an asteroid whose mass is 2.0 x 10^-4 times the mass of earth revolves in a circular orbit around the sun at a distance that is
twice the earth's distance from the sun calculate the period of revolution of the asteroid in years
1 answer:
Answer:
0.0002
Explanation:
First you have to solve exponents first, so you solve 10 to the power of negative 4 which is 0.0001 and now you have 2.0x0.0001 and that equals 0.0002
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