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Yuri [45]
3 years ago
9

A wooden block of dimensions of 6 cm ⨯ 6 cm ⨯ 15 cm floats with the long axis oriented horizontally and the of the square faces

are oriented so that they point up vertically, straight up out of the water like a diamond shape. If the wooden block is submerged to a depth 7.96 cm (the bottom point to the water level), then: (a) What is the submerged volume of the block?(b) What is the mass of the block
(c) What is the density of the block?
Physics
1 answer:
schepotkina [342]3 years ago
3 0

Explanation:

(a)  Formula to calculate volume of the submerged wooden block is as follows.

            V_{sub} = l \times w \times d

It is given data of the wooden block is as follows.

          depth = 7.96 cm,       length (l) = 6 cm

         width (w) = 4 cm,      

So, we will calculate the volume of the submerged wooden block as follows.

           V_{sub} = l \times w \times d

                       = 6 \times 6 \times 7.96

                       = 286.56  cm^{3}

Hence,  the submerged volume of the block is 286.56  cm^{3}.

(b)   Expression for the buoyant force acting on the wooden block is as follows.

            F_{B} = \rho_{w} g V_{sub}

And, expression for the force of gravity of the wooden block is as follows.

            F_{g} = m_{b}g

As the wooden block is floating on the water hence, buoyant force is balanced by the weight of the block.

                 F_{g} = F_{B}

Hence, mass of the wooden block will be calculated as follows.

         F_{g} = F_{B}

       m_{b}g = \rho_{w}gV_{sub}

          m_{b} = \rho_{w}V_{sub}

                      = 997 kg/m^{3} \times 286.56 cm^{3}

                      = 997 kg/m^{3} \times 286.56 \times 10^{-6} m^{3}

                      = 0.02857 kg

Therefore, mass of the given block is 0.02857 kg

(c)   Expression for the density of the block is as follows.

             \rho_{b} = \frac{m_{b}}{V_{b}}

Now, expression for the total volume of the wooden block is as follows.

             V_{b} = l \times w \times h

Hence, density of the given block is as follows.

              \rho_{b} = \frac{m_{b}}{V_{b}}

                         = \frac{m_{b}}{lwh}

                         = \frac{0.02857 kg}{4 \times 4 \times 15}

                         = 1.19 \times 10^{-4} kg/cm^{3}

Therefore, density of the given block is 1.19 \times 10^{-4} kg/cm^{3}.

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