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A small family home in Tucson, Arizona, has a rooftop area of 1967 square feet, and it is possible to capture rain falling on about \(56 \%\) of the roof. A typical annual rainfall is about 14 inches. If the family wanted to install a tank to capture the rain for an entire year, without using any of it, what would be the required volume of the tank in \(m^{3}\) and in gallons? How much would the water weigh when the tank was full (in \(\mathbf{N}\) and in \(\mathrm{Ib}_{\mathrm{f}}\) )?

Short Answer

Expert verified
The required volume of the tank is about \(36.36 m^{3}\) or 9605.3 gallons, and the water inside would weigh about \(356685.6 N\) or 2579946.3 lb_f when the tank is full.

Step by step solution

01

Convert units to SI

First convert all the given data into appropriate units. Converting the roof area from square feet to square meters gives \(1967 ft^{2} \times 0.092903 m^{2}/ft^{2} = 182.64 m^{2}\). The annual rainfall in meters is \(14 \times 0.0254 m/inch = 0.3556 m\).
02

Calculate the capturing area and the total rainfall volume

Based on the given information, \(56 \%\) of the roof area can collect the rain. So, the effective area is \(56/100 \times 182.64 m^{2} = 102.2784 m^{2}\). The total volume of water collected over a year can be calculated by multiplying this area by the height of the rainfall, giving \(0.3556 m \times 102.2784 m^{2} = 36.36 m^{3}\).
03

Convert to Gallons

To express this volume in gallons, we make use of the relation 1 gallon is \(0.00378541 m^{3}\), from which we get \(36.36 m^{3} / 0.00378541 m^{3}/gallon = 9605.3 gallons\).
04

Calculate the weight of the water

We know that the density of water is \(1,000 kg/m^{3}\). The mass of the water is therefore this density times the volume of the water, yielding \(1000 kg/m^{3} \times 36.36 m^{3} = 36360 kg.\) Since weight is mass times gravity, the weight in N is \(36360 kg \times 9.81 m/s^{2} = 356685.6 N.\) Turning to imperial units, 1 kg is \(2.20462 lb_f\), so the mass is \(36360 kg \times 2.20462 lb_f/kg = 80157 lb_f.\) and the weight is \(80157 lb_f \times 32.174 ft/s^2 = 2579946.3 lb_f\).

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Key Concepts

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

SI Units
Understanding the International System of Units (SI) is fundamental to scientific and educational contexts, as it provides a standardized method for measuring various physical quantities. One of the primary advantages of SI units is their global acceptance which facilitates the sharing and comparison of scientific data.

For example, when converting from feet to meters, as seen in the exercise, we use the conversion factor \( 1 ft^2 = 0.092903 m^2 \). Such conversion factors are critical when we need to communicate findings or perform calculations that require consistent units. Use of SI units is encouraged in scientific communities to ensure clarity and avoid confusion that can arise from different measurement systems.

To master SI units, familiarize yourself with the basic units of measure for length (meter - m), mass (kilogram - kg), and volume (cubic meter - m3), among others. Each of these can be converted into other units using standard conversion factors, such as converting inches to meters for rainfall measurement, where \( 1 inch = 0.0254 m \).
Volume Calculation
Volume calculation is an essential skill in disciplines ranging from engineering to environmental science. The volume of a three-dimensional space is the amount of space enclosed within it, and it can be measured in various units, such as cubic meters (\(m^3\)), cubic feet (\(ft^3\)), or gallons.

One must start by understanding the shape of the space and applying the corresponding mathematical formula. For example, the volume of a rectangular prism is calculated as length times width times height. In our exercise, the effective area for collecting rainwater is determined by the roof's total area that is capable of capturing rain, which then gets multiplied by the height of the rainfall – here represented by the annual rainfall depth.

\textbf{Calculating Volume}:
  • Determine the capturing area: \( Area = length \times width \)
  • Find the volume: \( Volume = Area \times height \)
It is important to match the units when multiplying area and height to get the volume in a consistent unit like cubic meters, which often involves converting from other units such as inches or feet.
Weight and Mass Conversion
The terms 'weight' and 'mass' are often used interchangeably in everyday language, but they have distinct meanings in physics. Mass is a measure of the amount of matter in an object, while weight is the force exerted on the mass by gravity.

\textbf{Converting Mass to Weight}:
  • To convert mass (in kilograms) to weight (in newtons), you multiply the mass by the acceleration due to gravity (\(9.81 m/s^2\)), since \(Weight = Mass \times Gravity\).
  • This results in the object's weight under Earth's gravity, which can be further converted to pounds-force (\(lb_f\)) using the conversion factor \( 1 kg = 2.20462 lb_f \).
Moreover, the mass of water can be found by multiplying its volume by its density. In the metric system, water has a convenient density of \(1,000 kg/m^3\), which simplifies calculations. For instance, \(36.36 m^3\) of water has a mass of \(36,360 kg\), which can then be converted to weight. To navigate these conversions, it is helpful to maintain a reference list of key conversion factors and practice converting back and forth to instill a deeper understanding.

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Most popular questions from this chapter

A rectangular block of solid carbon (graphite) floats at the interface of two immiscible liquids. The bottom liquid is a relatively heavy lubricating oil, and the top liquid is water. Of the total block volume, \(54.2 \%\) is immersed in the oil and the balance is in the water. In a separate experiment, an empty flask is weighed, \(35.3 \mathrm{cm}^{3}\) of the lubricating oil is poured into the flask, and the flask is reweighed. If the scale reading was \(124.8 \mathrm{g}\) in the first weighing, what would it be in the second weighing? (Suggestion: Recall Archimedes' principle, and do a force balance on the block.)

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