Answer:
Step-by-step explanation:
To determine which cube-shaped cell would be most efficient in removing waste by diffusion, we need to consider the surface area-to-volume ratio. The larger the surface area compared to the volume, the more efficient the cell will be in facilitating diffusion.
The surface area of a cube is given by the formula 6 * (side length)^2, and the volume is given by (side length)^3.
Let's calculate the surface area-to-volume ratios for each option:
Cube with side length of 10 μm:
Surface area = 6 * (10 μm)^2 = 600 μm^2
Volume = (10 μm)^3 = 1,000 μm^3
Surface area-to-volume ratio = 600 μm^2 / 1,000 μm^3 = 0.6 μm^-1
Cube with side length of 20 μm:
Surface area = 6 * (20 μm)^2 = 2,400 μm^2
Volume = (20 μm)^3 = 8,000 μm^3
Surface area-to-volume ratio = 2,400 μm^2 / 8,000 μm^3 = 0.3 μm^-1
Cube with side length of 30 μm:
Surface area = 6 * (30 μm)^2 = 5,400 μm^2
Volume = (30 μm)^3 = 27,000 μm^3
Surface area-to-volume ratio = 5,400 μm^2 / 27,000 μm^3 = 0.2 μm^-1
The cell with the highest surface area-to-volume ratio is the cube with a side length of 10 μm, with a ratio of 0.6 μm^-1. This means that this cube-shaped cell has the most efficient surface area for facilitating diffusion of waste compared to the other options provided.