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1 March, 15:59

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?

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Answers (1)
  1. 1 March, 16:06
    0
    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
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