Exam 1 2006 - Solutions

Exam 1 2006 - Solutions - Judy is 20 6=26 yr c distance for...

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1 PHYSICS 6 HOUR EXAM 1 SPRING 2006 SOLUTIONS I. 1d, 2a, 3c, 4d, 5e, 6b, 7c, 8b. II. a. distance it falls = (1/2) g t 2 = (1/2) × (9.8 m/s 2 ) × (6 s) 2 = 4.9 × 36 m = 176 m b. Ball takes 6 sec, so car travels horizontally a distance of 28 m/s × 6 s = 168 m. c. Path is parabolic starting at height 176 m and falling to 168 m to the left on the graph III. a. time for trip in Andy’s frame = 8c.yr / 0.8c = 10yr. Andy’s age =20+10=30 yr b. Judy’s time for trip is 10yr/gamma and gamma=1/sqrt(1-(0.8)^2)=1/sqrt(0.36)=1/0.6=5/3 So Judy’s time is 10/(5/3)=(3/5)10=6 yr and
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Unformatted text preview: Judy is 20+6=26 yr. c. distance for Judy = distance for Andy/gamma = 8 c yr/(5/3)=4.8 light years d. For Judy the distance was 4.8 c yr and the time was 6 yr, so the planet approached her at 4.8 c yr / 6 yr = 0.8 c. IV. a. 235 grams is one “gram molecular weight” or 1 mole b. 1 Mole has N(Avogadro)= 6.0x10^23 nuclei of U. c. E = mc 2 = 0.235 Kg × (3.0 × 10 8 m/s) 2 = 2.12 × 10 16 J d. Energy converted = E/1000 = 2.12 × 10 13 J (at 4.2 × 10 12 J/ kiloton that is equivalent to about 5 kilotons of TNT)...
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