Solve m₁g-t=m₁a₁, m₂g-½t=m₂a₂, m₃g-½t=m₃a₃, -2a₁=a₂+a₃, for t,a₁,a₂,a₃, where t is a tension of the cord of the pulley A

Solve
m₁g-t=m₁a₁,
m₂g-½t=m₂a₂,
m₃g-½t=m₃a₃,
-2a₁=a₂+a₃,
for t,a₁,a₂,a₃
where t is a tension of the cord of the pulley A pic.twitter.com/O9K1sCoidm

— Physics I (@PHY210) October 18, 2016

Solve
m₁g-t=m₁a₁,
m₂g-½t=m₂a₂,
m₃g-½t=m₃a₃,
-2a₁=a₂+a₃,
for t,a₁,a₂,a₃
where t is a tension of the cord of the pulley A

Physics Tutor

https://t.co/9OMwyeWNZe #phy210 #phy215 #GeneralPhysicsI #GeneralPhysics1 #GeneralPhysics #PhysicsI #Physics1 #Physics #PHYS1441 #физика pic.twitter.com/8BomlH4f0d

— Physics Tutor (@PhysicsTutor1) October 15, 2016

Calculus Based Physics, Chapter 7, Problem 42. Solution

Capter 7, Problem 42. Solution #Physics #problem #PhysicsProblem #solution #ProblemSolution

A photo posted by Vasiliy S. Znamenskiy (@znamenski) on Oct 5, 2016 at 4:49pm PDT

Power Output

#ground #energy #lost #friction #air #resistance #Derive #formula #intermsof #variables #FinalSpeed #PowerOutput #EnergyLost #AirResistance pic.twitter.com/bSkHZ2yv3r

— Physics I (@PHY210) October 1, 2016