This sort of question I'm more checking if I'm doing itcorrectly, its been awhile since I've done something like this,always need a refresher anyways the question(s) is...

E. Coli cells are about $2\mu $ m long and $0.8\mu $ m indiameter:

How many E. Coli cells laid end to end would fit across thediameter of the pinhead, assuming the pinhead diameter of 0.5 mm,and what would be the volume of an E.Coli cell assuming that its acylinder.

Ok, what I've done was simple I guess, the first part whichI've assumed that "long" ment length, and I've taken the longestpart of E. Coli and did the idea of $0.5\times {10}^{-3}m/2.0\times {10}^{-6}m=250$ Cellsgoing across.

[Wasn't sure if it was suppose to use the othernumber, just assuming it was talking about long ways]

For the second part of the question, I just used the formulaof $V=\pi {r}^{2}h$ in which I came up with

$V=(3.14)(0.4\times {10}^{-6}m)2(2.04\times {10}^{-6}m)=1.00\times {10}^{-18}m$

Which something doesn't seem correct, or atleast seems oddto me.

Also, I don't know if I'm pushing the amount of questions Ican ask in a single problem, but the next part its asking for the Surface area of an E. Coli cell, and what is the surface to volumeratio of an E. Coli cell?

Which I have no idea, what type of surface area I should usefor a cell, then do I just compare the volume to the surface area,that I got in the previous problem?

E. Coli cells are about $2\mu $ m long and $0.8\mu $ m indiameter:

How many E. Coli cells laid end to end would fit across thediameter of the pinhead, assuming the pinhead diameter of 0.5 mm,and what would be the volume of an E.Coli cell assuming that its acylinder.

Ok, what I've done was simple I guess, the first part whichI've assumed that "long" ment length, and I've taken the longestpart of E. Coli and did the idea of $0.5\times {10}^{-3}m/2.0\times {10}^{-6}m=250$ Cellsgoing across.

[Wasn't sure if it was suppose to use the othernumber, just assuming it was talking about long ways]

For the second part of the question, I just used the formulaof $V=\pi {r}^{2}h$ in which I came up with

$V=(3.14)(0.4\times {10}^{-6}m)2(2.04\times {10}^{-6}m)=1.00\times {10}^{-18}m$

Which something doesn't seem correct, or atleast seems oddto me.

Also, I don't know if I'm pushing the amount of questions Ican ask in a single problem, but the next part its asking for the Surface area of an E. Coli cell, and what is the surface to volumeratio of an E. Coli cell?

Which I have no idea, what type of surface area I should usefor a cell, then do I just compare the volume to the surface area,that I got in the previous problem?