Practice Exam 2 - Solutions

Practice Exam 2 - Solutions - PHYSICS 6 I II Practice Exam...

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PHYSICS 6 Practice Exam 2 (2003) SOLUTIONS I. 1d ; 2b; 3c; 4a; 5d; 6d II. E n = -13.6/n 2 (eV) a) E(photon)= E 4 – E 2 = 13.6(1/4 – 1/16) eV = 2.6 eV b) E(photon) = h f f = E(photon)/h = 2.6 eV/ 4.1 × 10 -15 eV sec = 6.3 × 10 14 sec -1 III. a) probability for one day to have precipitation P = 11/38 = 0.29 or 29% b) prob for no precipitation in one day = (1-P) = 0.71 or 71% So for 3 days have (1-P) 3 = (0.71) 3 = 0.36 c) prob for ONLY ONCE in 3 days? For (yes-no-no) it would be P × (1-P) 2 . Same for (no-yes-no) and (no-no-yes). So total prob for ONLY ONCE is 3 × P × (1-P) 2 =3 × 0.29 × (0.71) 2 = 0.44 IV. x p h/4 π So p(min) = h/(4 π x) = 6.6 × 10 -34 J sec / (4 π × 10 -7 m) = 5.3 × 10 -28 Kg m/sec (or units of J sec/m since 1 J=1 Kg m 2 /sec 2 ) V. For the essay the grade depends on evidence that you have read the play as demonstrated by your use of appropriate citations to the text. Each essay choice should refer to a section (or sections) of the play in which one or more of the characters address the issue. Along with those references a broader
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