4. (15 points) The object consists of two parts. The density of one of them is 7800 kg $/ \mathrm{m}^{3}$. In addition, it is known that this part occupies $30 \%$ of the volume of the entire object and its mass is $60 \%$ of the total mass. Determine the density of the second part.
Answer: $2229 \kappa / / \mathrm{m}^{3}$.

Solution. From the condition: $m_{1}=0.6 m$, and $V_{1}=0.3 V$. We get $m_{2}=0.4 m$, and $V_{2}=0.7 V$. The density of the first part $\rho_{1}=\frac{m_{1}}{V_{1}}=\frac{0.6 m}{0.3 V}$. The density of the second part:

$$
\rho_{2}=\frac{m_{2}}{V_{2}}=\frac{0.4 m}{0.7 V}=\frac{0.4}{0.7} \cdot \frac{0.3}{0.6} \rho_{1}=2229 \mathrm{\kappa} / \mathrm{M}^{3} .
$$