An ore sample weighs 17.50 N in air. When the sample is suspended by a light cord and totally immersed in water, the tension in

Question

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An ore sample weighs 17.50 N in air. When the sample is suspended by a light cord and totally immersed in water, the tension in

the cord is 11.20 N. Find the total volume and the density of the sample.

Physics

1 answer:

kicyunya [3.2K] · Answer 1 · 2026-04-07T13:03:54+03:00

Answer:

Estimate of the sample's volume: approximately $\rm 0.6422 \; L = 6.422 \times 10^{-4} \; m^{3}$ .

Mean density of the sample: approximately $\rm 2.77\; g \cdot cm^{3} = 2.778 \times 10^{3}\; kg \cdot m^{3}$ .

Assumption:

$\rm g = 9.81\; N \cdot kg^{-1}$ .
$\rho(\text{water}) = \rm 1.000\times 10^{3}\; kg \cdot m^{-3}$ .
The volume of the cord is considered negligible.

Explanation:

Overall volume of the sample

The magnitude of the buoyant force equals $\rm 17.50 - 11.20 = 6.30\; N$ .

This also corresponds to the weight (weight, $m \cdot g$ ) of the water displaced by the object. To determine the mass of the displaced water from its weight, apply the formula: divide weight by $g$ .

$\displaystyle m = \frac{m\cdot g}{g} = \rm \frac{6.30\; N}{9.81\; N \cdot kg^{-1}} \approx 0.642\; kg$ .

Assuming the density of water is $\rho(\text{water}) = \rm 1.000\times 10^{3}\; kg \cdot m^{-3}$ . To find the volume of the displaced water, use the formula: divide mass by density $\rho(\text{water})$ .

$\displaystyle V(\text{water displaced}) = \frac{m}{\rho} = \rm \frac{0.642\; kg}{1.000\times 10^{3}\; kg \cdot m^{-3}} \approx 6.42201 \times 10^{-4}\; m^{3}$ .

Assuming the cord's volume is negligible, since the sample is completely submerged in water, its volume should equal the volume of the displaced water.

$V(\text{sample}) = V(\text{water displaced}) \approx \rm 6.422\times 10^{-4}\; m^{3}$ .

Mean Density of the sample

Average density can be calculated by the mass divided by volume.

To compute the mass of the sample from its weight, utilize the formula: divide by $g$ .

$\displaystyle m = \frac{m \cdot g}{g} = \rm \frac{17.50\; N}{9.81\; N \cdot kg^{-1}} \approx 1.78389 \; kg$ .

The volume from the previous section can be utilized.

Lastly, divide mass by volume to find the average density.

$\displaystyle \rho(\text{sample, average}) = \frac{m}{V} = \rm \frac{1.78389\; kg}{6.42201 \times 10^{-4}\; m^{3}} \approx 2.778\; kg \cdot m^{-3}$ .