Question

Carry out the following conversions: (a) 0.105 in. to $\mathrm{mm},$ (b) 0.650 qt to $\mathrm{mL},$ (c) $8.75\mu \mathrm{m}/\mathrm{s}$ to $\mathrm{km}/\mathrm{hr}$, (d) $1.955{\mathrm{m}}^3$ to ${\mathrm{yd}}^3$, (e) $$\mathrm{\$}3.99/\mathrm{lb}$$ to dollars per $\mathrm{kg}$, (f) $8.75\mathrm{lb}/{\mathrm{ft}}^3$ to $\mathrm{g}/\mathrm{mL}$

Short answer

The short answer for each conversion is as follows: (a) 0.105 in. = 2.667 mm (b) 0.650 qt = 615.13 mL (c) 8.75 μm/s = 0.0000275 km/hr (d) 1.955 m³ = 2.257 yd³ (e) $3.99/lb = $8.80/kg (f) 8.75 lb/ft³ = 0.139 g/mL

Step-by-step solution

(a) 0.105 in. to mm

Conversion factor: 1 in. = 25.4 mm 0.105 in. x 25.4 mm = 2.667 mm

(b) 0.650 qt to mL

Conversion factor: 1 qt = 946.353 mL 0.650 qt x 946.353 mL = 615.13 mL

(c) 8.75 μm/s to km/hr

Conversion factor: 1 μm = 1 x 10⁻⁶ km and 1 s = 1/3600 hr 8.75 μm/s x (1 x 10⁻⁶ km / 1 μm) x (1 hr / 3600 s) = 0.0000275 km/hr

(d) 1.955 m³ to yd³

Conversion factor: 1 m³ = 1.30795 yd³ 1.955 m³ x 1.30795 yd³ = 2.257 yd³

(e) $3.99/lb to dollars per kg

Conversion factor: 1 lb = 0.453592 kg (3.99 $/lb)÷(0.453592kg/lb)=8.80$/kg

(f) 8.75 lb/ft³ to g/mL

Conversion factor: 1 lb = 453.592 g and 1 ft³ = 28316.8 mL (8.75 lb/ft³) x (453.592 g/lb) x (1 ft³ / 28316.8 mL) = 0.139 g/mL

Key concepts

Dimensional Analysis

Dimensional analysis is a mathematical technique widely used in science and engineering to convert one unit of measure to another. It's like a road trip where you map out your path using the specific directions and distances between points. In the classroom, it's an essential tool in understanding the relationship between different units and mastering unit conversions.

To carry out dimensional analysis, you start with the quantity you want to convert and multiply it by a series of conversion factors. These factors are fractions that represent the equality between units but are written in a form that allows canceling out the units you're converting from. It’s like exchanging currency when traveling abroad - you start with what you have and, using the exchange rate (a conversion factor), you end up with the currency you need.

For example, step 2 of the exercise given uses dimensional analysis to convert 0.650 quarts to milliliters. The solution uses the conversion factor that 1 quart is equivalent to 946.353 milliliters to navigate from quarts to milliliters.

Conversion Factors

Conversion factors are the exchange rates between units of measurement that you use during dimensional analysis. They allow you to convert quantities from one unit to another seamlessly. A conversion factor is created from any two quantities known to be equivalent. One key aspects when working with conversion factors is the principle of unity; it states that any number or quantity multiplied by one remains unchanged. Conversion factors are essentially a sophisticated form of the number one.

Imagine you're translating a book from one language to another. Each conversion factor is like a snippet of a bilingual dictionary that tells you exactly how a unit in one language translates to a unit in another language without changing the meaning - or in this case, the value. In our exercises, each step includes a specific conversion factor to translate the units correctly. For instance, to convert dollars per pound to dollars per kilogram, step 5 utilizes the conversion factor that 1 pound is equivalent to about 0.453592 kilograms.

Metric Conversion

Metric conversion involves changing measurements within the metric system or to different systems of measurement, like the imperial system used in the United States. The metric system is a decimal-based system of measurement, which makes it very straightforward to move between units - just like adding zeroes or moving decimal points.

Analogous to resizing photos for different platforms, where the image is still the same but fits into various contexts, metric conversions reshape measurements into new units without altering the quantity. For example, converting micrograms per second to kilometers per hour, as in step 3, requires understanding both the metric unit prefixes (micro and kilo) and the time unit relationships (seconds and hours). It's this versatility and ease of use that makes metric conversions a fundamental skill in many fields, particularly science and engineering.