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anonymous
 4 years ago
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.
anonymous
 4 years ago
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.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I know that delta P = PinPout = 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 * 1E3) = 9.42 mm Except the answer is 1.47 cm. Help please!

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.0Coulda used the equation editor...

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.0I might not be able to help much. All I can say is to check your calculations...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thanks 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.
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