Asolid steel sphere of density 7.86 g/cm3 and mass 1 kg spin on an axis through its center with a period of 2.3 s. given vsphere = 4 3 π r3 , what is its angular momentum? answer in units of kg m2 /s.

The solution for this problem is:

density = mass / volume 
7860 = 1 / ((4/3) pi r^3) 
r^3 = 1 / (7860 * 4/3*pi) 
r = (1 / (7860 * 4/3*pi))^(1/3)

= 0.067 m 

Inertia = (2/5)mr^2

= (2/5) x 1 x 0.067^2

= 0.0017956 kg-m^2 

1/2.3 = 0.4348 rev/s 

0.4348 x 2pi = 2.732 rad/s 

Angular momentum = Inertia x rad/s 

0.0017956 x 2.732 = 0.00490557 kg m^2/s

luster is not a good characteristic to use by itself because luster is the property of minerals that indicates how much the surface of a mineral reflects light. the luster of a mineral is affected by the brilliance of the light used to observe the mineral surface.

im guessing the west of station b