A steel ball bearing 8.00 mm in diameter is dropped into a cylinder of glycerin. The densities of steel and glycerin are 7.80 x 10^3 and 1.26 x 10^3 kg/m^3, respectively. What is the terminal speed of the ball bearing?
a. 153 m/s
b. 15.3 m/s
c. 1.53 m/s
d. 0.153 m/s
2006-08-07 03:31:07 · 2 answers · asked by Lou 1 in Science & Mathematics ➔ Physics
Use the terminal velociy formula:
Tv = sqrt(2*m(steel)*g / ρ(glycerin)*A*C)
see link below
You can find the mass of the steel ball by knowing the density and diameter (and hence the volume), so that you end up with:
Tv = sqrt(8*ρ(steel)*g*radius / ρ(glycerin)*3*C)
Now if you use a C of 2.75 you get 1.53m/s
Hope that helps
2006-08-07 04:21:23 · answer #1 · answered by 1,1,2,3,3,4, 5,5,6,6,6, 8,8,8,10 6 · 0⤊ 0⤋
Stoke's Law
mg - d*V*g= 6*pi*n*r*v
n = viscosity of fluid
r = radius of ball
m = mass of steel (density of steel * ball's vol)
V = ball's volume
d = density of glycerine
g = 9.8m/s^2
v = Terminal velocity.
2006-08-07 04:01:54 · answer #2 · answered by ag_iitkgp 7 · 0⤊ 0⤋