User:PuzzleMeister/sandbox

Help With Formulæ

$$f(x)=\frac{log_{10}n}{log_{10}(2006n-n^2)}$$

$$f(1), f(2), f(3), ... , f(2005)$$

$$\left \{ \sqrt[3]{27+\overbrace{3+57+99}} \right \}$$

$$\sqrt2$$

$$\Delta x=x_f-x_0$$

$$x=\bar{v}t$$

$$\bar{v}=\frac{\Delta x}{\Delta t}$$

$$v_f = v_0+at$$

$$\bar{v}=\frac{v_0+v_f}{2} \iff \bar{a} constant$$

$$x=v_0t+\frac{1}{2}at^2$$

$${v_f}^2={v_0}^2+2ax$$

$$\vec{v_x} = \vec{v_0}\cos{\Theta_0}$$

$$p_{final}$$ $$\vec{v_y} = \vec{v_0}\sin{\Theta_0}$$

$$\Delta x = v_{x0}t=t\left(v_0\cos{\Theta_0}\right)$$

$$v_y = v_{y0}-gt$$

$$\Delta y = v_{y0}t-\frac{1}{2}gt^2$$

$$\textrm{Time\ to\ top\ of\ parabola}=\frac{v_0\sin{\Theta_0}-v_y}{g}$$

$$t=\sqrt{\frac{2\Delta y}{g}}\iff v_0=0$$

$${v_y}^2 = {v_{y0}}^2-2g\Delta y = {v_{y0}}^2-2g\left(v_{y0}t-\frac{1}{2}gt^2\right)$$

$$\textrm{Magnitude\ of\ }\vec{v}=\sqrt{{v_x}^2+{v_y}^2}$$

$$\Theta\textrm{\ from\ positive\ }x\textrm{-axis}=\arctan{\frac{v_y}{v_x}}$$

$$\Sigma F = F_{net} = m\bar{a}=\frac{m\Delta v}{\Delta t} = F_a - F_f$$

$$Ft= m\Delta v\textrm{\ (impulse=momentum)}$$

$$T-F_f=ma$$

$$\Sigma F_x = 0 = T_{1x} - T_{2x}$$

$$\Sigma F_y = 0 = \left(T_{1y}+T_{2y}\right)-Newtons$$

$$\mu \left(\textrm{coefficient\ of\ friction}\right) = \frac{friction}{normal}$$

$$f_k=\mu_kn$$

$$f_s \leq \mu_sn, f_{s\textrm{\ max}} = \mu_sn$$

$$\textrm{Atwood\ Machines:\ }m_1<m_2$$

$$T_1-N_1 = m_1a$$

$$T_2+N_2 =m_2a$$

$$a^2 = b^2 + c^2 -2bc\cos{\alpha}$$

$$\frac{a}{\sin{\alpha}}=\frac{b}{\sin{\beta}}=\frac{c}{\sin{\gamma}}$$

$$1N\equiv1kg*m/s^2=0.225lb$$

$$g=9.8m/s_2=32ft/s^2$$

$$1lb=4.44N$$

$$1kg=2.2lbs\iff g=9.m/s^2$$

$$\vec{R}=\sqrt{37.65^2+22.9^2}\approx 44.1$$

$$2+p_{final}$$

$$\lim_{x\to\infty}f(x)=\mathbb{M}\textrm{\ where\ }\mathbb{M}\equiv\textrm{ madness.}$$