CCF08182010_00018

# CCF08182010_00018 - — Solutions for Exam I—Morning...

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Unformatted text preview: — Solutions for Exam I—Morning Section 7 3 no is the lowest common denominator. no 0.8563 a 2 6.3535 2 1 . 1.71 2—: = m = 1.997 m 2 5.1 2—: = m = 5.956 m 6 77:7: = \$ = 1.997 e 2 02H60012 Answer is C. 35. Each compound ionizes to form 1 mol of each ion. NaOH + HCl ——> NaCl + H20 Na+ + OH‘ + H+ + Cl‘ _3 Na+ + 01‘ + H20 Answer is C. 36. C—12 has a mass of 12 g/mol. mass mass per H101 molecule molecules per mol _ (12 g _) 1 mol _ mol 6.02 x 10—23 molecules = 1.99 x 10-23 g (2.0 X 10-23 g) Answer is B. 37. 252 +143 — 36 = (23 — 4)(s + 9) = 0 [at pole] —-> poles at s = —9,2 5 = 2 Hz Positive poles are unstable. Answer is B. Compare the system with the general system sketched. 012(8) = Go(8) : 53 G1(s) = 7/5 13(8) = 02(8) = 2 H(s) = % + \$411+ — 0(8) _ GC(5)GI(3)G2(5)GR(5) 12(5) ‘ 1 + GC(S)GI(S)02(8)H(5) <5s>(7)<1)<1> S 1+;(5s)(7,§)<1>(—:+ 2—} +1) 35 || ._. 39 2 s +3s+ 4 For a second-order control system7 C(s) _ kwi 11 12(3) _ s2 + ZCwns + 10721 k is the steady—state gain, C is the damping ratio7 and wn is the natural frequency. Comparing Eqs. I and II, .2. = 3.9/4 39 wn = I = 3.122 (3.12) Answer is B. 39. Compare Eqs. I and II from Sol. 38. Imi=35 35 35 k:—:—:1.42 wE. (5q>2 ”1 Answer is A. 40. Compare Eqs. I and II from 501. 38. 201)” : 3 3 3 C“ m ‘ <2><5q> ‘ 0'3” Answer is B. Professional Publications, Inc. ...
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