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Chapter 15: Problem 15

Viscosity is measured as follows. Write each in acceptable form using base units only: a) \(\frac{17}{9800} \mathrm{pa} \cdot \mathrm{s}\) b) \(\frac{32}{8741} \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}\) c) \(\frac{61}{34982} \mathrm{kN} \cdot \mathrm{h} / \mathrm{mm}^{2}\)

Short Answer

Expert verified
Question: Convert and simplify the following expressions into their base units: a) \(\frac{17}{9800} \mathrm{~Pa} \cdot \mathrm{s}\) b) \(\frac{32}{8741} \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}\) c) \(\frac{61}{34982} \mathrm{~kN} \cdot \mathrm{h} / \mathrm{mm}^{2}\) Answer: a) \(\frac{17}{9800} \mathrm{~Pa} \cdot \mathrm{s}\) b) \(\frac{32}{8741} \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}\) c) \(\frac{219600000}{0.34982} \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}\)

Step by step solution

01

Understanding the given quantities and their units

There are three quantities given, which are required to be written in the base units. The units are Pascal (Pa), second (s), Newton (N), meter (m), kilonewton (kN), hour (h), and millimeter (mm).
02

Convert the first quantity in base units

For the first quantity, we have the expression \(\frac{17}{9800} \mathrm{pa} \cdot \mathrm{s}\). As Pascal (Pa) is already a base unit, we don't need to perform any conversion for this term. Thus, the expression remains: \(\frac{17}{9800} \mathrm{pa} \cdot \mathrm{s}\)
03

Convert the second quantity in base units

The second quantity is given by \(\frac{32}{8741} \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}\). In this case, all units (Newton, second, and meter squared) are already in their base units. Therefore, we don't need to perform any conversion for this expression either. It remains: \(\frac{32}{8741} \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}\)
04

Convert the third quantity in base units

The third expression is \(\frac{61}{34982} \mathrm{kN} \cdot \mathrm{h} / \mathrm{mm}^{2}\). Here, we need to convert kilonewton (kN) to Newton (N), hour (h) to seconds (s), and millimeter squared (mm^2) to meter squared (m^2). We know the following conversion factors: 1 kN = 1000 N 1 h = 3600 s 1 mm = 0.001 m Applying these conversions: \(\frac{61}{34982} \cdot 1000 \mathrm{N} \cdot 3600 \mathrm{s} / (0.001 \mathrm{m})^{2} = \frac{61 \cdot 1000 \cdot 3600}{34982 \cdot (0.001)^{2}} \mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2}\)
05

Simplify the expressions

Now, we will simplify the expressions for all three quantities: a) \(\frac{17}{9800} \mathrm{pa} \cdot \mathrm{s}\) b) \(\frac{32}{8741} \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}\) c) \(\frac{61 \cdot 1000 \cdot 3600}{34982 \cdot (0.001)^{2}} \mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2} = \frac{219600000}{0.34982} \mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2}\) With these simplified expressions, we have converted and simplified all the given quantities in their base units.

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Most popular questions from this chapter

A jet plane is traveling at \(900 \mathrm{~km} / \mathrm{h}\) at an altitude where the temperature is such that the speed of sound is \(600 \mathrm{~km} / \mathrm{h}\). a) Determine the Mach number for the jet plane. b) How will the answer in part (a) be different if the speeds were given in miles per hour?

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