Question
A rectangular bar of platinum has a width of 5 centimeters, a height of 8 centimeters, and an unknown length. You measure the platinum bar and it is exactly 1 kilogram. Find the length of the bar if you know its density is 21.4 g/cm^3
. Round your answer to the nearest hundredth
. Round your answer to the nearest hundredth
Answers
GPT-4o mini
To find the length of the rectangular bar, we can use the formula for density, which is given by:
\[
\text{Density} = \frac{\text{Mass}}{\text{Volume}}
\]
From this relationship, we can rearrange the formula to solve for the volume:
\[
\text{Volume} = \frac{\text{Mass}}{\text{Density}}
\]
We are given:
- Mass = 1 kg = 1000 g (since 1 kg = 1000 g)
- Density = 21.4 g/cm³
Now we can calculate the volume:
\[
\text{Volume} = \frac{1000 \text{ g}}{21.4 \text{ g/cm}^3} \approx 46.73 \text{ cm}^3
\]
Next, we can express the volume of the rectangular bar using its dimensions:
\[
\text{Volume} = \text{length} \times \text{width} \times \text{height}
\]
Substituting the known values (width = 5 cm, height = 8 cm), we can write:
\[
\text{Volume} = \text{length} \times 5 \text{ cm} \times 8 \text{ cm}
\]
Setting the two expressions for volume equal gives us:
\[
46.73 \text{ cm}^3 = \text{length} \times 5 \text{ cm} \times 8 \text{ cm}
\]
Now, solve for the length:
\[
\text{length} = \frac{46.73 \text{ cm}^3}{5 \text{ cm} \times 8 \text{ cm}} = \frac{46.73 \text{ cm}^3}{40 \text{ cm}^2} \approx 1.16825 \text{ cm}
\]
Rounding to the nearest hundredth gives:
\[
\text{length} \approx 1.17 \text{ cm}
\]
Thus, the length of the bar is approximately **1.17 cm**.
\[
\text{Density} = \frac{\text{Mass}}{\text{Volume}}
\]
From this relationship, we can rearrange the formula to solve for the volume:
\[
\text{Volume} = \frac{\text{Mass}}{\text{Density}}
\]
We are given:
- Mass = 1 kg = 1000 g (since 1 kg = 1000 g)
- Density = 21.4 g/cm³
Now we can calculate the volume:
\[
\text{Volume} = \frac{1000 \text{ g}}{21.4 \text{ g/cm}^3} \approx 46.73 \text{ cm}^3
\]
Next, we can express the volume of the rectangular bar using its dimensions:
\[
\text{Volume} = \text{length} \times \text{width} \times \text{height}
\]
Substituting the known values (width = 5 cm, height = 8 cm), we can write:
\[
\text{Volume} = \text{length} \times 5 \text{ cm} \times 8 \text{ cm}
\]
Setting the two expressions for volume equal gives us:
\[
46.73 \text{ cm}^3 = \text{length} \times 5 \text{ cm} \times 8 \text{ cm}
\]
Now, solve for the length:
\[
\text{length} = \frac{46.73 \text{ cm}^3}{5 \text{ cm} \times 8 \text{ cm}} = \frac{46.73 \text{ cm}^3}{40 \text{ cm}^2} \approx 1.16825 \text{ cm}
\]
Rounding to the nearest hundredth gives:
\[
\text{length} \approx 1.17 \text{ cm}
\]
Thus, the length of the bar is approximately **1.17 cm**.