The dimensions are
"[\\rho] = \\frac{M}{L^3}"
"[v] = \\frac{L}{T}" hence
"\\bigg[\\frac{\\rho v^2}{2}\\bigg] = \\frac{M}{L^3} \\frac{L^2}{T^2} = M L^{-1} T^{-2}" On the other hand
"[p] = \\bigg[\\frac{F}{S}\\bigg] = \\frac{M L}{T^2} \\frac{1}{L^2} = M L^{-1} T^{-2}" The dimensions are the same.
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