The aluminum foil's thickness is estimated to be about 1819.4 millimeters (mm).
To determine the thickness of the aluminum foil, we can use the formula for density, which relates mass (m), volume (V), and density (ρ) as follows: density = mass / volume.
First, we need to convert the given measurements to consistent units. The weight of the foil is 9.0 oz, which can be converted to grams by using the conversion factor 1 oz = 28.3495 g. Therefore, the mass (m) of the foil is approximately 9.0 oz * 28.3495 g/oz ≈ 255.1455 g.
The volume (V) of the foil can be calculated using the given area (50. ft²) and the thickness (t) of the foil. The formula for volume is V = area * thickness.
First, convert the area to square centimeters: 1 ft² = 929.0304 cm². So, the area (A) is 50. ft² * 929.0304 cm²/ft² ≈ 46451.52 cm².
Now, rearrange the volume formula to solve for thickness: t = V / A. Substituting the values, we get t = 46451.52 cm² / 255.1455 g ≈ 181.94 cm.
However, the density of aluminum is given in g/cm³, and we want the thickness in millimeters (mm). To convert cm to mm, multiply by 10: t ≈ 181.94 cm * 10 mm/cm ≈ 1819.4 mm.
Therefore, the approximate thickness of the aluminum foil is around 1819.4 millimeters.
It's important to note that the calculated value is unusually large for a common household aluminum foil, which is typically around 0.016 mm thick. There might be a discrepancy in the provided values or units in the question.
To learn more about aluminum foil refer here: