Air Bubbles and Terminal Velocity--> why is my answer wrong? (all work shown) 1. The problem statement, all variables and given/known data 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) 2. Relevant equations r = 1.0mm = 1.0 x 10-3m T= 20C <--I have no idea what to do with this! [tex]\eta[/tex]= 1.0 x 10-3 (viscosity of water) [tex]\rho[/tex]= 1000kg/m3 im assuming density of water will be needed 3. 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[tex]\pi[/tex][tex]\eta[/tex]rv (v=velocity) ^This is known as Stoke's Law next i solved for vt v = vt vt= mass water(g)/ 6[tex]\pi[/tex][tex]\eta[/tex]r =4/3[tex]\pi[/tex]r3g([tex]\rho[/tex]water/(6[tex]\pi[/tex][tex]\eta[/tex]r) next i simplified it by taking out pi and =4/3r2g([tex]\rho[/tex]water/(6[tex]\eta[/tex]) 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 dont know why it does that-but it is not an exponent.