User:FeralWolf

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I am FeralWolf25. Located in Aberdeen, UK and currently attending a secondary school within Aberdeen. Will become an undergraduate student of the University of Edinburgh, UK, in Autumn 2007 to study Chemical Physics with Industrial Experience MChemPhys (Hons) and am set to graduate in Summer 2011.


My Sandbox

2 + 3 = Cats\,

http://en.wikipedia.org/wiki/Wikipedia:Editing_Math

\epsilon_0 = \dfrac{C d}{A}

C = \dfrac{Q}{V}

\Rightarrow \epsilon_0 = \dfrac{Qd}{VA}

F = mg\,

B = \mu_0 n I\,

F = B I L\,

\Rightarrow F = \mu_0 n I^2 L\,

mg = \mu_0 n I^2 L\,

\Rightarrow mg = \mu_0 n I^2 L\,

\Rightarrow \mu_0 = \dfrac{m g}{n I^2 L}

I^2 = \dfrac{g}{n \mu_0 L} m

\dfrac{g}{n\mu_0 L}

Centroid = \left ( \dfrac{5.2+6.4+7.8+4.9+5.6}{5},\dfrac{4.24+5.34+6.76+4.08+4.37}{5} \right )

\Rightarrow Centroid = (6.0,4.96)\,

\dfrac {g}{n \mu_0 L} = \frac {4.96}{6.0} = 0.82666667

\dfrac {g}{n \mu_0 L} = 0.82666667\mathbf{x}10^5

\Rightarrow \mu_0 = \dfrac {g}{n\mathbf{x}L\mathbf{x}8.2666667\mathbf{x}10^4}

g=9.8, n=4094, L=0.02319\,

\Rightarrow \mu_0 = \dfrac {9.8}{4094\mathbf{x}0.02319\mathbf{x}8.2666667\mathbf{x}10^4}

\mu_0 = 1.248668232\mathbf{x}10^{-6}\ Hm^{-1}

\mu_0 \approx 1.25\mathbf{x}10^{-6}\ Hm^{-1}

M_1 = 1.031034483\mathbf{x}10^5

M_2 = 0.6103448276\mathbf{x}10^5

\Delta M = \dfrac {M_1 - M_2}{2\sqrt{n - 2}} = \dfrac {(1.031034483 - 0.6103448276)\mathbf{x}10^5}{2\sqrt{5 - 2}}

= 12144.26429\,

M = 8.2666667\mathbf{x}10^4\pm12144.26429

M = 8.2666667\mathbf{x}10^4\pm14.7%

\Rightarrow \mu_0 \approx 1.25\mathbf{x}10^{-6}\pm14.7%\ Hm^{-1}

V = \dfrac{d}{\epsilon_0 A} Q

M = 2.397260274\mathbf{x}10^9\pm8.87850468\mathbf{x}10^7

M = 2.397260274\mathbf{x}10^9\pm3.7%

\Rightarrow \epsilon_0 \approx 1.14\mathbf{x}10^{-11}\pm3.7%\ Fm^{-1}

M = 3.191489362\mathbf{x}10^9\pm2.866753813\mathbf{x}10^7

M = 3.191489362\mathbf{x}10^9\pm0.9%

\Rightarrow \epsilon_0 \approx 3.21\mathbf{x}10^{-11}\pm0.9%\ Fm^{-1}

c = \dfrac {1}{\sqrt{\epsilon_0\mu_0}}

c = \dfrac {1}{\sqrt{1.25\mathbf{x}10^{-6}\ \mathbf{x}\ 1.14\mathbf{x}10^{-11}}}

c \approx 2.65\mathbf{x}10^8\ ms^{-1}

\Delta c = \sqrt{(\Delta\epsilon_0)^2 + (\Delta\mu_0)^2}

\Delta c = \sqrt{3.7%^2 + 14.7%^2}

\Delta c = 15.2%\,

c = 2.65\mathbf{x}10^8\pm15.2%\ ms^{-1}

c = 1.58\mathbf{x}10^8\pm14.7%\ ms^{-1}

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