fundamentals of electric circuits 3rd edition solutions manual chapter 1

Fundamentals of Corporate Finance 8th edition: Solutions Manual

Fundamentals of Corporate Finance 8th edition: Solutions Manual

Ngày tải lên : 01/07/2014, 14:22
... percentage of assets, fell by: 1 – .9850 = . 015 0 or 1. 50 percent. 16 . 2006 Sources/Uses 2007 Assets Current assets Cash $ 15 ,18 3 1, 002 U $ 16 ,18 5 Accounts receivable 35, 612 1, 514 U 37 ,12 6 ... 37 ,12 6 Inventory 62 ,18 2 2,6 71 U 64,853 Total $ 11 2,977 5 ,18 7 U $11 8 ,16 4 Fixed assets Net plant and equipment 327 ,15 6 31, 007 U 358 ,16 3 Total assets $ 440 ,13 3 36 ,19 4 U $476,327 ... Sales $1, 014 ,000 $1, 098,500 $1, 140,750 Costs 788,400 854 ,10 0 886,950 Other expenses 21, 000 22,750 23,625 EBIT $ 204,600 $ 2 21, 650 $ 230 ,17 5 Interest 12 ,500 12 ,500 12 ,500 Taxable income $ 19 2 ,10 0...
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fundamentals of electric circuits

fundamentals of electric circuits

Ngày tải lên : 08/05/2014, 15:37
... Filter † 14 .9 Scaling 619 14 .9 .1 Magnitude Scaling 14 .9.2 Frequency Scaling 14 .9.3 Magnitude and Frequency Scaling 14 .10 Frequency Response Using PSpice 622 † 14 .11 Applications 626 14 .11 .1 Radio Receiver 14 .11 .2 ... source v s i o 6i o + − Figure 1. 19 For Review Question 1. 10. Answers: 1. 1b, 1. 2d, 1. 3c, 1. 4a, 1. 5a, 1. 6b, 1. 7c, 1. 8c, 1. 9b, 1. 10d. | | | | ▲ ▲ e-Text Main Menu Textbook Table of Contents Problem Solving ... Student ix 1. 1 Introduction 4 1. 2 Systems of Units 4 1. 3 Charge and Current 6 1. 4 Voltage 9 1. 5 Power and Energy 10 1. 6 Circuit Elements 13 † 1. 7 Applications 15 1. 7 .1 TV Picture Tube 1. 7.2 Electricity...
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fundamentals of heat and mass transfer solutions manual phần 1 docx

fundamentals of heat and mass transfer solutions manual phần 1 docx

Ngày tải lên : 08/08/2014, 17:20
... P T. A ss SS 4 sss SS 1/ 4 s SS s s α αεσ α εσ ′′ −+ ′′ −+ ′′  + =   In the shade () q0, S ′′ = 10 00 W T 364 K. 1 m 1 5.67 10 W/m K 1/ 4 s 2824 == ìì ì < In the sun, 0.25 1 m 750 W/m 10 00 W T 380 K. 1 m 1 5.67 10 W/m K 1/ 4 22 s 2824 ìì + == ìì ì < COMMENTS: ... () () P 500 W elec m 0. 019 9 kg/s cT T po i 10 07 J/kg K 25 C == = − ⋅   m 0. 019 9 kg/s 3 0. 018 1 m /s 3 1. 10 kg/m ρ ∀= = =   < () 3 4 4 0. 018 1 m /s V 4.7 m/s o 22 A c D 0.07 ... 7.85m 10 W/m K 15 0 25 K 0.8 5.67 10 W/m K 423 298 K =+ìì ()() 22 q 7.85m 1, 250 1, 095 w/m 9 813 8592 W 18 ,405 W=+=+= < (b) The annual energy loss is 11 E qt 18 ,405 W 3600...
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fundamentals of heat and mass transfer solutions manual phần 2 docx

fundamentals of heat and mass transfer solutions manual phần 2 docx

Ngày tải lên : 08/08/2014, 17:20
... lens: () t,wo 23 2-3 3 311 1 Rm 4 0.35 W/m K 10 .2 12 .7 10 12 W/m K4 10 .2 10 m =+ ì ì () 2 2-3 3 19 1.2 K/W +13 .2 K/W+246.7 K/W=4 51. 1 K/W 6 W/m K4 12 .7 10 m += ì With lens: t,w 3 311 1 R 19 1.2 K/W +13 .2 K/W+ ... With the insulation, s ,1 2 w1 2 i2 3 3 11 1 11 1 1 TTq 4k r r 4k r r 4rh ππ π ∞     =+ −+ −+       () () 4 s ,1 2 11 1 1K T 25 C 489W 1. 84 10 4 0.04 0. 51 0.53 W 4 0.53 6 =+ ... insulation () () () s ,1 g 2 2 2 w1 2 2 TT 50 25 C Eq 11 1 1 111 1 417 W/mK 0.50m0.51m 4k r r 40.51m6W/mK 4rh π π π π ∞ − −° == = −+ −+ ⋅ ⋅        ( ) g 42 25 C E q 489W 1. 84 10 5 .10 10 K / W ==...
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fundamentals of heat and mass transfer solutions manual phần 3 pdf

fundamentals of heat and mass transfer solutions manual phần 3 pdf

Ngày tải lên : 08/08/2014, 17:20
... 75 10 0 12 5 15 0 17 5 200 225 250 275 300 0 18 0.7 18 0.2 17 8.4 17 5.4 17 1 .1 165.3 15 8 .1 149.6 14 0 .1 129.9 11 9.4 10 8.7 98.0 25 204.2 203.6 2 01. 6 19 8.2 19 3.3 18 6.7 17 8.3 16 8.4 15 7.4 14 5.6 13 3.4 12 1.0 50 ... isotherms. - 10 0 10 0 10 0 10 0 10 0 - 50 86.0 10 5.6 11 9 13 1.7 15 1.6 200 50 88.2 11 7.4 13 8.7 15 6 .1 174.6 200 50 99.6 13 7 .1 162.5 17 9.2 19 0.8 200 50 12 3.0 16 8.9 19 4.9 207.6 209.4 200 50 17 3.4 220.7 ... T 1 (°C) T 2 (°C) T 3 (°C) T 4 (°C) 0 18 5 18 5 18 5 18 5 ← initial estimate 1 186.3 18 6.6 18 6.6 10 5.8 2 18 7 .1 187.2 16 7.0 96.0 3 18 7.4 18 2.3 16 3.3 94.2 4 18 4.9 18 0.8 16 2.5 93.8 5 18 4.2 18 0.4 16 2.3...
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fundamentals of heat and mass transfer solutions manual phần 4 docx

fundamentals of heat and mass transfer solutions manual phần 4 docx

Ngày tải lên : 08/08/2014, 17:20
... PROBLEM 5 .11 2 (Cont.) () () ()() ( ) () p +1 p +1 p +1 p +1 mm m -1 m +1 p +1 p p +1 mm omp TT TT kd1 kd1 xx TT q x 1 2h x 1 T T x d 1 c t ρ ∞ −− ⋅+⋅ ∆∆ − ′′ + ∆⋅ + ∆⋅ − = ∆⋅⋅ ∆ () pp +1 mm T12Fo2FoBiT =+ ... 296.2 308.6 315 .8 320.4 322.5 9 16 2 276.7 294 .1 306.0 314 .3 319 .0 10 18 0 274.8 2 91. 3 304 .1 312 .4 Hence, find ( ) ( ) 10 10 03 T0, 18 0sT275C T45mm, 18 0sT 312 C.==== oo < COMMENTS: (1) The above ... 307.8 318 .9 322.5 324.5 325 325 325 5 90 287.6 305.8 315 .2 3 21. 5 323.5 324.5 325 325 6 10 8 285.6 3 01. 6 313 .5 319 .3 322.7 324.0 324.5 325 7 12 6 2 81. 8 299.5 310 .5 317 .9 3 21. 4 323.3 324.2 8 14 4...
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Fundamentals of Ecological Modelling, Third Edition pot

Fundamentals of Ecological Modelling, Third Edition pot

Ngày tải lên : 23/03/2014, 01:20
... Balance 11 1 3A.4 Energetic Factors 11 6 3A.5 Settling and Resuspension 12 3 3B .1 Chemical Reactions 12 9 3B.2 Chemical Equilibrium 13 6 3B.3 Hydrolysis 14 0 3B.4 Redox 14 1 3B.5 Acid-Base 14 5 3B.6 ... Sediment-N To Nitrate - 1 0 0 0 0 0 Ammonium 0 - 0 1 0 1 1 Phyt-N 1 1 - (I 0 0 0 Zoopl-N 0 0 1 - 0 0 0 Fish N 0 0 0 1 - 0 0 Detritus-N 0 0 1 1 1 - 0 Sediment-N 0 0 1 (~ 0 1 - The adjacency ... Growth 19 9 3C.7 Ecotoxicological Processes 2 01 Problems 208 4. Conceptual Models 211 4 .1 Introduction 211 4.2 Application of Conceptual Diagrams 211 4.3 Types of Conceptual Diagrams 214 4.4....
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unit operation of chemical engineering 7th ed, solutions manual

unit operation of chemical engineering 7th ed, solutions manual

Ngày tải lên : 01/04/2014, 11:53
... using this Manual, you are using it without permission. Page 26 - 1 PROPRIETARY MATERIAL. â The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may ... using it without permission. Page 13 - 1 PROPRIETARY MATERIAL. â The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed ... are using it without permission. Page 21 - 1 PROPRIETARY MATERIAL. â The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed...
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essentials of programming languages 3rd edition apr 2008

essentials of programming languages 3rd edition apr 2008

Ngày tải lên : 11/06/2014, 13:24
... 1 of definition 1. 1.4 or the first rule of definition 1. 1.5. 2. (14 . ()) is a list of integers, because of property 2 of definition 1. 1.4, since 14 is an integer and () is a list of integers. We ... is (2n 1 + 1) + (2n 2 + 1) + 1 = 2(n 1 + n 2 + 1) + 1 which is once again odd. This completes the proof of the claim that IH(k + 1) holds and therefore completes the induction. The key to the proof ... element of los is s,saylos = (ss 1 s n 1 ),thefirst occurrence of s is as the first element of los. So the result of removing it is just (s 1 s n 1 ). remove-first : Sym ì Listof(Sym) Listof(Sym) (define...
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