Physics

Here's a little bit of physics.(my favorite equations)
d=r(t) or x=v(t) and x=Xo+Vo(t)+1/2(A)(T**2)

Here's a physics problem.

Find the acceleration of the system for m1>m2, m2>m1. Assume negligible friction in the pulley. The incline is a right triangle <(theta). The coefficient of friction of the surface=*m that is supposed to be a meew, a little squiggly m.
W=m1g W'=m2g Wx=m1g(cos(theta2)) Ffr=*m(Fn)=*m(m1g)(sin(theta2)) Wy=m1g(sin(theta2))=Fn

for m1>m2:

sum of forces in the Y direction=Wx-W'-Ffr

(m1+m2)a=m1g(cos(theta2))-m2g-*m(m1g)(sin(theta2))

a =m1g(cos(theta2))-m2g-*m(m1g)(sin(theta2))/(m1+m2)

for m2>m1:

sum of forces in the x direction=W'-Wx-Ffr

(m1+m2)a=m2g-m1g(cos(theta2))-*m(m1g)(sin(theta2))

a =m2g-m1g(cos(theta2))-*m(m1g)(sin(theta2))/(m1+m2)

There you have it, something you've always wanted to know.

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Physics

Here's a little bit of physics.(my favorite equations) d=r(t) or x=v(t) and x=Xo+Vo(t)+1/2(A)(T**2)

Here's a physics problem.

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Here's a little bit of physics.(my favorite equations)
d=r(t) or x=v(t) and x=Xo+Vo(t)+1/2(A)(T**2)