Find the diameter of a gold coin that has a weight of 1.00 (troy) oz and a thickness of 3.00 mm.
mass= density*volume= density*thickness*PI*radius squared.
solve for diameter (twice radius)
they wanted the answer in cm to ---------> cm. (1.00 oz = 28.35 g).
so do i still solve it the same way
You have to work in consistent units. I have no idea what you are using for units in density. Make consistent units, then convert the answer to cm.
To find the diameter of a gold coin given its weight and thickness, you'll need to use the density and volume of the coin. Here's how you can do it step by step:
1. Determine the density of gold: The density of gold is approximately 19.30 grams per cubic centimeter (g/cm³). Since 1 troy ounce is equal to 31.1035 grams, the density of gold can be converted to 19.30 g/cm³ ÷ 31.1035 g/oz ≈ 0.6215 g/cm³.
2. Calculate the volume of the gold coin: Since density is mass per unit volume, you can rearrange the formula to solve for volume: volume = mass ÷ density. In this case, the mass of the coin is 1.00 oz, which can be converted to grams by multiplying it by 31.1035 g/oz. So, the volume of the coin is 31.1035 g ÷ 0.6215 g/cm³ ≈ 50.03 cm³.
3. Use the volume of a cylinder formula: Since a coin is often shaped like a cylinder, you can use the formula for the volume of a cylinder to solve for the diameter. The formula is volume = π * radius² * height. In this case, the height of the coin is given as 3.00 mm, which can be converted to centimeters by dividing it by 10. Rearranging the formula, we get the radius² = volume ÷ (π * height). Applying the values, we find that the radius² is approximately 50.03 cm³ ÷ (π * 0.3 cm) ≈ 53.12 cm².
4. Solve for the radius: Take the square root of the radius² to find the radius. In this case, the radius is approximately √53.12 cm² ≈ 7.29 cm.
5. Calculate the diameter: Finally, double the radius to get the diameter. In this case, the diameter of the gold coin is approximately 2 * 7.29 cm ≈ 14.58 cm.
Therefore, the diameter of the gold coin is approximately 14.58 cm.