Question

Find initial speed and max acceleration

s The Energy Stored in × Lakehead University , pHTS 1 x earning.ca/ibiscms/mod/ficn/vilew.php?ld-429903 Upgrade to macos Mojave umpto. S 1010/1133-Winter19-MACKAY Activities and Due Dates >HW:Electric Potential nment Score: 1927/2500 Resourcesvp Feedback on 12 of 25> Attempt 1 Two red blood cells each have a mass of 9.05 x 10-4 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 -2.30 pC and the other-3.10 pC, and each cell can be modeled as a sphere 3.75 x 10* m in radius. If the red blood cells start very far apart and move directly toward each other with the same speed, what initial spoed would each need so that they get clone encugh to just barely touch? Assume that there is no viscous drag from any of the surrounding liquid initial speed: 326.5 What is the maximum acceleration of the cells as they move toward each other and just barely touch? maximum acceleration: 1.42 x101 30 5 3

Answer 1

the initial speed of blood cells is , v1 = v

the final speed of blood cells is , v2 = 0

initial distance = very far off (infinity)

final speed = 2*radius = 2*3.75*10^-6 = 7.5*10^-6 m

The conserving energy is ,

q1*q2/4*pi*$\varepsilon$o*r + 1/2*m*v² = constant

(q1*q2/4*pi*$\varepsilon$o*$\infty$) + 2*(1/2*m*v²) = q1*q2/4*pi*$\varepsilon$o*2r + 0

calculate the velocity of the cell ,

m*v² = q1*q2/4*pi*$\varepsilon$o*2r

v = sqrt[q1*q2/4*pi*$\varepsilon$o*2r*m]

v = sqrt[(-2.30*10^-12)*(-3.10*10^-12)/4*3.14*8.85*10^-12*7.5*10^-6*9.05*10^-14]

v = 307.41 m/s

These are experiences the maximum acceleration when they just barely touch each other .

so,

m*a = q1*q2/4*pi*$\varepsilon$o*(2r)²

a =  q1*q2/4*pi*$\varepsilon$o*(2r)²*m

a = (-2.30*10^-12)*(-3.10*10^-12)/4*3.14*8.85*10^-12*(7.5*10^-6)²*9.05*10^-14

a = 1.26*10¹⁰ m/s²