In a pollution control experiment, minute solid particles (mass 1.46x10-12 kg) a

Question

In a pollution control experiment, minute solid particles (mass 1.46x10-12 kg) are dropped in air. The terminal speed of the particles is measured to be -0.061 m/s. The drag of these particles is given by Fd = kV, where V is the instantons particle speed. Neglect the buoyancy force. a. Find the value of the constant k. b. Find the time required to reach 99% of terminal speed. c. Find the distance traveled before reaching 99% of terminal speed. d. Plot the distance as a function of time.

(please do all parts)

Explanation / Answer

Terminal speed , S = -0.061 m /s,  Mass of solid particle , m = 1.46 X 10-12 kg,

Fd = K*V where, V = instantaneous particle speed .

V = m4 / 2g2 ( 1 - g2 / m2 )2

          = 1.46 X 10-12 )4 / 2 x 102 ( 1 - 102 / X 10-24 )2 = 4.91 m/s

a) From the formula , Fd = K V

     K = Fd / V

     Fd =?

Fd = 2W / v * V2 * A = 2 / 1.1 X 0.0612 X 56

= 8.66

   K = 8.66 / 4.91

     = 1.8

b) The time required to reach the 99 % of terminal speed.

    V = U + gt

-0.061 = 0.06 + 10 t

    t = 1 x 10-4 s

c)   S = ut + 1/2 gt2

        = -0.06 X 10-4 + 1/2 X 10 X 10-4

        = 4.94 X 10-4 m

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