Question

The volume of an iceberg is 308,000ft³ .(a) What is its weight (in Ib), assuming it is pure ice? (Round your answer to at least four significant figures.) Ib (b) What is the volume of seawater ( in ft³ ) it displaces when floating? (Hint: You know what the welght of the seawate is. Round your answer to at least three significant figures.) X ft³ (c) What is the volume of the part of the iceberg out of the water ( in ft³ )?

Answer 1

The weight of the iceberg is 17,671,336.82 lb. The volume of seawater displaced when the iceberg is floating is approximately 8,009.02 m³. The volume of the part of the iceberg out of the water is 728.92 m³.

Step-by-step explanation:

To find the weight of the iceberg, we need to use the formula:
Weight = Volume × Density

Given:
Volume = 308,000 ft³
Density of ice = 917 kg/m³

Converting the units:
Volume = 308,000 ft³ × (0.3048 m/ft)³ = 8736.94 m³

Weight = 8736.94 m³ × 917 kg/m³ = 8,009,747.38 kg

Since 1 kg = 2.20462 lb, the weight of the iceberg is:
8,009,747.38 kg × 2.20462 lb/kg = 17,671,336.82 lb

To find the volume of seawater displaced when the iceberg is floating, we need to use Archimedes' principle:
Volume of seawater displaced = Volume of iceberg × (Density of ice / Density of seawater)

Given:
Density of seawater = 1000 kg/m³

Volume of seawater displaced = 8736.94 m³ × (917 kg/m³ / 1000 kg/m³) ≈ 8,009.02 m³

To find the volume of the part of the iceberg out of the water, we subtract the volume of seawater displaced from the total volume of the iceberg:
Volume out of water = Volume of iceberg - Volume of seawater displaced

Volume out of water = 8736.94 m³ - 8009.02 m³ = 728.92 m³