Respuesta :
Answer: 20 m³
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
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
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!