Question

Show all work. (a) Calculate the pressure at a depth of 16.0 m below the surface of the ocean. Assume the density of sea water is 1037 kg/m, and atmospheric pressure at sea level is 1.02 x 10 Pa (b) A submersible of mass 67.0 kg, and volume 66.3 litres, is at this depth. Will it rise to the surface or sink? (c) If you could change the mass of the submersible, so that it would have neutral buoyancy, and remain at the same depth, what would the new mass be? G

Answer 1

Date depth (d)= 16 m. density of water s = 1037 kg/m3 I a). Pressure due to water 1. Pw aggh - 1037 x 9.8 x 16 = 1.626 X 105 N/m2 Total pressure = atomospheric Pressure fpw = 2.646 x 105 N/m² = 2.646X 105 Pa e.

Carlin Page Date I volume of submersible = 66.3 lit V = 66.3 X 10-3 m3 weight of submersible = 67g N W = 656,6 N. Buoyancy force Fur FB = 8gv | = |234 66.3 Xe kr@g 8 = 673.78 N Here for > W buoyancy force is greater than weight of submersible hence it will float i.e. will rise to the surface c) for at sa submersible to be remain same depth W =Fo mg = sqV m = 1037X 66.3 X 10-3 = 68.75 kg.