1. Two red blood cells each have a mass of 9.05 × 10-14 kg and carry a negative
ID: 1647491 • Letter: 1
Question
1. 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 -2.50 pC and the other -2.60 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.
??? m/s
2. What is the maximum acceleration of the cells as they move toward each other and just barely touch?
???m/s^2
Explanation / Answer
Q1.
mass of each cell=m=9.05*10^(-14) kg
charge on first cell=q1=-2.5 pC
charge on second cell=q2=-2.6 pC
radius of each sphere=r=3.75*10^(-6) m
let initial speed be v m/s.
as both are initially very far apart, initial electrical potential energy=0
as they barely touch in the end and we are looking for minimum initial speed final kinetic energy also 0.
then using energy conservation principle:
initial kinetic energy=final potential energy
==>0.5*m*v^2+0.5*m*v^2=k*q1*q2/(2*r)
where k=coloumb's constant=9*10^9
==>m*v^2=k*q1*q2/(2*r)
==>v=sqrt(k*q1*q2/(2*r*m))
=sqrt(9*10^9*2.5*10^(-12)*2.6*10^(-12)/(2*3.75*10^(-6)*9.05*10^(-14)))
=293.58 m/s
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