Two red blood cells each have a mass of 9.0×10-14 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 -2.30 pC of charge and the other -3.30 pC, and each cell can be modeled as a sphere 7.5 μμm in diameter.

1)Using conservation of energy, calculate the speed each cell would 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 and that each cell starts with the same speed aimed directly at the other cell. (Express your answer to two significant figures.)

320 m/s

Explanation:

Parmeters given:

Mass of red blood cells = 9 * 10^(-14) kg

Charge on first red blood cell = -2.3 pC = - 2.3 * 10^(-12) C

Charge on second red blood cell = -3.3 pC = -3.3 * 10^(-12) C

Diameter of red blood cells = 7.5 * 10^(-6) m

Radius of red blood cells 3.75 * 10^(-6) m

When both red blood cells come to closest to each other, the total kinetic energy of both red blood cells will convert into potential energy of the cells (using conservation of energy) , hence:

0.5mv² + 0.5mv² = k(q1)(q2) / 2r

=> mv² = k(q1)(q2) / 2r

v² = k(q1)(q2) / 2mr

v² = ( 9 * 10^9 * - 2.3 * 10^(-12) * - 3.3 * 10^(-12)) / (2 * 9 * 10^(-14) * 3.75 * 10^(-6))

v² = 1.012 * 10^5

=> v = 318.12 m/s

Approximating to 2 significant figures, v = 320 m/s. This is the speed each cell would need.

f=ma newton

f=100x2.5=250n

the right answer is 0.3 meters per second