Asked by Mike
During heavy rain, a section of a mountainside measuring 2.5 km horizontally, 0.80 km up along the slope, and 2.0m 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 0.40km x 0.40km and that mud has a density of 1900kg/m3 (cubic meters). What is the mass of the mud sitting above 4.0m2 (squared meters)area of the valley floor?
Answers
Answered by
nullpointer
1) get initial volume (2500x800x2)
2) divide that volume by the surface area (400x400)
3) you have depth from equation 2
4) (4x4xdepthx1900)
Visualize it as this. You first find the cube. Then they want a cube inside that cube, the base being a 2x2 which is 4 meters squared.
If Depth is 25 then 25x4x1900 = 1.9x10^5kg
2) divide that volume by the surface area (400x400)
3) you have depth from equation 2
4) (4x4xdepthx1900)
Visualize it as this. You first find the cube. Then they want a cube inside that cube, the base being a 2x2 which is 4 meters squared.
If Depth is 25 then 25x4x1900 = 1.9x10^5kg