EBK BUSINESS DRIVEN INFORMATION SYSTEMS
6th Edition
ISBN: 8220106796986
Author: BALTZAN
Publisher: YUZU
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Chapter 1, Problem 3CCO
To determine
To explain:
System thinking, and the ways MIS solves the issue with the information silos throughout Costco’s entire worldwide organization.
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A three-phase transmission line supplies power to three loads at a voltage
408 Vrms (line to line). The loads are as follows:
Load 1:
S₁ = 100+ j50 VA
Load 2:
S₂ = 40-j20 VA
Load 3:
S3 = 10 + j0 VA
Find the magnitude of the line current | Line and the combined power factor of
the loads.
Hint:
Steral \= √3 | Vime |× | Ime |
line
line
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Chapter 1 Solutions
EBK BUSINESS DRIVEN INFORMATION SYSTEMS
Ch. 1 - Prob. 1OCQCh. 1 - Prob. 2OCQCh. 1 - Prob. 3OCQCh. 1 - Prob. 4OCQCh. 1 - Prob. 5OCQCh. 1 - Prob. 6OCQCh. 1 - Prob. 1RQCh. 1 - Prob. 2RQCh. 1 - Prob. 3RQCh. 1 - Prob. 4RQ
Ch. 1 - Prob. 5RQCh. 1 - Prob. 6RQCh. 1 - Prob. 7RQCh. 1 - Prob. 8RQCh. 1 - Prob. 9RQCh. 1 - Prob. 10RQCh. 1 - Prob. 11RQCh. 1 - Prob. 12RQCh. 1 - Prob. 13RQCh. 1 - Prob. 14RQCh. 1 - Prob. 15RQCh. 1 - Prob. 1CCOCh. 1 - Prob. 2CCOCh. 1 - Prob. 3CCOCh. 1 - Prob. 4CCOCh. 1 - Prob. 5CCOCh. 1 - Prob. 6CCOCh. 1 - Prob. 1CCTCh. 1 - Prob. 2CCTCh. 1 - Prob. 3CCTCh. 1 - Prob. 4CCTCh. 1 - Prob. 5CCTCh. 1 - Prob. 6CCTCh. 1 - Prob. 1CBTCh. 1 - Prob. 2CBTCh. 1 - Prob. 3CBTCh. 1 - Prob. 4CBTCh. 1 - Prob. 5CBTCh. 1 - Prob. 6CBTCh. 1 - Prob. 7CBTCh. 1 - Prob. 8CBTCh. 1 - Prob. 9CBTCh. 1 - Prob. 10CBTCh. 1 - Prob. 11CBTCh. 1 - Prob. 12CBTCh. 1 - Prob. 13CBTCh. 1 - Prob. 14CBTCh. 1 - Prob. PIAYKBPCh. 1 - Prob. PIIAYKBPCh. 1 - Prob. PIIIAYKBPCh. 1 - Prob. PIVAYKBPCh. 1 - Prob. PVIAYKBPCh. 1 - Prob. PVIIAYKBPCh. 1 - Prob. PVIIIAYKBPCh. 1 - Prob. PIXAYKBPCh. 1 - Prob. PXAYKBPCh. 1 - Prob. PXIAYKBP
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- Chapter 12 - Lecture Notes.pptx: (MAE 272-01) (SP25) DY... Scoresarrow_forwardA square flexible foundation of width B applies a uniform pressure go to the underlying ground. (a) Determine the vertical stress increase at a depth of 0.5B below the center using Aσ beneath the corner of a uniform rectangular load given by Aσ Variation of Influence Value I m n 0.5 0.6 0.8 1.0 0.2 0.4 0.2 0.01790 0.03280 0.03866 0.04348 0.05042 0.05471 0.4 0.03280 0.06024 0.07111 0.08009 0.09314 0.10129 0.5 0.03866 0.07111 0.08403 0.09473 0.11035 0.12018 0.6 0.04348 0.08009 0.09473 0.10688 0.12474 0.13605 0.8 0.05042 0.09314 0.11035 0.12474 0.14607 0.15978 1.0 0.05471 0.10129 0.12018 0.13605 0.15978 0.17522 (Enter your answer to three significant figures.) Ασ/90 = Activity Frame (b) Determine the vertical stress increase at a depth of 0.5B below the center using the 2 : 1 method equation below. 90 x B x L Aσ = (B+ z) (L+ z) (Enter your answer to three significant figures.) Δσ/90 = (c) Determine the vertical stress increase at a depth of 0.5B below the center using stress isobars in…arrow_forwardNeed help!!!arrow_forwardCan you show why the answer to this question R = 199 ohm?arrow_forward2 A flexible circular area is subjected to a uniformly distributed load of 450 kN/m² (the figure below). The diameter of the load area is 2 m. Estimate the average stress increase (Aσay) below the center of the loaded area between depths of 3 m and 6 m. H₂ 1.0 H₂ B 0.8 CHI HD DV 0.6 C 1.0 1.5 0.4 0.2 6.0 8.0. 10.0 2.0 2.5 3.0 4.0 5.0 H₁ (Enter your answer to two significant figures.) Δσαν τ kN/m² 6arrow_forwardRefer to the figure below. Using the procedure outlined in your textbook, determine the average stress increase in the clay layer below the center of the foundation due to the net foundation load of 45 tons. Use the equations: Aσ = and qo x B x L (B+ z)(L+ z) Aσ av (H2/H₁) Δσι +44 + Δσο net load 6 4:5 ft 10 ft 5ft x 5ft Sand Sand y=100 lb/ft³ Ysat 122 lb/ft³:" Ysat 120 lb/ft³: 0.7 C=0.25 Groundwater table C=0.06 Preconsolidation pressure = 2000 lb/ft² (Enter your answer to three significant figures.) Ασαν = lb/ft²arrow_forwardRefer to the figure below, which shows a flexible rectangular area. Given: B₁ = 4 ft, B₂ = 6 ft, L₁ = 8 ft, and L2 = 10 ft. If the area is subjected to a uniform load of 4100 lb/ft², determine the stress increase at a depth of 10 ft located immediately below point O. Use the table below. T B(1) 3 B(2) 2 L(1) * 4 L2) Table 1 Variation of Influence Value I n m 0.8 0.9 1.0 1.2 1.4 0.1 0.02576 0.02698 0.02794 0.02926 0.03007 0.2 0.05042 0.05283 0.05471 0.05733 0.05894 0.3 0.07308 0.07661 0.07938 0.08323 0.08561 0.4 0.09314 0.09770 0.10129 0.10631 0.10941 0.5 0.11035 0.11584 0.12018 0.12626 0.13003 0.6 0.12474 0.13105 0.13605 0.14309 0.14749 0.7 0.13653 0.14356 0.14914 0.15703 0.16199 0.8 0.14607 0.15371 0.15978 0.16843 0.17389 0.9 0.15371 0.16185 0.16835 0.1766 0.18357 1.0 0.15978 0.16835 0.17522 0.18508 0.19139 1.1 0.16843 0.17766 0.18508 0.19584 0.20278 (Enter your answer to three significant figures.) Aσ = lb/ft²arrow_forwardPoint loads of magnitude 100, 200, and 380 kN act at B, C, and D, respectively (in the figure below). Determine the increase in vertical stress at a depth of 6 m below point A. Use Boussinesq's equation. B 6 m A 6 m с 3 m D (Enter your answer to three significant figures.) Δαχτ kN/m²arrow_forwardTwo line loads q₁ = 30 kN/m and 92 = 44 kN/m of infinite lengths are acting on top of an elastic medium, as shown in the figure below. Find the vertical stress increase at A. 92 91 6 m 3 m 3 m Δσ A (Enter your answer to three significant figures.) Vertical stress increase at A = kN/m²arrow_forward2.5. Find the half-power beamwidth (HPBW) and first-null beamwidth (FNBW), in radians and degrees, for the following normalized radiation intensities: (a) U(9) cos θ cos(20) (b) U(θ)-cos2 θ cos2(26) (c) U(θ) = cos(θ) cos(30) (0 < θ < 900,0 < φ 360) (d) U(t) = cos2(9) cos2(39) (e) U(9) = cos(29) cos(39) (f) U (ecos (20) cos (30)arrow_forwardStart with the initial angles within the integration and just integrate them without mapping them to specific quadrants. Use python and radiansarrow_forwardA flexible circular area is subjected to a uniformly distributed load of 144 kN/m² (see the figure below). The diameter of the load area is 2 m. Estimate the average stress increase (Aσay) below the center of the loaded area between depths of 3 m and 6 m. Use the equations: 1 Ασ = go 1 [1 + (2) ² ³/2 and Aσ av (H2/H1) Δσι + 41ση + Ασο 6 9 B/2 krark do Δε Aσ (Enter your answer to three significant figures.) Ασαν = kN/m²arrow_forwardarrow_back_iosSEE MORE QUESTIONSarrow_forward_ios
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