ok lets say when doing e=mc^2
Mass 1kg
C (speed of light) is 3.0 x 10^8 m s^-1

e = 1 kg x (3.0 x 10^8 m s^-1)^2

e = 1 kg x 9.0 x 10^8 m^2 s^-2

result is 9 x 10^16 kg m^2 s^-2

its the bits about kg m^2 s^-2 that im talking ablout changing, since when the m s^-1 was squared it changed to m^2 s^-2.

So im needing to know what happens to my km^3 s when I find the square roof of my equation.

Reply 5

Zhen Lin

12

Well, looking at your units should tell you that taking the square root of this gives you an unphysical quantity. So why worry about it?

Original post by gibbyni

ok lets say when doing e=mc^2
Mass 1kg
C (speed of light) is 3.0 x 10^8 m s^-1

e = 1 kg x (3.0 x 10^8 m s^-1)^2

e = 1 kg x 9.0 x 10^8 m^2 s^-2

result is 9 x 10^16 kg m^2 s^-2

its the bits about kg m^2 s^-2 that im talking ablout changing, since when the m s^-1 was squared it changed to m^2 s^-2.

So im needing to know what happens to my km^3 s when I find the square roof of my equation.

Ok I understand dimensions - I want to know why you think you need to take

\sqrt{km^3 s}

because, if this is a physics question, you have certainly made a mistake. As Zhen Lin said the answer you obtain is not a physical quantity.

work out the root of a unit

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