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Chapter 1: Problem 8

If \(p\) is a pressure, \(V\) a velocity, and \(\rho\) a fluid density, what are the dimensions (in the \(M L T\) system) of (a) \(p / \rho\) (b) \(p V \rho,\) and (c) \(p / \rho V^{2} ?\)

Short Answer

Expert verified
The dimensions in M L T system are: (a) \(L^{2} T^{-2}\), (b) \(M^{2} L^{-4} T^{-3}\), and (c) \(M^{0} L^{0} T^{0}\) (which is dimensionless)

Step by step solution

01

Identify the Dimensions of the given Parameters

Let's start by identifying the dimensions in the \(M L T\) system: - Pressure (\(p\)), has units of force per area, which is \(Mass \cdot Acceleration / Area\). Thus, \(p\) has dimensions \(M L^{-1} T^{-2}\). - Velocity (\(V\)), is distance over time, which gives us dimensions of \(L T^{-1}\)- Density (\(\rho\)), is mass over volume, so its dimensions are \(M L^{-3}\)
02

Evaluate the Dimensions for p/ρ

For part (a), we want the dimensions of \(p / \rho = (M L^{-1} T^{-2}) / (M L^{-3})\). Canceling terms, we find that the units of \(p / \rho\) are \(L^{2} T^{-2}\)
03

Evaluate the Dimensions for pVρ

For part (b), we want the dimensions of \(p V \rho = (M L^{-1} T^{-2}) \cdot (L T^{-1}) \cdot (M L^{-3})\). Multiplying these together, we find that the units of \(p V \rho\) are \(M^{2} L^{-4} T^{-3}\)
04

Evaluate the Dimensions for p/ρV²

For part (c), we want the dimensions of \(p / (\rho V^{2}) = (M L^{-1} T^{-2}) / ((M L^{-3}) \cdot (L T^{-1})^{2})\). Dividing these terms, we find that the units of \(p / (\rho V^{2})\) are \(M^{0} L^{0} T^{0}\), or dimensionless

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