Question

1. ninety - eight cut boulders are being brought in to help prevent further erosion on a portion of a coastline. each boulder has been roughly cut into a cube with sides of length 0.8 meters. the stone in the boulders has an average density of 3000 kg/m³. a. find the average weight (mass) of each boulder in pounds (round to the nearest ten pounds). (3 points)

Explanation:

Step1: Calculate the volume of the cube

The volume \(V\) of a cube is given by \(V = s^{3}\), where \(s\) is the side - length of the cube. Here, \(s=0.8\space m\), so \(V=(0.8)^{3}=0.8\times0.8\times0.8 = 0.512\space m^{3}\)

Step2: Calculate the mass of each boulder using the density formula

The density formula is \(
ho=\frac{m}{V}\), where \(
ho\) is the density, \(m\) is the mass, and \(V\) is the volume. We know that \(
ho = 3000\space kg/m^{3}\) and \(V = 0.512\space m^{3}\). Rearranging the formula for \(m\), we get \(m=
ho V\). Substituting the values, \(m = 3000\times0.512=1536\space kg\)

Step3: Convert kilograms to pounds

We know that \(1\space kg\approx2.20462\space pounds\). So, if \(m = 1536\space kg\), then the mass in pounds \(M\) is \(M=1536\times2.20462\)

$$M = 1536\times2.20462=(1500 + 36)\times2.20462=1500\times2.20462+36\times2.20462$$

$$1500\times2.20462 = 3306.93$$

$$36\times2.20462=79.36632$$

$$M=3306.93+79.36632 = 3386.29632\approx3386.3$$

Answer:

\(3386.3\) pounds