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Unformatted text preview: ASTRONOMY 294Z: The History of the Universe Professor Barbara Ryden SOLUTIONS TO PROBLEM SET # 5 1) [20 points] Einstein showed that mass ( M ) and energy ( E ) are interchange able: E = Mc 2 , where c is the speed of light. This implies, for instance, that 1 kilogram of matter is equivalent to an energy E = (1 kg) × (3 × 10 8 m / sec) 2 = 9 × 10 16 kg m 2 / sec 2 . An energy of 1 kg m 2 / sec 2 is known as 1 joule, for short. The joule is not a unit of energy that is used much in everyday life. To give you a sense of scale, burning one gallon of gasoline releases 1 . 3 × 10 8 joules (130 million joules) of energy. Okay, enough background. Here’s the question: If you were capable of converting mass to energy with 100% efficiency, how much mass M would you need to produce an energy E = 1 . 3 × 10 8 joules ? If E = Mc 2 , then M = E/c 2 . Thus, to produce E = 1 . 3 × 10 8 joules = 1 . 3 × 10 8 kg m 2 / sec 2 , the amount of mass required is M = E c 2 = 1 . 3 × 10 8 kg m 2 / sec 2 (3 × 10 8 m / sec) 2 = 1 . 44 × 10 9 kg . (1) This mass can also be written as 1.44 micrograms; it’s roughly equivalent toThis mass can also be written as 1....
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 Winter '08
 RYDEN
 Astronomy, Energy, Mass, Kilogram, energy consumption

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