"_ .;...........'.1......,.. __..... 11 ........ •_ _........... _ti(i$l,..,;.S! !l'!lIgical, and s ooial a nalys!s. F urthermore, a s ttJ!y Q f n arrative t echnique, i n\:ludinq 1 :he u se o f n arrative v oiee' b rings t ogether· t he , t.hree p reviously e onsidered a spects.
| 6.99V | 0.14A | 7. | 8.00V | 0.16A | 8. | 9.03V | 0.18A | 9. | 9.98V | 0.20A | 10. | 18.01V | 0.36A | 50-ohm Resistor | V drop | I through | 1.
Data Table #1 Trial Mass of Stopper Hanging Mass Radius T1 T2 T3 Revolutions 1 0.014kg 0.05 kg 0.595m 12.93s 13.10s 14.6s 20 2 0.014kg 0.1 kg 0.595m 11.53s 11.78s 10.8s 20 3 0.014kg 0.15 kg 0.595m 8.59s 9.37s 9.35s 20 4 0.014kg 0.2 kg 0.595m 8.0s 8.03s 7.72s 20 5 0.014kg 0.1 kg 0.765m 12.97s 13.81 ------- 20 6 0.014kg 0.1 kg 0.50m 10.78s 9.72s ------- 20 7 0.014kg 0.1 kg 0.60m 10.0s 10.25s ------- 20 *For the last 3 trials, we only had time to take 2 time intervals. IV. Analysis and Interpretation 1. Organizing Data- Calculate the weight of the hanging mass for each trial. This weight is the force that maintains circular motion, Fc.
Apoi deodata se ncinge ca focul si izbucneste ragusit: Ho, cioara, fi ti ar neamul de rs, ho!. Si nchide pliscul ca gold swtor te pocnesc de ti sar maselele tocmai n curtea bisericii!. Briceag a mai patit o cu Ion; tace. DUBUISSON (les). Nous empruntons au dernier numro du Bulletin de la librairie Morgand et Fatout, page 329, une excellente notice sur cette famille de relieurs du dix huitime sicle.
One can check that v2 (16, 28) = 8, v2 (4, 16) = 1.25, v2 (4, 10) = 0.25 and v2 (1, 7) = 0. (iii) δ(s, y) = [vn+1 (2s, y + 2s) − vn+1 (s/2, y + s/2)]/[2s − s/2]. −→ 1.9: (i) Vn (w) = 1/(1+rn (w))[pn (w)Vn+1 (wH)+qn (w)Vn+1 (wT )] where pn (w) = (1 + rn (w) − dn (w))/(un (w) − dn (w)) and qn (w) = 1 − pn (w). (ii) ∆n (w) = [Vn+1 (wH) − Vn+1 (wT )]/[Sn+1 (wH) − Sn+1 (wT )] (iii) p = q = 1/2. V0 = 9.375.
| Wgt. | Prod. | Pack. | A | R | Action | Sign | 12/1/11 | 1pm | Topcutmeat | meat | 3 deg.c | ok | 20kg. | ok | ok | yes | | n/a | Rk.
1. 7588x 1011 Coulombs Kg-1 C. 9. 11x 10-31 Coulombs Kg-1 D. 1. 0974x 107 Coulombs Kg-1
T.M J.B. T.C A.W M.E. Z.G T.S. E.R. A.W .S.K K.W Total Total Draw Individ. Draws Yellow (75%) 13
Pre-Lab Questions: 1. Absorbance = - Log10T = 2-Log10 (%T) a) 2 – Log (89.95) = .0460 f) 2 – Log (28.18) = .5501 b) 2 – Log (80.91) = .0920 g) 2 – Log (20.51) = .6880 c) 2 – Log (59.02) = .2290 h) 2 – Log (17.38) = .7600 d) 2 – Log (47.75) = .3210 i) 2 – Log (14.45) = .8401 e) 2 – Log (34.83) = .4580 2. Molar Concentration 1.213 g Cu (1 mol Cu/ 63.546 g Cu) (1 mol [Cu (H2O6)2+]/ 1 mol Cu) (1/ .100 L) = .191 M [Cu (H2O6)2+] a) M2 = (M1V1)/V2 M2 = (.191 M [Cu (H2O6)2+])( 1.0 mL)/ (50.0 mL) = .0038 M b) (.191 M [Cu (H2O6)2+]) (2.0 mL)/ (50.0 mL) = .0076 M c) (.191 M [Cu (H2O6)2+]) (5.0 mL)/ (50.0 mL) = .0191 M d) (.191 M [Cu (H2O6)2+]) (7.0 mL)/ (50.0 mL) = .0267 M e) (.191 M [Cu (H2O6)2+]) (10.0 mL)/ (50.0 mL) = .0382 M f) (.191 M [Cu (H2O6)2+]) (12.0 mL)/ (50.0 mL) = .0458 M g) (.191 M [Cu (H2O6)2+]) (15.0 mL)/
Solve ENEE 644 2 Example: Quine-McCluskey Method Example: Quine f(w,x,y,z) = x’y’ + wxy + x’yz’ + wy’z wxy x’y’ x’z’ wx’y’z’ 1 1 w’x’y’z 1 w’x’y’z’ 1 wxyz 1 wxyz’ wxz wyz’ wy’z 1 1 1 {x’y’, x’z’,wxy, wxz}, {x’y’, x’z’,wxy, wy’z}, {x’y’, x’z’,wxz, wyz’}. 1 1 w’x’yz’ wx’y’z 1 1 wx’yz’ wxy’z minimum cover(s): 1 1 1 ENEE 644 1 3 Two-Level Logic Synthesis Two -- Unate Covering Problem Unate and Binate nf > A function f(x1,•••xi•••xn) is positive unate iin xi iif function f(x positive its cofactor f x includes f x ' its i i Negative unate iis defined in a similar way. If a function is s neither positive unate nor negative unate in a variable, it is called binate in this variable. binate A function is positive/negative unate iif it is so for all f function positive/negative variables,