Question

Evaluate the following conversion. Will the answer be correct? Explain. $$\text { rate }=\frac{75 \mathrm{m}}{1 \mathrm{s}} \times \frac{60 \mathrm{s}}{1 \mathrm{min}} \times \frac{1 \mathrm{h}}{60 \mathrm{min}}$$

Short answer

The converted rate is \(\frac{5}{4}\mathrm{h}\), and the answer will be correct because we used valid conversion factors and properly canceled out the units. The initial given rate of 75 meters per second is equal to \(\frac{5}{4}\) hours.

Step-by-step solution

Identify the given rate

The given rate is: $$\text{rate} = \frac{75\mathrm{m}}{1\mathrm{s}}$$ This means that the object is moving at a speed of 75 meters per second.

Analyze the conversion factors

The conversion factors given are: $$\frac{60\mathrm{s}}{1\mathrm{min}}$$ This factor will convert seconds to minutes. The other conversion factor is: $$\frac{1\mathrm{h}}{60\mathrm{min}}$$ This factor will convert minutes to hours.

Multiply the given rate by the conversion factors

Now, we will multiply the given rate by the conversion factors to convert meters per second to the desired unit: $$\text{converted rate} = \frac{75\mathrm{m}}{1\mathrm{s}} \times \frac{60\mathrm{s}}{1\mathrm{min}} \times \frac{1\mathrm{h}}{60\mathrm{min}}$$

Cancel out the units and simplify the expression

Upon multiplying, we cancel out the same units from the numerator and the denominator: $$\text{converted rate} = \frac{75\cancel{\mathrm{m}}}{1\cancel{\mathrm{s}}} \times \frac{60\cancel{\mathrm{s}}}{1\mathrm{min}} \times \frac{1\mathrm{h}}{60\cancel{\mathrm{min}}}$$ Now we can simplify the numerical expression: $$\text{converted rate} = \frac{75 \times 1\mathrm{h}}{ 1\times 1\times 60}$$ $$\text{converted rate} = \frac{75}{60} \mathrm{h}$$

Simplify the final fraction and determine the correct answer

Simplify the fraction by dividing the numerator and the denominator by their greatest common divisor (15): $$\text{converted rate} = \frac{75}{60} = \frac{5}{4}$$ So the converted rate is: $$\text { rate }= \frac{5}{4}\mathrm{h}$$ Since we have used valid conversion factors and properly canceled out the units, our answer should be correct. The initial given rate of 75 meters per second is equal to \(\frac{5}{4}\) hours.

Key concepts

Metric System Conversion

The metric system is the most widely used system of measurement in the world and is based on the powers of ten for its units of length, mass, and volume, among others. Converting within the metric system involves shifting the decimal point to the right or left, depending on whether you need to convert to a smaller or larger unit, respectively. In this context, we are interested in converting a rate of speed from meters per second to another unit of speed.

For example, the exercise requires converting a speed from meters (the base unit for length in the metric system) per second into a speed given in terms of hours. To perform this conversion properly, we need to use conversion factors that relate the different units of time—seconds to minutes, and minutes to hours. The conversion rates are universally accepted: 60 seconds in one minute and 60 minutes in one hour. These rates become our conversion factors when we seek to change the unit of time while maintaining the same physical quantity of speed.

Dimensional Analysis

Conversion Factors and Unit Cancellation

Dimensional analysis, also known as the factor-label method or unit factor method, is a technique used to convert one set of units to another. It involves multiplying the quantity by conversion factors, which are fractions that represent the equivalence between two units. An essential component of this technique is the cancellation of units to ensure the correct conversion.

In our exercise, dimensional analysis is exemplified when the speed given in meters per second is multiplied by the conversion factors 60 seconds per minute and 1 hour per 60 minutes. The correct conversion hinges on canceling out the original units (in this case, seconds and minutes) so that we are left only with the desired unit (hours). This methodical approach not only makes the conversion process systematic but also provides a clear visual representation that helps to prevent errors during calculations.

Rate and Speed Calculations

Understanding Rate and Speed

Rate and speed are often used interchangeably but can, in fact, refer to slightly different concepts. Rate can refer to any kind of quantity measured against another, such as liters per hour or miles per hour, whereas speed specifically refers to the distance traveled per unit of time. In chemistry, as well as in physics, understanding and calculating rates and speeds are crucial for discussing reaction rates, diffusion, and other time-dependent phenomena.

In the context of the original exercise, the rate being referred to is the speed of an object. Calculating this speed involves using the information provided about distances and times, and converting them as necessary to achieve the desired units. In the solution, the initial rate gets multiplied sequentially by valid conversion factors, and thorough simplification provides the final speed. Being adept at these calculations is essential for solving a vast array of problems in scientific disciplines.