The problem describes a solid object in the shape of a cylinder. The diameter of the cylinder is 2.8 m, and the height is 4 m. The mass of the object is 28 kg. We are asked to approximate the density of the object to the nearest hundredth.

1. Problem Description

The problem describes a solid object in the shape of a cylinder. The diameter of the cylinder is 2.8 m, and the height is 4 m. The mass of the object is 28 kg. We are asked to approximate the density of the object to the nearest hundredth.

2. Solution Steps

To find the density, we need to divide the mass by the volume. First, we calculate the volume of the cylinder.

The formula for the volume of a cylinder is:

$V = \pi r^2 h$

where $r$ is the radius and $h$ is the height.

The diameter is given as 2.8 m, so the radius is half of that:

$r = \frac{2.8}{2} = 1.4$ m.

The height is given as 4 m.

Now we can calculate the volume:

$V = \pi (1.4)^2 (4)$

$V = \pi (1.96)(4)$

$V = 7.84\pi$ cubic meters.

Using the $\pi$ button on a calculator, we have $V \approx 24.6300864$ cubic meters.

Now we can calculate the density. Density is defined as mass divided by volume:

$Density = \frac{Mass}{Volume}$

$Density = \frac{28}{7.84\pi}$

$Density \approx \frac{28}{24.6300864}$

$Density \approx 1.13682$ kg/m$^3$

Rounding to the nearest hundredth, we get 1.14 kg/m$^3$.

3. Final Answer

1. 14 kg/m$^3$