Question

12 formula 4 points a floating swimming platform in a lake is a rectangular solid with a length of 2.2 m and width 1.1 m. the bottom face of the platform is 11 cm below the surface of the water. find the mass in kilograms of the platform. round your answer to the nearest kilogram. the density of water is 1000 kg/m³. answer

Explanation:

Step1: Calculate the volume of the displaced water

The platform is a rectangular solid. The formula for the volume of a rectangular solid is \(V = l\times w\times h\).
Given \(l = 2.2\space m\), \(w = 1.1\space m\), and \(h=0.11\space m\) (since \(11\space cm=0.11\space m\)).

$$V = 2.2\times1.1\times0.11=\frac{22}{10}\times\frac{11}{10}\times\frac{11}{100}=\frac{22\times11\times11}{10\times10\times100}=\frac{2662}{10000}= 0.2662\space m^{3}$$

Step2: Use the density formula to find the mass

The density formula is \(
ho=\frac{m}{V}\), where \(
ho\) is density, \(m\) is mass, and \(V\) is volume.
We know \(
ho = 1000\space kg/m^{3}\) and \(V=0.2662\space m^{3}\).
Rearranging the formula for \(m\) gives \(m=
ho\times V\).

$$m = 1000\times0.2662=266.2\space kg$$

Answer:

\(266\space kg\)