AI Math Grader for Handwritten Work & Partial Credit

Methane bond angle

109°

IUPAC name for isobutane

2-methylpropane

Gauche butane interaction

0.9 kcal/mol

Ethane rotation barrier

3.0 kcal/mol

Ethane pKa

50

Initial angular momentum

$L_0=I_{cm}{\omega }_0=(4.0\times 10^{-5}kg\cdot m^2)(200\frac{rad}{sec})=8.0\times 10^{-3}\frac{kg\cdot m^2}{sec}$

Initial kinetic energy

$k_0=\frac{1}{2}{I}_{cm}{\omega }_0^2=\frac{1}{2}(4.0\times 10^{-5}kg\cdot m^2)(2.0\times 10^2\frac{rad}{sec}{)}^2=8.0\times 10^{-1}J$

Frictional torque magnitude

$\tau =I_{cm}|\frac{\Delta \omega }{\Delta t}|=(4.0\times 10^{-5}kg\cdot m^2)|\frac{140\frac{rad}{sec}-200\frac{rad}{sec}}{6sec}|=4\times 10^{-4}J$

Angular velocity at 6.2 sec

$-\tau \Delta t=\Delta L=2I_{cm}{\omega }_f-I_{cm}{\omega }_0$ ${\omega }_f=\frac{-\tau \Delta t+I_{cm}{\omega }_0}{2I_{cm}}=\frac{(-4\times 10^{-4}J)(0.2sec)+(4.0\times 10^{-5}kg\cdot m^2)(140\frac{rad}{sec})}{(2)(4.0\times 10^{-5}kg\cdot m^2)}$ $w_f=69\frac{rad}{sec}$

Average torque exerted by dropped washer

$-{\tau }^{total}\Delta t=I_{cm}(w_f-w_b)$ $-{\tau }^{total}=I_{cm}(w_f-w_b)=(4\times 10^{-5}kg-m^2)(69\frac{rad}{sec}-140\frac{rad}{sec})=-1.42\times 10^{-2}J$ $\Delta t=2sec$ ${\tau }^{total}={\tau }_{frictional}+{\tau }_{dropperwasher}$ ${\tau }_{dropperwasher}={\tau }^{total}-{\tau }_{fr}=-1.42\times 10^{-2}J-4\times 10^{-4}J=-1.38\times 10^{-2}J$

Derivative

$(x^3{e}^x{)}^{\prime }=3x^2{e}^x+x^3{e}^x=e^x(3x^2+x^3).$

Seventh derivative

$f^{(7)}(x)=-128\cos (2x)$