A = 4\pi r^2
A = 4\pi (2\mu m /2)^2 (10^{-6}m/1\mu m)^2 (1mm/10{-3})^2
A = 1.33*!0^{-5}MM^2
Answer: W =
J
Explanation: Since the potassium ion is at the outside membrane of a cell and the potential here is lower than the potential inside the cell, the transport will need work to happen.
The work to transport an ion from a lower potential side to a higher potential side is calculated by
![W=q.\Delta V](https://tex.z-dn.net/?f=W%3Dq.%5CDelta%20V)
q is charge;
ΔV is the potential difference;
Potassium ion has +1 charge, which means:
p =
C
To determine work in joules, potential has to be in Volts, so:
![\Delta V=65.10^{-3}V](https://tex.z-dn.net/?f=%5CDelta%20V%3D65.10%5E%7B-3%7DV)
Then, work is
![W=1.6.10^{-19}.65.10^{-3}](https://tex.z-dn.net/?f=W%3D1.6.10%5E%7B-19%7D.65.10%5E%7B-3%7D)
![W=1.04.10^{-20}](https://tex.z-dn.net/?f=W%3D1.04.10%5E%7B-20%7D)
To move a potassium ion from the exterior to the interior of the cell, it is required
J of energy.
It looks blue as it is only reflecting blue light
Answer:
Energy. They need energy.
Explanation: