My Philosophy of Ministry Essay

My philosophy of ministry is to teach believers, through teaching the Bible, and training leaders. So, that they may, in turn, teach others to be rooted and built up in Him as seen in Colossians 2:6-7. The first part of my philosophy of ministry is to teach believers in the knowledge and the ways of Jesus Christ. Every member in the church has gifts. They need to understand their gifts and if you have the gift to teach, be trained in the work of the ministry. Teaching God’s people to do the work of the ministry takes place through the teaching of Scripture (2 Tim. 3:16-17). The second part of my philosophy of ministry is to train leaders. A leader in the church is a godly servant who is influencing others in the church. Building leadership development in the local church is vital to the continual growth of the ministry (Acts 6: 3-4; 2 Tim. 2:2). I want to always be in the process of training individuals and preparing them to step into a leadership position. The leadership training process needs to include not only theological study and â€Å"hands-on† involvement, but community and relationship development as well. My philosophy of ministry desires to see believers become mature disciples of Christ that are equipped to reach out and show Christ to a lost and dying world. Realizing that the first step in any philosophy of ministry must be establishing priorities, I have established the following personal priorities. The personal priorities must be firmly established and followed through on. Personal Priorities: (1) Seek God first as the central priority in life. (2) Model a character of Christ. Jesus Christ set an example that all Christians should follow. (3) Model the great commission by doing the work of an evangelist (4) Lead others by establishing loving relationships. People make life changing decisions during unique occasions of challenge and reflection. 5) Bible study, prayer, and spiritual preparation for teaching. Lack of fruitfulness and lack of vision in the church are due to lack of spiritual leadership. I define myself as a leader and a teacher in my local church. I teach the people of my church old Jewish traditions, so we understand more of why Jesus did some of what he did. Also, the Old Testament was written to show history of Jesus’s genealogy, but also what was foretold of Him. I am the Director of the woman’s ministry, and the Awana program for the children to help them learn Bible verses at a young age.

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

One Foot in Eden Essay

It is impossible to go through life without making any. What we choose can define us, can close off a part of our life that, had we chosen differently, could have led to something completely different. Many things can influence our choices, from morals, to peers, to experience. Usually, it is our own morals and opinions that decide what we do. How we were raised, what we were taught, and what we have picked up along the way. Family plays a large part in our decisions. Many people think about what someone in their family would do when faced with a difficult decision. In One Foot in Eden, Amy weighs the pros and cons of sleeping with Holland to get pregnant. She carefully thinks about what her husband would do if he found out. This may be a bad example, as she decides to sleep with him anyways. Also, simple things that we learn in school allow us to make informative judgments. Peers also play a significant role. It is almost human nature to please others, or to fit in. Something we would not normally do, we do to make others like us. We might also refrain from acting a certain way that might prevent others from accepting us. This is why it is common that teens begin smoking, or doing some other dangerous habit, all to â€Å"be cool. † Our experiences shape many of our choices. If we know from experience that we will get a bad or painful response to an action, we avoid that action. We will also repeatedly make choices that produce a good or pleasant effect. We face many obstacles by trial and error, and if we have done something before, we know what actions elicit what responses. In conclusion, our choices, whether they are made from our experiences, our peers, or our morals, immensely affect our lives. The factors that influence our choices, be they good or bad, are directly related to the outcome of those decisions.

Business of Events Essay Example | Topics and Well Written Essays - 3000 words

Business of Events - Essay Example The idea is to raise consciousness among students of University of Bedfordshire through a collective effort that will bring together community groups, local businesses, educators and artists to an event where green innovations are highlighted. To ensure the event’s success, students of Travel and Tourism Management will be assigned to organize the Earth Fair, making sure that event plans are actualized and well-coordinated. The entire event plan will be divided into four categories – event research, event design, event planning, and event coordination. Each category will be handled by a committee which will consist of one committee chair and five members. As a head of the planning committee, I will be assigned to present this paper detailing the plans for the University of Bedfordshire - Earth Fair 2013. The challenge is to attract local groups to participate at the Fair and invite approximately students at the university’s Park Square. With a budget of ?10,000, it is important to ensure effective event promotion, proper coordination with local groups, and efficient management of the events team that will guarantee of the event’s success. As such, this paper presents a carefully designed plan to successfully gather students, community organizations, and local establishments to increase awareness about how to care for the environment. Event Research As an annual global event, Earth Day celebration has evolved from bringing environmental issues on the political arena to increasing public awareness through various green activities (Nelson, 1993). One of the well-known activities during Earth Day include holding an Earth Fair which features innovative works and products that promote environmental sustainability. Its aim is to bring together different sectors, from businesses to government agencies, to showcase their works and innovations in order to help people understand more about environmental conservation. All over the world, Earth D ay organizers showcase different ways and themes of holding and preparing for an environmental fair. For instance, in Plymouth, Michigan, organizers for the Earth Day have conceptualized the â€Å"Green Street Fair† which promotes green products and services on the streets of Downtown Plymouth. Also, for this year, the Piedmont Earth Day Fair organizers are challenging themselves to host a zero-waste environmental event that welcomes more than 10,000 people in the area (Piedmont Environmental Alliance, 2012). Here in the country, Greenpeace UK (2011) has initiated an environmental fair which features children’s’ activities, plant swaps, rainbow face painting, yoga, and performances to create a positive display of green lifestyle. Moreover, the London Green Fair, which is the largest environmental fair in London, has been holding the event at the Regent’s Park where they give away trees to visitors for free to offset the carbon used during the event. It is from the concepts and organization of these events that the planning for the University of Bedfordshire – Earth Day Fair will be based on. The themes and manner of presentation will be guided on the previous management and organization of green fair events. Needs Assessment As an events management tool, needs assessment is conducted to assess the preparation, intention, and responsibilities of organizers in an event project. Basically, it involves answering the 5 W’s (Who, What, Where, When, and Why) to develop how the event will be

Negotiating Strategy on XYZ Airport Services Essay

Negotiating Strategy on XYZ Airport Services - Essay Example Under this stage, XYZ should get to know the other party (Airport Authority) by initially setting an appointment. Before making any attempt to establish a wide-range of the settlement with the Airport Authority, the top management of XYZ should carefully plan on some strategies that could enable both parties to come up with a win-win situation or conduct negotiations on neutral ground (Guirdham 2002, pp. 400 – 404). Since negotiation process is dynamic by nature, Shell (2001) suggests the need to be careful when choosing the best bargaining style. In line with this, negotiator such as in the case of XYZ company should consider not only the culture but also the personality of the people behind the Airport Authority. For instance: The personality of the people behind the Airport Authority is outgoing. Therefore, it is necessary on the part of XYZ company to satisfy the personality, needs and wants of the people behind the Airport Authority. Since most of them are outgoing people, XYZ management should invite and treat these people to have lunch or dinner as a group meeting or play golf while discussing the issue involved. This strategy will enable XYZ management to easily win the trust and sympathy of the people behind Airport Authority. Becoming familiar with the entry phase of negotiation is a crucial factor that determines whether or not the outcome of a given negotiation process can be successful. Right timing is very important in the negotiation process. As explained by Zartmann (2002) and Rubin (1989), the process of knowing the phases of negotiation could somehow enable the negotiator to know whether or not it is the right time to begin the pre-negotiation stage, the formal negotiation stage which normally takes place after the preparation stage, and ends the bargaining process which is the post-negotiation stage.

Security Cameras in Building Essay Example | Topics and Well Written Essays - 1000 words

Security Cameras in Building - Essay Example Some of the justifications given by stakeholders within one of the test schools claim that face recognition cameras will protect their children against known criminal offenders and other types of dangerous persons. Another safety consultant in the schools that adopted the security system claims that they adopted the use of face recognition cameras in order to prevent another saga that occurred in a school in Columbia during the year 1999. Advantages of using cameras in buildings have also been put forward in the GSA (1997) article. Although this has been mentioned shortly, they assert that security cameras in public places and government buildings are essential in detecting assailants and criminals who may have intentions of causing harm. There are no such advantages mentioned by Lyon (2001). Keen (2006) also points out to some advantages of the cameras. She describes how bars in Chicago will be required to install cameras in buildings. She affirms that some business owners feel that this enhances their patron's confidence and security. In the article, the major of Chicago claims that security cameras in business premises such as bars have worked wonders for other cities that have done the same in the prevention of crime. He cites examples like London where they were able to prevent a bomb attack by a potential terrorist through their cameras. Keen (2006) affirms that security cameras in public places will help catch some seemingly passive crimes such as drivers who pass red lights or those who leave bomb packages in public places. The article also claims that placing cameras in buildings is easy and cheap consequently denying residents any excuses. Disadvantages of using cameras in buildings Three out of the four articles also examine some of the mishaps, inefficiencies and misgivings about the use of cameras in buildings. GSA (1997) is the only article that does not look at the disadvantages of using security cameras in buildings. Frank (2007) states that the cost that come with this new technology will only place undue burdens on the consumer of that technology. He claims that for a school to employ face recognition camera systems, they require finances reaching the tune of 30, 000 dollars. This is something that is rather costly considering the fact that those schools still have to go about their daily operations. Keen (2006) also points out to this same problem. She says that many business owners who in Chicago who are required to place security cameras in their premises will have to deal with the additional expenses that come with the system. This means that there will be greater costs for bar owners who may not necessarily see the direct benefits that come with the installation. Keen (2006) also says that security cameras in buildings do not necessary bring about positive effects. In her articles, one of the stakeholders (a president of the Chicago restaurants Association) claim that this is just another form of intrusion from the government since no tangible results are visible. In this same article, some representatives from the American Civil liberties Union claim that security cameras in buildings should only be allowed unless there is a direct link to the September eleven attacks. Since this is not possible, then the idea should be abandoned. The Union claims that The American

Does the Japanese state deserve most of the credit for Japan's Essay

Does the Japanese state deserve most of the credit for Japan's development - Essay Example During Japan's postwar economic miracle, it was the Japanese state that deserves most of the credit for successful industrial development. The state's use of industrial policy was the single most important cause of the transformation and growth of the economy. All factors, including external environment, political leadership, and the role played by the private sector, are insignificant when compared to industrial policy. Without the industry-specific interventionist policies followed by the MITI, the economy would not have developed at the pace or in the direction it ultimately did. At the outset, those who answered the question affirmatively can point out that the roots of Japan's successful post-war industrialization and economic development can be traced to efforts of the Japanese state as early as the period of the restoration of imperial rule in Japan. Prior to the restoration of Japanese imperial rule, the Tokugawa Shogunate after its experience with Commodore Matthew Perry's gunboat diplomacy in 1853 had accepted many unequal treaties leading to dissatisfaction among the country's samurais and feudal clans. For example, Japanese tariff rates were kept low and a system of extraterritoriality was established. Restoration of imperial rule through the installation of the political rule of Emperor Meiji became the rallying point of a significant portion of the country's ruling elite and leading warlords in expressing opposition to foreign encroachments. Japan is one of countries of Asia that started early in modernizing their banks. As early as 1872, Emperor Meiji established four national banks in Tokyo and other cities of Japan.

Strategic marketing 2 Essay Example | Topics and Well Written Essays - 1750 words

Strategic marketing 2 - Essay Example Presently the company operates with 40 staffs. The company started competing with branded glass manufacturers from early 1990s and cemented their position as premier glass manufacturers in Singapore, Malaysia and Hong Kong. Sales growth for the company slowed down during the period of 2003 due to external factors. In the initial years Henrik Skagen used two types of marketing strategy to increase sales revenue. Henrik Skagen used trade shows to promote glass product and increase brand equity among customers. Internal sales force was used to explore retail channel sales. New generation of Skagen family have changed the traditional product strategy of the company in order to fillip the growth of the company. Sandra and Lars Skagenby extended product portfolio by including items such as reading sun-glasses, non-prescription reading glasses, sports goggles and glass cases. Marketing Overview Product Reading sun-glasses, Non-prescription reading glasses, Sports goggles and Glass cases. Th e company uses Nordic style of strong lines and bold colours in their offering. Price Retail price of reading glasses is between $35-$85 while retailer sells a pair of sunglasses at $35-$140 Place The company sells their product through distribution channel complemented with accessory shops, department stores and sports outlets Promotion The company uses trade shows and in store sales promotion to create awareness among customers Target Market Seventy percent of sales are contributed by consumers over the age group of 40 while sports products are targeted for young people Target Country Singapore (company owned retail shops), Malaysia & Hong Kong (franchise business model) Marketing Challenge The company wants to expand their business in South Asian Market and Vietnam has been selected for their future business expansion Financial Overview Sales Revenue (2011) $58.6m Operating Profit $5m Sales Revenue (Country & Product category wise) Singapore Malaysia Hong Kong Sunglasses $7.7m $1 2.4m $5.1m Spectacles $5.3m $11.7m $6.7m External Analysis Bright Eyes needs to conduct macro environment audit such as PESTLE of the country in order to explore business opportunities of the country (Elearn, 2012, p. 75). PESTLE PESTLE analysis helps companies to get a picture in terms of macro environmental perspective (Henry, 2008, pp. 51-56). Political CPV or Communist Party of Vietnam has recently changed their industrial policy and has taken progressive approach for industrial development. The government is also supporting foreign players to invest in the country hence Bright Eyes will get support from government to expand their business. Country trend suggests that CPV has not faced any major threat from opposition party in recent times hence from the view point of political stability the country is going strong. Economic The country is suffering from high inflation rate hence overall GDP growth is slow for Vietnam (Tucker, 2010, pp. 194-195). Foreign players are investing in developing industrial park in the country. Vietnam has attracted foreign players to invest $271m in industrial projects last year. Government has sanctioned nine FDI projects worth of $112bn for next three years (Pham, 2004, pp. 69-97). Economic situation of the country is positive for companies like

Artefacts (Idenitifying Materials) Essay Example | Topics and Well Written Essays - 1750 words

Artefacts (Idenitifying Materials) - Essay Example Quality assurance plays an active role in materials testing. Quality assurance carries out calibration test and manufacturing test. Calibration test is carried out to determine the workability of the measuring properties. Manufacturing test is carried out to determine the adherence to standardization. Some of the tests through which engineering materials are subjected to can be categorized as follows; Mechanical test These tests are carried out on the engineering materials to determine their strength, elastic constants, material properties and performance properties. Mechanical test involves testing the test specimen. The test specimen is obtained through the breaking of the original sample materials. Tensional test These are quasi-static test carried out on the engineering materials to determine the properties of these materials. This test is carried out by exposing the sample material to uniaxial loading conditions. Hardness test The resistances to penetration by most engineering m aterials are determined by hardness test. Hardness test is carried out using three different methods that are scratch test, rebound test, indentation test. Scratch test involves progressive scratching of a hard material. Rebound test involves the ability to determine the material resilience. Material resilience is determined by measuring the potential energy of the material. Indentation test involves production of a permanent impression on the surface of the materials. The size and the force of the impression determine the material hardness. Hardness test can further be categorized as macro hardness test or micro hardness test. The macro hardness test includes Brinell, Vickers and Rockwell tests. Micro hardness test includes Knoop and Tukon test. Torsion test This is a test that is carried out to obtain the stress-strain relationship for the metal. Torsion test is able to generate both the shear stress and shear strain of the material. Impact test This test is conducted to determine the static properties and the mechanical behavior of the engineering material. Heat treatment and the stress concentrations of the material are carried out under impact test. When the drill (Osaki CD 1202 12V) was stripped down, the following material components were identified; Polystyrene polymer Polystyrene is a vinyl polymer structurally made up of very long hydrocarbon chain. The polystyrene polymer has a phenyl group attached to its carbon atom. This polymer is hard and it is a clear plastic. Polystyrene is made from free radical vinyl polymerization of the monomer styrene. This polymer is hard and it is used for making of the drill handle. Due to its hard nature, it is able to withstand a lot of heat. Nylon This is a thermoplastic polymer which is used as a fiber. Nylons are made from diacid chlorides and the diamines.One of the nylon polymer that is nylon 6, 6 is produced from the combination of two monomers. The two monomers are adipoyl chloride and hexamethylene diamine. Another form of nylon is the nylon 6. Nylon 6 is made from a ring opening polymerization. These nylons are used in making the casing for the ball bearing. It is preferred because of its ability to be fabricated into any shape. Diamond This is the hardest engineering material. They are gem-quality carbon crystals. They are highly refractive crystalline carbon types and are used in abrasives, cutting tools and drilling operations. They have slight impurities in their structures, but they are

Importance of Internal Relations Case Study Example | Topics and Well Written Essays - 250 words

Importance of Internal Relations - Case Study Example Generally it is the role of the school to form the best and most appropriate way of informing the public on their school system and the operations of the institution. The community which is part of the stakeholders has the massive role of bringing up and sustaining collaborative measures and means which lead to the success of the institutions and the eventual improvements of the institution. The result of the collaborative actions of the public and the school should yield students success and the improvement of the institution in all round perspective without compromising any aspect of an institution. The administrator of an institution should have the knowledge and understanding of building of consensus effectively and negotiation techniques, effective and good communication skills, he should believe in values and have an inclusion of all the members of the community. Most importantly the administration should ensure that all the visions and missions of the institution are effectively and clearly communicated to the parents, students, staff and the community members .The vision of the institution should be developed with among the community members and all the stakeholders. Generally the administrator is the head of the institution hence should nurture and promote the success of the learning process. This can be achieved through the advocating for effective learning mechanisms and friendly environment for the students and the staff to perform to their best potentials. Internal communication is very critical and determines the depiction of the staff and the students on how they perceive the ability of the institution to meet its goals and targets. Effective leads to the success of an institution due to the constant flow and sharing of information that will eventually lead to

Angel in the House Virgina Woolf Essay Example for Free

Angel in the House Virgina Woolf Essay Virginia Woolf was an English author. She was a feminist, publisher, essayist and critic. Woolf commonly acquired female authors Jane Austen and Charlotte Bronte. Woolf analyses women and their struggles as artists, their position in literary history and need for independence in her works of literature. Woolf’s short story â€Å"Angel in the House† has a deeper meaning then just a female author sharing pointers and stories on how she succeeded in her career to another woman trying to become a successful professional. These personal encounters and struggles the author in the story discusses with the other women really expose how different women’s jobs were compared to men’s and recognizes that different approaches needed to be made by a woman in order to do well in her occupation. Women were not to be perceived more superior then men. The speaker of the story is clearly Virginia in Angel in the House. The fact that the woman in the story is a female author and continuously compares how being a professional woman is much harder than people consider it to be makes it obvious to the reader that Virginia is the speaker. She begins by sarcastically discusses how easy being a writer is and that she is unsure why someone recommended her to give advice on being successful because her job is so easy. The author says â€Å"to show how little I deserve to be a professional women how little I know of the struggles and difficulties of such lives, I have to admit instead of spending that sum upon bread, butter, shoes, stockings or butcher bills I went out and bought a cat†(Woolf 109). She starts off speaking about her profession in this manner because that was how men felt about being a professional author or critic or essayist. They believed that job was stress free. If this was an author who was a man though he would of never spoke down about his job like that, but it was strictly because it was a women doing it that they felt this way about the work. After talking down about being a professional writer, the author starts to tell the other women about inner struggles she has had to deal with in her profession. Inner struggles that men were not aware of. She speaks about a phantom that she named the Angel in the house. This phantom angel is referred to by the author as a she; she was pure, charming, sympathetic and selfless. This angel in the house made the author feel guilty for having her own opinion and personality. This phantom wanted every woman to have the same morals and characteristics as her. She was what a woman was supposed to act and think like especially in a society that men ruled. The narrator states she had to kill the angel in the house in order to become successful in her profession. This angel wanted to pluck the heart out of the authors writing, basically wanting it to be average and appropriate incase a man were to read it. Women who obeyed the angel and acted as she did were not living up to their fullest potential, allowing the men to outshine them. A woman does not discover herself until she kills that phantom, forgets what men expect and follow her own intuition no matter what.

Cultural Tasks Essay Example for Free

Cultural Tasks Essay The main idea of postmodernism is that individuals from different cultures have diversified values, beliefs, interests and perspectives, which are not the result of natural human nature, but have been formed by human history and culture. Since at a particular time knowledge is socially constructed in different ways, it changes through time from one context to another. As a result, there is no universal truth about anything; it should rather be perceived as multiple truths about different issues and concepts, which are unique and applicable for particular situations and cultures. The current chapter discusses the principles of postmodernism as applied to educational changes offered by modern grand schemes and programs. According to postmodernists, the best way to achieve changes is to combine collaborative and individual efforts. Though, since changes are usually unpredictable, there is no guarantee whether those efforts and actions can lead to real success and progress. Moreover the perception of the progress differs among nations and cultures: while one culture can consider change as positive leading to progress, the other might consider such change as negative or undesired. If to apply postmodernism ideas toward the educational programs and changes they can bring, it is possible to state that there is little guarantee that contemporary grand schemes and programs such as No Child Left Behind can bring positive change to schools and educational changes. On the contrary, the changes within the school premises are usually caused and influenced by the local context, which might include the school culture, the environment of the community the school is located and other aspects. To determine what can bring educational change might require certain research by gathering and analyzing the related information. Since self-study, according to postmodernism, is socially constructed knowledge, it cannot be applied without the external data of multiple values and experiences. Nowadays many schools try to implement uniformity in cloths, lectures, teaching methods, attitude to students, etc. From one side, it is a good way of making all students feel as they all have equal chances to learn and show their knowledge. However, according to postmodernists, such uniformity might suppress the personality and opportunities of some more progressive students. That is why, it is very important to identify what should be the same in the schools and what should be different. To ensure positive and effective changes in schools it is important to organize the combined work between school members by applying innovative and latest teaching methods in order to change teaching and learning for all the students. Change in the school teaching methods can be achieved if every teacher realizes the importance of personal professional learning, the results of which can mature the learning process of the students. It is not mandatory that such learning should be professional and obtained from some learning center or university. It can start from everyday desire to learn something new and apply this new knowledge into the teaching process. Moreover, every teacher should try to understand each students way of perceiving information, recognize and respect diversity among students and apply the teaching methods, which will be the most effective for all the students in the classroom. Finally, collaborative learning stimulates deep thinking about teaching and learning while ensuring that the students are prepared to be active participants in a global world. There can be large number of different educational programs, teaching styles, and advices, but the most important thing is to realize and understand the individual nature of each student and his/her demands in the education. Friendly, trustful and respectful teachers/students relationships are the most important factor in the teaching process. Works Cited: Part 6: Cultural Tasks of Supervision, pp. 414-456.

Affect The Flight Time Of A Balsa Wood Glider Engineering Essay

Affect The Flight Time Of A Balsa Wood Glider Engineering Essay The purpose of the study was to obtain the relation between the flight time and launch height of a balsa glider of different wing area. Also the effect on flight time was recorded by changing the position of wings of the gliders. In the first part of the experiment the gliders were launched from four different height and the flight time was recorded. The experiment was conducted in a closed area in order to avoid the effect of wind. The gliders used were of different size and shape. The net acceleration acting on the gliders was also calculated by plotting a graph between square of flight time and launch height. In the second part of the experiment the wing positions for three different gliders were changed and the flight time was recorded in each of the three gliders. The launch height of the gliders was kept same. It was observed that as the launch height was increased the flight time also increased. The amount of lift could only be analysed by calculating the net acceleration in each of the gliders. The variation in the flight time due to the change in the position of wings was interpreted in terms of the increase in the loops made by the gliders i. e. the instability of the gliders or the moments of wings. Word Count- 221 Table of Contents- Introduction _______________________________________ 4 Theory _______________________________________ 5 Experimental Set up ___________________________________ 9 Method ________________________________________ 10 Result _________________________________________ 12 Conclusion _________________________________________ 22 Limitations ________________________________________ 23 Unresolved Questions _________________________________ 24 Bibliography ______________________________________ 25 Appendix ______________________________________ 26 Introduction: The question about aerodynamics has great importance in todays times and the various factors that affect aerodynamics of an aircraft or a glider is necessary in order to improve the efficiency. By taking up this experiment I have tried to analyse the acting forces on the flight of a glider such as the lift, the drag, and the weight also the theory of projectile motion plays an important role in determining the flight time of a glider. Also keeping in mind the laws of physics that relate to flight and also checking whether they are in accordance with the following experiment. Research Question- How the Wing area and position along with the launch height affect the flight time of a Balsa Wood Glider? The aim of this experiment is to study how the wing area, wing position and the launch height have an impact on the flight time of a balsa glider. In order to establish the above relations balsa gliders of different wing areas were hand launched from different heights and the flight time was recorded. As for the relation with the wing position and flight time they were hand launched as well from a fixed height with different wing positions. Theory The experiment is based on the theory of projectile motion, and fluid mechanics. Projectile Motion- When a body is projected from a certain height with a certain velocity, the acceleration on the body acts only in the downward (along y-axis) direction and the acceleration along the horizontal (x-axis) direction is zero. Since the air resistance also affects the horizontal motion of the body there might be some deceleration, however it is of a very small magnitude hence it can be neglected in the case of my experiment. The same explanation can be taken for the wind; there might be some component of the force (due to wind) which might affect the horizontal motion as well. This is also taken care of in the experiment as it is conducted in closed area. As the initial velocity is in the horizontal direction its vertical component is zero, by taking the vertical motion the flight time of the glider should depend on the height from which it has been launched and the net acceleration in the vertical direction. By varying the height in the experiment the flight time should also vary proportionately. As the height is increased the flight time should also increase. Due to the change in the forces acting on the glider the net acceleration can also change which in turn will affect the flight time. Forces There are three types of forces acting on a glider which are: Weight The weight of the body always acts in the downward direction. The weight of a body dependent on the mass of the glider and gravitational acceleration which can be taken constant for a given space. By Newtons 2nd law of motion, the weight is given by (F = mg) It  is the  force  due to the gravitational attraction of the earth on the  glider. But this force weight, which is the gravitational force, is different from the aerodynamic forces,  lift  and  drag.  The lift and the drag are mechanical forces that will act on the glider only when it is in physical contact with air which generates these forces. The gravitational force or weight is a field force; and is a non-contact force. The gravitational force (weight) between two objects depends on the masses of the two objects and the inverse of the square of the distance between these objects. The more the masses of the objects the stronger is the attraction, and closer the object are the stronger the attraction. Lift Lift is the  force  that acts on a glider upward and helps the glider to stay in the air. The lift on the glider is mainly generated by the wings.  It is an  aerodynamic  force produced by the motion of air flowing through the glider. Lift acts through the  centre of the pressure  of the glider and is in the direction which is normal to the flow of air. Lift occurs when a flow of air is  turned  by the glider. According to Newtons 3rd law of motion the flow of air is turned in one direction, and the lift is generated in the opposite direction. Since air is a  gas  and its molecules are free to move about, any solid can deflect its flow. For an  air foil, both the upper and lower surfaces plays part in turning the flow of air. There can be two types of lift static and dynamic lift. Static lift according to the Archimedes principle whenever a body is immersed in a fluid it experiences an upward force called the buoyant force. The factors on which the buoyant force depends are: The volume of the fluid displaced and; the density of the fluid. If the area of the wings increases the magnitude of the lift should also increase and the net force acting on the glider should also increases which in turn should decrease the net acceleration and hence increase the flight time. Dynamic Lift According to Bernoullis principle; the dynamic lift is due to the difference in pressure on the two sides of the body which is due to the difference in the speed of the air on the two sides of the body. There are various factors that affect lift, these are: Aircraft  wing geometry  has a large effect on the amount of lift generated. The shape and size of the wing will have a significant impact on the amount of lift generated. In order to generate lift there must be some velocity; hence if the object is moved in air then lift will be generated. Lift also depends on the  mass  of the flow.it also depends in a major way on the viscosity and compressibility of air. Viscous force or Drag Viscous force is a mechanical force. The drag, like lift, is also produce by the interaction and contact of a solid body with air. For drag to be produce, the solid body must be in contact with the air. If there is no air, there is no drag. Drag is generated by the difference in the speeds of the solid object and the air. There must be relative motion between the object and the air. If there is no relative motion, there is no drag. Viscous force always opposes the motion, hence it will be opposite to the motion of the glider. The most of the factors affecting drag is same as that affecting the lift. The viscous force can be taken as the aerodynamic friction, and one of the sources of this force is the  skin friction  between the molecules of the air and the solid surface of the glider. Since the skin friction is an interaction between a glider and the air, the magnitude of the skin friction depends on properties of both glider surface and air. The smooth, waxed surface of glider will produces less skin friction than a rough surface. And for the air, the magnitude of skin friction depends on the  viscosity  of the air. The relative magnitude of the viscous forces to the motion of the flow of air is called the  Reynolds number. Also the drag can be taken as the aerodynamic resistance to the motion of the object through the air. This source of drag depends on the shape of the glider and is called  form drag. When air flows around a body, the velocity and  pressure around the glide  are changed. The pressure is a measure of the momentum of the air molecules and a change in momentum results in a  force, a change in pressure will produce a force on the body. This component of the aerodynamic force that is opposed to the motion is the drag. Viscous force directly depends on the  mass  of the air flow going past the glider. Effect of Height on Flight time In this experiment there will be 4 heights taken but the height intervals will not be uniform in order to check the trend and see if there is a clear distinction in readings of flight time. As the height increases the flight time should increase as there is more distance to cover for the glider and since there are no forces acting as mentioned above, it is only the height that acts as a factor to change flight time. Wing area In this experiment the wing area should effect the flight time of the gliders as seen earlier, lift has a direct connection with wing area. As the lift increases the glider goes higher in the air thus increasing the flight time. Every glider has a different wing area and this makes a clear distinction between the flight times for the gliders. Wing position The wing position whether towards the front or back determines the stability of the glider while its flight. The more it is to the front of the socket it tends to do be more unstable and has a very loopy flight which increases the flight time. The further behind the wing is in the socket the more stable the glider is as the weight is more towards the centre of mass, making it more stable and also increasing the flight time however whilst a straight and balanced there are other forces acting on the flight that might pull it down to the ground. Experimental Set up Gliders In the initial stages the glider used were made by hand using a template to cut out the parts of the gliders. However these gliders lacked perfect stability and the edges had to be rubbed and smoothened in order to use make them completely aerodynamic. Even after doing so they lacked perfect specification and the material used to make them was not the right material hence they did not glide as required to. The gliders used in the later stages were bought online from amazon.com. These gliders are laser cut and ready to fly. They are made out of balsa wood. There were 3 types of gliders that were used, the parts of the gliders were precisely cut and well balanced in order to obtain a decent flight. The 3 gliders varied differently in shape, size, and weight and wing area. Area The experiment was conducted in a closed environment. The length of the place was approximately 35meters and width approximately 12meters. The wind factor was controlled as all doors and windows were shut and it was an enclosed area. Method: There were various steps that are involved in this experiment. The glider was first flown in an outdoor environment to check the flight. Since the wind factor cannot be controlled in an outdoor environment it was not possible to conduct the experiment outside as this affects the flight time. Hence an indoor area was chosen. Measuring Wing Area The gliders that will be used which each have different shape, size, and wing area. The first variable wing area cannot be measured by a given formula as it is not a uniform shape and cannot be broken into smaller shapes. The wing area will be measured by keeping the entire wing on a graph sheet whose each square area is known and the outline of the wing shape will be sketched out on the graph sheet. After having done this the number of complete squares of the known area that are enclosed by the outline of the wing will be counted. After this the number of incomplete squares and an approximation will have to be made as it is not feasible to calculate area of a fraction of the square. This will be done for each of the three gliders and will be noted down. In order to make sure that the gliders, glide properly without any hindrance and technical difficulty a test flight will be done. If there seem to be any technical problems with any of the three gliders they shall be fixed at first in order to provide accurate and legit readings. If any parts seem to be broken they will be fixed by the special adhesive which is used to stick balsa wood. Launching and Measuring Flight Time Keeping in mind that there are no holders or launching devices provided with these gliders they will have to be launched by the free hand as it is not possible to devise a launch method. This is so because making any alterations to the glider might distort its stability and will cause unequal weight balance. Although launching from the hand will have uncertainties such as different launch force and height it will be controlled as much as possible. In order to make sure the height is constant a plumb line will be taken and held from the comfortable launch height. This plumb line will then be measured in accordance with a metre scale. Every time a glider is launched it will be launched from the same height as the thread will be held at that height while launching the glider. The second height from which it will be launched will be after standing on top of a dining table. The height of the table will be measured by using another plumb line and will be measured in accordance to the metre scale. This height of the table will be added to the initial launch height and then a plumb line of that height will be held while the launch. The fourth height will be from approximately the first floor. The vast interval difference is taken so that there can be a clear trend that can be observed for the flight time. The height of the wall will be measured and then the initial height of the launch will be added to this. The fourth height will be from above a table on the first floor, in order to obtain this height the same step as the one for the table height on the lower level will be used. Taking the readings of the flight time will be done by using a stopwatch. It is not possible for me to do this alone as starting the stopwatch and launching the glider is not possible at the same time hence a little assistance will be required to measure the flight time. The assistant will start recording the flight time as soon as the glider leaves contact with my hand and will stop as soon as the glider touches the ground. These readings will then be recorded in a table. There will be 5 readings taken for each glider at each height. There are two wing positions possible either at the front of the socket or the back. Each gliders wings will be adjusted as front and back and for each position there will be 5 readings taken. Result Flight time for different launch height (Experiment1) Glider A Height (meters) Flight time ( ± 0.01s) Average( ±0.13s) t2(sec2) 1.5 1.64 1.78 1.84 1.81 1.74 3.03 2.24 2.21 2.45 2.53 2.28 2.38 5.68 5.3 3.83 3.94 4.04 3.72 3.91 15.29 6.04 4.81 4.63 4.79 4.84 4.75 22.54 As the height was increased the flight time also increased. Since the glider used was the same and the speed with which it was projected also remained same the lift experienced by the glider did not change. The increase in the time was only due to the increase in height from which the glider was projected. Glider B Height (meters) Flight time ( ± 0.01s) Average ( ±0.13s) t2(sec2) 1.5 1.25 1.5 1.28 1.3 1.32 1.75 2.24 1.54 1.68 1.61 1.59 1.61 2.60 5.3 2.19 1.94 2.4 2.33 2.24 5.02 6.04 2.78 3.01 2.92 2.84 2.87 8.26 The trend between the time of flight and the height of launch was same as in the case of glider A. Glider C Height (meters) Flight time ( ± 0.01s) Average ( ±0.04s) t2(sec2) 1.5 0.93 0.90 1.01 0.94 0.94 0.88 2.24 1.16 1.21 1.24 1.19 1.19 1.43 5.3 1.57 1.59 1.55 1.6 1.57 2.45 6.04 2.1 2.12 2.11 2.15 2.11 4.47 The trend between the time of flight and the height of launch was same as in the case of glider A and B. hence in all the 3 cases it was found that the lift experienced by the gliders did not depend upon the height of projection or the height at which the glider was flying. As seen above in the graph the flight time of the glider is directly proportional to the height from which it has been released i.e. as the height increases the time taken by the glider to touch the ground also increases. According to the equation of motion 2 If we are considering the vertical motion then the initial speed in the vertical direction will be taken as zero then 2. The three forces acting on the glider that is the weight (mg), the drag and the lift are all constant. Since the drag and the lift depend on the speed of the glider it is not changing as the speed in all the cases are constant. The above relation can be made linear by plotting a graph between t2 and h from this graph the net acceleration acting on the glider can be calculated by measuring the slope of the graph. Since: 2. Net acceleration = 2/slope of the curve. Measurement of net acceleration of the glider. Glider A Calculation Net acceleration = 2/slope of the curve. Net acceleration in glider A= = 0.51 m/s2 Glider B Calculation Net acceleration = 2/slope of the curve. Net acceleration in glider B= = 1.62 m/s2 Glider C Calculation Net acceleration = 2/slope of the curve. Net acceleration in glider B= = 3.1 m/s2 The Glider C has the maximum acceleration and Glider A has the least. This also means that Glider C comes to rest much more quickly than Glider A, Glider A also has a longer glide time and this is because of the light weight and wing span that makes it more stable while in flight. Whereas when compared to Glider C the weight is much more and the glider isnt stable enough to stay in air for a long time. The wing area also has a significant impact on this as the more wing area means more lift however when we check Glider C has more wing area but it yet doesnt get enough lift; this explains that this glider needs more thrust when launched in order to stay in air longer than the others. Wing Position (Experiment2) Glider A Wing Position for A Position Flight time ( ± 0.01s) Average ( ±0.11secs) Front 1.72 1.85 1.9 1.68 1.77 1.78 Back 1.56 1.59 1.72 1.6 1.52 1.60 As seen above when the wings of the glider are moved forward the flight time of the glider increases. Glider B Wing Position for B Position Flight time ( ± 0.01s) Average ( ±0.11secs) Front 2.13 2.06 2.24 2.17 2.26 2.17 Back 1.66 1.59 1.63 1.82 1.74 1.69 As in the case of Glider A when the wings of the glider are moved forward the flight time of the glider increases. Glider C Wing Position for C Position Flight time ( ± 0.01s) Average ( ±0.12secs) Front 1.47 1.59 1.61 1.53 1.42 1.52 Back 0.84 0.79 0.94 0.75 1.03 0.87 As in the case of Glider A and B when the wings of the glider are moved forward the flight time of the glider increases. As we can see in the graphs above the time taken for the flight when the wings are in front is more than the flight time for when the wings are pushed backward. This is because the forward wings make the flight of the glider much more unstable and cause it to loop more. The looping increases the time to touch the ground as it causes sudden immediate lift and this also increases the horizontal gliding time. Whereas on the other hand when the wings are pulled back in order to make the flight more stable but there are still other factors like weight, lift and drag that are constantly affecting the flight and causing it to descend. Conclusion: In the first part of the experiment it was found that the flight time increases as the height from where the glider was launched increases. I each of the 3 gliders the lift acting on the glider did not change as the area of the glider remained same. The change in the flight time was only due to the change in the height of the launch. The graph between the height of launch and the square of flight time gave the measure of the net acceleration in case of the 3 gliders which in turn could be interpreted in terms of net force acting on the glider. The measure of net acceleration showed that the lift produced in case of the gliders increased with the wing area. The lift in the case of Glider A was found to be maximum, as it had the maximum wing area. In the second experiment the flight time increased as the wing position was shifted towards the front of the glider. The trend obtained in all the three gliders was the same. The increase in the flight time was due to the increase in lift that made the glider shoot up and loop in the air. As the glider looped in the air the height also increased which in turn increased the flight time. The increase in the loop made by the glider was due to the instability produced by shifting the weight of the wing to the front of the plane. Limitations: There were several limitations while doing this experiment. The first and what can be considered as the most important is the launching technique. This is the most crucial part of this experiment and all the readings depend on this. Due to the fragile bodies of the glider there could be no launching technique devised that would make sure the force on each launch is the same. The lift acting on the glider depends on the speed of the glider; if the launching speed varies it can affect the lift experienced by the glider. Since it is not possible to neither control the force used for each launch nor measure it, it was the biggest drawback. When the experiment was conducted in the open the wind factor had a great impact on the flight of the gliders. The gliders being light and very fragile, the wind outdoor was drifting the gliders into different directions and also slowing them down. This change in path and time was random and unpredictable. Since it was not possible to control the wind the experiment was carried out in a closed environment, yet there was some wind that affected the flight and caused slight deviation in path which could have possibly increased the flight time or even decreased it. Since the gliders were launched several times and had not landing mechanism as well, the rough landing chipped quite a few parts of the gliders that made the flight for the later readings relatively unstable and defective. Unresolved Questions: The effect of projection velocity on the flight time was not clear as the gliders were launched with almost same velocity and force. The speed with which the glider moves effects both the lift on the glider and the drag acting on it. The two forces are very important in deciding the flight of the gliders. By designing a proper launch mechanism the effect of speed or the launch force would had been studied. Also the efficiency of the flight could have been studied by measuring glide ratio i.e. the ratio of the horizontal distance travelled and the loss of height travelled in a given time. Bibliography Books: Giancoli, Douglas C., Physics Principles with Applications, 6th Edition, Pearson Education Limited, NJISBN: 0-13 -184661-2-1. Nelkon, M., and Parker, P., Advanced level Physics, London, Heinemann Educational Books Limited, ISBN 0-435-68636-4 Tsokos, K., A., Physics for the IB Diploma,5th Edition, Cambridge University Press, ISBN 978-0-521-13821-5 Websites: http://en.wikipedia.org/wiki/Gliding_flight http://www.grc.nasa.gov/WWW/k-12/airplane/glider.html http://www.skysailing.com/pages/theory.htm http://web.mit.edu/16.00/www/aec/flight.html