Answer:
The density is [tex]\rho_{block} = 356 \ kg/m^3 [/tex]
Explanation:
From the question we are told that
The proportion of the plastic under water is p = 0.356
Generally at equilibrium the buoyant force is equal to the weight of the block i.e
[tex]B = mg[/tex]
Generally the Buoyant force is mathematically represented as
[tex]B = \rho * V * g[/tex]
Here [tex]\rho[/tex] density of the water with value
[tex]\rho = 1000 kg/m^3[/tex] ,
V is the volume of water displaced by the block which is [tex] p * V_{block [/tex] ,
So
[tex]1000 * 0.356 V_{block} * g = m * g[/tex]
Here m is the mass of the block which is mathematically represented as
[tex]m = \rho_{block} * V_{block}[/tex]
So
[tex]1000 * 0.356 V_{block} * g = \rho_{block} * V_{block}* g[/tex]
[tex]1000 * 0.356 V_{block} = \rho_{block} * V_{block}* g[/tex]
=> [tex]\rho_{block} = 1000 * 0.356[/tex]
=> [tex]\rho_{block} = 356 \ kg/m^3 [/tex]
