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

[afg_91320]

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 F_net = 0
and its in the y directions, so
F_y= F_D + F_B=0
F_D= 6$$\pi$$$$\eta$$rv (v=velocity)
^This is known as Stoke's Law

next i solved for v_t v = v_t

v_t= mass water(g)/ 6$$\pi$$$$\eta$$r
=4/3$$\pi$$r³g($$\rho$$water/(6$$\pi$$$$\eta$$r)

next i simplified it by taking out pi and
=4/3r²g($$\rho$$water/(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.

 

[ideasrule]

Check your calculations. Using your final equation, I do get 2.2 m/s.

 

[afg_91320]

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

how did u put in ur calc?

 

[ideasrule]

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.

 

[afg_91320]

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! =)

 

[mk88]

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).