Help With Formulæ
f ( x ) = l o g 10 n l o g 10 ( 2006 n − n 2 ) {\displaystyle f(x)={\frac {log_{10}n}{log_{10}(2006n-n^{2})}}}
f ( 1 ) , f ( 2 ) , f ( 3 ) , . . . , f ( 2005 ) {\displaystyle f(1),f(2),f(3),...,f(2005)}
{ 27 + 3 + 57 + 99 ⏞ 3 } {\displaystyle \left\{{\sqrt[{3}]{27+\overbrace {3+57+99} }}\right\}}
2 {\displaystyle {\sqrt {2}}}
Δ x = x f − x 0 {\displaystyle \Delta x=x_{f}-x_{0}}
x = v ¯ t {\displaystyle x={\bar {v}}t}
v ¯ = Δ x Δ t {\displaystyle {\bar {v}}={\frac {\Delta x}{\Delta t}}}
v f = v 0 + a t {\displaystyle v_{f}=v_{0}+at}
v ¯ = v 0 + v f 2 ⟺ a ¯ c o n s t a n t {\displaystyle {\bar {v}}={\frac {v_{0}+v_{f}}{2}}\iff {\bar {a}}constant}
x = v 0 t + 1 2 a t 2 {\displaystyle x=v_{0}t+{\frac {1}{2}}at^{2}}
v f 2 = v 0 2 + 2 a x {\displaystyle {v_{f}}^{2}={v_{0}}^{2}+2ax}
v x → = v 0 → cos Θ 0 {\displaystyle {\vec {v_{x}}}={\vec {v_{0}}}\cos {\Theta _{0}}}
p f i n a l {\displaystyle p_{final}} v y → = v 0 → sin Θ 0 {\displaystyle {\vec {v_{y}}}={\vec {v_{0}}}\sin {\Theta _{0}}}
Δ x = v x 0 t = t ( v 0 cos Θ 0 ) {\displaystyle \Delta x=v_{x0}t=t\left(v_{0}\cos {\Theta _{0}}\right)}
v y = v y 0 − g t {\displaystyle v_{y}=v_{y0}-gt}
Δ y = v y 0 t − 1 2 g t 2 {\displaystyle \Delta y=v_{y0}t-{\frac {1}{2}}gt^{2}}
Time\ to\ top\ of\ parabola = v 0 sin Θ 0 − v y g {\displaystyle {\textrm {Time\ to\ top\ of\ parabola}}={\frac {v_{0}\sin {\Theta _{0}}-v_{y}}{g}}}
t = 2 Δ y g ⟺ v 0 = 0 {\displaystyle t={\sqrt {\frac {2\Delta y}{g}}}\iff v_{0}=0}
v y 2 = v y 0 2 − 2 g Δ y = v y 0 2 − 2 g ( v y 0 t − 1 2 g t 2 ) {\displaystyle {v_{y}}^{2}={v_{y0}}^{2}-2g\Delta y={v_{y0}}^{2}-2g\left(v_{y0}t-{\frac {1}{2}}gt^{2}\right)}
Magnitude\ of\ v → = v x 2 + v y 2 {\displaystyle {\textrm {Magnitude\ of\ }}{\vec {v}}={\sqrt {{v_{x}}^{2}+{v_{y}}^{2}}}}
Θ \ from\ positive\ x -axis = arctan v y v x {\displaystyle \Theta {\textrm {\ from\ positive\ }}x{\textrm {-axis}}=\arctan {\frac {v_{y}}{v_{x}}}}
Σ F = F n e t = m a ¯ = m Δ v Δ t = F a − F f {\displaystyle \Sigma F=F_{net}=m{\bar {a}}={\frac {m\Delta v}{\Delta t}}=F_{a}-F_{f}}
F t = m Δ v \ (impulse=momentum) {\displaystyle Ft=m\Delta v{\textrm {\ (impulse=momentum)}}}
T − F f = m a {\displaystyle T-F_{f}=ma}
Σ F x = 0 = T 1 x − T 2 x {\displaystyle \Sigma F_{x}=0=T_{1x}-T_{2x}}
Σ F y = 0 = ( T 1 y + T 2 y ) − N e w t o n s {\displaystyle \Sigma F_{y}=0=\left(T_{1y}+T_{2y}\right)-Newtons}
μ ( coefficient\ of\ friction ) = f r i c t i o n n o r m a l {\displaystyle \mu \left({\textrm {coefficient\ of\ friction}}\right)={\frac {friction}{normal}}}
f k = μ k n {\displaystyle f_{k}=\mu _{k}n}
f s ≤ μ s n , f s \ max = μ s n {\displaystyle f_{s}\leq \mu _{s}n,f_{s{\textrm {\ max}}}=\mu _{s}n}
Atwood\ Machines:\ m 1 < m 2 {\displaystyle {\textrm {Atwood\ Machines:\ }}m_{1}<m_{2}}
T 1 − N 1 = m 1 a {\displaystyle T_{1}-N_{1}=m_{1}a}
T 2 + N 2 = m 2 a {\displaystyle T_{2}+N_{2}=m_{2}a}
a 2 = b 2 + c 2 − 2 b c cos α {\displaystyle a^{2}=b^{2}+c^{2}-2bc\cos {\alpha }}
a sin α = b sin β = c sin γ {\displaystyle {\frac {a}{\sin {\alpha }}}={\frac {b}{\sin {\beta }}}={\frac {c}{\sin {\gamma }}}}
1 N ≡ 1 k g ∗ m / s 2 = 0.225 l b {\displaystyle 1N\equiv 1kg*m/s^{2}=0.225lb}
g = 9.8 m / s 2 = 32 f t / s 2 {\displaystyle g=9.8m/s_{2}=32ft/s^{2}}
1 l b = 4.44 N {\displaystyle 1lb=4.44N}
1 k g = 2.2 l b s ⟺ g = 9. m / s 2 {\displaystyle 1kg=2.2lbs\iff g=9.m/s^{2}}
R → = 37.65 2 + 22.9 2 ≈ 44.1 {\displaystyle {\vec {R}}={\sqrt {37.65^{2}+22.9^{2}}}\approx 44.1}
2 + p f i n a l {\displaystyle 2+p_{final}}
lim x → ∞ f ( x ) = M \ where\ M ≡ madness. {\displaystyle \lim _{x\to \infty }f(x)=\mathbb {M} {\textrm {\ where\ }}\mathbb {M} \equiv {\textrm {madness.}}}