I'm trying to rearange this equation for Mu s That is the greek letter Mu subscript s... the static coeficent of friction m2 = m1 sin theta - Mu s m1 cos theta ok I got this Mu s = -((m1 cos theta)^-1 (m2 - m1 sin theta)) ok I know that sin over cosine is tangent but I can...
a wire has a resistance of 10 ohms. The length of the wire is doubled, and its radius is also doubled. What is the new resistance?
a block (mass m1) on a smooth horizontal surface, connected by a thin cord that passes over a pulley to as second block (m2), which hangs vertically. Ignore friction and the masses of the pulley and cord. If m1 = 13.0 kg and m2 = 6.0 kg determine the accleeration of each block...
I took algebra two and there not really verticle asymptotes threr limits you have been lied to which you should be used to if you are going to a public high school All I remeber is you try to get them to cancel out and get removable discontinuties or something this was a long ...
I think you mean the limits
thank you so much
m2a=m2g-m1a m2a+m1a=m2g a= m2g/(m1+m2) I got this m2a=m2g-m1a m2a + m1 a = m2g-m1a + m1 a m2 a + m1 a = m2g (m2 a + m1 a = m2g(m2 + m1)^-1 = a + a = (m2 + m1)^-1 m2g a + a = (m2 + m1)^-1 m2g 2a = (m2 + m1)^-1 m2g (2a = (m2 + m1)^-1 m2g)2^-1 = a = (2(m2 + m1))^-1 m2g a = (2(m2 ...
I was taking upwards in the y direction positve and downwards in the y dierction negetive my bad
a = m2^-1 (m1 a) -g solve for acceleartion it's the same post below they cancel out
y^2 + 6y -12 y -72 simplify