Please Check my work and help me to answer the questions i could not get.
Measure the cube that you are given. Complete the chart on the next page for your cube, then fill in the spaces for two other cubes, one measuring 1 cm on an edge and the other measuring 0.5 cm on a side. The ratio of surface area volume should be divided so that it is expressed as square cm per cubic cm.
Your Cube:
(1)Length of 1 Edge- 2.5
(2)Area of One Side- 6.25
(3)Total Surface Area- 37.5
(4)Volume- 2.5 squared
(5)Ratio of SA/Volume (reduced to /1 cc)-
1 cm cube:
(1) 1
(2) 1
(3) 6
(4)1 squared
(5)?
.5 CM cube:
(1) .5
(2) .25
(3) 18
(4) .5 squared
(5) ?
Please check the above math and help. Also:
1. If a cell is made smaller and smaller, what happens to the ratio of surface area to volume. (It Decreases?)
2. Of what value is it to a cell to have a large surface area to volume ratio?
3. How does having small cells affect the metabolic rate that is possible in these cells?
Thanks
Let's check the calculations and complete the missing spaces.
Your Cube:
(1) Length of 1 Edge - 2.5 cm
(2) Area of One Side - (2.5 cm) * (2.5 cm) = 6.25 cm^2
(3) Total Surface Area - 6.25 cm^2 * 6 faces = 37.5 cm^2
(4) Volume - (2.5 cm)^3 = 15.625 cm^3
(5) Ratio of SA/Volume (reduced to /1 cc) - 37.5 cm^2 / 15.625 cm^3 = 2.4 cm^-1
1 cm cube:
(1) Length of 1 Edge - 1 cm
(2) Area of One Side - (1 cm) * (1 cm) = 1 cm^2
(3) Total Surface Area - 1 cm^2 * 6 faces = 6 cm^2
(4) Volume - (1 cm)^3 = 1 cm^3
(5) Ratio of SA/Volume (reduced to /1 cc) - 6 cm^2 / 1 cm^3 = 6 cm^-1
0.5 cm cube:
(1) Length of 1 Edge - 0.5 cm
(2) Area of One Side - (0.5 cm) * (00.5 cm) = 0.25 cm^2
(3) Total Surface Area - 0.25 cm^2 * 6 faces = 1.5 cm^2
(4) Volume - (0.5 cm)^3 = 0.125 cm^3
(5) Ratio of SA/Volume (reduced to /1 cc) - 1.5 cm^2 / 0.125 cm^3 = 12 cm^-1
Now let's address your additional questions:
1. If a cell is made smaller and smaller, what happens to the ratio of surface area to volume?
As a cell decreases in size, the ratio of surface area to volume increases. This is because the volume of a shape increases more rapidly with size compared to its surface area. Therefore, the smaller the cell, the larger the surface area-to-volume ratio.
2. Of what value is it to a cell to have a large surface area to volume ratio?
Having a large surface area to volume ratio is advantageous for a cell because it allows for efficient exchange of nutrients, gases, and waste products with its environment. A larger surface area facilitates a faster rate of diffusion, which is crucial for cell survival and proper functioning.
3. How does having small cells affect the metabolic rate that is possible in these cells?
Having small cells with a high surface area to volume ratio enables a higher metabolic rate. This is because a larger surface area allows for more efficient absorption and distribution of necessary resources, such as oxygen and nutrients, facilitating faster metabolic processes. Small cell size also permits a larger number of cells to be packed into a given space, increasing the overall metabolic capacity of the tissues or organisms.