2003 Practice Final Exam - Solutions

# 2003 Practice Final Exam - Solutions - PHYSICS 6 I...

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PHYSICS 6 Practice Final Exam (2003) SOLUTIONS I. 1e ; 2c; 3a; 4b; 5c; 6b; 7c; 8d; 9e; 10d II. a) distance = v(horiz) t so t = d/v(horiz) = 200 Km/ 1 (Km/s) = 200 sec b) reach max ht in half the total time or 100 sec c) max ht = ½ g t 2 = 0.5 × 9.8(m/s 2 ) × (100 s) 2 = 4.9 × 10 4 m =49 Km d) Consider the clock on the ground at the time of impact. It reads 200 sec. According to the rocket, that clock is moving and runs slow. The corresponding rocket clock will read t × γ = t ÷ (1-v 2 /c 2 ) =200 ÷ (1-(0.8) 2 ) = 200/0.6 =330 sec e) distance measured by the rocket is contracted = d/ γ = 200 Km × 0.6 = 120 Km III. a) E n = -13.6 eV / n 2 so E 3 = - 13.6/9 eV and E 1 = -13.6 eV Then E(photon) = -13.6 × (1/9 – 1) = 12.1 eV b) E(photon) = h f , so f = E(photon)/h = 12.1eV/4.1x10 -15 eV·s = 2.95 × 10 15 s -1 c) 1 Watt=1 J/s. To get 1 Joule of photon energy note that 1 Joule=(1/1.6 × 10 -19 )eV = 6.25 × 10 18 eV Then have number of photons is 6.25 × 10 18 eV/(12.1 eV/photon) = 5.17 × 10 18 photons per sec. IV. a) Dow went up 27 times in 53 days, or prob=27/53=0.509 to go up in one day.
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