Answer: 0.439488 N
Explanation:
The apparent weight of the metal is computed as
apparent weight = weight - weight of the displaced fluid
To compute the weight of the metal, we use the following formula:
[tex]\text{weight} = \rho gV[/tex]
where
[tex]\rho = \text{density of the metal = 6,600 kg/m}^3[/tex]
[tex]g = \text{gravitational acceleration = 9.81 m/s}^2[/tex]
[tex]V = \text{volume of the metal}[/tex]
Note that the volume is unknown but we can compute this because the metal is a cube with edge = 2 cm = 0.02 m. So, the volume of the metal is given by
[tex]\text{Volume} = \text{edge}^3
\\ = (0.02)^3
\\ \boxed{\text{Volume = 0.000008 m} ^3} [/tex]
Thus, the weight of the metal is computed as
[tex]\text{weight} = \rho gV
\\ = (\text{6,600 kg/m}^3)(\text{9.81 m/s}^2)(\text{0.000008 m}^3) \\ = \text{0.517968 kg \(\cdot \) m/s}^2
\\ \boxed{\text{weight of the metal} = \text{0.517968 N}}[/tex]
Next, we compute the displaced weight or buoyancy, which has the following formula
[tex]\text{weight of the displaced fluid} = \rho' gV'[/tex]
where
[tex]\rho' = \text{density of the fluid (water) = 1,000 kg/m}^3
\\ g = \text{gravitational acceleration = 9.81 m/s}^2
\\ V' = \text{displaced volume}[/tex]
Note that the displaced volume is equal to the volume of the submerged metal. Since the metal has a volume of [tex]\text{0.000008 m} ^3[/tex], the displaced volume is [tex]\text{0.000008 m} ^3[/tex].
Thus, the weight of the displaced fluid is calculated as
[tex]\text{weight of the displaced fluid} = \rho' gV'
\\ = (\text{1,000 kg/m}^3)(\text{9.81 m/s}^2)(\text{0.000008 m}^3) \\ = \text{0.07848 kg \(\cdot \) m/s}^2 \\ \boxed{\text{weight of the displaced fluid} = \text{0.07848 N}}[/tex]
Therefore,
apparent weight of the metal
= weight of the metal - weight of the displaced fluid
= 0.517968 N - 0.07848 N
apparent weight of the metal = 0.439488 N
