1) A. float
There are two forces acting on an object in the water: the weight of the object (downward) and the buoyancy (upward), which is equal to the weight of displaced water. If the weight of displaced water is greater than the weight of the object, it means that there is a net force directed upward, so the object will float.
2) A. buoyant force is equal to the weight of the object
As stated in the previous question, there are only these two forces acting on an object in the water (buoyant force and weight of the object), so if the two forces are equal, then the object is in equilibrium, so it will float.
3) C. 0.59 N
The buoyancy force is given by:
\(B=\rho_L V_d g\)
where
\(\rho_L = 1000 kg/m^3\) is the density of the liquid (water)
\(V_d=60 mL=6 \cdot 10^{-2} L=6 \cdot 10^{-5} m^3\) is the volume of displaced water
\(g=9.8 m/s^2\) is the acceleration of gravity
Substituting numbers into the formula, we find
\(B=(1000 kg/m^3)(6 \cdot 10^{-5} m^3)(9.8 m/s^2)=0.59 N\)
4) C. 100.0 g
The density of the object is 20 g/cm^3, which is greater than the density of the water (1 g/cm^3): this means that the object will sink, so its volume is equal to the volume of displaced water.
Therefore, we have:
- object's density: \(\rho=20 g/cm^3\)
- object's volume: \(V=5 mL=5 cm^3\)
so, the mass of the object is
\(m=dV=(20 g/cm^3)(5 cm^3)=100.0 g\)
5) A. 2.0 g
The specific gravity of an object is given by the ratio between its density (\(\rho\)) and the density of a reference substance (\(\rho_W\)), in this case water:
\(SG=\frac{\rho}{\rho_W}\)
whe can rewrite each density as the ratio between mass and volume:
\(SG=\frac{m_o /V_o}{m_w/V_w}\)
where the suffix o refers to the object, while the suffix w refers to the water. However, if we assume that the object is completely in the water, the two volumes are equal, so we can simplify the formula:
\(SG=\frac{m_o}{m_w}=\frac{16.2 g}{8.1 g}=2.0\)
