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2 Pages

ITGS SL P1

Course: IB 2002, Fall 2000
School: InterAmerican Barranquitas
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BACCALAURAT INTERNATIONAL BACHILLERATO BACCALAUREATE INTERNATIONAL INTERNACIONAL M01/390/S(1)M MARKSCHEME May 2001 INFORMATION TECHNOLOGY IN A GLOBAL SOCIETY Standard Level Paper 1 2 pages 2 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. A D C B D B C D A C B B D B B/C 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. C D C B A/D A A C D C B A/B A A 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41....

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BACCALAURAT INTERNATIONAL BACHILLERATO BACCALAUREATE INTERNATIONAL INTERNACIONAL M01/390/S(1)M MARKSCHEME May 2001 INFORMATION TECHNOLOGY IN A GLOBAL SOCIETY Standard Level Paper 1 2 pages 2 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. A D C B D B C D A C B B D B B/C 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. C D C B A/D A A C D C B A/B A A 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42. 43. 44. 45. D C C A B B C A A 46. 47. 48. 49. 50. 51. 52. 53. 54. 55. 56. 57. 58. 59. 60. M01/390/S(1)M -
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InterAmerican Barranquitas - IB - 2002
InterAmerican Barranquitas - IB - 2002
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Brandon Ivey Sex and Controversy Final Draft I choose to compare "The Eye of the Beholder: Parents View on Sex of Newborns", and "The Scientist in Society". I liked the way both of those articles addressed issues in society that not everybody might b
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Is Natural Selection Natural 1 Brandon Ivey Writing 2 Susan Cook 11-6-06 Is Natural Selection Natural? America has always been an advocate for human equality and considers its borders to be a safe haven of tolerance and understanding. This is ironic
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Asset PricingJohn H. CochraneJune 12, 20001AcknowledgmentsThis book owes an enormous intellectual debt to Lars Hansen and Gene Fama. Most of the ideas in the book developed from long discussions with each of them, and trying to make sense of
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CHAPTER 1The Role of the Public Accountant in the American EconomyReview Questions 1-1 The "crisis of credibility" largely arose from the number of companies that restated their previously issued financial statements as a result of accounting irr
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CHAPTER 2Professional StandardsReview Questions 2-1 The Sarbanes-Oxley Act of 2002 created the PCAOB and gave this body authority to develop auditing standards for the audits of public companies. The AICPA has the authority, based on general acce
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24. The metric prefixes (micro (), pico, nano, . . .) are given for ready reference on the inside front cover of the textbook (see also Table 1-2). The surface area A of each grain of sand of radius r = 50 m = 5010-6 m is given by A = 4(50 10-6 )2 =
UniFor - PHY - 025
25. From the Figure we see that, regarding differences in positions x, 212 S is equivalent to 258 W and 180 S is equivalent to 156 Z. Whether or not the origin of the Zelda path coincides with the origins of the other paths is immaterial to considera
UniFor - PHY - 026
26. The first two conversions are easy enough that a formal conversion is not especially called for, but in the interest of practice makes perfect we go ahead and proceed formally: (a) (11 tuffet) (b) (11 tuffet) (c) (5.5 bushel) 36.3687 L 1 bushel
UniFor - PHY - 027
27. We make the assumption that the clouds are directly overhead, so that Figure 1-3 (and the calculations that accompany it) apply. Following the steps in Sample Problem 1-4, we have t = 360 24 h which, for t = 38 min = 38/60 h yields = 9.5 . We o
UniFor - PHY - 028
28. In the simplest approach, we set up a ratio for the total increase in horizontal depth x (where x = 0.05 m is the increase in horizontal depth per step) x = Nsteps x = 4.57 0.19 (0.05) = 1.2 m .However, we can approach this more carefully by no
UniFor - PHY - 029
29. Abbreviating wapentake as "wp" and assuming a hide to be 110 acres, we set up the ratio 25 wp/11 barn along with appropriate conversion factors: (25 wp)100 hide 1 wp 110 acre 1 hide 110-28 m2 1 barn 4047 m2 1 acre(11 barn) 1 1036 .
UniFor - PHY - 030
30. It is straightforward to compute how many seconds in a year (about 3 107 ). Now, if we estimate roughly one breath per second (or every two seconds, or three seconds it won't affect the result) then to within an order of magnitude, a person tak
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Cinco de Mayo!The Sun! nuclear fusion (producing energy) energy transport in the sun hydrostatic equilibrium sun's atmosphereMidterm Info Will be returned to your slots either today or Wednesday note that the scantrons will have the correct answ
UniFor - PHY - 031
31. A day is equivalent to 86400 seconds and a meter is equivalent to a million micrometers, so (3.7 m)(106 m/m) = 3.1 m/s . (14 day)(86400 s/day)
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5/7/08Sun's corona Determining distances to stars (parallax) Flux and luminosity Determining stellar masses Watch out for all the crazy units we'll use today! (And for all the equations) Uppermost layer of sun's atmosphere: 104 105 km above photo
UniFor - PHY - 032
32. The mass in kilograms is (28.9 piculs) 100 gin 1 picul 16 tahil 1 gin 10 chee 1 tahil 10 hoon 1 chee 0.3779 g 1 hoonwhich yields 1.747 106 g or roughly 1750 kg.
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5/9/08Calculating Stellar Masseswatch out for units and equationsMass-Luminosity Relation Colors and Temperatures of Stars Main Sequence Sizes of StarsDEMO Two light bulbs have same luminosity Their flux (apparent brightness) changes dependin
UniFor - PHY - 033
33. (a) In atomic mass units, the mass of one molecule is 16 + 1 + 1 = 18 u. Using Eq. 1-9, we find (18 u) 1.6605402 10-27 kg 1u = 3.0 10-26 kg .(b) We divide the total mass by the mass of each molecule and obtain the (approximate) number of wate
UniFor - PHY - 034
34. (a) We find the volume in cubic centimeters (193 gal) 231 in3 1 gal 2.54 cm 1 in3= 7.31 105 cm3and subtract this from 1 106 cm3 to obtain 2.69 105 cm3 . The conversion gal in3 is given in Appendix D (immediately below the table of Volume
UniFor - PHY - 035
35. (a) When is measured in radians, it is equal to the arclength divided by the radius. For very large radius circles and small values of , such as we deal with in this problem, the arcs may be . . approximated as . R Sun .. . . . . . . . . . . . .
UniFor - PHY - 036
36. (a) For the minimum (43 cm) case, 9 cubit converts as follows: (9 cubit) 0.43 m 1 cubit = 3.9 m .And for the maximum (43 cm) case we obtain (9 cubit) 0.53 m 1 cubit = 4.8 m .(b) Similarly, with 0.43 m 430 mm and 0.53 m 530 mm, we find 3.9
UniFor - PHY - 037
37. (a) Squaring the relation 1 ken = 1.97 m, and setting up the ratio, we obtain 1 ken2 1.972 m2 = = 3.88 . 2 1m 1 m2 (b) Similarly, we find 1.973 m3 1 ken3 = = 7.65 . 1 m3 1 m3 (c) The volume of a cylinder is the circular area of its base multiplie
UniFor - PHY - 038
38. Although we can look up the distance from Cleveland to Los Angeles, we can just as well (for an order of magnitude calculation) assume it's some relatively small fraction of the circumference of Earth which suggests that (again, for an order of
UniFor - PHY - 039
39. Using the (exact) conversion 2.54 cm = 1 in. we find that 1 ft = (12)(2.54)/100 = 0.3048 m (which also can be found in Appendix D). The volume of a cord of wood is 8 4 4 = 128 ft3 , which we convert (multiplying by 0.30483 ) to 3.6 m3 . Therefo
UniFor - PHY - 040
40. (a) When is measured in radians, it is equal to the arclength s divided by the radius R. For a very large radius circle and small value of , such as we deal with in Fig. 1-9, the arc may be approximated as the straight line-segment of length 1 A
UniFor - PHY - 041
41. We reduce the stock amount to British teaspoons: 1 breakfastcup 1 teacup 6 tablespoons 1 dessertspoon = 2 8 2 2 = 64 teaspoons = 8 2 2 = 32 teaspoons = 6 2 2 = 24 teaspoons = 2 teaspoonswhich totals to 122 teaspoons which corresponds (s
UniFor - PHY - 042
42. (a) Megaphone. (b) microphone. (c) dekacard ("deck of cards"). (d) Gigalow ("gigalo"). (e) terabull ("terrible"). (f) decimate. (g) centipede. (h) nanonannette. ("No No Nanette"). (i) picoboo ("peek-a-boo"). (j) attoboy ("at-a-boy"). (k) Two hect
UniFor - PHY - 043
43. The volume removed in one year is V = 75 104 m2 (26 m) 2 107 m3 which we convert to cubic kilometers: V = 2 107 m3 1 km 1000 m3= 0.020 km3 .
UniFor - PHY - 006
6. (a) Using the fact that the area A of a rectangle is widthlength, we find Atotal = = = (3.00 acre) + (25.0 perch)(4.00 perch) (40 perch)(4 perch) (3.00 acre) + 100 perch2 1 acre 580 perch2 .We multiply this by the perch2 rood conversion factor
UniFor - PHY - 007
7. The volume of ice is given by the product of the semicircular surface area and the thickness. The semicircle area is A = r2 /2, where r is the radius. Therefore, the volume is V = 2 r z 2where z is the ice thickness. Since there are 103 m in 1
UniFor - PHY - 008
8. The total volume V of the real house is that of a triangular prism (of height h = 3.0 m and base area A = 20 12 = 240 m2 ) in addition to a rectangular box (height h = 6.0 m and same base). Therefore, V = 1 hA + h A = 2 h +h 2 A = 1800 m3 .(a)
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UniFor - PHY - 023
23. We introduce the notion of density (which the students have probably seen in other courses): = m Vand convert to SI units: 1000 g = 1 kg, and 100 cm = 1 m. (a) The density of a sample of iron is therefore = 7.87 g/cm31 kg 1000 g100 cm 1m
UniFor - PHY - 009
9. We use the conversion factors found in Appendix D. 1 acre ft = (43, 560 ft2 ) ft = 43, 560 ft3 . Since 2 in. = (1/6) ft, the volume of water that fell during the storm is V = (26 km2 )(1/6 ft) = (26 km2 )(3281 ft/km)2 (1/6 ft) = 4.66 107 ft3 .
UniFor - PHY - 022
22. The volume of the water that fell is V = = = = (26 km2 )(2.0 in.) (26 km2 ) 1000 m 1 km2(2.0 in.)0.0254 m 1 in.(26 106 m2 )(0.0508 m) 1.3 106 m3 .We write the mass-per-unit-volume (density) of the water as: = m 3 = 1 103 kg/m . VThe
UniFor - PHY - 021
21. We introduce the notion of density (which the students have probably seen in other courses): = and convert to SI units: 1 g = 1 10-3 kg. (a) For volume conversion, we find 1 cm3 = (1 10-2 m)3 = 1 10-6 m3 . Thus, the density in kg/m3 is 1g cm3
UniFor - PHY - 020
20. To organize the calculation, we introduce the notion of density (which the students have probably seen in other courses): m = . V (a) We take the volume of the leaf to be its area A multiplied by its thickness z. With density = 19.32 g/cm3 and m
UniFor - PHY - 014
14. The time on any of these clocks is a straight-line function of that on another, with slopes = 1 and y-intercepts = 0. From the data in the figure we deduce tC tB = = 2 594 tB + 7 7 33 662 tA - . 40 5These are used in obtaining the following res
UniFor - PHY - 018
18. We denote the pulsar rotation rate f (for frequency). f= 1 rotation 1.55780644887275 10-3 s(a) Multiplying f by the time-interval t = 7.00 days (which is equivalent to 604800 s, if we ignore significant figure considerations for a moment), we
UniFor - PHY - 017
17. The last day of the 20 centuries is longer than the first day by (20 century)(0.001 s/century) = 0.02 s . The average day during the 20 centuries is (0 + 0.02)/2 = 0.01 s longer than the first day. Since the increase occurs uniformly, the cumulat
UniFor - PHY - 015
15. We convert meters to astronomical units, and seconds to minutes, using 1000 m = 1 km 1 AU = 1.50 108 km 60 s = 1 min . Thus, 3.0 108 m/s becomes 3.0 108 m s 1 km 1000 m AU 1.50 108 km 60 s min = 0.12 AU/min .
UniFor - PHY - 010
10. The metric prefixes (micro (), pico, nano, . . .) are given for ready reference on the inside front cover of the textbook (also, Table 1-2). 1 century = = 10-6 century 52.6 min . 100 y 1 century 365 day 1y 24 h 1 day 60 min 1hThe percent differ
UniFor - PHY - 011
11. We use the conversion factors given in Appendix D and the definitions of the SI prefixes given in Table 12 (also listed on the inside front cover of the textbook). Here, "ns" represents the nanosecond unit, "ps" represents the picosecond unit, an
UniFor - PHY - 019
19. If ME is the mass of Earth, m is the average mass of an atom in Earth, and N is the number of atoms, then ME = N m or N = ME /m. We convert mass m to kilograms using Appendix D (1 u = 1.661 10-27 kg). Thus, ME 5.98 1024 kg N= = = 9.0 1049 . m
UniFor - PHY - 012
12. The number of seconds in a year is 3.156107 . This is listed in Appendix D and results from the product (365.25 day/y)(24 h/day)(60 min/h)(60 s/min) . (a) The number of shakes in a second is 108 ; therefore, there are indeed more shakes per secon
UniFor - PHY - 013
13. None of the clocks advance by exactly 24 h in a 24-h period but this is not the most important criterion for judging their quality for measuring time intervals. What is important is that the clock advance by the same amount in each 24-h period. T
UniFor - PHY - 016
16. Since a change of longitude equal to 360 corresponds to a 24 hour change, then one expects to change longitude by 360 /24 = 15 before resetting one's watch by 1.0 h.
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