Ask Question
27 July, 13:44

A cube of wood with a density of 0.780 g/cm 3 is 10.0 cm on each side. When the cube is placed in water, what buoyant force acts on the wood

+4
Answers (1)
  1. 27 July, 14:08
    0
    The buoyant force on the wood = 7.652 N

    Explanation:

    According to the principle of flotation, a body floats when the upthrust exerted upon it by the fluid n which it floats equals the weight of the body.

    W = U ... Equation.

    Where W = weight of the wood, U = Upthrust or buoyant force.

    Recall That,

    Density = mass/volume

    Mass = Density*volume

    m = D*V ... Equation 2

    Where m = mass of the wood, V = Volume of the wood, D = Density of the wood.

    But

    Volume of a cube = a³

    V = a³ where a = length of the cube.

    V = 10³ = 1000 cm³.

    Given: V = 1000 cm³ D = 0.780 g/cm³

    Substituting these values into equation 2,

    m = 1000 (0.780)

    m = 780 g

    m = 0.78 kg.

    But W = mg

    Where m = 0.78 kg, g = 9.81 m/s²

    W = 0.78 (9.81)

    W = 7.652 N.

    Since W = U = 7.652 N.

    U = 7.652 N

    Therefore the buoyant force on the wood = 7.652 N
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “A cube of wood with a density of 0.780 g/cm 3 is 10.0 cm on each side. When the cube is placed in water, what buoyant force acts on the wood ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers