Question

When the weight of an object equals the weight of the water displaced (or pushed away) by the object, the object is neutrally buoyant and easy to lift to the surface. a tanker made of 2.5\times10^3\text{ m}^32.5×10 3 m 3 2, point, 5, times, 10, start superscript, 3, end superscript, space, m, start superscript, 3, end superscript of steel sank to the bottom of the ocean. the density of steel is approximately 8\times10^3\text{ kg/m}^38×10 3 kg/m 3 8, times, 10, start superscript, 3, end superscript, space, k, g, slash, m, start superscript, 3, end superscript. the density of seawater is approximately 1\times10^3\text{ kg/m}^31×10 3 kg/m 3 1, times, 10, start superscript, 3, end superscript, space, k, g, slash, m, start superscript, 3, end superscript. because gravity is pretty much the same everywhere on earth, the weight of the tanker can be assumed to be approximately equal to its mass. in cubic meters, what total volume of seawater needs to be displaced in order to lift the tanker to the surface?

Answer 1

Actualy, the answer to this question is 2x10^4m^3. This is the answer that Khan Academy wants, so I hope you already knew how to do this... Thanks!

Answer 2

Answer: 20 m³

Step-by-step explanation:


1) Data:
Tanker:
Vt = 2.5 × 10³ m³
Dt = 8 × 10³ kg / m³

Sea water:
Vs = ?

Ds = 1 × 10³ kg / m³

2) Physical principles and formulas

Buoyancy: weight of the object equals the weight of the water displaced (or pushed away) by the object.

Weight of water = Weight of tanker

weight = mass × g ..... g = acceleration due to gravity ≈ constant
density = mass / volume ⇒ mass = density × volume


3) Solution

Weight of water = Weight of tanker
Ds × Vs × g = Dt × Vt × g


1×10³ kg/m³ × Vs × g = 2.5 × 10³ m³ × 8 × 10³ kg / m³ × g


⇒ Vs = 20 × 10⁶ kg / ( 1 × 10³ kg/m³) = 20 m³

Answer: 20 m³