Practice Exam 2 - Solutions

Practice Exam 2- - PHYSICS 6 I II Practice Exam 2(2003 SOLUTIONS 1d 2b 3c 4a 5d 6d En =-13.6/n2(eV a E(photon)= E4 E2 = 13.6(1/4 1/16 eV = 2.6 eV b

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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
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This note was uploaded on 03/10/2008 for the course PHY 6 taught by Professor Garyr.goldstein during the Spring '06 term at Tufts.

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