Question

Perform the following conversions: (a) \(5.00\) days to s, (b) \(0.0550 \mathrm{mi}\) to \(\mathrm{m}\), (c) \(\$ 1.89 / \mathrm{gal}\) to dollars per liter, (d) \(0.510\) in. \(/ \mathrm{ms}\) to \(\mathrm{km} / \mathrm{hr}\), (e) \(22.50 \mathrm{gal} / \mathrm{min}\) to \(\mathrm{L} / \mathrm{s}\), (f) \(0.02500 \mathrm{ft}^{3}\) to \(\mathrm{cm}^{3}\)

Short answer

The short answers for the conversions are: (a) 432000 s (b) 88.514 m (c) 0.499 dollars per liter (d) 1.839 km/hr (e) 14.157 L/s (f) 707.921 cm³

Step-by-step solution

Identify the conversion factor

We know that there are 24 hours in a day, 60 minutes in an hour, and 60 seconds in a minute. So, we have to first convert days to hours, hours to minutes, and finally, minutes to seconds.

Perform the conversion

To perform the conversion, we multiply the given quantity by the conversion factors: \[5.00~\text{days} \times \frac{24~\text{hours}}{1~\text{day}} \times \frac{60~\text{minutes}}{1~\text{hour}} \times \frac{60~\text{seconds}}{1~\text{minute}} = 432000~\text{s}\] (b) 0.0550 mi to m ---

Identify the conversion factor

We need to convert miles to meters. We know that 1 mile = 1609.34 meters.

Perform the conversion

Now, multiply the given quantity by the conversion factor: \[0.0550~\text{mi} \times \frac{1609.34~\text{m}}{1~\text{mi}} = 88.514~\text{m}\] (c) \(1.89 / \mathrm{gal}\) to dollars per liter ---

Identify the conversion factor

We need to convert gallons to liters. We know that 1 gallon = 3.78541 liters.

Perform the conversion

Multiply the given quantity by the conversion factor: \[\frac{1.89~\text{dollars}}{1~\text{gal}} \times \frac{1~\text{gal}}{3.78541~\text{L}} = 0.499~\text{dollars per liter}\] (d) 0.510 in. \(/ \mathrm{ms}\) to \(\mathrm{km} / \mathrm{hr}\) ---

Identify the conversion factors

We need to convert inches to kilometers and milliseconds to hours. We know that 1 inch = 0.0254 meters, and 1 kilometer = 1000 meters, 1 millisecond = 0.001 second, and 1 hour = 3600 seconds.

Perform the conversion

Multiply the given quantity by the conversion factors: \[\frac{0.510~\text{in}}{1~\text{ms}} \times \frac{0.0254~\text{m}}{1~\text{in}} \times \frac{1~\text{km}}{1000~\text{m}} \times \frac{0.001~\text{s}}{1~\text{ms}} \times \frac{1~\text{hr}}{3600~\text{s}} = 1.839~\frac{\text{km}}{\text{hr}}\] (e) \(22.50 \mathrm{gal} / \mathrm{min}\) to \(\mathrm{L} / \mathrm{s}\) ---

Identify the conversion factors

We need to convert gallons to liters and minutes to seconds. We know that 1 gallon = 3.78541 liters and 1 minute = 60 seconds.

Perform the conversion

Multiply the given quantity by the conversion factors: \[\frac{22.50~\text{gal}}{1~\text{min}} \times \frac{3.78541~\text{L}}{1~\text{gal}} \times \frac{1~\text{min}}{60~\text{s}} = 14.157~\frac{\mathrm{L}}{\mathrm{s}}\] (f) \(0.02500 \mathrm{ft}^{3}\) to \(\mathrm{cm}^{3}\) ---

Identify the conversion factor

We need to convert cubic feet to cubic centimeters. We know that 1 ft = 30.48 cm, so 1 cubic foot (ft³) = (30.48 cm)³ = 28316.85 cm³.

Perform the conversion

Multiply the given quantity by the conversion factor: \[0.02500 ~\mathrm{ft}^{3} \times \frac{28316.85 ~\mathrm{cm}^{3}}{1 ~\mathrm{ft}^{3}} = 707.921 ~\mathrm{cm}^{3}\]

Key concepts

Conversion Factors

Conversion factors are essential tools in the world of unit conversion. They act as a bridge between two different units of measure, allowing us to convert from one unit to another. These factors are numerical values that relate the amount of one unit to the equivalent value in another unit.

For example, when converting from days to seconds, as mentioned in the textbook solution, you need to understand how many of the smaller unit (seconds) equal one of the larger unit (days). Knowing that one day equals 24 hours, one hour equals 60 minutes, and one minute equals 60 seconds, you can create a conversion factor that when multiplied by the initial value, gives you the result in the desired unit. It is a chain of simple multiplications when several conversions are involved, like converting days to seconds through hours and minutes.

In any conversion, clarity about what you want to convert (distance, volume, time, etc.) and using the correct conversion factor are crucial. Remember, the unit you want to get rid of should be in the denominator, and the unit you're converting to should be in the numerator.

Dimensional Analysis

Dimensional analysis is a systematic way of checking the consistency within equations or calculations involving measurements. It is a powerful technique often used for unit conversions. The process uses the fact that you can multiply by a form of one (such as 24 hours / 1 day) without changing the value, just the units.

Let's take a closer look at the example of converting miles to meters. Starting with the known quantity of 0.0550 miles, we apply the conversion factor 1609.34 meters per mile. In dimensional analysis, we ensure that the units align so that unwanted units cancel out, leaving us with the desired units. This makes calculations not only accurate but also helps in understanding the relationship between different units.

Dimensional analysis is not just a tool for simple conversions; it can also help solve more complex problems and ensure that equations make sense dimensionally - that's to say, the units on one side of an equation match the units on the other side.

The Metric System

The metric system is an international system of measurement based on multiples of ten, which makes it very easy to use and understand. Most of the world uses this system, and it includes units like meters for length, liters for volume, and grams for mass.

In the metric system, prefixes like milli-, centi-, and kilo- indicate the scale of measurement and are used to describe units smaller or larger than the base unit. For example, 1 kilometer is 1000 meters, and 1 milliliter is 0.001 liters. When solving unit conversion problems, it's essential to know these prefixes and use them correctly. This is especially important when converting between systems, like from the metric system to the imperial system used predominantly in the United States, as it often requires precise conversion factors to maintain accuracy.

Understanding the metric system's simplicity can help students grasp the concept of unit conversions more quickly. It reduces the complexity of remembering numerous distinct units and concentrates on the changes in magnitude signified by the different prefixes.