(3) The speed of light in vacuum, c, is 3 X 105 km/s and 1 kilometer = 1000 meters. How many Joules of energy are released if 2 kg of matter are converted into energy at 0.1% efficiency? First convert c to units of meters/second: c = 3 x 105 km/s x 1000 m/km = 3 x 108 m/s. Then express efficiency (eff) in decimal terms: eff = 0.001. In this type of problem, the mass/ energy equivalence equation is written as: E = (eff) mc2 Joules,

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Chapter1: Units, Trigonometry. And Vectors
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(3) The speed of light in vacuum, c, is 3 X 105 km/s and 1 kilometer = 1000 meters. How many
Joules of energy are released if 2 kg of matter are converted into energy at 0.1% efficiency?
First convert c to units of meters/second: c = 3 x 105 km/s x 1000 m/km = 3 x 108 m/s.
Then express efficiency (eff) in decimal terms: eff = 0.001. In this type of problem, the mass/
energy equivalence equation is written as:
E = (eff) mc2 Joules,
Transcribed Image Text:(3) The speed of light in vacuum, c, is 3 X 105 km/s and 1 kilometer = 1000 meters. How many Joules of energy are released if 2 kg of matter are converted into energy at 0.1% efficiency? First convert c to units of meters/second: c = 3 x 105 km/s x 1000 m/km = 3 x 108 m/s. Then express efficiency (eff) in decimal terms: eff = 0.001. In this type of problem, the mass/ energy equivalence equation is written as: E = (eff) mc2 Joules,
Expert Solution
Step 1

Given:

Speed of light: c = 3×105 km/s  =3×105×1000 m/s =3×108 m/s 

Mass of the matter: m = 2 kg

Efficieny of conversion: eff = 0.1%=0.001

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