91影视

The given data can be listed below as,

• The weight of an ore is, $\mathrm{W}=\mathrm{mg}=17.50\mathrm{N}$.
• The tension in the cord is, $\mathrm{T}=11.20\mathrm{N}$.

Whenever the force is imbalanced, Newton's second law applies. It means that the total force used on the system depends on its volume, density, and rate of change velocity.

From Newton鈥檚 second law, the force equation can be expressed as,

$鈭�F=ma$ 鈥� (i)

Here, m is the mass and a is the acceleration.

The buoyant force can be expressed as,

$$\begin{gathered} B=p_w{V}_{obj}g \\ {V}_{obj}=\frac{B}{p_{w}g} \end{gathered}$$ 鈥� (ii)

Here, $p_w$is the density of water whose value is $\left(1000kg/m^3\right)$, $V_{obj}$is the volume of the object and $g$is the acceleration due to gravity.

The total force on the system can be given by,

$$\begin{gathered} T+mg+B=0 \\ B=mg-T \end{gathered}$$ 鈥� (iii)

Here, T is the tension in string, mgis the weight of the ore and Bis the buoyant force.

Substitute all the values in above equation.

$$\begin{gathered} \mathrm{B}=17.50\mathrm{N}-11.20\mathrm{N} \\ =6.30\mathrm{N} \end{gathered}$$

Substitute this value in equation (ii).

$$\begin{gathered} V_{obj}=\frac{6.30N}{\left(1脳{10}^{3}kg/m^3\right)\left(9.8m/s^2\right)}\left(\frac{kg.m/s^2}{1N}\right) \\ {V}_{obj}=6.43脳{10}^{-4}{m}^3 \end{gathered}$$

Thus, the total volume of the ore sample is $6.43脳{10}^{-4}{m}^3$

The density of the ore can be expressed as,

$p=\frac{m}{V_{obj}}$

Here, mis the mass of ore and$V_{obj}$is volume of the ore.

Substitute all the values in above equation.

$$\begin{gathered} p=\frac{1.79kg}{6.43脳{10}^{-4}{m}^3} \\ =2.78脳{10}^{3}kg/m^3 \end{gathered}$$

Thus, the density of the ore sample is $2.78脳{10}^{3}kg/m^3$.