A hollow palladium (rhoPd = 1.202 ✕ 104 kg/m3) spherical shell of mass m = 0.9020 kg floats in water with its entire volume below the surface. (Assume the temperature of the water is 4°C. Enter your answers to at least four significant figures.) (a) What is the radius of the sphere?

59.90 mm Correct

(b) What is the thickness of the shell wall?

1.3 mm Incorrect

please help with part B)

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

hi there!

your question asks at which segment does freezing occur.

since we're talking about water, the freezing point of water will be 32 degrees fahrenheit or below.

hope this ! ~