2pts: a student scientist has a block of wood and a block of aluminum that have the same length and width…

Explanation:

Step1: Analyze volume

Since the wood and aluminum blocks have the same length and width, and assuming they have the same height (not stated otherwise but for equal - cross - sectional area and three - dimensional objects), their volumes $V$ are equal. Volume is calculated as $V = l\times w\times h$, and with $l_{wood}=l_{aluminum}$, $w_{wood}=w_{aluminum}$, and assuming $h_{wood}=h_{aluminum}$, we have $V_{wood}=V_{aluminum}$.

Step2: Analyze mass

The aluminum block sinks and the wood block floats. According to Archimedes' principle, the buoyant force $F_b=\rho_{water}gV_{submerged}$. For a floating object, the buoyant force equals the weight of the object, $F_b = m_{wood}g$, and for a sinking object, the weight of the object is greater than the maximum buoyant force it can experience when fully submerged, $m_{aluminum}g>\rho_{water}gV_{aluminum}$. Since $V_{wood}=V_{aluminum}$, we know that $m_{aluminum}>m_{wood}$.

Step3: Analyze density

Density is defined as $\rho=\frac{m}{V}$. Since $V_{wood}=V_{aluminum}$ and $m_{aluminum}>m_{wood}$, using the density formula, we get $\rho_{aluminum}=\frac{m_{aluminum}}{V_{aluminum}}$ and $\rho_{wood}=\frac{m_{wood}}{V_{wood}}$, so $\rho_{aluminum}>\rho_{wood}$. Also, since the wood floats, $\rho_{wood}<\rho_{water}$, and since the aluminum sinks, $\rho_{aluminum}>\rho_{water}$.

Answer:

• Volume: The volumes of the wood and aluminum blocks are equal, $V_{wood} = V_{aluminum}$.
• Mass: The mass of the aluminum block is greater than the mass of the wood block, $m_{aluminum}>m_{wood}$.
• Density: The density of the aluminum block is greater than the density of the wood block. The density of the wood block is less than the density of water ($\rho_{wood}<\rho_{water}$), and the density of the aluminum block is greater than the density of water ($\rho_{aluminum}>\rho_{water}$), $\rho_{aluminum}>\rho_{wood}$.