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Miles per Kilogram. The density of gasoline is 737 \(\mathrm{kg} / \mathrm{m}^{3}\) . If your new hybrid car gets 45.0 miles per gallon of gasoline, what is its mileage in miles per kilogram of gasoline? (See Appendix E.)

Short Answer

Expert verified
The car achieves approximately 16.14 miles per kilogram of gasoline.

Step by step solution

01

Convert Gallons to Cubic Meters

To find out how much gasoline is there in gallons in terms of cubic meters, use the conversion factor: 1 gallon = 0.00378541 cubic meters. So for 1 gallon, we compute:\[1 ext{ gallon} = 0.00378541 ext{ m}^3\]
02

Determine the Mass of Gasoline

Now we need to find the mass of one gallon of gasoline using the given density. We use the formula \( \text{Mass} = \text{Density} \times \text{Volume} \). The density of gasoline is given as 737 kg/m³. Therefore, the mass of one gallon of gasoline is:\[ \text{Mass} = 737 \times 0.00378541 = 2.7888 \text{ kg} \]
03

Calculate Mileage in Miles per Kilogram

In this step, we'll calculate how many miles the car can travel per kilogram of gasoline. Since the car travels 45.0 miles per gallon, and we've calculated that one gallon equals 2.7888 kg, we can find the mileage in miles per kilogram by dividing the mileage per gallon by the mass of one gallon:\[\text{Miles per kilogram} = \frac{45.0}{2.7888} \approx 16.14\text{ miles/kg} \]

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Key Concepts

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

Density
Density is a fundamental concept in physics and chemistry that describes how compactly mass is distributed in a given volume. It is an intrinsic property of materials, and it's calculated as mass divided by volume. The formula for density is given by \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \). The units often used are kilograms per cubic meter (kg/m³). For substances like gasoline, understanding density is essential for converting volumes into masses, especially since gasoline density is typically listed as 737 kg/m³.
  • Because density relates mass to volume, it helps in determining how much of a substance is present in a specific space.
  • It is a critical factor when performing conversions in chemistry and physics, such as in determining the mass of liquids from their volumes.
The use of density is especially important when needing to convert between measures such as gallons and kilograms, as it allows us to transition from a volume-based measure to a mass-based measure.
Mass Calculation
Mass calculation involves determining the weight or mass of a substance when given its volume and density. The equation used for this is \( \text{Mass} = \text{Density} \times \text{Volume} \). In the context of gasoline, if you know the density and the volume that you have (like in cubic meters), you can calculate how much it weighs in kilograms.
  • This calculation is essential for converting between units of volume and mass, a common requirement in scientific and engineering contexts.
  • It allows for precise measurements and comparisons of different materials, making it easier to understand their properties and behavior.
By calculating the mass of a gallon of gasoline using its known density, one can facilitate further calculations like determining how far a vehicle can travel.
Volume Conversion
Converting between different volume units is often necessary in calculations involving liquids, such as gasoline. For example, we often convert gallons to cubic meters due to varying measurement systems. The conversion factor for gallons to cubic meters is 1 gallon = 0.00378541 m³. This conversion allows precise calculations using the metric system.
  • Volume conversion is necessary due to the widespread use of different unit systems worldwide, like liters and gallons.
  • Understanding how to convert between units allows for universal understanding and calculations in scientific, engineering, and everyday contexts.
By converting the volume to a consistent unit measure, it sets the stage for additional calculations, such as density and mass, to be applied consistently.
Mileage Calculation
Mileage calculation in vehicles typically refers to how many miles a car can travel on a specific quantity of fuel, often expressed in miles per gallon. However, it can also be expressed in alternative units such as miles per kilogram when dealing with different measurements.
  • Mileage is a measure of a vehicle's fuel efficiency, offering insights into fuel consumption and cost-effectiveness.
  • Expressing mileage in terms like miles per kilogram can provide a more direct understanding of fuel efficiency relative to the weight of the fuel used.
By converting mileage from miles per gallon to miles per kilogram, you gain insights into how fuel consumption relates directly to the fuel's mass, offering a different perspective on efficiency and usage patterns.

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Most popular questions from this chapter

Untethered helium balloons, floating in a car that has all the windows rolled up and outside air vents closed, move in the direction of the car's acceleration, but loose balloons filled with air move in the opposite direction. To show why, consider only the horizontal forces acting on the balloons. Let \(a\) be the magnitude of the car's forward acceleration. Consider a horizontal tube of air with a cross-sectional area \(A\) that extends from the windshield, where \(x=0\) and \(p=p_{0},\) back along the \(x\) -axis. Now consider a volume element of thickness \(d x\) in this tube. The pressure on its front surface is \(p\) and the pressure on its rear surface is \(p+d p\) . Assume the air has a constant density \(\rho\) . (a) Apply Newton's second law to the volume element to show that \(d p=\rho a d x\) . (b) Integrate the result of part (a) to find the pressure at the front surface in terms of \(a\) and \(x\) . (c) To show that considering \(\rho\) constant is reasonable, calculate the pressure difference in atm for a distance as long as 2.5 \(\mathrm{m}\) and a large acceleration of 5.0 \(\mathrm{m} / \mathrm{s}^{2}\) . (d) Show that the net horizontal force on a balloon of volume \(\dot{V}\) is \(\rho V a\) . (e) For negligible friction forces, show that the acceleration of the balloon (average density \(\rho_{\text { bal }} )\) is \(\left(\rho / \rho_{\text { bal }}\right) a,\) so that the acceleration relative to the car is \(a_{n e 1}=\left[\left(\rho / \rho_{\text { bel }}\right)-1\right] a\) (f) Use the expression for \(a_{\mathrm{rel}}\) in part (e) to explain the movement of the balloons.

Dropping Anchor. An iron anchor with mass 35.0 \(\mathrm{kg}\) and density 7860 \(\mathrm{kg} / \mathrm{m}^{3}\) lies on the deck of a small barge that has vertical sides and floats in a freshwater river. The area of the bottom of the barge is 8.00 \(\mathrm{m}^{2}\) . The anchor is thrown overboard but is suspended above the bottom of the river by a rope; the mass and volume of the rope are small enough to ignore. After the anchor is overbound and the barge has finally stopped bobbing up and down, has the barge risen or sunk down in the water? By what vertical distance?

At one point in a pipeline the water's speed is 3.00 \(\mathrm{m} / \mathrm{s}\) and the gauge pressure is \(5.00 \times 10^{4} \mathrm{Pa}\) . Find the gauge pressure at a second point in the line, 11.0 \(\mathrm{m}\) lower than the first, if the pipe diameter at the second point is twice that at the first.

(a) Calculate the difference in blood pressure between the feet and top of the head for a person who is 1.65 \(\mathrm{m}\) tall. (b) Consider a cylindrical segment of a blood vessel 2.00 \(\mathrm{cm}\) long and 1.50 \(\mathrm{mm}\) in diameter. What additional outward force would such a vessel need to withstand in the person's feet compared to a similar vessel in her head?

A tall cylinder with a cross-sectional area 12.0 \(\mathrm{cm}^{2}\) is partially filled with mercury; the surface of the mercury is 5.00 \(\mathrm{cm}\) above the bottom of the cylinder. Water is slowly poured in on top of the mercury, and the two fluids don't mix. What volume of water must be added to double the gauge pressure at the bottom of the cylinder?

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