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Consider an e. coli to be a cylinder with a diameter of 1 micrometer (um) and with a length of 2 micrometer (um).

Part 1. calculate the volume of the e. coli cell and express the answer in um^3.

Part 2. using the same cylindrical model, calculate the surface area of an e. coli cell in um^2. (round to 2 decimal places.

Respuesta :

Answer:

[tex]A=6.28\ \mu m^2[/tex]

Part 1

[tex]V=1.57 \ \mu m^3[/tex]

Part 2

[tex]A=6.28\ \mu m^2[/tex]

Explanation:

Given that

Diameter,d=1 μm

Length ,l=2 μm

As we know that volume of cylinder given as

[tex]V=\pi r^2l[/tex]

[tex]V=\pi \times 0.5^2\times 2 \ \mu m^3[/tex]

[tex]V=1.57 \ \mu m^3[/tex]

Surface area,A

A=π d l

[tex]A=\pi \times 1 \times 2\ \mu m^2[/tex]

[tex]A=6.28\ \mu m^2[/tex]

Part 1

[tex]V=1.57 \ \mu m^3[/tex]

Part 2

[tex]A=6.28\ \mu m^2[/tex]

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