Answer:
Volume of the sample: approximately [tex]\rm 0.6422 \; L = 6.422 \times 10^{-4} \; m^{3}[/tex].
Average density of the sample: approximately [tex]\rm 2.77\; g \cdot cm^{3} = 2.778 \times 10^{3}\; kg \cdot m^{3}[/tex].
Assumption:
Explanation:
The size of the buoyant force is equal to [tex]\rm 17.50 - 11.20 = 6.30\; N[/tex].
That's also equal to the weight (weight, [tex]m \cdot g[/tex]) of water that the object displaces. To find the mass of water displaced from its weight, divide weight with [tex]g[/tex].
[tex]\displaystyle m = \frac{m\cdot g}{g} = \rm \frac{6.30\; N}{9.81\; N \cdot kg^{-1}} \approx 0.642\; kg[/tex].
Assume that the density of water is [tex]\rho(\text{water}) = \rm 1.000\times 10^{3}\; kg \cdot m^{-3}[/tex]. To the volume of water displaced from its mass, divide mass with density [tex]\rho(\text{water})[/tex].
[tex]\displaystyle V(\text{water displaced}) = \frac{m}{\rho} = \rm \frac{0.642\; kg}{1.000\times 10^{3}\; kg \cdot m^{-3}} \approx 6.42201 \times 10^{-4}\; m^{3}[/tex].
Assume that the volume of the cord is negligible. Since the sample is fully-immersed in water, its volume should be the same as the volume of water it displaces.
[tex]V(\text{sample}) = V(\text{water displaced}) \approx \rm 6.422\times 10^{-4}\; m^{3}[/tex].
Average density is equal to mass over volume.
To find the mass of the sample from its weight, divide with [tex]g[/tex].
[tex]\displaystyle m = \frac{m \cdot g}{g} = \rm \frac{17.50\; N}{9.81\; N \cdot kg^{-1}} \approx 1.78389 \; kg[/tex].
The volume of the sample is found in the previous part.
Divide mass with volume to find the average density.
[tex]\displaystyle \rho(\text{sample, average}) = \frac{m}{V} = \rm \frac{1.78389\; kg}{6.42201 \times 10^{-4}\; m^{3}} \approx 2.778\; kg \cdot m^{-3}[/tex].
