What are the dimensions of a and b coefficients in the relation of P=ar^-1+bsqrt (x) where Fis force, xis position and tis -3 time [b]=[M][L]^1/2[T]^2[TII]M[M]=[a] Ca 1/2[T]^-1/2[b]=[M][L]^1/2[T]^-4[L][M]=[a] (b Ma[(I)]^2[b]=[M][L][C]^2(M)=[a] 12[L][C]^2[b]=[M][L][T]M[M]=[a] II_(2)[I]^2[D]=[M][M^1/2[T]^2[L][M]=[a]-(e

The dimensions of the coefficients a and b in the relation can be determined by analyzing the dimensions of each term in the equation.In this case, the force (F) is given as a function of position (x) and time (t). The dimensions of force are [M][L][T]^{-2}, where [M] represents mass, [L] represents length, and [T] represents time. The dimensions of position (x) are [L], and the dimensions of time (t) are [T].Let's analyze the dimensions of each term in the equation:1. : The dimensions of this term are [M][L][T]^{-2} / [L] = [M][T]^{-2}. Therefore, the dimensions of a must be [M][T]^{-2}.2. : The dimensions of this term are [M][L][T]^{-2} / [L]^{1/2} = [M][L]^{1/2}[T]^{-2}. Therefore, the dimensions of b must be [M][L]^{1/2}[T]^{-2}.Therefore, the dimensions of the coefficients a and b in the relation are [M][T]^{-2} and [M][L]^{1/2}[T]^{-2}, respectively.