Exercise PhysicsWiki

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$$a=b$$

(posted by TerriC, Group 2)

(Zach McDargh)

(posted by KimberlyWynne 19:05, 29 August 2009 (EDT))

(Sandy Simmons)

(Steve Honeywell)



(Andrew Wray)


(Ryan Taylor)

<math>\nabla^2\psi = \frac{1}{v^2}\frac{\partial^2\psi}{\partial t^2}