Consider a semi-spherical bowl machined out of wood.

Question

Consider a semi-spherical bowl machined out of wood. The outer radius of the bowl is 20.0 cm and it is uniformly 2.0cm thick. The bowl is filled to the top with water. What is the total mass (wood+water) of the water-filled bowl (in units of kg)? The mass densities of water and the wood are, respectively, 1000.0 kg/m3 and 500.0 kg/m3.

Answer 1

 

Please check the densities of the wood and water.

Density of most wood is a single digit decimal number of kg/m^3

Here's a link:

http://www.engineeringtoolbox.com/wood-density-d_40.html

 

The density of water is 1 g/ cm^3 = 1 g x   1 kg / 1000g  x 100^3 cm^3/ 1m^3 = 1000 kg/m^3

 

The volume of a half ball is 1/2 x 4/3 x pi x radius^2

 

Volume of the whole thing is 1/2 x 4/3 x 22/7 x 20 x 20

                                      = 2/3 x 22/7 x 20/1 x 20/1

                                        = 17600/21

                                       

The volume of the empty wooden bowl is 1/2 x 4/3 x 22/7 x 18 x 18

 

                                             = 2/3 x 22/7 x 18/1 x 18/1

 

                                             = 14256/21

 

                                             = 4752/7

 

The volume of the water is 17600/21 - 14256/21 = 3344/21

 

density = mass / volume

 

mass = density x volume

 

M = D* V

 

For the wooden bowl, the mass is 4752/7 x 1000/1 = 4752000/7 = 14256000/21

 

For the water the mass is  3344/21 x 500/1 = 1672000/21

 

If you want to add them together, they already have common denominator of 21.

Just add the top numbers.

 

You can now rewrite the problem with the correct densities, once you find them.