Respuesta :
Answer:
0.0166 mm
Explanation:
Solution:-
- We are given a hollow palladium of material density ( ρpd ).
- The object is given to be spherical in shape with shell thickness ( t ).
- The radius of the spherical shell was determined in the first part ( R = 59.90 mm ).
- We are to determine the thickness of the shell wall.
- We will use the simple mass - density - volume relationship as follows:
m = ρpd*V
Where,
V: The volume of spherical shell
m: The mass of the object = 0.9020 kg
- The volume ( V ) of the hollow sphere can be written down in differential form as follows:
dV = As.dt
Where,
The thickness of shell ( t ) is uniform; hence,
V = As.dt
Where,
As = the surface area of sphere with median radius ( R )
- Therefore the formulation becomes:
m = ρpd*As*t
t = [ m / 4*ρpd*π*R^2 ]
t = [ 0.902 / (4*1.202*10^4*π*0.599^2) ]
t = 0.0166 mm ... Answer
