Castings 2 Episode 4 docx

Castings 2 Episode 4 docx

Castings 2 Episode 4 docx

... 4 8 121 620 2 42 8 323 640 444 8 525 660 Copper/weight per cent (a) I < 12. 5 x 14. 5 CUAI, + Maxima n 4, “0 10 20 30 40 50 60 Copper/weight per cent (b) 40 0 300 E E 5 . Ln 2 ... by the ratio: = (0 .46 0}’ x {0.867}’ x (4. 65}’ = 0 .21 x 0.75 x 22 (3.9b) = 3 .4 (3.9c) Thus from Equation 3.9b, pure Si would have been expected to have 2...

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Castings 2 Episode 8 docx

Castings 2 Episode 8 docx

... (am) (am) - - Water 0.0 72 1 320 16 Mercury 0.5 0.30 I6 700 22 300 20 0 Aluminium 0.9 0 .29 31 000 30 000 360 Copper I .3 0 .26 50 000 50 000 600 Iron I .9 0 .25 76 000 70 000 850 packed ... travelled nR: t = (3~~* /2( q/P)(R/h)* = 15(q/P)(R/h )2 (6 .4) For viscosity q = 1 .4 x lo-’ Nsm -2, and reasonable figures for P of about 0 .2 atmosphere (0...

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Hazardous Chemicals Handbook 2 Episode 4 docx

Hazardous Chemicals Handbook 2 Episode 4 docx

... — 24 00 0.15 Metaldehyde 1000 — — Morestan 1800 — — Naphthalene 24 00 25 00 50 Nicotine Sulphate 83 28 5 — Ovex 20 50 — — Paris Green 100 24 00 — Pyrethrum 1500 1800 — Rotenone 50–75 940 5 Ryania 120 0 ... 0.5 SK Chlorobenzilate 1 040 – 122 0 <5000 — DDT 113–118 25 10 1.0 Dichloropropane-Dichloropropene 140 21 00 — Dicofol (Kelthane) 1000–1100 1000– 123 0 — Dieldrin 46 60–90 0 .25...

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SAT II success literature Episode 2 Part 4 docx

SAT II success literature Episode 2 Part 4 docx

... E 21 . C 22 . B 23 . D 24 . B 25 . A 26 . B 27 . C 28 . E 29 . B 30. B 31. A 32. E 33. B 34. E 35. C 36. E 37. D 38. A 39. B 40 . E 41 . C 42 . D 43 . A 44 . B 45 . C 46 . E 47 . B 48 . A 49 . D 50. E 51. B 52. B 53. ... O A O B O C O D O E 41 O A O B O C O D O E 42 O A O B O C O D O E 43 O A O B O C O D O E 44 O A O B O C O D O E 45 O A O B O C O D O E 46 O A O...

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New SAT Math Workbook Episode 2 part 4 docx

New SAT Math Workbook Episode 2 part 4 docx

... coordinates. 62 2 26 2 44 ++ ,,       = () 2. (C) O is the midpoint of AB. x xx y yy + + + + 4 2 244 0 6 2 1 6 24 === === , , − A is the point (0, 4) . 3. (A) d = ()() = === 84 63 4 3 16 9 25 5 22 22 − ... < 3 2 (B) b > 3 2 (C) b <− 3 2 (D) b < 2 3 (E) b > 2 3 7. If x 2 < 4, then (A) x > 2 (B) x < 2 (C) x > 2 (D) 2 <...

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Engineering Tribology Episode 2 Part 4 docx

Engineering Tribology Episode 2 Part 4 docx

... shown in Figure 7 .20 [7]. 100 20 0 500 1000 20 00 5000 10 4 2 × 10 4 5 × 10 4 10 5 2 × 10 5 5 × 10 5 10 6 2 × 10 6 5 × 10 6 10 7 20 0 500 1000 20 00 5000 10 4 2 × 10 4 5 × 10 4 10 5 2 × 10 5 5 × 10 5 10 6 100 50 20 10 Piezoviscous-elastic Piezoviscous-rigid Lubrication ... p max = 3W 2 ab = 3 × 50 2 (2. 32 × 10 4 ) × (1.75 × 10 4 ) = 588.0 [MPa] p...

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Handbook of Mechanical Engineering Calculations ar Episode 2 Part 4 docx

Handbook of Mechanical Engineering Calculations ar Episode 2 Part 4 docx

... con- ditions is: ͚M ϭ 0 O 2 ϪF ϫ 14 ϩ F ϫ 12 Ϫ F ϫ 2 ϭϪF ϫ 14 ϩ 510 ϫ 12 Ϫ 1, 140 ϫ 2 ϭ 0 Bs2 B F ϭ 27 4 lb B ͚M ϭ 0 O 3 P ϫ 31. 82 Ϫ F ϫ 2. 18 ϭ P ϫ 31. 82 Ϫ 27 4 ϫ 2. 18 ϭ 0 B from which Solving for ... / 2) ␲ ϭ e ϭ e ϭ 5 .20 F 2 and F ϭ 5 .20 F 12 Solving the equations for force simultaneously, we find F ϭ 5 940 lb (26 42 1 N) F ϭ 1 140 lb (5071 N) 12 Then, ͚M ϭ...

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Atomic Force Microscopy in Cell Biology Episode 2 Part 4 docx

Atomic Force Microscopy in Cell Biology Episode 2 Part 4 docx

... cells in phosphate-buffered saline (PBS) 1 mM CaCl 2 , 3 mM KCl, 1 mM K 2 HPO 4 ; 2 mM MgCl 2 , 140 mM NaCl, 8 mM Na 2 HPO 4 ; pH 7 .4 (2 × 5 min). 2. Depending upon the epitope tag used, incubate ... Membr. Biol. 1 32, 24 3 25 2. 20 . Marsh, D. J., Jensen, P. K., and Spring, K. R. (1985) Computer-based determina- tion of size and shape in living cells. J. Microsc. 137, 28...

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Engineering Statistics Handbook Episode 2 Part 4 docx

Engineering Statistics Handbook Episode 2 Part 4 docx

... Type I Distribution http://www.itl.nist.gov/div898/handbook/eda/section3/eda366g.htm (2 of 12) [5/1 /20 06 9:58 :22 AM] Percent Point Function The formula for the percent point function of the power ... Type I Distribution http://www.itl.nist.gov/div898/handbook/eda/section3/eda366g.htm (4 of 12) [5/1 /20 06 9:58 :22 AM] ... Tukey-Lambda Distribution http://www.itl.nist.gov/div898/handboo...

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Diffusion Solids Fundamentals Diffusion Controlled Solid State Episode 2 Part 4 docx

Diffusion Solids Fundamentals Diffusion Controlled Solid State Episode 2 Part 4 docx

... Ta V H D 0 /m 2 s −1 5 × 10 −8 (T> ;27 3 K) 4. 4 × 10 −8 2. 9 × 10 −8 0.9 × 10 −8 (T< ;27 3 K) ∆H/eV 0.106 (T> ;27 3 K) 0. 140 0. 043 0.068 (T< ;22 3 K) D D 0 /m 2 s −1 5 .4 × 10 −8 4. 9 × 10 −8 3.7 ... (19.6), and (19 . 24 ), b 1 can be expressed as a function of the ratio D ∗ B (0)/D ∗ A (0): b 1 = −18 + 4 D ∗ B (0) D ∗ A (0) f 1 − f 2 4 1 (ω 1 /ω 3 )+ 14 2 f (4 1...

Ngày tải lên: 06/08/2014, 15:21

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