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(D) Given dimensional formulas:$[a] = \alpha \beta^{-2} = LT^{-2}$$[v] = \alpha \beta^{-1} = LT^{-1}$$[\ell] = \alpha \gamma = L$From $[v] = \alpha \b

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$\alpha^0 \beta^0 \gamma^0$

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(D) Given dimensional formulas:$[a] = \alpha \beta^{-2} = LT^{-2}$$[v] = \alpha \beta^{-1} = LT^{-1}$$[\ell] = \alpha \gamma = L$From $[v] = \alpha \beta^{-1} = LT^{-1}$ and $[a] = \alpha \beta^{-2} = LT^{-2}$,we can divide the two:$\frac{[v]}{[a]} = \frac{\alpha \beta^{-1}}{\alpha \beta^{-2}} = \frac{LT^{-1}}{LT^{-2}} = T$Thus,$[\beta] = T$.Substituting $[\beta] = T$ into $[v] = \alpha \beta^{-1} = LT^{-1}$:$\alpha T^{-1} = LT^{-1} \Rightarrow [\alpha] = L$.Substituting $[\alpha] = L$ into $[\ell] = \alpha \gamma = L$:$L \gamma = L \Rightarrow [\gamma] = 1$ (dimensionless).The coefficient of friction is a ratio of two forces (frictional force and normal force),so it is a dimensionless quantity,represented by $[M^0 L^0 T^0]$.Checking the options:Option $(D)$ is $\alpha^0 \beta^0 \gamma^0 = 1$,which is dimensionless. Therefore,the correct option is $(D)$.

The dimensional formulas for acceleration,velocity,and length are $\alpha \beta^{-2}$,$\alpha \beta^{-1}$,and $\alpha \gamma$ respectively. What is the dimensional formula for the coefficient of friction?

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