1/7
Looks like no tags are added yet.
Name | Mastery | Learn | Test | Matching | Spaced | Call with Kai |
|---|
No analytics yet
Send a link to your students to track their progress
Kinematic Equations
\overrightarrow{v_{f}}=\overrightarrow{v_{i}}+\overrightarrow{a}\cdot\Delta t
\overrightarrow{v}_{f}^2=\overrightarrow{v}_{i}^2+2\overrightarrow{a}\cdot\overrightarrow{\Delta x}
\overrightarrow{\Delta x}=\frac12\left(\overrightarrow{v_{i}}+\overrightarrow{v_{f}}\right)\cdot\Delta t
\overrightarrow{\Delta x}=\overrightarrow{v}_{i}\cdot\Delta t+\frac12\overrightarrow{a}\cdot\Delta t^2
Forces (Linear Dynamics)
\overrightarrow{F}=m\cdot\overrightarrow{a}
\overrightarrow{F_{s,\max}}=\mu_{s}\cdot F_{N}
0<\overrightarrow{F_{s}}<\overrightarrow{F_{s,\max}}
\overrightarrow{F_{k}}=\mu_{k}\cdot F_{N}
Work, Energy, Power
W=F\cdot d\cdot\cos\theta
W_{net}=\Delta KE=\frac12mv_{f}^2-\frac12mv_{i}^2
KE=\frac12mv^2
PE_{g}=mgh (“h” is the height from the chosen zero level)
PE_{elastic}=\frac12k\Delta x^2 (“k” is the spring constant; “x” is the distance the object is from the rest position
F_{g},F_{elastic,}F_{electrostatic} —Conservative forces
KE_1+PE_1=ME_1=ME_2=KE_2+PE_2 —Use the Conservation of Mechanical Energy to solve if work is only done by conservative forces.
W_{NC}=\Delta ME=\Delta KE+\Delta PE —Use this if work is done by non-conservative forces
P_{ave}=\frac{\Delta E}{\Delta t}=\frac{\Delta W}{\Delta t} —Average power
P_{Inst}=F\cdot v\cdot\cos\theta —Instant power
Linear Momentum
\overrightarrow{P}=m\cdot\overrightarrow{v} (“P”=Linear Momentum)
\overrightarrow{J}=\overrightarrow{F}\cdot\Delta t (“J”=Linear Impulse)
\Sigma\overrightarrow{P}_{i}=\Sigma\overrightarrow{P_{f}} —Conservation of Linear momentum
m_1\cdot\overrightarrow{v_{1,i}}+m_2\cdot\overrightarrow{v_{2,i}}=m_1\cdot\overrightarrow{v_{1,f}}+m_2\cdot\overrightarrow{v_{2,f}} —For two objects
\Sigma\overrightarrow{P}_{i}=\Sigma\overrightarrow{P_{f}} —In inelastic collisions, only momentum is conserved
\Sigma\overrightarrow{P}_{i}=\Sigma\overrightarrow{P_{f}} —In elastic collisions, momentum is conserved.
\Sigma KE_{i}=\Sigma KE_{f} —And kinetic energy is conserved
\overrightarrow{v_{1,i}}+\overrightarrow{v_{1,f}}=\overrightarrow{v_{2,i}}+\overrightarrow{v_{2,f}}
Circular Motion
\theta=\frac{s}{r} “s”=arc length, “r”=radius
\omega=\frac{\theta}{t} —angular velocity
v=\frac{s}{t}=\frac{r\cdot\theta}{t}=r\cdot\omega —linear velocity
\alpha=\frac{\omega}{t} —angular acceleration
\overrightarrow{\omega_{f}}=\overrightarrow{\omega_{i}}+\overrightarrow{\alpha}\cdot\Delta t
\left(\overrightarrow{\omega_{f}}\right)^2=\left(\overrightarrow{\omega_{i}}\right)^2+2\cdot\overrightarrow{\alpha}\cdot\overrightarrow{\Delta\theta}
\overrightarrow{\Delta\theta}=\frac12\left(\overrightarrow{\omega_{f}}+\overrightarrow{\omega_{i}}\right)\cdot\Delta t
\overrightarrow{\Delta\theta}=\overrightarrow{\omega_{i}}\cdot\Delta t+\frac12\cdot\overrightarrow{\alpha}\cdot\Delta t^2
v_{T}=r\cdot\omega —tangential velocity
a_{T}=r\cdot\alpha —tangential acceleration
a_{c}=\frac{v_{T^{}}^2}{r}=r\cdot\omega^2 —centripetal acceleration
\overrightarrow{F_{c}}=m\cdot\overrightarrow{a_{c}} —centripetal force (net force required to maintain circular motion)
v=\sqrt{rg\tan\theta} —Banked curve formula
\overrightarrow{F_{g}}=G\cdot\frac{m_1\cdot m_2}{r^2} —universal gravitational force. “r”=distance between the centers of the objects, “G” is gravitational constant (6.674×10^-11 m³/kg*s² or N*m²*kg^-2)
U_{g}=-G\cdot\frac{m_1\cdot m_2}{r}
\char"03A4 =\frac{4\pi^2}{GM}\cdot r^3 “T”=orbital period, “M”=Mass of orbited body, “r”=distance between orbited and orbiting body.
v_{esc}=\sqrt{\frac{2GM}{r}} —Escape velocity. “M'“=Mass of orbited body
Torque (Rotational Dynamics)
\overrightarrow{\tau}=\overrightarrow{F}\cdot r\cdot\sin\theta
\Sigma\overrightarrow{\tau}=I\cdot\overrightarrow{\alpha} (“I”=moment of inertia in kg*m², “alpha”=angular acceleration)
KE_{R}=\frac12I\omega^2 (“omega”=angular speed)
I=I_{\operatorname{cm}}+mr^2 —Parallel Axis Theorem
\overrightarrow{L}=I\cdot\overrightarrow{\omega} —Angular Momentum (“L”=angular momentum)
\overrightarrow{\Delta L}=\overrightarrow{\tau}\cdot\Delta t —Angular Impulse
Simple Harmonic Motion
\overrightarrow{F_{rest}}=-k\cdot\overrightarrow{\Delta x} —Hooke’s Law
v_{x}=\pm\sqrt{\frac{k}{m}\left(A^2-x^2\right)}
\char"03A4 =\frac{1}{f}
f=\frac{1}{\char"03A4 }
\char"03A4 _{spring}=2\pi\sqrt{\frac{m}{k}} —Period of mass-spring system. “m”=Mass; “k”=spring constant
f_{spring}=\frac{1}{2\pi}\sqrt{\frac{k}{m}} —Frequency of mass-spring system
\char"03A4 _{pendulum}=2\pi\sqrt{\frac{L}{g}} —Period of simple pendulum system. “L”=Length
f_{pendulum}=\frac{1}{2\pi}\sqrt{\frac{g}{L}} —Frequency of simple pendulum system
\char"03A4 =\frac{2\pi}{\omega}
f=\frac{\omega}{2\pi}
x=A\cos\left(2\pi ft\right) (“x”=position)
v=-A\cdot2\pi f\sin\left(2\pi ft\right) (“v”=speed)
a=-A\cdot4\pi^2f^2\cos\left(2\pi ft\right) (“a”=acceleration)
Fluids
\rho=\frac{m}{v} (“rho”=density)
P=\frac{F}{A} (“P”=pressure)
P=P_0+\rho gh —Absolute pressure; Atmospheric pressure is 101.3 kPa; “h” is distance below the surface
P_{gauge}=\rho gh —Gauge pressure
F_{b}=\rho_{fluid}\cdot V_{displaced}\cdot g
\frac{F_1}{A_1}=\frac{F_2}{A_2} —Pascal’s principle
Q=\frac{Volume}{Time}=\frac{A\cdot d}{\Delta t}=Av —Flow Rate. A=cross sectional area, v=speed
A_1v_1=A_2v_2 —Equation of Continuity
P_1+\frac12\rho v_1^2+\rho gh_1=P_2+\frac12\rho v_2^2+\rho gh_2 —Bernoulli’s equation. “P”=static pressure, “rho”=density of fluid, “v”=velocity of fluid, “h”=height from zero level.
v=\sqrt{2gh} —Torricelli’s Theorem.