Exercise PhysicsWiki: Difference between revisions

Latest revision as of 16:12, 31 August 2009

$$a=b$$

$\nabla \cdot {\overrightarrow {B}}=0$ (posted by TerriC, Group 2)

$e^{i\theta }=cos(\theta )+isin(\theta )$ (Zach McDargh)

$f\lambda =c$

$r={\frac {p}{1+\varepsilon \cdot \cos \theta }}$ (posted by KimberlyWynne 19:05, 29 August 2009 (EDT))

$x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}$ (Sandy Simmons)

$e^{i\pi }+1=0$ (Steve Honeywell)

\psi (x,\ t)=Ae^{i\left(kx-\omega t\right)}\ ,

(Michael Brandow)

$f=\langle P,Q,R\rangle$

$\nabla f=\langle {\partial f \over \partial P}+{\partial f \over \partial Q}+{\partial f \over \partial R}\rangle$ (Andrew Wray)

$U=0.5kx^{2}$ (Ryan Taylor)

$\nabla ^{2}\psi ={\frac {1}{v^{2}}}{\frac {\partial ^{2}\psi }{\partial t^{2}}}$ (Matthew Denagy)

$F=G{m1m2 \over r^{2}}$ (August Larson)

${\frac {-\hbar ^{2}}{2m}}\nabla ^{2}\psi (r)+V(r)\psi (r)=E\psi (r)$ (Brandon Bryant)

$E=mc^{2}$ Cheyvonne Perry

$F=ma$ (Jamie Kalb)

@@ Line 19: / Line 19: @@
+:<math> \psi (x, \ t) = A e^{i \left( kx - \omega t \right)} \ , </math> (Michael Brandow)
 <math>f=\langle P, Q, R \rangle</math>
-<math>\nabla f = {\partial f \over \partial x} P + {\partial f \over \partial y} Q + {\partial f \over \partial z} R</math> (Andrew Wray)
+<math>\nabla f = \langle {\partial f \over \partial P} + {\partial f \over \partial Q} + {\partial f \over \partial R} \rangle</math> (Andrew Wray)
 <math>U=0.5kx^2</math> (Ryan Taylor)
-<math>\nabla^2\psi = \frac{1}{v^2}\frac{\partial^2\psi}{\partial t^2}
+<math>\nabla^2\psi = \frac{1}{v^2}\frac{\partial^2\psi}{\partial t^2} </math> (Matthew Denagy)
+<math>F = G{m1m2 \over r^2 } </math> (August Larson)
+<math>\frac{-\hbar^2}{2m}\nabla^2\psi(r) + V(r)\psi(r) = E\psi(r)</math> (Brandon Bryant)
+<math>E = mc^2</math> Cheyvonne Perry
+<math>F = ma</math> (Jamie Kalb)