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Course: PHY 028, Spring 2008
School: UniFor
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In 28. 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 noting if that there are N = 4.57/.19 24 rises then under normal circumstances we would expect N - 1 = 23 runs (horizontal pieces) in that staircase. This would yield (23)(0.05) = 1.15 m, which to two significant figures agrees with our first result.
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UniFor - PHY - 029
29. Abbreviating wapentake as &quot;wp&quot; 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
UCSB - ASTRO - 1
5/16/08Globular Clusters Stellar lifetimes Learning from star clusters Life after the main sequenceHe burning Red giantsOpen Clusters Note the blue stars Note the haze: blue light scattered by dust still in the open clusterGlobular Clusters
UCSB - ASTRO - 1
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)
UCSB - ASTRO - 1
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.
UCSB - ASTRO - 1
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 (&quot;deck of cards&quot;). (d) Gigalow (&quot;gigalo&quot;). (e) terabull (&quot;terrible&quot;). (f) decimate. (g) centipede. (h) nanonannette. (&quot;No No Nanette&quot;). (i) picoboo (&quot;peek-a-boo&quot;). (j) attoboy (&quot;at-a-boy&quot;). (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)
Texas San Antonio - ACCT - 4013
CHAPTER 3Professional EthicsReview Questions 3-1 An ethical dilemma is a situation that an individual faces involving a decision about appropriate behavior. Ethical dilemmas generally involve situations in which the welfare of one or more other i
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, &quot;ns&quot; represents the nanosecond unit, &quot;ps&quot; 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.
Texas San Antonio - ACCT - 4013
CHAPTER 4Legal Liability of CPAsReview Questions 4-1 There are several reasons why the potential legal liability of CPAs for professional &quot;malpractice&quot; exceeds that of physicians and other professionals. One reason is the vast number of people wh
Texas San Antonio - ACCT - 4013
CHAPTER 5Audit Evidence and DocumentationReview Questions 5-1 Audit risk is the possibility that the auditors may unknowingly fail to appropriately modify their opinion on financial statements that are materially misstated. It is composed of the p
Rutgers - STATS - 580
Introduction to Probability Models Ninth EditionThis page intentionally left blankIntroduction to Probability ModelsNinth EditionSheldon M. RossUniversity of California Berkeley, CaliforniaAMSTERDAM BOSTON HEIDELBERG LONDON NEW YORK OXFO
Texas San Antonio - ACCT - 4013
CHAPTER 21Internal, Operational and Compliance AuditingReview Questions 21-1 Internal auditing may be defined as an independent, objective assurance and consulting activitydesigned to add value and improve an organization's operations. It helps
Texas San Antonio - ACCT - 4013
CHAPTER 6Planning the Audit; Linking Audit Procedures to RiskReview Questions 6-1 In their investigation of a prospective client, the CPAs should assess the backgrounds and reputations of the prospect and its major shareholders, directors, and of
Texas San Antonio - ACCT - 4013
CHAPTER 7Internal ControlReview Questions 7-1 Internal control is a process, effected by the entity's board of directors, management and other personnel, designed to provide reasonable assurance regarding the achievement of objectives in the cate
Texas San Antonio - ACCT - 4013
CHAPTER 8Consideration of Internal Control in an IT EnvironmentReview Questions 8-1 System software monitors and controls hardware and provides other support to application programs. The computer's operating system and utility programs are import
Texas San Antonio - ACCT - 4013
CHAPTER 9Audit SamplingReview Questions 9-1 Nonstatistical sampling is an audit sampling technique in which the risk of sampling error is estimated by the auditors using professional judgment rather than by the laws of probability. Statistical sa
Texas San Antonio - ACCT - 4013
CHAPTER 10Cash and Financial InvestmentsReview Questions 10-1 The following circumstances might cause a client to understate assets: (1) (2) (3) 10-2 Management of a privately held company may be motivated to understate assets so as to minimize i
Texas San Antonio - ACCT - 4013
CHAPTER 11Accounts Receivable, Notes Receivable, and RevenueReview Questions 11-1 The term &quot;customer's order&quot; refers to the purchase order received from a customer. The term &quot;sales order&quot; refers to the document created upon receipt of a customer'
Texas San Antonio - ACCT - 4013
CHAPTER 12Inventories and Cost of Goods SoldReview Questions 12-1 Substantiation of the figure for inventories is an especially challenging task because of the variety of acceptable methods of valuation. In addition, the variety of materials foun
Texas San Antonio - ACCT - 4013
CHAPTER 13Property, Plant, and Equipment: Depreciation and DepletionReview Questions 13-1 Factors that facilitate the auditors' verification of plant and equipment but are not applicable to audit work on current assets include the following: (1)
Texas San Antonio - ACCT - 4013
CHAPTER 14 Accounts Payable and Other LiabilitiesReview Questions 14-1 Overstated earnings are associated with understated liabilities. To overstate earnings causes an overstatement of owners' equity. An overstatement of owners' equity must be acco
Texas San Antonio - ACCT - 4013
CHAPTER 15Debt and Equity CapitalReview Questions 15-1 A trust indenture is drawn to protect the position of bondholders by imposing restrictions upon the borrowing corporation. One of the most common of these restrictions is that the company mus
Texas San Antonio - ACCT - 4013
CHAPTER 16Auditing Operations and Completing the AuditReview Questions 16-1 Revenue accounts that are verified during the audit of balance sheet accounts are the following (only three required): Balance Sheet Item Accounts Receivable Notes Receiv
Texas San Antonio - ACCT - 4013
CHAPTER 17Auditors' ReportsReview Questions 17-1 The three paragraphs of the standard audit report for a nonpublic company are (1) introductory paragraph, (2) scope paragraph, and (3) opinion paragraph. The function of notes to financial statemen
Texas San Antonio - ACCT - 4013
CHAPTER 18Integrated Audits of Public CompaniesReview Questions 18-1 Section 404a requires that each annual report filed with the Securities and Exchange Commission include an internal control report prepared by management in which management ack
Texas San Antonio - ACCT - 4013
CHAPTER 19Additional Assurance Services: Historical Financial InformationReview Questions 19-1 This statement is incorrect. An audit can be a significant expense to a small company. The audit fee must be justified by the benefits received from th
Texas San Antonio - ACCT - 4013
CHAPTER 20Additional Assurance Services: Other InformationReview Questions 20-1 Assurance services are independent professional services that improve the quality of information, or its context for decision makers; attestation services are those a
NJIT - CS - 661
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Texas San Antonio - ACCT - 4013
Chapter 17 PracticeMultiple Choice Identify the letter of the choice that best completes the statement or answers the question. 1. Tests of details, rather than analytical procedures are appropriate when all of the following are true except: a. Tra
Texas San Antonio - ACCT - 4013
Audit Chapter 1pMultiple Choice Identify the letter of the choice that best completes the statement or answers the question. 1. Assurance services a. Include tax services, compliance audits, and review engagements. b. Are contracts in which the ass
Texas San Antonio - ACCT - 4013
Chapter 2Multiple Choice Identify the letter of the choice that best completes the statement or answers the question. 1. An auditor strives to achieve the appearance of independence in order to a. Become independent in fact with respect to a client