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?

Question

08-04-2020
Physics

contestada

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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Xema8 Xema8 · Answer 1 · 2020-04-08T06:10:04+02:00

Answer:

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

lhmarianateixeira lhmarianateixeira · Answer 2 · 2022-01-18T12:34:26+01:00

In this exercise we have to use the knowledge of force to calculate the requested alternatives, in this way we have:

A) [tex]2658.55 \ Kg[/tex]

B)Downward direction as weight is higher

C)[tex]13.5 \ kg[/tex]

D)Down wards

So from the knowledge of force we have that:

A ) By the force formula, we have:

[tex]weight\ of\ air + weight\ of\ tank = 013.5 + 3.2\\= 16.7 kg[/tex]

By the volume of air formula, we have:

[tex]volume\ of\ water\ displaced = 3.2 / 1.225 = 2.61 m^3 \\weight\ of\ water\ displaced = 2.61 * 1025 = 2675.25 kg\\buoyant\ force = weight\ of\ displaced\ sea\ water = volume \ of\ displaced\ sea water * density\\= 15.7 x 10^{-3} * 1025\\= 16.0925 kg\\net\ force = buoyant\ force - weight\\= 2675.25 - 16.7\\= 2658.55 kg[/tex]

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

C ) By the force formula, we have:

[tex]weight\ of\ tank + weight\ of\ water\ in \ it = 13.5 + 15.7 * 10^{-3} * 1025= 13.5 + 16.0925= 29.5925 kg[/tex]

D ) net force down wards

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