Final answer:
To determine the density of a piece of driftwood floating in the ocean with 35% of its volume above the water, we use the principle of floatation, which results in finding that the density of the driftwood is 669.5 kg/m³.
Step-by-step explanation:
The student's question involves calculating the density of a piece of driftwood floating in the ocean with 35% of its volume above water, given the density of the ocean water. To find the density of the driftwood, one can apply the principle of floatation that states that the fraction of an object submerged in a fluid is equal to the ratio of the density of the object to the density of the fluid.
We know the density of the ocean water is 1,030 kg/m³ and 65% (100% - 35%) of the driftwood's volume is submerged. Therefore, we can set up a ratio where the unknown density of the driftwood (ρdriftwood) over the known density of ocean water (ρwater) is equal to the volume submerged (Vsubmerged) over the total volume of the driftwood (Vtotal).
ρdriftwood / ρwater = Vsubmerged / Vtotal
This equation simplifies to ρdriftwood = ρwater * (Vsubmerged / Vtotal). By inserting the given values, we calculate the density of the driftwood:
ρdriftwood = 1,030 kg/m³ * (65/100) = 669.5 kg/m³
Therefore, the density of the driftwood is 669.5 kg/m³.