2003 Practice Final Exam - Solutions 

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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.

Ask a homework question - tutors are online