Lab Answers

converted to a more homogeneous form. Solution: Clearly the formula cannot be dimensionally homogeneous, because B and H do not contain the dimension time. The formula would be invalid for anything except English units (ft, sec). By comparing with the answer to Prob. 1.14 just above, we see that the constant “3.3” hides the square root of the acceleration of gravity. 1.16 Test the dimensional homogeneity of the boundary-layer x-momentum equation: ρu ∂u ∂u ∂p ∂τ + ρv =− + ρ gx +…

discusses the steps needed to establish a goaldriven software metrics program. The important points are to choose your business goals and to determine what you expect to learn from the metrics program. The measures and derived indicators used will need to answer questions related to the attainment of these goals. It is also important to keep in mind that the modern view of software quality assurance includes customer satisfaction goals as well as product quality and process improvement…

analisar e discutir sua disposição, pois é nesse ambiente que estamos formando professores e as atividades de formação ocorrem nele. Fotografia 06 – Laboratório de línguas da Fundação Comunitária do Ensino Superior de Itabira. Fonte: www.funcesi.brlabacesso em 14 de nov. 2005 73 A fotografia 06 mostra a evolução do laboratório de línguas em ambiente equipado com as novas tecnologias. Revela que o ambiente agora está mais aberto, as cabines mais baixas e o professor mais…

caught up onthat west wall. I had to pull out...4500:04:32,000 --> 00:04:34,300War is a team sport.4600:04:34,400 --> 00:04:37,600Those AAA sites were thick.Gave extra time to the SAM sites.4700:04:37,800 --> 00:04:40,600See, that's your answer. There reallywas no room for improvement.4800:04:40,800 --> 00:04:43,000Actually, what Lieutenant Wadedid was right.4900:04:43,200 --> 00:04:46,000She took out their communicationsand exposed their SAM sites...5000:04:46,200 --> 00:04:48

its simple sequencer predecessors. While Babbage's machine had to be constructed from mechanical building blocks, it took almost 100 years before his ideas were realized with more advanced technology such as electromechanical relays e.g., the Bell Labs Model 1 in 1940 and vacuum tubes ENIAC in 1946. 5 The Birth of Electronic Computers ENIAC, the Electronic Numerical Integrator And Computer, is considered to be the rst modern, electronic computer. It was built from 1944 through…

following: “A valid password is issued to an authorised user to allow the user to logon. The user must logon to obtain access to the system. Access is granted when a valid password is entered to complete a logon” into modal logic as shown below, and answers queries about the policies: ∀x (password(x) → (∃ s (∀c (character(c) ∧ part_of(c,x)) → member(c,s))) → (∃n (size(s,n) ∧ n ≥ 8) )))) 2.2.2 High-level Policy Languages for Security Although most of the efforts on security policy…

A2 g2 ; L12 = µ0 N1 N2 A2 g2 Problem 1.23 RA = part (a): L11 = lA ; µAc R1 = l1 ; µAc R2 = l2 ; µAc Rg = g µ0 Ac 2 2 N1 µAc N1 = R1 + R2 + Rg + RA /2 l1 + l2 + lA /2 + g (µ/µ0 ) 10 LAA = LBB = part (b): LAB = LBA = lA + l1 + l2 + g (µ/µ0 ) N 2 µAc N2 = RA + RA ||(R1 + R2 + Rg ) lA lA + 2(l1 + l2 + g (µ/µ0 )) N 2 µAc N 2 (R1 + R2 + Rg ) l1 + l2 + g (µ/µ0 ) = RA (RA + 2(R1 + R2 + Rg )) lA lA + 2(l1 + l2 + g (µ/µ0 )) −N N1 µAc −N N1 = RA +…

Problem 1.23 RA = part (a): L11 = lA ; µAc R1 = l1 ; µAc R2 = l2 ; µAc Rg = g µ0 Ac 2 2 N1 µAc N1 = l1 + l2 + lA /2 + g (µ/µ0 ) R1 + R2 + Rg + RA /2 WWW.COMUNIDADEELETRICIDADE.BLOGSPOT.COM 10 LAA = LBB = part (b): LAB = LBA = lA + l1 + l2 + g (µ/µ0 ) N 2 µAc N2 = RA + RA ||(R1 + R2 + Rg ) lA lA + 2(l1 + l2 + g (µ/µ0 )) N 2 µAc N 2 (R1 + R2 + Rg ) l1 + l2 + g (µ/µ0 ) = RA (RA + 2(R1 + R2 + Rg )) lA lA + 2(l1 + l2 + g (µ/µ0 )) −N N1 µAc −N N1 = RA…

part (a): L11 = 2 2 N1 µAc N1 = R1 + R2 + Rg + RA /2 l1 + l2 + lA /2 + g (µ/µ0 ) A2 g2 I2 10 LAA = LBB = lA + l1 + l2 + g (µ/µ0 ) N 2 µAc N2 = RA + RA ||(R1 + R2 + Rg ) lA lA + 2(l1 + l2 + g (µ/µ0 )) part (b): LAB = LBA = N 2 µAc N 2 (R1 + R2 + Rg ) l1 + l2 + g (µ/µ0 ) = RA (RA + 2(R1 + R2 + Rg )) lA lA + 2(l1 + l2 + g (µ/µ0 )) LA1 = L1A = −LB1 = −L1B = −N N1 µAc −N N1 = RA + 2(R1 + R2 + Rg ) lA + 2(l1 + l2 + g (µ/µ0 )) part (c): v1 =

A2 g2 ; L12 = µ0 N1 N2 A2 g2 Problem 1.23 RA = part (a): L11 = lA ; µAc R1 = l1 ; µAc R2 = l2 ; µAc Rg = g µ0 Ac 2 2 N1 µAc N1 = R1 + R2 + Rg + RA /2 l1 + l2 + lA /2 + g (µ/µ0 ) 10 LAA = LBB = part (b): LAB = LBA = lA + l1 + l2 + g (µ/µ0 ) N 2 µAc N2 = RA + RA ||(R1 + R2 + Rg ) lA lA + 2(l1 + l2 + g (µ/µ0 )) N 2 µAc N 2 (R1 + R2 + Rg ) l1 + l2 + g (µ/µ0 ) = RA (RA + 2(R1 + R2 + Rg )) lA lA + 2(l1 + l2 + g (µ/µ0 )) −N N1 µAc −N N1 = RA +…