Answer:
[tex]0.5\; \rm m^{3}[/tex].
Explanation:
Calculate the mass of this object:
[tex]\begin{aligned}& m(\text{object}) \\ &= \rho(\text{object}) \cdot V(\text{object}) \\ &= 500\; \rm kg \cdot m^{-3} \times 1\; \rm m^{3} = 500\; \rm kg\end{aligned}[/tex].
Multiple the mass of this object by the gravitational field strength, [tex]g[/tex], to find the weight of this object: [tex]W(\text{object}) = m(\text{object}) \cdot g[/tex].
Since this object is floating in water, the buoyancy force on it should be equal to its weight:
[tex]F(\text{buoyancy}) = W(\text{object}) = m(\text{object}) \cdot g[/tex].
By Archimedes' Principle, the weight of the water that this object displaces would be equal to [tex]F(\text{buoyancy})[/tex], the size of buoyancy force on this object,
Hence, the weight of water displaced would be [tex]W(\text{object}) = m(\text{object}) \cdot g[/tex].
Divide this weight by [tex]g[/tex] to find the mass of water displaced:
[tex]\begin{aligned} & m(\text{water displaced}) \\ &= \frac{W(\text{water displaced})}{g} \\ &= \frac{m(\text{object}) \cdot g}{g} = m(\text{object}) = 500\; \rm kg\end{aligned}[/tex].
Assume that the density of water is [tex]\rho(\text{water}) = 1000\; \rm kg \cdot m^{-3}[/tex]. The volume of water displaced would be:
[tex]\begin{aligned}& V(\text{water displaced}) \\ &= \frac{m(\text{water displaced})}{\rho(\text{water})} \\ &= \frac{500\; \rm kg}{1000\; \rm kg \cdot m^{-3}} = 0.5\; \rm m^{3}\end{aligned}[/tex].
