Recent content by budd99

Ok, I used 0.5 = 0 + 0.5a12 to find a, which gives 1. So, if F = ma, then the F on the 3kg block is 3(1) which is 3. On the block suspended in air, the F = mg. Since we already found F = 3, then 3 = m(9.8), and finally m = .31 This still doesn't make sense, maybe I'm forgetting a T...

Homework Statement The system shown (attached image) is released from rest and moves 50 cm in 1.0 s. What is the value of M? All surfaces are frictionless. a. 0.42 kg b. 0.34 kg c. 0.50 kg d. 0.59 kg e. 0.68 kg Homework Equations Since the incline is 90 degress, I am trying...

Ok, I think I figured it out... I took mv2/r (M(L/T)(L/T))/r I assumed r = L since both are length measures so... (M(LL/TT))/L (M(LL/TTL) M(L/TT) M(L/T2) I don't know if I should be equating r to L though...

I'm working on it, but what is the r in the answers? Also, why does the m become a capital M?

Which one of the quantities below has dimensions equal to ML/T2 a) mv b) mv2 c) mv2/r d) mrv e) mrv2/r2 I tried splitting it up in M*(L/T)*1/T which would lead me to mv*T-1 and then I get stuck Please help!