Part A Data Analysis

Water

1. -0.02746 N/cm

2. (0.111 N) / [(9.8 m/s^2) (2.81x10^-4 m^2) (0.04 m) = 1,007.7 kg/m^3

3. (1,007.7 - 1,000) / 1,000 = 0.077 = 0.77 %

Oil

1.  -0.02422 N/cm

2. (0.103 N) / [(9.8 m/s^2) (2.81x10^-4 m^2) (0.04 m) = 935.072 kg/m^3

3. ρoil = .464 kg / 500 mL = 928 kg/m^3; (935.072 - 928) / 928 = .0076 = .76 %

Part B Data Analysis

Block 1 Dimensions: h = 7.42 cm; w = 4.31 cm; l = 4.31 cm; submerged h = 6.92 cm; mass = 129.3 g

Block 2 Dimensions: h = 3.8 cm; w = 7.6 cm; l = 7.6 cm; submerged h = 2.8 cm; mass = 162.3 g

Volumes: block 1 = (7.42 cm) (4.31 cm) (4.31 cm) = 137.835 cm^3 = 1.378x10^-4 m^3; block 2 = 219.488 cm^3 = 2.195x10^-4 m^3

Volumes Submerged (and dispaced): block 1 = 1.286x10^-4 kg/m^3; block 2 = 1.617x10^-4 kg/m^3

Fractions Submerged: block 1 = 0.933; block 2 = 0.7367

Density: block 1 = 938.316 g/m^3; block 2 = 739.08 g/m^3

Discussion Questions

1. By filling the submarine's ballasts with water or air to submerge or rise.

2. No. When the block is vertical there is less surface area than when it is horizontal.

3. No. The measured weight would be the same. While the water is exerting a buoyant force on the block, the block is exerting an equal and opposite force down on the water.

4.  a) ρliq = (ρball) / [weight / (weight - apparent immersed weight) = (2.8x10^3 kg/m^3) / [ (2.8 kg x 9.8 m/s/s) / ( (2.8 kg x 9.8 m/s/s) - (1.9 kg x 9.8 m/s/s) ) ] = 317 kg/m^3

5. No. The buoyant force is given as Fbuoy = ρVg. The hollow steel ball has less density and therefore a smaller buoyant force.