Air Bubbles and Terminal Velocity-> why is my answer wrong? (all work shown)

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Homework Help Overview

The discussion revolves around the terminal velocity of air bubbles rising in water, specifically focusing on a scenario involving a scuba diver. The problem includes calculations related to the viscosity of water and the implications of pressure changes for divers.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply Stoke's Law to determine the terminal velocity of air bubbles, expressing confusion over their calculations and the expected answer. Other participants check calculations and discuss potential calculator settings. Another participant questions the origin of a specific term in the equations used.

Discussion Status

Participants are actively engaging with the problem, with some offering guidance on calculation methods and settings. There is a mix of verification of approaches and clarification of concepts, but no consensus has been reached regarding the calculations or the underlying principles.

Contextual Notes

Participants note the importance of understanding the relationship between the density of water and the mass of the bubbles, as well as the implications of pressure changes for divers. There is also mention of potential confusion regarding calculator usage and settings.

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Air Bubbles and Terminal Velocity--> why is my answer wrong? (all work shown)

Homework Statement


Air bubbles of 1.0mm radius are rising from a scuba diver to the surface of the sea. Assume a
water temp of 20C.
a) If viscosity of water = 1.0 X 10-3 Pa.s, what is the terminal
velocity of the bubbles
b) What is the largest rate of the pressure change tolerable for the diver
according to this rule?
(Rule is divers cannot rise faster than their air bubbles when riding to the surface
* helps avoid rapid pressure changes that cause the bends)


Homework Equations


r = 1.0mm = 1.0 x 10-3m
T= 20C <--I have no idea what to do with this!
\eta= 1.0 x 10-3 (viscosity of water)
\rho= 1000kg/m3
im assuming density of water will be needed


The Attempt at a Solution



With terminal velocity you produce the right drag so the net force is 0.
And with the air bubbles i know that the terminal velocity is upward

first i set Fnet = 0
and its in the y directions, so
Fy= FD + FB=0
FD= 6\pi\etarv (v=velocity)
^This is known as Stoke's Law

next i solved for vt v = vt

vt= mass water(g)/ 6\pi\etar
=4/3\pir3g(\rhowater/(6\pi\etar)

next i simplified it by taking out pi and
=4/3r2g(\rhowater/(6\eta)

and i got .22
but the answer is suppose to be 2.2 m/s upward-WHAT did i do wrong?

please help me!
thanks.

NOTE the greek letters ARE NOT in subsrcipt. i don't know why it does that-but it is not an exponent.
 
Last edited:
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Check your calculations. Using your final equation, I do get 2.2 m/s.
 


i have been trying it all day
and just did it again.
now I am getting 2177777

how did u put in ur calc?
 


I put 1E-3 for r, 9.8 for g, 1000 for rho, and 1E-3 for eta. If you don't know how to use your calculator, I can't help you there; my calculator is likely different from yours, and all I can suggest is to watch how the calculator responds to each key you press to make sure it's doing what you want it to.
 


hmm i did the same thing-ill check my settings then.
oh well.
the important thing is that my setup for the prblm was right! =)
 


I'm working on the same problem, but I don't understand where the 4/3(Pi)(r^3) came from. I'm not sure how to sub density for the masses (which aren't given).
 

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