Question

1. icebergs are hazardous to shipping because so much of their volume is below the water level. if the density of seawater is 1025 kg/m³ and the density of ice is 919 kg/m³, what percentage of an iceberg is below the water level?

Explanation:

Step1: Recall Archimedes' Principle

When an iceberg floats in seawater, the buoyant force equals the weight of the iceberg. So, \( F_{buoyant}=G_{iceberg} \), which means \(
ho_{seawater}gV_{submerged}=
ho_{ice}gV_{iceberg} \). The \( g \) (acceleration due to gravity) cancels out, giving \(
ho_{seawater}V_{submerged}=
ho_{ice}V_{iceberg} \).

Step2: Solve for the ratio of volumes

We want to find the percentage of the iceberg below the water level, which is \( \frac{V_{submerged}}{V_{iceberg}}\times100\% \). From the equation above, \( \frac{V_{submerged}}{V_{iceberg}}=\frac{
ho_{ice}}{
ho_{seawater}} \).

Step3: Substitute the given densities

Given \(
ho_{seawater} = 1025\space kg/m^3 \) and \(
ho_{ice}=919\space kg/m^3 \). Substitute these values into the ratio: \( \frac{V_{submerged}}{V_{iceberg}}=\frac{919}{1025} \).

Step4: Calculate the percentage

First, calculate \( \frac{919}{1025}\approx0.896576 \). Then multiply by 100% to get the percentage: \( 0.896576\times100\%\approx89.7\% \).

Answer:

The percentage of the iceberg below the water level is approximately \( 89.7\% \) (or more precisely, using the exact fraction \( \frac{919}{1025}\times100\%\approx89.66\% \) which can be rounded as needed).