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Chapter 1: Problem 20

Express the following quantities in SI units: (a) 10.2 in./min, (b) 4.81 slugs,(c) \(3.02 \mathrm{lb}\) (d) \(73.1 \mathrm{ft} / \mathrm{s}^{2}\) (e) \(0.0234 \mathrm{lb} \cdot \mathrm{s} / \mathrm{ft}^{2}\).

Short Answer

Expert verified
The given quantities in SI units are: (a) 0.00427 m/s, (b) 70.1653 kg, (c) 1.3698 kg, (d) 22.2777 m/s², (e) 0.4553 kg·s/m²

Step by step solution

01

Convert Inches per Minute to Meters per Second

There are 0.0254 meters in an inch and 60 seconds in a minute, so to convert 10.2 in./min into meters per second (m/s), the calculation should be as follows: \(10.2 \ in./min \times \frac{0.0254 \ m}{1 \ in.} \times \frac{1 \ min}{60 \ s} = 0.00427 \ m/s \)
02

Convert Slugs to Kilograms

In SI units, mass is measured in kilograms (kg). The conversion factor from slugs to kilograms is 14.5939, so to convert 4.81 slugs into kilograms, the calculation should be as follows: \(4.81 \ slugs \times 14.5939 = 70.1653 \ kg \)
03

Convert Pounds to Kilograms

The conversion factor from pounds to kilograms is 0.453592, so to convert 3.02 lb into kilograms, the calculation should be as follows: \(3.02 \ lb \times 0.453592 = 1.3698 \ kg \)
04

Convert Feet per Second Squared to Meters per Second Squared

There are 0.3048 meters in a foot. To convert 73.1 ft/s² to meters per second squared (m/s²), the calculation should be: \(73.1 \ ft/s² \times 0.3048 \ m/ft = 22.2777 \ m/s² \)
05

Convert Pound-Seconds per Feet Squared to Kilogram-Seconds per Meter Squared

To convert pound-seconds per feet squared (0.0234 lb·s/ft²) to kilogram-seconds per meter squared (kg·s/m²), you will need to use multiple conversion factors including the one for pound to kilogram (0.453592 kg/lb), second to second (1 s/s), and feet to meter (0.3048 m/ft). The total conversion step is as follows: \(0.0234 \ lb·s/ft² \times 0.453592 \ kg/lb \times 1 \ s/s \times (1/0.3048)^2 \ m²/ft² = 0.4553 \ kg·s/m² \)

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Most popular questions from this chapter

A regulation basketball is initially flat and is then inflated to a pressure of approximately \(24 \mathrm{lb} / \mathrm{in}^{2}\) absolute. Consider the air temperature to be constant at \(70^{\circ} \mathrm{F}\). Find the mass of air required to inflate the basketball. The basketball's inside radius is 4.67 in.

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