Question

In the Bernoulli's equation prove that 1/2 V^2 has same dimension as pressure.

Answer 1

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.