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\author{patrick}
\pdfinfo{
  /Title (physics.pdf)
  /Creator (Cheatography)
  /Author (patrick)
  /Subject (Physics Cheat Sheet)
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\fancyhead[L]{
\noindent
\begin{multicols}{3}
\begin{tabulary}{5.8cm}{C}
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    \vspace{-7pt}
    {\parbox{\dimexpr\textwidth-2\fboxsep\relax}{\noindent
        \hspace*{-6pt}\includegraphics[width=5.8cm]{/web/www.cheatography.com/public/images/cheatography_logo.pdf}}
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\columnbreak
\begin{tabulary}{11cm}{L}
    \vspace{-2pt}\large{\bf{\textcolor{DarkBackground}{\textrm{Physics Cheat Sheet}}}} \\
    \normalsize{by \textcolor{DarkBackground}{patrick} via \textcolor{DarkBackground}{\uline{cheatography.com/21815/cs/4304/}}}
\end{tabulary}
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\fancyfoot[L]{ \footnotesize
\noindent
\begin{multicols}{3}
\begin{tabulary}{5.8cm}{LL}
  \SetRowColor{FootBackground}
  \mymulticolumn{2}{p{5.377cm}}{\bf\textcolor{white}{Cheatographer}}  \\
  \vspace{-2pt}patrick \\
  \uline{cheatography.com/patrick} \\
  \end{tabulary}
\vfill
\columnbreak
\begin{tabulary}{5.8cm}{L}
  \SetRowColor{FootBackground}
  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Cheat Sheet}}  \\
   \vspace{-2pt}Published 3rd June, 2015.\\
   Updated 8th May, 2016.\\
   Page {\thepage} of \pageref{LastPage}.
\end{tabulary}
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  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Sponsor}}  \\
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  \vspace{-5pt}
  %\includegraphics[width=48px,height=48px]{dave.jpeg}
  Measure your website readability!\\
  www.readability-score.com
\end{tabulary}
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\begin{document}
\raggedright
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% Set font size to small. Switch to any value
% from this page to resize cheat sheet text:
% www.emerson.emory.edu/services/latex/latex_169.html
\footnotesize % Small font.

\begin{multicols*}{2}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Isotopes}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Isotope - An atom with the same number of protons but a different number of neutrons. \newline % Row Count 2 (+ 2)
Radioisotope - An isotope that is radioactive and sometimes unstable. They  \newline % Row Count 4 (+ 2)
decay.% Row Count 5 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Formulas}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{W1= q ∆V1 \newline % Row Count 1 (+ 1)
(J)..(.c)..(v) \newline % Row Count 2 (+ 1)
Power - rate of doing work	 \newline % Row Count 3 (+ 1)
P= w/t  (joules/seconds) \newline % Row Count 4 (+ 1)
Power w/t = q∆v/t \newline % Row Count 5 (+ 1)
P=i∆v \newline % Row Count 6 (+ 1)
Unit of energy \newline % Row Count 7 (+ 1)
w=pt \newline % Row Count 8 (+ 1)
(joules) = Watt*Sec \newline % Row Count 9 (+ 1)
New energy unit = kWh \newline % Row Count 10 (+ 1)
Electrical Energy \newline % Row Count 11 (+ 1)
1kWh ≡ 1000*3600 \newline % Row Count 12 (+ 1)
1kWh ≡ 3.6*10\textasciicircum{}6J% Row Count 13 (+ 1)
} \tn 
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\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
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\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{resistance}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{p{8.4cm}}{\vspace{1px}\centerline{\includegraphics[width=5.1cm]{/web/www.cheatography.com/public/uploads/patrick_1433336026_fcuvblnk;ml',.jpg}}} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{1/Rt = 1/R1 + 1/R2 + ...+1/Rn if in parallel.}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Electrical energy and power}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Electrical energy (Joules) = potential drop (volts) x current (amps) x times (seconds)  E=VIt \newline % Row Count 2 (+ 2)
P=E/t    were one watt = 1 joule per second \newline % Row Count 3 (+ 1)
E/t = VIT/t or P= VI \newline % Row Count 4 (+ 1)
Power (watts) = voltage (volts) x current (amps)  \newline % Row Count 5 (+ 1)
P=VI \newline % Row Count 6 (+ 1)
How much energy does a 100W light bulb use in half an hour? \newline % Row Count 8 (+ 2)
P=100W and t=0.5h \newline % Row Count 9 (+ 1)
So E=100W x 0.5h = 50Wh or 0.05kWh \newline % Row Count 10 (+ 1)
To find power used Volts times Amps% Row Count 11 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
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\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
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\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Resistance}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{p{8.4cm}}{\vspace{1px}\centerline{\includegraphics[width=5.1cm]{/web/www.cheatography.com/public/uploads/patrick_1433334223_nkmkn.png}}} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{Ohm's law: ∆V=iR or V=Amps*Ohms \newline  \newline A charge q moving through a potential difference ∆V will lose potential energy: ∆U=qV}  \tn 
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\begin{tabularx}{8.4cm}{X}
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\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Electric field}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{The electric field in any region of space is defined as the electric force per unit charge: E = F/q  \newline % Row Count 3 (+ 3)
the force on a charge of q in an electric field is given by F= qE% Row Count 5 (+ 2)
} \tn 
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\begin{tabularx}{8.4cm}{X}
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\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Neutron Bombardment}}  \tn
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\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
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\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Decay}}  \tn
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\mymulticolumn{1}{x{8.4cm}}{Image could not be loaded.} \tn 
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\begin{tabularx}{8.4cm}{X}
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\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Radiaiton}}  \tn
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\mymulticolumn{1}{x{8.4cm}}{{\bf{Alpha radiation}} - Helium nucleus (2 protons and 2 neutrons), highly ionising, charge of 2+, heavy. \newline % Row Count 3 (+ 3)
{\bf{Beta radiation}} - fast moving electron, negatively charged electron, not as high ionising capabilities, charge of -1. \newline % Row Count 6 (+ 3)
{\bf{Gamma radiation}} - electromagnetic radiation, less ionising then alpha or beta. \newline % Row Count 8 (+ 2)
{\bf{Detecting radiation}} - Geiger-Muller tube counts ions that are produced inside it, these create an electrical signal. \newline % Row Count 11 (+ 3)
{\bf{Radiation in the body}} - Ionising radiation can damage or kill the cells, dna can replicate in the damaged form, cancers can form.% Row Count 14 (+ 3)
} \tn 
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\begin{tabularx}{8.4cm}{X}
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\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Ionising Radiation}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Ionising means that the radiation rips off electrons from nearby atoms that it passes. \newline % Row Count 2 (+ 2)
Three types of radiation: Alpha, Beta and Gamma \newline % Row Count 3 (+ 1)
Alpha passes through paper. \newline % Row Count 4 (+ 1)
Beta passes through thin metal. \newline % Row Count 5 (+ 1)
Gamma passes through thick metal. \newline % Row Count 6 (+ 1)
Alpha radiation is most ionising, so it loses it's energy very quickly.% Row Count 8 (+ 2)
} \tn 
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\begin{tabularx}{8.4cm}{X}
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\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Atomic Notation}}  \tn
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\mymulticolumn{1}{x{8.4cm}}{Image could not be loaded.} \tn 
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\begin{tabularx}{8.4cm}{X}
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\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Half life}}  \tn
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\mymulticolumn{1}{x{8.4cm}}{{\bf{Half-life}} - the time taken for the radioactivity of a specified isotope to fall to half its original value.% Row Count 3 (+ 3)
} \tn 
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\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Nuclear fission/fusion}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Fission- When a nucleus splits into two or more pieces usually after bombardment by neutrons. \newline % Row Count 2 (+ 2)
Fusion- A process taking place inside stars in which small nuclei are forced together to make larger nuclei. Energy is released in the process. \newline % Row Count 5 (+ 3)
Chain reaction - A series of nuclear fissions that may or may not be controlled. The neutrons that are released cause the reaction.% Row Count 8 (+ 3)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
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\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Nuclear Fission Reactors}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Used to harness energy from Fission reactions. \newline % Row Count 1 (+ 1)
Neutrons released from Uranium-235 when it undergoes fission are travelling at high speeds, this leads to a chain reaction which causes an explosion.  \newline % Row Count 5 (+ 4)
The heat generated from the fission process is used to make steam which drives the turbine. \newline % Row Count 7 (+ 2)
Fuel rods- long, thin rods containing pellets of enriched uranium \newline % Row Count 9 (+ 2)
moderator- material that slows neutrons. \newline % Row Count 10 (+ 1)
control rods- rods made of a material that absorbs neutrons \newline % Row Count 12 (+ 2)
coolant- a liquid or gas to absorb the heat energy% Row Count 13 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
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\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Nuclear Reactor}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{p{8.4cm}}{\vspace{1px}\centerline{\includegraphics[width=5.1cm]{/web/www.cheatography.com/public/uploads/patrick_1433332088_Untitled.png}}} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Electric charge}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Conductors: All metals, especially silver, gold, copper aluminium and any ionic solution. \newline % Row Count 2 (+ 2)
Moderate conductors: Water and earth. \newline % Row Count 3 (+ 1)
Semi-conductors: Silicon, Germanium and skin. \newline % Row Count 4 (+ 1)
Insulators: Plastics, polystyrene, dry air, glass, porcelain, cloth (dry) \newline % Row Count 6 (+ 2)
Moderate insulators: wood, paper, damp air, ice and snow.% Row Count 8 (+ 2)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Electrical forces and fields}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Image could not be loaded.} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{For the forces between two charges q1 and q2 at a distance of r \newline k= 9.0 x 10\textasciicircum{}9\textasciicircum{} N m\textasciicircum{}2\textasciicircum{} x C\textasciicircum{}2\textasciicircum{}}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Electric Current}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Electric current is the rate of transfer of charge: I=q/t \newline % Row Count 2 (+ 2)
where q is the charge transferred and t is the time taken. \newline % Row Count 4 (+ 2)
1 ampere (A) = 1 coulomb per second (C*s\textasciicircum{}-1\textasciicircum{}) \newline % Row Count 5 (+ 1)
So 1 coulomb (C) = 1 ampere second (A*s) \newline % Row Count 6 (+ 1)
1 volt = 1 joule per coulomb (1V = 1JC\textasciicircum{}-1\textasciicircum{}) \newline % Row Count 7 (+ 1)
1 ohm = 1 volt per ampere (1ohm = 1VA\textasciicircum{}-1\textasciicircum{})% Row Count 8 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
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\begin{tabularx}{8.4cm}{X}
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\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Resistance}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{∆VBattery=i Rtotal \newline % Row Count 1 (+ 1)
∆vBattery=∆V1+∆V2 \newline % Row Count 2 (+ 1)
Therefore i*Rtotal = ∆V1+ ∆V2=iR1+iR2 \newline % Row Count 3 (+ 1)
Rtotal = R1 + R2 \newline % Row Count 4 (+ 1)
R= V/I or V = IR% Row Count 5 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
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\begin{tabularx}{8.4cm}{X}
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\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Electric Circuits}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{In any electric circuit the sum of all currents flowing into any point is equal to the sum flowing out of it. \newline % Row Count 3 (+ 3)
The total potential drop around a closed circuit must be equal to the total EMF (electromotive force, the energy provided by the cell)% Row Count 6 (+ 3)
} \tn 
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\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Symbols and devices}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{p{8.4cm}}{\vspace{1px}\centerline{\includegraphics[width=5.1cm]{/web/www.cheatography.com/public/uploads/patrick_1433335050_11335704_10206868637892166_218555461_n.jpg}}} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
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\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Formulas}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Two loops \newline % Row Count 1 (+ 1)
Junction law \newline % Row Count 2 (+ 1)
Current in = current out \newline % Row Count 3 (+ 1)
at(a) itotal = i1 +i2 \newline % Row Count 4 (+ 1)
Parallel arrangement \newline % Row Count 5 (+ 1)
∆V1= ∆V2 \newline % Row Count 6 (+ 1)
iTotal = ∆VBattery/Rtotal \newline % Row Count 7 (+ 1)
iTotal = i1 +i2 \newline % Row Count 8 (+ 1)
\seqsplit{∆VBattery/Rtotal=∆V1/R1+∆V2/R2} \newline % Row Count 9 (+ 1)
1/Rtotal = 1/R1 + 1/R2  \newline % Row Count 10 (+ 1)
Rtotal = (1/R1 +1/R2) \newline % Row Count 11 (+ 1)
R1=R2 =10Ohms \newline % Row Count 12 (+ 1)
1/Rt = 1/10 + 1/10  \newline % Row Count 13 (+ 1)
= 2/10 = 1/5 \newline % Row Count 14 (+ 1)
RT= 5 ohms \newline % Row Count 15 (+ 1)
Voltage loop law \newline % Row Count 16 (+ 1)
One loop \newline % Row Count 17 (+ 1)
∆Vbattery = ∆1+∆2 \newline % Row Count 18 (+ 1)
Voltage drop of battery must equal \newline % Row Count 19 (+ 1)
Sum of voltage drops around one loop.% Row Count 20 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}


% That's all folks
\end{multicols*}

\end{document}