arning 4Jump to... Madison Cade 7/6/2018 10:59 PM 4 80.2/100 7/5/2018 03:22 PM P
ID: 1864010 • Letter: A
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arning 4Jump to... Madison Cade 7/6/2018 10:59 PM 4 80.2/100 7/5/2018 03:22 PM PrintCaloulatorPeriodic Table Gradeb estion 27 of 32 Mape presented by Sapling Learning Two red blood cells each have a mass of 9.05 x10* 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 and the other -3.30 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 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. Number m/s What is the maximum acceleration of the celis as they move toward each other and just barely touch? Number m/s O Previous G ve Up & View Solution check Answer 0 Next Ext about us careers privacy policy terms of use contact us help re to searchExplanation / Answer
Given,
Mass, m = 9.05 x 10^-14 kg
Charge q1 = -2.1*10^-12 C
Charge q2 = -3.3*10^-12 C
a) Assume that the speed is v, then when they are close enough the speed will be zero and when they are very far the potential energy will be zero, so the initial kinetic energy will be converted into the potential energy completely.
So, 0.5*m*v^2 = kq1q2/r
2*0.5*( 9.05^-14)*v^2 = 9*10^9*(2.10*10^-12)*(3.3*10^-12)/(2 x 3.75 *10^-6)
9.05*10^-14*v^2 = 8.316*10^-9
v = 303.13 m/s
b) maximum acceleration will be when they are closest, the force on any one of them will be,
F = kq1q2/r^2 = 9*10^9*(2.1*10^-12)*(3.3*10^-12)/(2 x 3.75*10^-6)^2 = 0.0011 N
acceleration, a = F/m = 0.0011/(9.05*10^-14) = 1.225*10^10 m/s^2
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