Answered

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When an object is put into a liquid, it experiences a buoyant force that is equal to the weight of the liquid the object displaces. The force on the wire is given as the block is slowly lowered into the liquid (position is given in "centimeters", which you have to change to "meters") and force is given in "newtons"). Choose a mass of the block between 0.125 kg and 0.375 kg and a density of the liquid between 500 kg/m^3 and 1000 kg/m^3. The object is in static equilibrium when the clock stops.

Required:
a. What is the weight of the block and the tension, F, in the string when the block is in the liquid?
b. What is the volume of the block in the liquid—either the submerged part of the block if the block is partially submerged when you paused it or the entire block if it is completely submerged (the dimension of the block that is into the screen is 5.00 cm = 0.0500m)?
c. What is the volume of the water that is displaced by the block (the dimension of both water containers into the screen is 10.00 cm = 0.100m)?

Sagot :

Answer:

A) F = 1.8375 N, B) V = 1.25 10⁻⁴ m³, C) V= 1.25 10⁻⁴ m³

Explanation:

This exercise should select some values ​​such as the weight of the block m=0.250 kg and the density of the liquid ρ = 500 kg / m³ (water); with these values ​​answer the question, for this we use the static equilibrium relations.

          Σ F = 0

          F + B - W = 0

the expression for thrust is

          B = ρ g V_liquid

A) the weight of the block is

          W = m g

          W = 0.250 9.8

          W = 2.45 N

Let's look for the maximum push

           V_body = V_liquid

            B = 500 9.8 0.05³

           B = 0.6125 N

we substitute

          F = W - B

          F = 2.45 - 0.6125

          F = 1.8375 N

B) As the body is totally submerged, the volume of the liquid and the body are equal

           V = l³

           V = 0.05³

           V = 1.25 10⁻⁴ m³

C) the volume of water displaced is equal to the volume of the body

           V_liquid = 1.25 10⁻⁴ m³

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