The largest and the smallest balls used in the experiment are with diameter 9. 52 mm, and 2. 38 mm respectively. For a glycerin with viscosity 1. 0 pa. S, what is the time necessary for each ball to reach a velocity 95% of the terminal velocity? density of the ball material is given in the text. Round the result to three decimal places.

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Raync1977 Raync1977

25-10-2022
Physics

Answered

The largest and the smallest balls used in the experiment are with diameter 9. 52 mm, and 2. 38 mm respectively. For a glycerin with viscosity 1. 0 pa. S, what is the time necessary for each ball to reach a velocity 95% of the terminal velocity? density of the ball material is given in the text. Round the result to three decimal places.

Answer :

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tutorconsortium102 tutorconsortium102 · Answer 1 · 2022-10-30T04:35:46-04:00

Here the answer = 0.0080 .

Solution:

Based on stokes law,

v = g * D^2 *(d p - d m) / (18 V)

v = terminal velocity

D = diameter of particle

V = Viscosity

dp = density of particle

dm = density of medium.

Here density of ball = 1.42 gm/cc

By substituting,

v = 9.81 x 9.52 x 10^-3 ^2 (1420 - 1300) / 18 x 1.0

=5.92x10^-4

v(t) =0 .99 x Vtrm

=0 .99 x 0.0059

= 0.00059 m/s

v(t)/Vterm =1-e^(-t/r)

0.99 = 1 - e ^ (-t / 9.52 x 10^-3)

e ^ (-t / 9.52 x 10^-3) = 0.01

Taking log

-t / 9.52 x 10^-3 = -4.6

t = 0.0438 s

Then smaller ball

v = 9.81 x 2.38 x 10^-3 ^2 (1420 - 1300) / 18 x 1.0

= 0.000037 m/s

v(t) / Vterm = 1-e^(-t/r)

0.99 = 1 - e ^ (-t / 2.38 x 10^-3)

e ^ (-t / 2.38 x 10^-3) = 0.01

Taking log

-t / 2.38 x 10^-3 = -4.6

t = 0.0109 s

Here Reynolds number is,

R= rho vd / mu

= 1420 x 0.00059 x 9.52 x 10^-3 x / 1.0

= 0.0080

To learn more about stroke's law refer to :

https://brainly.com/question/18240039

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