A brick is completely submerged in SALTWATER (D = 1.03 kg/L). The brick experiences a buoyant force of 193 oz. What is the weight of the SALTWATER displaced by the brick?

I believe the answer is 193 because, wt. of water displaced = buoyant force

Let me know if this is correct. Thanks!

correct.

Correct.

Actually, the weight of the saltwater displaced by the brick is not equal to the buoyant force acting on the brick. The buoyant force is the force exerted by the saltwater on the brick, which is equal to the weight of the volume of saltwater that the brick displaces.

To find the weight of the saltwater displaced by the brick, you need to use the concept of buoyancy and the density of the saltwater.

The buoyant force (F_b) acting on the brick is equal to the weight of the displaced saltwater. Mathematically, we can represent it as:

F_b = ρ_fluid * V_displaced * g

Where:
ρ_fluid is the density of the saltwater (1.03 kg/L, or you can convert it to kg/m^3 by multiplying by 1000),
V_displaced is the volume of the saltwater displaced by the brick,
and g is the acceleration due to gravity (approximately 9.8 m/s^2).

Now, rearranging the equation to solve for V_displaced:

V_displaced = F_b / (ρ_fluid * g)

Substituting the given value of the buoyant force (193 oz) as F_b:

V_displaced = (193 oz) / (ρ_fluid * g)

To convert ounces to kilograms, you can use the conversion factor 1 oz ≈ 0.0283495 kg.

So the weight of the saltwater displaced by the brick is:

Weight_displaced = ρ_fluid * V_displaced * g

Now, you have the volume of the saltwater displaced by the brick (V_displaced), and you can calculate the weight by substituting the values into the equation.