A scuba tank, when fully submerged, displaces15.7 Lof seawater. The tank itself has a mass of13.5kgand, when "full," contains 3.20kgof air. The density of seawater is 1025 kg/m3. Assume that only its weight and the buoyant force act on the tank.

Part A
Determine the magnitude of the net force on the fully submerged tank at the beginning of a dive (when it is full of air).
Part B
Determine the direction of the net force on the fully submerged tank at the beginning of a dive (when it is full of air).
Part C
Determine the magnitude of the net force on the fully submerged tank at the end of a dive (when it no longer contains any air).
Part D
Determine thedirection of the net force on the fully submerged tank at the end of a dive (when it no longer contains any air).

Upward or Downward?

Explanation:

A ) weight of air + weight of tank = 013.5 + 3.2

= 16.7 kg

volume of air

= 3.2 / 1.225 = 2.61 m³

= volume of water displaced = 2.61 m³

weight of water displaced = 2.61 x 1025

= 2675.25 kg

buoyant force = 2675.25 kg

buoyant force = weight of displaced sea water

= volume of displaced sea water x density

= 15.7 x 10⁻³ x 1025

= 16.0925 kg

net force =  buoyant force - weight

= 2675.25 - 16.7

= 2658.55 kg

B ) net force will be in downward direction as weight is higher .

C ) weight of tank + weight of water in it

= 13.5 + 15.7 x 10⁻³ x 1025

= 13.5 + 16.0925

= 29.5925 kg

buoyant force = 16.0925 kg

net force = 29.5925 - 16.0925

= 13.5 kg

D ) net force down wards

current and voltage can both be different in different parts of a circuit so it matters where you place the ammeter and voltmete