During heavy rain, a section of a mountainside measuring 2.5 km horizontally, 0.80 km up along the slope, and 2.0 m deep slips into a valley in a mud slide. Assume that the mud ends up uniformly distributed over a surface area of the valley measuring 0.40 km × 0.40 km and that mud has a density of. What is the mass of the mud sitting above a 4.0 m² area of the valley floor?

Dimensions of section of mountainside are $2.5\text{km}\times 0.80\text{km}\times 2.0\text{m}$

The dimensions of surface area of valley $0.40\text{km}\times 0.40\text{km}$

Density of mud ${\text{ρ=1900kg/m}}^3$

The horizontal distance is A = 2.5 km = 2500 m , the height is,B = 0.80 km = 800m and the depth is, C = 2.0 m

Area of the valley floor is

The density of a material is given by mass per unit volume.A volume is the amount of space occupied by an object in three-dimensional space.

The expression for density is given as:
$\rho =\frac{m}{V}$… (i)

Here, $\rho$is the density, $m$is the mass and$V$ is the volume

The volume of the section is calculated as:

Thus, the volume of the section is$4\times {10}^6{\text{m}}^3$

Let d be the thickness of the mud after it has uniformly distributed in the valley.

So, the volume of the mud is,

$V=400\text{m}\times 400\text{m}\times d$

Since the mud is distributed uniformly, therefore,

$$\begin{gathered} 400\text{m}\times 400\text{m}\times d=4\times {10}^6{\text{m}}^3 \\ d=25\text{m} \end{gathered}$$

Calculate the volume of mud over area of$\text{4.0}{\text{m}}^2$

$$\begin{gathered} V_{\text{m}}=4.0{\text{m}}^{\text{2}}\times d \\ =4.0m^2\times 25\text{m} \\ =100.0{\text{m}}^{\text{3}} \end{gathered}$$

As each cubic meter corresponds to a mass of 1900 kg, the mass of that part of the mud can be calculated using the formula for the density.

Thus, the mass of that part of mud is$1.9\times {10}^5\text{kg}$