Answer:
A) W = 103.55 N
B) mass of displaced water = 4186 g
C) W_displaced water = 41.06 N
D) Buoyant force = 41.06 N.
E) ZERO
F) 62.54 N
Explanation:
We are given;
mass of statuette;m = 10,566 g = 10.566 kg
volume = 4,064 cm³
Density of seawater;ρ = 1.03 g/mL = 1.03 g/cm³
A) The dry weight of the statuette can be calculated as;
W = mg
So;
W = 10.556 × 9.81
W = 103.55 N
B) Mass of displaced water is calculated from;
Density = mass/volume
So, mass = Density × Volume
m = 1.03 × 4,064 = 4186 g
C) Weight of displaced water is given by;
W_displaced water = (m_displaced water) × g
W_displaced water = 4.186 kg × 9.81 m/s^2 = 41.06 N
D) The buoyant force is the same as the weight of the displaced water.
Thus, Buoyant force = 41.06 N.
E) The apparent weight of the statuette is calculated from;
Apparent weight = Dry weight - Weight of displaced water
Apparent weight = 103.6 N - 41.06 N = 62.54 N. It is sitting on the bottom of the sea, so the sea floor is providing an opposite force that is equal but opposite the weight so that the net force on the statuette is zero. Since It has zero acceleration, in any direction, hence the net force on it is zero.
F. From E above, The Force required to lift the statuette = 62.54 N
