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A rectangular wooden block of weight W floats with exactly one-half of its volume below the waterline.Masses are stacked on top of the block until the top of the block is level with the waterline. This requires 20 g of mass. What is the mass of the wooden block?A.) 40 gB.) 20 gC.) 10 g

Respuesta :

Answer:

b) M=20g

Explanation:

For this exercise we must use the Archimedes principle that states that the thrust that a body receives is equal to the weight of the dislodged liquid.

         B = ρ g V

Let's use balance healing for this case

Initial.

          B - W = 0

The weight of the body can be related to its density

          W = ρ V_body g

         ρ_liq g (½ V_body) = m g

Final

   Some masses were added

          M = 20 g = 0.020 kg

        B - W - W₂ = 0

        ρ_liq  g  V_Body = m g + M g

Let's replace and write the system of equations

     ½ ρ_liq  V_body = m

         ρ  V_body = m + M

We solve the equations

        2 m = m + M

        m = M

         m = 20 g

The answer is b

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