A certain person had a brain that weighed 1.30 kg and contained \( 6.71 \times 10^{10} \) cells. Part: \( 0 / 2 \) Part 1 of 2 Assuming that each cell was completely filled with water (density \( \left.=1.00 \frac{\mathrm{~g}}{\mathrm{~mL}}\right) \), calculate the liength of one side of such a cell if it were a cube. Round your answer to 3 significant digits.

To determine the length of one side of a cubic cell, we first need to find the total volume occupied by all the cells. Given that there are \( 6.71 \times 10^{10} \) cells and the density of water is \( 1.00 \, \text{g/mL} \), we can convert the brain's weight into grams: \[ 1.30 \, \text{kg} = 1300 \, \text{g} \] Now, since the density of water is \( 1.00 \, \text{g/mL} \), the total volume \( V \) of the brain (assuming it is all water) is: \[ V = \frac{\text{mass}}{\text{density}} = \frac{1300 \, \text{g}}{1.00 \, \text{g/mL}} = 1300 \, \text{mL} \] Now, we can find the volume of each individual cell by dividing the total volume by the number of cells: \[ V_{\text{cell}} = \frac{1300 \, \text{mL}}{6.71 \times 10^{10}} \approx 1.9375 \times 10^{-8} \, \text{mL} \] Since \( 1 \, \text{mL} = 1 \, \text{cm}^3 \), we can say: \[ V_{\text{cell}} \approx 1.9375 \times 10^{-8} \, \text{cm}^3 \] Now that we have the volume of one cell, and knowing that the volume of a cube is given by \( V = a^3 \), where \( a \) is the length of a side of the cube, we can solve for \( a \): \[ a = \sqrt[3]{V_{\text{cell}}} = \sqrt[3]{1.9375 \times 10^{-8}} \] Calculating the cube root: \[ a \approx 0.00257 \, \text{cm} \] Converting to micrometers (1 cm = 10,000 micrometers): \[ a \approx 25.7 \, \mu m \] Rounding to three significant digits, we find that the length of one side of a cubic cell is: \[ \boxed{25.7 \, \mu m} \]