anonymous
  • anonymous
Fluid Mechanics. An experimental fluid is used to create a spherical bubble with a diameter of 0.25 cm. WHen in contact with a surface made of plastic, it has a contact angle of 30 degrees. The pressure inside the bubble is 101453 Pa and outside the bubble the pressure is atmospheric. In a particular experiment, you are asked to calculate how high this experimental fluid will rise in a capillary tube made of the same plastic as used in the surfae described above. The diameter of the capillary tube is 0.2 cm. The density of the experimental fluid used in this experiment is 750 k/m^3.
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
I know that delta P = Pin-Pout = 4 (sigma) /r Pin = 101453 and Pout = 102325 Pa, so delta P is 218 Pa. You plug that into the first equation to find sigma, the surface tension, and get 0.04 N/m. Then to find the height, you use the equilibrium equation: (2pi r) (sigma) cos(theta) - rho g pi r^2 h= 0, so h = (2 * sigma * cos (theta)) / (rho * g* r) = (2*0.04*cos(30))/(750 * 9.81 * 1E-3) = 9.42 mm Except the answer is 1.47 cm. Help please!
IsTim
  • IsTim
Coulda used the equation editor...
IsTim
  • IsTim
I might not be able to help much. All I can say is to check your calculations...

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
thanks for the advice, but I've already checked everything by myself and with several other students and we're collectively trying to solve the problem so some of us posted questions to look for help. thanks anyway, IsTim.
IsTim
  • IsTim
oh well...

Looking for something else?

Not the answer you are looking for? Search for more explanations.