etc. For example, many organizations use the digital signature format, which certifies and timestamps a document. If the document has been modified, the signatureShow MoreRelatedFirst Managers Needed Skils1242 Words   |  5 Pageshow the managers become more efficient, many researcher devoted themselves to develop theories on what exactly skills or abilities does managers need. In 1955, Robert L. Katz developed three essential managerial skills or competencies: technical, human and conceptual. He also believed that different levels of managers need different skills within the organization. More specially, Katz thought that top managers need conceptual skills, middle managers need human skills and first line managers need technicalRead MoreOrganizational Behavior and Leadership Essay1204 Words   |  5 PagesThe success of a corporation has been traditionally tied to how well the managers execu te the roles of leading, planning and controlling. One of the key components of the controlling aspect of management is the evaluation of the people that they are managing. The function of this process involves monitoring performance goals and has management taking corrective actions when necessary. Performance evaluations can be described as the measurement of the performance of an employee related to definedRead MoreContingency Theories in Management1657 Words   |  7 PagesThis essay sets out to show where the four popular management contingency variables of organisational size, routineness of task technology, environmental uncertainty and individual differences are reflected in the work of the manager that was interviewed. Using classical theories of Fayol, Mintzberg and Katz along practical examples from the managers’ day-to-day routine, this essay sets out to explain how these theories and functions impact upon how the manager applies the situational approach toRead M oreWhat is Management? Essay920 Words   |  4 PagesEssay 1: What is Management? The purpose of this paper is to analyze the role and function of management in society and the workforce and to look into how it is implemented. Three main questions will be discussed, outlining, †What is management? aWhat do managers do? And what skills do managers need?† We will define management as a profession, look into what managers undertake and deliver in their daily output of work, different levels of management, and what skills managers need in order to produceRead MoreTaylors Principles of Management Essay1472 Words   |  6 Pagestheir job, compare with the previous time when people would have been doing their job without applying the knowledge of management. Likewise, management can be considered as a science in how managers will be able to manage people and help them to involve further and to be more effective. The term of manager here refers to the person who responsible for planning and directing the work of a group of individuals, monitoring their work, and taking corrective action when it necessary (Robbins, BergmanRead MoreCorporate Trends And Accounting Scandals946 Words   |  4 Pages(2007) suggest that the roles of a management accountant have indeed changed and in turn they have turned more into ‘hybrid accountants’ with their roles becoming more ‘exiting’ and ‘consulting-based’ (Burns, J. Baldvinsdottir, G. 2007). In this essay I intend to show that although key drivers such as globalisation, advances in technology, corporate trends and accounting scandals have cause a shift in the roles of management accountants, moving away from just ‘score-keepers’, the fundamental jobRead MoreAn Advanced Risk Management Method1270 Words   |  6 PagesAbstract This essay aims to debate an advanced Risk Management method while slightly in comparison to other advanced or not-so-advanced processes to deduct the importance on an effective mitigation phase. The Risk Management method that is examined in this essay will be used to compare and to make a conclusion on the mitigation’s effectiveness with the help of a detailed assessment phase. -- Introduction Terminologically, risk is known to be the possibility of an action having complications whileRead MoreWhat skills does a manager need This essay gives a general overview of the different skills managers require at the different levels of management1660 Words   |  7 PagesESSAY TOPIC: WHAT SKILLS DOES A MANGER NEED? A managers job is complex and multidimensional, certain skills are required in order to effectively run an organisation. As used here management is the process of coordinating work activities so that they are completed efficiently and effectively with and through other people (Robbins, S., Bergman, R, Stagg, J. Coulter, M. 2006). A manager is someone who works with and through other people by coordinating their work activities in order to accomplishRead MoreSelf Introduction975 Words   |  4 PagesSelf-Introduction Essay What is the objective of a self-introduction essay? The objective of a self-introduction essay is to provide a short, concise introduction to others. A self-introduction essay can be useful for different reasons such as employment, graduate school, or professional activities. We will focus on the use of a self-introduction essay solely for the use of employment opportunities. Employers may request a self introduction essay to provide a ‘summary’ of each candidate. ThisRead MoreThe Leadership Skills And Behaviours For Successful Implementation Of Human Resource Management Plan1737 Words   |  7 Pages1. INTRODUCTION This essay aims to discuss and evaluate the leadership skills behaviours for successful implementation of human resource management plan in a special case study of Offshore Gas Project (CSOGP). Identification of Leadership behaviours categories that are relevant and meaningful for all leaders is subject to controversy. For this essay discussion three main leadership categories have been considered as follows : 1) Task Oriented Leadership (TOL) which is a behaviour that organizes

Financial Analysis Japans Financial Markets - 1191 Words

INTRODUCTION In March 2015, Japan’s Financial Markets Agency for the first time in its history set out Corporate Governance Code and a year earlier Stewardship Code. Even though some efforts towards corporate governance and transparency have been made in Japan previously, specifically introduction of dual system in 2003, they did not gain popularity. Only 40 out of 3,000 firms adopted this system immediately rising to 112 five years later. However, these codes were necessary due to the pressure from foreigners investing and doing business in Japan, several scandals such as Olympus 2011-2012 accounting scandal and ineffective, high cash holdings of Japanese companies (Eberhart, 2012). They are both addressing different aspects, but principal-based Governance code is binding on all non-foreign companies listed on securities exchanges in Japan and mirrors the UK’s approach to Corporate Governance while Stewardship Code is on voluntary basis (Freshfileds Bruckhaus Deringer, 2015). This essay seeks to determine how the reforms from 2014, 2015 and dual system from 2003 influenced companies’ performance and provides the evidence that transparent corporate governance leads to higher performance of the firm. In order to respond to the question, it is first necessary to examine collected empirical evidence from Tokyo Stock Exchange and further analyze using Stata. LITERATURE BACKGROUND AND METHODOLOGY The analytical section considers panel data on Japanese companies listedShow MoreRelatedWhat Is The Link Between The Expansion Of Bank Loans In Gapan1035 Words   |  5 Pagesto the developments in M3 and bank loans to the non-financial private sector. These discoveries are consistent with economic theory and the findings of Fisher and Seater (1993) and McCandless and Weber (1995). The EG co-integration test does not approve a statistically significant connection between GDP and M3 or GDP and bank loans to the non-financial private sector. Nevertheless, the link between the volume of bank loans to the non-financial private sector and GDP relates with the findings ofRead MoreThe NSI And Japans National System Of Innovation Model921 Words   |  4 Pagesexamining an overview of the NSI and Japan’s NSI, a triad relationship which consists of industry, academia, and government. Additionally, its strengths, weaknesses, and to what extent is this approach useful toward understanding differences in Japan’s rates of innovation between other countries. Japan’s National System of Innovation NSI focuses on the flows and relationships of knowledge and technology among people, enterprises, and institutions, especially for analysis on â€Å"knowledge based economiesRead MoreJapan s International Business Environment Essay1352 Words   |  6 Pages2015). In this report the international business environment of Japan is analyzed from economical, political and legal, financial, and technological perspectives. In particular, the focus is on economic indicators, trade, Japan’s political system, the rule of law, the Japanese stock market, debt, the technological infrastructure and intellectual property. Subsequently, the analysis of the Japanese business environment is linked to international business. Here, implications of the business environmentRead MoreMacroeconomic Analysis Japan3941 Words   |  16 Pages| |Japan: Macroeconomic Analysis | | | | |Juan Pablo Giraudo Read MoreAmerican Dollar vs Japanese Yen910 Words   |  4 PagesThis paper aims to compare the Japanese Yen against the US Dollar over a five year period starting from 2005 till 2010. The exchange traded fund for Japanese Yen shall also be discussed in the paper and afterwards an analysis of both the currencies shall be presented. There are different factors that influence the exchange rate differences between any two chosen currencies. The effects produced by these different exchange rates can be of quite different intensity. The most common elements that haveRead MoreThe Monetary Policy Of Japan2450 Words   |  10 PagesThis essay will illustrate an analysis of the monetary policy of Japan and the impact of the global crisis of 2008-11 with the measures taken to recover from this crisis. The conduct and the effectiveness of a country’s monetary policy depend on the structure of its financial mark ets. According to Takatoshi Ito in Japan, governments have regulated the financial markets heavily. The Securities Transaction Law, article 65 creates barriers between banking and security dealings. Within the banking sectorRead MoreEssay about Japanese Malaise793 Words   |  4 PagesTHE IMPACT OF SUPPLY-SIDE FACTORS ON JAPANESE ECONOMIC STAGNATION Japan’s ongoing economic stagnation for decades has provoked wider debate in the US. Along with the raise in unemployment rate and weak industrial production relative to other major industrial countries, the economic malaise in Japan was described as a lost decade. Studies have shown that Japanese economy suffers of severe problems that are not only cyclical but also structural in natureRead MoreThe Bubble Economy. The Direct Cause Of The Japanese Banking834 Words   |  4 Pagesmacroeconomic environment in Japan. This condition caused the upward growth expectations of asset prices, uncontrolled credit expansion and financial deregulation. At the same period, the United States has a substantial increase in the current account deficit and a sharp increase in the foreign trade deficit. In order to depreciate the U.S. dollar by intervening in currency markets, five countries including France, West Germany, Japan, the United States and the United Kingdom signed the Plaza Accord on SeptemberRead MoreExternal Global Environment of Louis Vuitton in Japan Case Study Analysis1280 Words   |  6 PagesGlobal Environment of Louis Vuitton in Japan Case Study Analysis Kaplan University School of Business MT460 Management Policy and Strategy Author: Professor: Ernest Norris Date: June 9, 2014 External Global Environment of Louis Vuitton in Japan Introduction This case study is on the external and global environment of Louis Vuitton (LV) in Japan. For many years, Japan has been Louis Vuitton’s most profitable market but the economic crisis has had a negative effect onRead MoreThe Japanese Yen Currency System1505 Words   |  7 Pagescurrency and moved Japan onto the Gold Standard. As time passed, the yen was pegged at 1 USD =  ¥360. This remained true until 1971 when the Bretton Woods system collapsed. The yen became undervalued for exports were costing too little in international markets, and imports from abroad were costing the Japanese too much. Reluctantly, the currency system was switched a regime of a floating exchange rate, along with major currencies in the world, instead of being based on stable and adjustable exchange rates

College Physics 9e Free Essays

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. We will write a custom essay sample on College Physics 9e or any similar topic only for you Order Now 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), an d (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 second 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. Thus, 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 shou ld 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 ? 10 ? 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 t o 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 ( KE ) where the numeral 2 is a 2m dimensionless constant. Dimensional analysis gives the units of mome ntum 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 z eros 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 Introduction 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://helpyo ustudy. 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 Vba cteria 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 ft, 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 hypotenuse of the shaded triangle and is gi ven 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://helpyoustud y. 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 function to the large right triangle cont aining 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 de? nition of the tangent function to t he 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, t he 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 or ~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 n eutron 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 Dimension QUICK 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. If 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 to 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 appr oximately 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 18 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 them 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 time 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 directed 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 case. 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 ) wit h 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 true. 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 on 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 accele ration 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 farther 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 acceleration, 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 = ? 3 4. 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. 3 Distances 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, and 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 bei ng 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 min Finally, 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 = v t 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. 278 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 t ime 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 athlete 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. 00 ? 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 How to cite College Physics 9e, Essay examples