A 55 -kg swimmer climbs onto a Styrofoam block of density \(160 \mathrm{kg} / \mathrm{m}^{3} .\) If the water level comes right to the top of the Styrofoam, what's the block's volume?

The weight W_s of the swimmer can be calculated using the formula W_s = m_s g, where m_s = 55 kg is the swimmer's mass, and g = 9.8 m/s^2 is the acceleration due to gravity. So, W_s = 55 * 9.8 = 539 N.

The weight of the styrofoam block W_b can be calculated by the formula W_b = \(\rho_b\)Vg, where \(\rho_b\) = 160 kg/m^3 is the density of the styrofoam and V is the volume of the styrofoam block. But we do not know the volume of the styrofoam block yet, so we cannot calculate W_b at this step.

According to Archimedes' principle, the total weight W of the styrofoam block (when the swimmer climbs onto it) is equal to the weight of the water displaced by the styrofoam block. The weight of the water displaced by the styrofoam block can be calculated by the formula W = \(\rho_{water}\)Vg, where \(\rho_{water}\) = 1000 kg/m^3 is the density of water, V is the volume of the styrofoam block, and g = 9.8 m/s^2 is the acceleration due to gravity. However, since the total weight W is also the sum ofthe swimmer's weight and the volume of the styrofoam block, we have W = W_s + W_b = \(\rho_{water}\)Vg. We already calculated W_s in step 1, so we can use this information to calculate V as follows: V = (W_s + W_b) / (\(\rho_{water}\)g). As we noticed in step 2, we can not calculate the weight of the styrofoam block W_b without the volume of the styrofoam block, so the expression can be simplified to: V = W_s / (\(\rho_{water}\)g).

Now we can calculate the volume V of the styrofoam block by substituting the values into the formula V = W_s / (\(\rho_{water}\)g), so V = 539 / (1000gn). V = 0.055 m^3.