If z = xP + G,where P is pressure and G is th | vedclass

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(C) According to the principle of homogeneity of dimensions,the dimensions of each term in an equation must be the same.Given the equation $z = xP + G

Vedclass · Accepted Answer

$M^{-2} L^4 T^0, M^{-1} L^3 T^{-2}$

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(C) According to the principle of homogeneity of dimensions,the dimensions of each term in an equation must be the same.Given the equation $z = xP + G$,it implies that $[z] = [xP] = [G]$.First,find the dimensions of $G$ (universal gravitational constant):$[G] = \frac{[F][r^2]}{[m_1][m_2]} = \frac{[MLT^{-2}][L^2]}{[M][M]} = [M^{-1} L^3 T^{-2}]$.Since $[z] = [G]$,the dimension of $z$ is $[M^{-1} L^3 T^{-2}]$.Next,find the dimensions of $P$ (pressure):$[P] = \frac{[Thrust]}{[Area]} = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1} T^{-2}]$.Since $[z] = [xP]$,we have $[x] = \frac{[z]}{[P]} = \frac{[M^{-1} L^3 T^{-2}]}{[ML^{-1} T^{-2}]} = [M^{-2} L^4 T^0]$.Therefore,the dimensions of $x$ and $z$ are $[M^{-2} L^4 T^0]$ and $[M^{-1} L^3 T^{-2}]$ respectively.

If $z = xP + G$,where $P$ is pressure and $G$ is the universal gravitational constant; then the dimensional formulas for $x$ and $z$ respectively are (here,$G = \frac{Fr^2}{m_1 m_2}$,$P = \frac{\text{Thrust}}{\text{Area}}$).

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