In physics, it is essential to use the International System of Units (SI) for consistency and accuracy in calculations. When faced with non-SI units, the first step is to convert them into SI units before applying them in equations. This process involves a systematic approach to unit conversions, which can be broken down into clear steps.
To illustrate the conversion process, consider converting 22 pounds into kilograms. Start by identifying the given unit (22 pounds) and the target unit (kilograms). The next step is to use conversion factors, which are relationships between different units, expressed as fractions. For instance, the conversion factor between pounds and kilograms can be represented as:
$$\frac{2.2 \text{ pounds}}{1 \text{ kilogram}} \quad \text{or} \quad \frac{1 \text{ kilogram}}{2.2 \text{ pounds}}$$
To determine which fraction to use, ensure that the units cancel appropriately. Since pounds are in the numerator, the conversion factor should have pounds in the denominator:
$$\frac{1 \text{ kilogram}}{2.2 \text{ pounds}}$$
Now, multiply the given value by the conversion factor:
$$22 \text{ pounds} \times \frac{1 \text{ kilogram}}{2.2 \text{ pounds}} = \frac{22}{2.2} \text{ kilograms} = 10 \text{ kilograms}$$
This means that 22 pounds is equivalent to 10 kilograms.
Next, let’s convert 67.5 miles per hour to meters per second. Start with the given unit (67.5 miles/hour) and the target unit (meters/second). This conversion requires two steps: converting miles to kilometers and then kilometers to meters. The conversion factors are:
1 mile = 0.621 kilometers and 1 kilometer = 1000 meters.
Set up the conversion as follows:
$$67.5 \text{ miles/hour} \times \frac{0.621 \text{ kilometers}}{1 \text{ mile}} \times \frac{1000 \text{ meters}}{1 \text{ kilometer}} \times \frac{1 \text{ hour}}{3600 \text{ seconds}}$$
When you multiply these together, the units will cancel appropriately, leaving you with meters per second. The calculation yields:
$$\frac{67.5 \times 1000}{0.621 \times 3600} \approx 30.2 \text{ meters/second}$$
Finally, consider converting 100 square feet to square meters. Since the unit is squared, the conversion factor must be applied twice. The conversion factor is:
1 foot = 0.305 meters.
Set up the conversion as follows:
$$100 \text{ ft}^2 \times \left(\frac{0.305 \text{ meters}}{1 \text{ foot}}\right)^2 = 100 \text{ ft}^2 \times \frac{0.305^2 \text{ m}^2}{1 \text{ ft}^2}$$
Calculating this gives:
$$100 \times 0.305 \times 0.305 \approx 9.3 \text{ m}^2$$
Through these examples, it is clear that mastering unit conversions is crucial for solving physics problems accurately. By following a structured approach and using appropriate conversion factors, you can confidently convert between different units and ensure your calculations are correct.
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