Question

A rectangular block of plastic, measuring 2.27 cm by 2.43 cm by 2.11 cm, floats in the middle of the Pacific Ocean. The volume of sea water displaced by the plastic block is 8.24 cm3. Assuming the density of the sea water is 1.025 g/mL, what is the density of the plastic?

Answer 1

Answer on Question #45446, Physics, Mechanics – Kinematics – Dynamics

A rectangular block of plastic, measuring $2.27\mathrm{cm}$ by $2.43\mathrm{cm}$ by $2.11\mathrm{cm}$, floats in the middle of the Pacific Ocean. The volume of sea water displaced by the plastic block is $8.24\mathrm{cm}^3$. Assuming the density of the sea water is $1.025\mathrm{g/mL}$, what is the density of the plastic?

The sum of all forces on the block should be 0, because the block is in equilibrium position:

$$m _ {b} g - F _ {\cdot \cdot} = 0$$

Where $F_{\cdot \cdot}$ is the Archimedes' force, $m_b$ is the mass of block

$$m _ {b} g = \rho_ {w} g V _ {d}$$

Where $V_{d}$ – displaced volume

This equation could be transformed:

$$m _ {b} = \rho_ {w} V _ {d}$$ $$\rho_ {b} V _ {b} = \rho_ {w} V _ {d} \rightarrow \rho_ {b} = \frac {\rho_ {w} V _ {d}}{V _ {b}} = \frac {\rho_ {w} V _ {d}}{a b c}$$ $$\rho_ {b} = \frac {1 . 0 2 5 \frac {g}{c m ^ {3}} \cdot 8 . 2 4 c m ^ {3}}{2 . 2 7 c m \cdot 2 . 4 3 c m \cdot 2 . 1 1 c m} \approx 0. 7 2 6 \frac {g}{c m ^ {3}}$$

Answer: the density of the plastic is: $\rho_{b} = 0.726\frac{g}{cm^{3}}$

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