Two red blood cells each have a mass of 9.05 × 10-14 kg and carry a negative cha
ID: 1655471 • Letter: T
Question
Two red blood cells each have a mass of 9.05 × 10-14 kg and carry a negative charge spread uniformly over their surfaces. The repulsion arising from the excess charge prevents the cells from clumping together. One cell carries -3.00 pC and the other -3.50 pC, and each cell can be modeled as a sphere 3.75 × 10-6 m in radius. If the red blood cells start very far apart and move directly toward each other with the same speed, what initial speed would each need so that they get close enough to just barely touch? Assume that there is no viscous drag from any of the surrounding liquid.
Explanation / Answer
electric PE = k q1 q2 / d
they will approach until their kinetic energy becomes zero.
Applying energy conservation,
PEi + KEi = PEf + KEf
0 + 2 (m v^2 / 2) = 2 ( k q1 q2 / 2r) + 0
(9.05 x 10^-14) v^2 = (9 x 10^9) (3.50 x 10^-12)(3 x 10^-12) / (3.75 x 10^-6)
(9.05 x 10^-14) v^2 = 2.52 x 10^-8
v = 528 m/s .......Ans
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.