Answer:
[tex]\displaystyle m=\frac{2}{3},\ n=\frac{4}{3}[/tex]
Explanation:
Dimensional Analysis
It's given the relation between quantities A, B, and C as follows:
[tex]\displaystyle A=\frac{3}{2}B^mC^n[/tex]
and the dimensions of each variable is:
[tex]A=L^2T^2[/tex]
[tex]B=LT^{-1}[/tex]
[tex]C=LT^2[/tex]
Substituting the dimensions into the relation (the coefficient is not important in dimension analysis):
[tex]\displaystyle L^2T^2=\left(LT^{-1}\right)^m\left(LT^2\right)^n[/tex]
Operating:
[tex]L^2T^2=\left(L^mT^{-m}\right)\left(L^nT^{2n}\right)[/tex]
[tex]L^2T^2=L^{m+m}T^{-m+2n}[/tex]
Equating the exponents:
[tex]m+n=2[/tex]
[tex]-m+2n=2[/tex]
Adding both equations:
[tex]3n=4[/tex]
Solving:
[tex]n=4/3[/tex]
[tex]m=2-4/3=2/3[/tex]
Answer:
[tex]\mathbf{\displaystyle m=\frac{2}{3},\ n=\frac{4}{3}}[/tex]
