Question

A adult with a mass of 30kg jumps into a swimming pool tucked into a ball-like shape, with a radius of 30 cm, a velocity of 0.11 m/s (was calculated to find buoyancy), buoyancy force of 1078 N and a gravity force of -294 N.

With this information, calculate the terminal velocity of the adult still balled up as he sinks in the pool

*density: 1000 kg/m^3

*viscosity: 8.9*10^-4 kg/ms

*gravity: 9.8 m/s^2

*Please answer this to the best of your ability, I've been having trouble with it

Answer 1

IPLEASE NOTE THAT IF YOU SOLVE ABOVE EQUATION NUMBER 1 BY PUTTING VALUES OF Fb, W and Fv YOU'll GET THIS DIRECT VALUE SHOWN BELOW AND ALSO NOTICE THAT VISCOSITY IS DYNAMIC OR ABSOLUTE NOT KINEMATICSO KEE IN MIND WHILE SOLVING PROBLEMS

-fv 17 FB > FV + FB رن) دا = w viscous twtajebje ㅏ drag Force of buoyancy GIRO + 1078 = -294 6tnrv = -1372 -1372 -1372 6.nr 6X3.94 x 8.9x104x0.30 v=-27.26x109 m/s

Fv+ FB W 6 tmou + Po gxgror? = pengam? 38th th= 2 h (P-Podg 2 72 (P-Pog tva 3 on terminal velocity

rachin beet (ball Spherican Demotygoyect Density of liquod Va 2T (P-PG Terminal velocity gn Loviscosity