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Use dimensional analysis to change a. 50 meters per second to kilometers per hour. (Ti) b. \(0.025\) day to seconds. c. 1200 ounces to tons \((16 \mathrm{oz}=1 \mathrm{lb}\); \(2000 \mathrm{lb}=1\) ton).

Short Answer

Expert verified
a. 180 km/hr; b. 2160 sec; c. 0.0375 tons.

Step by step solution

01

Convert meters per second to kilometers per hour

To convert from meters per second to kilometers per hour, use the conversion factors: 1 kilometer = 1000 meters and 1 hour = 3600 seconds.The conversion is:\[50 \text{ m/s} = 50 \times \frac{1 \text{ km}}{1000 \text{ m}} \times \frac{3600 \text{ s}}{1 \text{ hr}} = 50 \times \frac{3600}{1000} \text{ km/hr} = 180 \text{ km/hr}\]
02

Convert days to seconds

To convert days to seconds, use the conversion factors:1 day = 24 hours, 1 hour = 60 minutes, and 1 minute = 60 seconds.The conversion is:\[0.025 \text{ day} = 0.025 \times 24 \text{ hr/day} \times 60 \text{ min/hr} \times 60 \text{ sec/min}\]\[0.025 \times 24 \times 60 \times 60 = 2160 \text{ seconds}\]
03

Convert ounces to tons

To convert from ounces to tons, use the conversion factors:16 ounces = 1 pound and 2000 pounds = 1 ton.The conversion is:\[1200 \text{ oz} = 1200 \times \frac{1 \text{ lb}}{16 \text{ oz}} \times \frac{1 \text{ ton}}{2000 \text{ lb}}\]\[1200 \times \frac{1}{16} \times \frac{1}{2000} = 0.0375 \text{ tons}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Unit Conversion
Unit conversion is the process of changing one unit of measurement to another. This is essential in many fields, such as science and engineering, where precise measurements are crucial. When performing unit conversions, it's important to understand the relationship between the two units involved. This ensures accuracy and consistency.

To begin a unit conversion, identify the given unit and the unit you want to convert to. Use conversion factors, which are numerical values that represent the relationship between these units. For example, if you need to convert meters to kilometers, remember that 1 kilometer is equal to 1000 meters. Using these factors allows for accurate calculations of the conversion process.

Here are some tips for successful unit conversion:
  • Always write down the units clearly to avoid confusion.
  • Ensure that the conversion factor is set up correctly to cancel out the unwanted unit.
  • Double-check the calculation to ensure that the final unit matches the desired outcome.
Exploring the Metric System
The metric system is a decimal-based system of measurement used worldwide. It is simple and consistent, making it a highly preferred choice for scientific calculations and daily use across multiple countries.

The base units of the metric system include:
  • Meter (m) for length
  • Kilogram (kg) for mass
  • Second (s) for time
One of the core advantages of the metric system is its use of powers of ten, which simplifies conversions among units, such as from kilometers to meters, or from grams to kilograms. For instance, converting 50 meters per second to kilometers per hour entails multiplying by a conversion factor that represents the difference in scale between meters and kilometers.

Understanding the metric system's unit hierarchy is essential for performing seamless conversions. Familiarity with common conversions, such as those mentioned in the original exercise, will enhance your proficiency in working within this system.
Different Measurement Units
Measurement units are essential as they provide a standard way of communicating the size, quantity, or extent of something. Units play a pivotal role in various aspects like trade, science, and daily life.

Measurement units can be broadly categorized into different types, including:
  • Length: These include meters and kilometers in the metric system, and feet and miles in the imperial system.
  • Mass: In the metric system, mass is measured in grams and kilograms, whereas the imperial system uses ounces, pounds, and tons.
  • Time: Seconds, minutes, hours, and days are standard units of time measurement.
It's crucial to be aware of these differences, especially when converting from one unit system to another. The ability to convert from units like ounces to tons as encountered in the example problem, illustrates the importance of using correct conversion factors, ensuring clarity and precision in reporting measurements.
Problem Solving in Mathematics with Dimensional Analysis
Dimensional analysis is a mathematical technique used to analyze and convert different units of measurement within problem-solving. This approach is particularly helpful in physics and engineering problems, where precise unit interpretation is necessary.

The process involves the use of conversion factors to change one set of measurement units to another. For example, when converting speed from meters per second to kilometers per hour, the necessary conversion factors link the desired and given units until the unwanted units are canceled out, leaving only the desired units. This structured step-by-step method helps in achieving accurate results.

Using dimensional analysis effectively requires understanding:
  • How to identify and apply the correct conversion factors.
  • The importance of unit consistency and cancellation.
  • The necessity of logical reasoning when organizing calculations for clarity and accuracy.
Applying dimensional analysis in mathematics encourages critical thinking and precision, which are vital traits for tackling complex problems with multiple units, as demonstrated in the exercise.

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Most popular questions from this chapter

Find five points that satisfy the inverse variation equation \(y=\frac{20}{x}\). Graph the equation and the points to make sure the coordinates of your points are correct.

Write a proportion and answer each question using the conversion factor 1 inch \(=2.54\) centimeters. a. A teacher is \(62.5\) inches tall. How many centimeters tall is she? (a) b. A common ceiling height is 96 inches ( 8 feet). About how high is this in centimeters? c. The diameter of a \(\mathrm{CD}\) is 12 centimeters. What is its diameter in inches? (a) d. The radius of a typical soda can is \(3.25\) centimeters. What is its radius in inches?

APPLICATION Portia drove her new car 308 miles on \(10.8\) gallons of gasoline. a. What is the car's rate of gasoline consumption in miles per gallon? b. If this is the typical mileage for Portia's car, how much gas will it take for a 750 -mile vacation trip? c. If gas costs \$2.35 per gallon, how much will Portia spend on gas on her vacation? d. The manufacturer advertised that the car would get 30 to 35 miles per gallon. How does Portia's mileage compare to the advertised estimates?

APPLICATION To squeeze a given amount of air into a smaller and smaller volume, you have to apply more and more pressure. Boyle's law describes the inverse variation between the volume of a gas and the pressure exerted on it. Suppose you start with a \(1 \mathrm{~L}\) open container of air. If you put a plunger at the top of the container without applying any additional pressure, the pressure inside the container will be the same as the pressure outside the container, or 1 atmosphere (atm). a. What will the pressure in atmospheres be if you push the plunger down until the volume of air is \(0.5 \mathrm{~L}\) ? (A) b. What will the pressure in atmospheres be if you push the plunger down until the volume of air is \(0.25 \mathrm{~L}\) ? c. Suppose you exert enough pressure so that the pressure in the container is \(10 \mathrm{~atm}\). What will the volume of the air be? (a) d. What would you have to do to make the pressure inside the container less than \(1 \mathrm{~atm}\) ? e. Graph this relationship, with pressure (in atmospheres) on the horizontal axis and volume (in liters) on the vertical axis.

Find the value of \(x\) in each proportion. a. \(\frac{1 \text { meter }}{3.25 \text { feet }}=\frac{15.2 \text { meters }}{x \text { feet }}\) b. \(\frac{1.6 \text { kilometers }}{1 \text { mile }}=\frac{x \text { kilometers }}{25 \text { miles }}\) c. \(\frac{0.926 \text { meter }}{1 \text { yard }}=\frac{200 \text { meters }}{x \text { yards }}\) d. \(\frac{1 \text { kilometer }}{0.6 \text { mile }}=\frac{x \text { kilometers }}{350 \text { miles }}\)

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