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. Round your
answer to the nearest
hundredth. (1 point)

Answer 1

To find the length of the platinum bar, we can use the relationship between mass, density, and volume. The formulas we will use are:

\[
\text{Density} = \frac{\text{Mass}}{\text{Volume}}
\]
and
\[
\text{Volume} = \text{Width} \times \text{Height} \times \text{Length}
\]

Given:
- Density of platinum = 21.4 g/cm³
- Mass of the bar = 1 kg = 1000 g
- Width = 5 cm
- Height = 8 cm
- Length = \(L\) (unknown)

First, we will rearrange the density formula to find the volume:

\[
\text{Volume} = \frac{\text{Mass}}{\text{Density}} = \frac{1000 \, \text{g}}{21.4 \, \text{g/cm}^3} \approx 46.733 \, \text{cm}^3
\]

Now we substitute this into the volume formula:

\[
\text{Volume} = \text{Width} \times \text{Height} \times \text{Length} = 5 \, \text{cm} \times 8 \, \text{cm} \times L
\]

Setting the two expressions for volume equal gives us:

\[
5 \, \text{cm} \times 8 \, \text{cm} \times L = 46.733 \, \text{cm}^3
\]

Calculating the left side gives:

\[
40 \, \text{cm}^2 \times L = 46.733 \, \text{cm}^3
\]

Now, solve for \(L\):

\[
L = \frac{46.733 \, \text{cm}^3}{40 \, \text{cm}^2} = 1.168325 \, \text{cm}
\]

Rounding to the nearest hundredth:

\[
L \approx 1.17 \, \text{cm}
\]

Therefore, the length of the platinum bar is approximately **1.17 centimeters**.