Two red blood cells each have a mass of 9.0 times 10^-14 kg and carry a negative

Question

Two red blood cells each have a mass of 9.0 times 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 -2.10 pC of charge and the other-3.10 pc, and each cell can be modeled as a sphere 7.5 mu m in diameter. What speed would they need when very far away from each other to get close enough to just touch? Assume that there is no viscous drag from any of the surrounding liquid. (Express your answer to two significant figures.) m/s You currently have 1 submissions for this question. Only 10 submission are allowed. You can make 9 more submissions for this question. What is the maximum acceleration of the cells in the previous part? (Express your answer to two significant figures.) times 10^10 m/s^2 You currently have 3 submissions for this question. Only 10 submission are allowed. You can make 7 more submissions for this question.

Explanation / Answer

Given that

mass m=9*10^-14 kg

charge q1=-2.1pC

charge q2=-3.1pC

distance d=7.5*10^-6 m

now we find the speed

1/2mv^2=kq1q2/d

1/2*9*10^-14*v^2=9*10^9*-2.1*0^-12*-3.1*10^-12/(7.5*10^-6)

v^2=1.74*10^5

speed v=4.2*10^2 m/s

now we find the acceleration

ma=kq1q2/d^2

9*10^-14*a=9*10^9*-2.1*-3.1*10^-24/(7.5*10^-6)^2

a=0.116*10^11=1.16*10^10 m/s^2

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.