91影视

The apparent weight of a body submerged in a fluid is the difference between its actual weight and buoyant force. It is expressed in terms of Newton.

The mass of the metal sample, when measured in air, is m_metal =63.5 g.

The apparent mass of the metal sample when measured in water is m_apparent = 55.4g

As discussed above,

\begin{aligned}{W_{{\rm{apparent}}}}={W_{{\rm{metal}}}}-{F_{\rm{B}}}\\{m_{{\rm{apparent}}}}\cdot g={m_{{\rm{metal}}}}\cdot g-{F_{\rm{B}}}\end{aligned}

Here, F_B is the buoyant force, is the apparent mass when submerged in water, is the mass of metal in the air, and g is the acceleration due to gravity.

But we know that the buoyant force is the weight of a mass of water occupying the volume of the metal sample.

Therefore,

Here, 蟻_H2Ois the density of water and 蟻_metal is the density of the metal.

Substitute the values of densities in the above formula.

\begin{aligned}{\rho_{{\rm{metal}}}}=\frac{{63.5\;{\rm{g}}}}{{\left({63.5\;{\rm{g}}-55.4\;{\rm{g}}}\right)}}\times{\rho_{{{\rm{H}}_{\rm{2}}}{\rm{O}}}}\end{aligned}

Now we know that the density of water is 1000kg/m³. Thus,

\begin{aligned}{\rho_{metal}}=\frac{{63.5\;{\rm{g}}}}{{\left({63.5\;{\rm{g}}-55.4\;{\rm{g}}}\right)}}\times\left({1000\;{{{\rm{kg}}}\mathord{\left/{\vphantom{{{\rm{kg}}}{{{\rm{m}}^3}}}}\right\{{{\rm{m}}^3}}}}\right)\\{\rho_{metal}}=7840\;{{{\rm{kg}}}\mathord{\left/{\vphantom{{{\rm{kg}}}{{{\rm{m}}^3}}}}\right\{{{\rm{m}}^3}}}\end{aligned}

Hence, based on the density, the metal is probably iron or steel (after referring to Table 10-1).