Certainly! Let's break down the process step-by-step to find the density of the liquid:
### Step 1: Determine the mass of the cube
Given:
- Volume of the cube [tex]\( V_{\text{cube}} = 5.50 \times 10^{-3} \, \text{m}^3 \)[/tex]
- Density of the cube [tex]\( \rho_{\text{cube}} = 7.50 \times 10^3 \, \text{kg/m}^3 \)[/tex]
The mass [tex]\( m \)[/tex] of the cube can be found using the formula:
[tex]\[ m = \rho_{\text{cube}} \times V_{\text{cube}} \][/tex]
[tex]\[ m = 7.50 \times 10^3 \, \text{kg/m}^3 \times 5.50 \times 10^{-3} \, \text{m}^3 \][/tex]
[tex]\[ m = 41.25 \, \text{kg} \][/tex]
### Step 2: Calculate the weight of the cube
The weight [tex]\( W \)[/tex] of the cube is given by:
[tex]\[ W = m \times g \][/tex]
where [tex]\( g \)[/tex] is the acceleration due to gravity [tex]\( \left( 9.8 \, \text{m/s}^2 \right) \)[/tex].
[tex]\[ W = 41.25 \, \text{kg} \times 9.8 \, \text{m/s}^2 \][/tex]
[tex]\[ W = 404.25 \, \text{N} \][/tex]
### Step 3: Determine the volume of the cube that is submerged
Since only the lower half of the cube is submerged, the submerged volume [tex]\( V_{\text{submerged}} \)[/tex] is half of the total volume of the cube:
[tex]\[ V_{\text{submerged}} = \frac{V_{\text{cube}}}{2} \][/tex]
[tex]\[ V_{\text{submerged}} = \frac{5.50 \times 10^{-3} \, \text{m}^3}{2} \][/tex]
[tex]\[ V_{\text{submerged}} = 2.75 \times 10^{-3} \, \text{m}^3 \][/tex]
### Step 4: Apply the principle of buoyancy
The buoyant force [tex]\( F_b \)[/tex] acting on the submerged part of the cube is given by:
[tex]\[ F_b = \rho_{\text{liquid}} \times V_{\text{submerged}} \times g \][/tex]
### Step 5: Establish the equilibrium condition for the forces
In equilibrium (when the cube is hanging), the tension [tex]\( T \)[/tex] in the cable plus the buoyant force balances the weight of the cube:
[tex]\[ T + F_b = W \][/tex]
Given that the tension [tex]\( T \)[/tex] in the cable is 375 N:
[tex]\[ 375 \, \text{N} + \rho_{\text{liquid}} \times V_{\text{submerged}} \times g = 404.25 \, \text{N} \][/tex]
### Step 6: Solve for the density of the liquid [tex]\( \rho_{\text{liquid}} \)[/tex]
Rearrange the equation to solve for [tex]\( \rho_{\text{liquid}} \)[/tex]:
[tex]\[ \rho_{\text{liquid}} = \frac{W - 375 \, \text{N}}{V_{\text{submerged}} \times g} \][/tex]
Substitute the known values into the equation:
[tex]\[ \rho_{\text{liquid}} = \frac{404.25 \, \text{N} - 375 \, \text{N}}{2.75 \times 10^{-3} \, \text{m}^3 \times 9.8 \, \text{m/s}^2} \][/tex]
[tex]\[ \rho_{\text{liquid}} = \frac{29.25 \, \text{N}}{2.75 \times 10^{-3} \, \text{m}^3 \times 9.8 \, \text{m/s}^2} \][/tex]
[tex]\[ \rho_{\text{liquid}} = \frac{29.25}{0.02695} \][/tex]
[tex]\[ \rho_{\text{liquid}} \approx 1085.34 \, \text{kg/m}^3 \][/tex]
Thus, the density of the liquid is approximately [tex]\( 1085.34 \, \text{kg/m}^3 \)[/tex].
