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% Document Info
\author{Jaco (brandenz1229)}
\pdfinfo{
  /Title (physics-midterm-2.pdf)
  /Creator (Cheatography)
  /Author (Jaco (brandenz1229))
  /Subject (Physics MidTerm 2 Cheat Sheet)
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    {\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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\begin{tabulary}{11cm}{L}
    \vspace{-2pt}\large{\bf{\textcolor{DarkBackground}{\textrm{Physics MidTerm 2 Cheat Sheet}}}} \\
    \normalsize{by \textcolor{DarkBackground}{Jaco (brandenz1229)} via \textcolor{DarkBackground}{\uline{cheatography.com/138824/cs/29649/}}}
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\noindent
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  \mymulticolumn{2}{p{5.377cm}}{\bf\textcolor{white}{Cheatographer}}  \\
  \vspace{-2pt}Jaco (brandenz1229) \\
  \uline{cheatography.com/brandenz1229} \\
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\begin{tabulary}{5.8cm}{L}
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  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Cheat Sheet}}  \\
   \vspace{-2pt}Published 3rd November, 2021.\\
   Updated 3rd November, 2021.\\
   Page {\thepage} of \pageref{LastPage}.
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  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Sponsor}}  \\
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  %\includegraphics[width=48px,height=48px]{dave.jpeg}
  Measure your website readability!\\
  www.readability-score.com
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\begin{document}
\raggedright
\raggedcolumns

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% from this page to resize cheat sheet text:
% www.emerson.emory.edu/services/latex/latex_169.html
\footnotesize % Small font.

\begin{multicols*}{3}

\begin{tabularx}{5.377cm}{x{2.43873 cm} x{2.53827 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Chapter 6: Work and Kinetic Energy}}  \tn
% Row 0
\SetRowColor{LightBackground}
m = mass & g = 9.8 m/s \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
F = Weight (N) & F = m ⋅ g \tn 
% Row Count 2 (+ 1)
% Row 2
\SetRowColor{LightBackground}
s = distance & K`E` = Kinetic Energy \tn 
% Row Count 4 (+ 2)
% Row 3
\SetRowColor{white}
\mymulticolumn{2}{x{5.377cm}}{W = Workdone} \tn 
% Row Count 5 (+ 1)
% Row 4
\SetRowColor{LightBackground}
Power = P (Watts) & x = cos \{\{nl\}\} y = sin \tn 
% Row Count 7 (+ 2)
% Row 5
\SetRowColor{white}
1 km = 1000m & 1kg = 1000g \tn 
% Row Count 8 (+ 1)
% Row 6
\SetRowColor{LightBackground}
ΔK = K`f` - K`i` & Friction = always negative \tn 
% Row Count 10 (+ 2)
% Row 7
\SetRowColor{white}
g = -9.8 (decreasing) & g = 9.8 (normal) \tn 
% Row Count 12 (+ 2)
% Row 8
\SetRowColor{LightBackground}
a = g (gravitational acceleration) & Θ = Angle between F and s \tn 
% Row Count 14 (+ 2)
% Row 9
\SetRowColor{white}
|| = Component of F parallel to dl & v = velocity \tn 
% Row Count 16 (+ 2)
% Row 10
\SetRowColor{LightBackground}
W = (∫P`2` to P`1`) F⋅dl & W = F`||` dl \tn 
% Row Count 18 (+ 2)
% Row 11
\SetRowColor{white}
W = (∫P`2` to P`1`) F⋅cosΘ⋅dl & W = F ⋅ s (Joules) \tn 
% Row Count 20 (+ 2)
% Row 12
\SetRowColor{LightBackground}
P`av` = ΔW / Δt & P = lim Δt \textgreater{} 0 (ΔW / Δt) = dW / dt \tn 
% Row Count 22 (+ 2)
% Row 13
\SetRowColor{white}
V`f` = V`i`\textasciicircum{}2\textasciicircum{} + 2 ⋅ a ⋅ s & P = (W/t) \tn 
% Row Count 24 (+ 2)
% Row 14
\SetRowColor{LightBackground}
Constant Speed : (a = 0) & F = force \{\{nl\}\} P = F ⋅ v \tn 
% Row Count 26 (+ 2)
% Row 15
\SetRowColor{white}
\mymulticolumn{2}{x{5.377cm}}{Friction (opposite) = cos(180\textasciicircum{}o\textasciicircum{})} \tn 
% Row Count 27 (+ 1)
% Row 16
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{W`x` = F (cosΘ)⋅s    ||    W`y` = F (sinΘ)⋅s} \tn 
% Row Count 28 (+ 1)
% Row 17
\SetRowColor{white}
\mymulticolumn{2}{x{5.377cm}}{a = (V`f`\textasciicircum{}2\textasciicircum{} - V`i`\textasciicircum{}2\textasciicircum{}) / (2⋅ s)} \tn 
% Row Count 29 (+ 1)
% Row 18
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{F`s` = (1/2)⋅m⋅V`f`\textasciicircum{}2\textasciicircum{} - (1/2)⋅m⋅V`i`\textasciicircum{}2\textasciicircum{}} \tn 
% Row Count 30 (+ 1)
\end{tabularx}
\par\addvspace{1.3em}

\vfill
\columnbreak
\begin{tabularx}{5.377cm}{x{2.43873 cm} x{2.53827 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Chapter 6: Work and Kinetic Energy (cont)}}  \tn
% Row 19
\SetRowColor{LightBackground}
W`grav` = m⋅g⋅h & P = (W/t) \tn 
% Row Count 1 (+ 1)
% Row 20
\SetRowColor{white}
K`E` = (1/2)⋅m⋅V\textasciicircum{}2\textasciicircum{} & P`E` = m⋅g⋅h \tn 
% Row Count 3 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
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\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{2.4885 cm} x{2.4885 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Chapter 7: Potential Energy, Energy Conservation}}  \tn
% Row 0
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Potential Energy = U, P`E` & ΔK  = -ΔU`grav` \tn 
% Row Count 2 (+ 2)
% Row 1
\SetRowColor{white}
K = Kinetic Energy & R = Radius \tn 
% Row Count 3 (+ 1)
% Row 2
\SetRowColor{LightBackground}
s = y`f` - y`i` & U`grav` =  m ⋅ g ⋅ y \tn 
% Row Count 5 (+ 2)
% Row 3
\SetRowColor{white}
Δs = Δxî + Δyĵ & `cm` = circular motion \tn 
% Row Count 7 (+ 2)
% Row 4
\SetRowColor{LightBackground}
 & k = constant of spring \tn 
% Row Count 9 (+ 2)
% Row 5
\SetRowColor{white}
\mymulticolumn{2}{x{5.377cm}}{P`E` = (1/2)⋅k⋅x\textasciicircum{}2\textasciicircum{}} \tn 
% Row Count 10 (+ 1)
% Row 6
\SetRowColor{LightBackground}
W`grav` = w-vector ⋅ Δs-vector & Diameter = 2 ⋅ Radius \tn 
% Row Count 12 (+ 2)
% Row 7
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W`f` = Work Done by Friction & if elastic... \{\{nl\}\} K`E` = P`E` \tn 
% Row Count 14 (+ 2)
% Row 8
\SetRowColor{LightBackground}
W`grav` = F × s & W`grav` = m ⋅ g ⋅ y`i` - m ⋅ g ⋅ y`f` \tn 
% Row Count 17 (+ 3)
% Row 9
\SetRowColor{white}
K`i` + U`i` = K`f` + U`f` & (1/2)⋅m⋅V`i`\textasciicircum{}2\textasciicircum{} + m⋅g⋅y`i` = (1/2)⋅m⋅V`f`\textasciicircum{}2\textasciicircum{} + m⋅g⋅y`f` \tn 
% Row Count 21 (+ 4)
% Row 10
\SetRowColor{LightBackground}
if gravity does work.... \{\{nl\}\} E = K + U`grav` & W`total` = K`f` - K `i` \{\{nl\}\}\{\{nl\}\} W`total` = W`grav` + W`el` + W`other` \tn 
% Row Count 25 (+ 4)
% Row 11
\SetRowColor{white}
W`other` + U`i` - U`f` = K`f` - K`i` \{\{nl\}\} arrange to.... \{\{nl\}\} K`i` + U`i` + W`other` = K`f` + U`f` & Work done on a spring \{\{nl\}\} W = (1/2)K`E`⋅X`f`\textasciicircum{}2\textasciicircum{} - (1/2)K`E`⋅X`i`\textasciicircum{}2\textasciicircum{} \{\{nl\}\} \{\{nl\}\} Work done by a spring  \{\{nl\}\} W = (1/2)K`E`⋅X`i`\textasciicircum{}2\textasciicircum{} - (1/2)K`E`⋅X`f`\textasciicircum{}2\textasciicircum{} \tn 
% Row Count 34 (+ 9)
\end{tabularx}
\par\addvspace{1.3em}

\vfill
\columnbreak
\begin{tabularx}{5.377cm}{x{2.4885 cm} x{2.4885 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Chapter 7: Potential Energy, Energy Conservation (cont)}}  \tn
% Row 12
\SetRowColor{LightBackground}
U`cm` = m ⋅ g ⋅ R & Elastic Potential Energy \{\{nl\}\} U`el` = (1/2)⋅K`E`⋅x\textasciicircum{}2\textasciicircum{}  \{\{nl\}\}\{\{nl\}\} Work Done by Elastic Force \{\{nl\}\} W`el` = (1/2)⋅K`E`⋅x`i`\textasciicircum{}2\textasciicircum{} - (1/2)⋅K`E`⋅x`f`\textasciicircum{}2\textasciicircum{} \tn 
% Row Count 9 (+ 9)
% Row 13
\SetRowColor{white}
\mymulticolumn{2}{x{5.377cm}}{if elastic force does work, and mechanical energy is conserved \{\{nl\}\} K`i` + U`el`, `i` = K`f` + U`el`, `f`} \tn 
% Row Count 12 (+ 3)
% Row 14
\SetRowColor{LightBackground}
Work Done by Friction:  \{\{nl\}\} W`f` = -W`fric`\{\{nl\}\} W`f` = -(-f`k`⋅ s) \{\{nl\}\} W`f` = μ`k`⋅m⋅g⋅s & Law of Conservation of Energy \{\{nl\}\} ΔK + ΔU + ΔU`int` = 0 \tn 
% Row Count 18 (+ 6)
% Row 15
\SetRowColor{white}
F = F`x` + F`y` + F`z`\{\{nl\}\} F`x`(x) = -m⋅g \{\{nl\}\} F`y`(y) = -m⋅g \{\{nl\}\} F`z`(z) = -m⋅g & F`x` = (1/2)⋅K⋅x\textasciicircum{}2\textasciicircum{} \tn 
% Row Count 23 (+ 5)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{2.4885 cm} x{2.4885 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Chapter 8: Momentum, Impulse, Collisions}}  \tn
% Row 0
\SetRowColor{LightBackground}
p = momentum & J = Impulse \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
m = mass & v = velocity \tn 
% Row Count 2 (+ 1)
% Row 2
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{P = m ⋅ v (kg ⋅ m/s)} \tn 
% Row Count 3 (+ 1)
% Row 3
\SetRowColor{white}
F = d`p` / d`t` \{\{nl\}\}\{\{nl\}\} J`y` = (∫t`f` to t`i`) ΣFy dt \{\{nl\}\} J`y`= (F`av`)`y` (t`f` - t`i`) \{\{nl\}\} J`y` = P`fy` - P`iy` \{\{nl\}\} J`y` = (m⋅V`fy`) - (m⋅V`iy`) & J = ΣF (t`f` - t`i`) \{\{nl\}\} J = ΣF⋅Δt \{\{nl\}\} J = (∫t`f` to t`i`) ΣF dt \{\{nl\}\}\{\{nl\}\} J`x` = (∫t`f` to t`i`) ΣFx dt \{\{nl\}\} J`x`= (F`av`)`x` (t`f` - t`i`) \{\{nl\}\} J`x` = P`fx` - P`ix` \{\{nl\}\} J`x` = (m⋅V`fx`) - (m⋅V`ix`) \tn 
% Row Count 15 (+ 12)
% Row 4
\SetRowColor{LightBackground}
ΣF = (P`f` - P`i`) / (t`f` - t`i`) & J = F`av`(t`f`- t`i`) \tn 
% Row Count 17 (+ 2)
% Row 5
\SetRowColor{white}
\mymulticolumn{2}{x{5.377cm}}{J = (P`f` - P`i`) = \{F\} : Change in Momentum} \tn 
% Row Count 18 (+ 1)
% Row 6
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{P = P`A` + P`B` = |P`A` + P`B`|} \tn 
% Row Count 19 (+ 1)
% Row 7
\SetRowColor{white}
\mymulticolumn{2}{x{5.377cm}}{Assuming m`1` and m`2` don't change} \tn 
% Row Count 20 (+ 1)
% Row 8
\SetRowColor{LightBackground}
\seqsplit{m`1`⋅v`1`+m`2`⋅v`2`} = constant & (P`1`+P`2`)`i` = (P`1`+P`2`)`f` \tn 
% Row Count 22 (+ 2)
% Row 9
\SetRowColor{white}
P`1`+P`2` = constant & P`i` = P`f` \tn 
% Row Count 23 (+ 1)
% Row 10
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{V`f` = \seqsplit{(m`1`⋅v`1`+m`2`⋅v`2`)} / (m`1`++m`2`)} \tn 
% Row Count 24 (+ 1)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}


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

\end{document}