For the following dimensional equations, find the base dimensions of the paramet
ID: 1919075 • Letter: F
Question
For the following dimensional equations, find the base dimensions of the parameter k: a) MLt^-2 = kM(L^-1)(t^-2) b) ML(t^-2)(L^-1) = kLt^-3 c) (L^2)(t^-2) = k(M^4)(T^2) d) M(L^2)(t^-3) = k LT e) nL(L^3)k = (T^2)(M^-2)L f) M(I^2)k = nT(M^-3)(L^-1) g) I(L^2)t = (k^2)(M^4)(t^2) h) (k^3)(T^6)(M^3)(L^-5) = (T^-3)(t^-6)L I) (T^(-1/2))(L^-1)(I^2) = (k^(-1/2))(t^4)(T^(-5/2))(L^-3) J) ML(t^-2) = ML(t^-2) sin(k(L^-2)(M^-1)) K) (T^2)n = (T^2)n ln(kn(T^-1)) Please show some work!! this one really got me so thanks for any help!! =]Explanation / Answer
a) L^-2t^0M^0 = k b)ML^0(t^1)(L^-1) = k c) (M^-4)(L^2)(t^-4) = k d) M(L^1)(t^-4) = k e) k = (T^2)(M^-2)nL^-1(L^-3) f) k = nT(M^-4)(L^-1)(I^-2) g) I^(1/2)(L^1)(M^-2)(t^-1/2) = (k) h) (k) = (T^-5)(t^-2)(M^-1)(L^2) I) (T^(4))(L^4)(I^4)(t^-8) = (k) J) M^0L^0(t^0) = sin(k(L^-2)(M^-1)) DEGREE unit UNDER SIN SO K MUST CANCELL OUT THE MULTIPLYING UNITS : K = L^2M^1 K) (T^0)n^0 = ln(kn(T^-1)) SAME HERE : K = n^-1T^1