Density:

$\overline{){\mathbf{\rho}}{\mathbf{=}}\frac{\mathbf{m}}{\mathbf{V}}}$

**Part A)**

The weight of the block = the weight of the displaced water.

The volume of the block:

$\begin{array}{rcl}\mathbf{V}& \mathbf{=}& \mathbf{(}\mathbf{10}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathbf{)}\mathbf{(}\mathbf{10}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathbf{)}\mathbf{(}\mathbf{10}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathbf{)}\end{array}$

V = 1.0 × 10^{-3} m^{3}

From the density equation, mass of the wood block is:

$\begin{array}{rcl}\mathbf{m}& \mathbf{=}& \mathbf{\rho}\mathbf{V}\\ & \mathbf{=}& \mathbf{\left(}\mathbf{670}\mathbf{\right)}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{)}\end{array}$

A 10cm×10cm×10cm woodblock with a density of 670 kg/m^{3} floats in water.

Part A) What is the distance from the top of the block to the water if the water is fresh? Express your answer to two significant figures and include the appropriate units.

Part B) If it's seawater? Suppose that ρ=1030kg/m^{3}. Express your answer to two significant figures and include the appropriate units.

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