Recent content by dtfw

right M; a+b=0 because there is no number value L;-a-3b+c however from the left there is still LT^-1 therefore there is 1L T^-1 T; -2a-2c 1L left over if I am doing this wrong please advise , the example I have used before was a lot simpler than this where...

M; a+b=0 L; -a-3b+c=1T^-1 T; -2a-2c= 1L correct?

so LT^-1=M^a+b L^-a-3b+c T-2a-2c M; a+b=0 L; -a-3b+c=? T; -2a-2c=? Is that correct other than the = part?

dimensional analysis is completely new to me and I am going through the steps correctly , its literally how it ends up with a T^_1 being on one side. The only thing I can determine is that the equation doesn't work at all. If not I am utterly clueless, basically for the equation to work , the L...

so it is ML^-1T^-2 . thanks for your help however, this still honestly doesn't help where I am stuck at, it is still going to be as follows M^(a+b) L^(-1a+3b+c) T^(-2a+2c) M; a+b=0 L; -1a+3b+c=1 +T^-1and still there are too many unknowns to do anything with

sorry that's a type error it MLT^-2/L^2 from p=f/a rearranging is ML^-2T^-2 Is that correct or do you mean it is the wrong formula to insert ?

u=prg LT^-1 = MLT^-2 ML^-3 LT^-2 = (ML^-2T^-2)^a (ML^-3)^b (LT^-2)c L^2 LT^-1 = M^a L^-2a T^-2a M^b L^-3b L^c T^-2c =M^(a+b) L^(-2a+3b+c) T(-2a+2c) M; a+b=0 L;-2a+3b+c=1 + T^-1 Either the formula is mixed up or...