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$v=484m/s$

Giancoli 6th Edition, Chapter 6, Problem 15

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Chapter 6, Problem 15 is solved.

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Transcript for this Giancoli solution

Knowing that kinetic energy is one half 'mv' squared we can solve for the speed of this oxygen molecule by solving for 'v'. We'll multiply both sides by two and divide by 'm' canceling those. So times 2 divide by 'm' and we'll take the square root of both sides and we'll solve for 'v'. So 'v' is the square root of 2 times the kinetic energy divided by 'm'. And substituting numbers we're told that the molecule has an energy of 6.21 times 10 to the negative 21 joules. And that's divided by its mass of 5.31 times 10 to the negative 26 kilograms and this gives ANSWER meters per second.