Question

A cube of Iron, 0.0800 m on a side,has a density of 7874 kg/m^3.What is its mass?Skip(Unit = kg)

Answer 1

ANSWER:

4.03 kilograms.

STEP-BY-STEP EXPLANATION:

We have that the density is given by the following formula:

$\begin{gathered} d=\frac{m}{v} \\ \text{ we solve for m:} \\ m=d\cdot v \end{gathered}$

We can calculate the raisin by multiplying the voulem by the density, we can calculate the volume since it is a cube, the volume of the cube is calculated like this:

$\begin{gathered} v=s^3 \\ \text{ replacing} \\ v=(0.0800)^3 \\ v=0.000512 \end{gathered}$

Now if we can calculate the mass, like this:

$\begin{gathered} m=7874\cdot0.000512 \\ m=4.03\text{ kg} \end{gathered}$

The mass of the iron cube is 4.03 kilograms.