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Chapter 1: Problem 6

If \(u\) is a velocity, \(x\) a length, and \(t\) a time, what are the (a) \(\partial u / \partial t\) dimensions (in the \(M L T\) system) of (b) \(\partial^{2} u / \partial x \partial t,\) and (c) \(\int(\partial u / \partial t) d x ?\)

Short Answer

Expert verified
The dimensions of (a) \(\frac{\partial u}{\partial t}\) are \([L][T]^{-2}\), (b) \(\frac{\partial^2 u}{\partial x \partial t}\) are \([L]^{-1}[T]^{-2}\), and (c) \(\int(\partial u / \partial t) dx\) are \([L]^2[T]^{-2}\) in the MLT system.

Step by step solution

01

Calculate dimensions of \(\frac{\partial u}{\partial t}\)

Velocity (\(u\)) is length per time, so its dimension is \([L][T]^{-1}\). Thus, \(\frac{\partial u}{\partial t}\) would add an extra \(1/T\) to the dimensions of \(u\), making the dimension of \(\frac{\partial u}{\partial t}\) to be \([L][T]^{-2}\).
02

Calculate dimensions of \(\frac{\partial^2 u}{\partial x \partial t}\)

From the dimensions of \(\frac{\partial u}{\partial t}\) calculated in Step 1, \(\frac{\partial^2 u}{\partial x \partial t}\) would add an extra \(1/L\) to the dimensions, since \(\partial x\) on the denominator indicates division by length. Therefore, the dimension of \(\frac{\partial^2 u}{\partial x \partial t}\) is \([L]^{-1}[T]^{-2}\).
03

Calculate dimensions of \(\int(\partial u / \partial t) dx\)

Integration will add length to the dimensions of \(\frac{\partial u}{\partial t}\), since \(\int dx\) indicates multiplication by length. Thus, the dimension of \(\int(\partial u / \partial t) dx\) would be \([L][T]^{-2} \times [L]\), which equals \([L]^2[T]^{-2}\).

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Most popular questions from this chapter

A stick of butter at \(35^{\circ} \mathrm{F}\) measures 1.25 in. \(\times 1.25\) in. \(x\) 4.65 in. and weighs 4 ounces. Find its specific weight.

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