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That would depend on the components of the soil, i.e. clay would be heavier than sand.
Assuming the standard size of a red clay brick 4"x8"x2-1/4", (1) CY of bricks would weight 6,912 lbs, or 3.46 tons.
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He's gained weight.
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He is a lightweight fighter.
Assuming the "dirt" has some organic material in it, figure about 100lbs per cubic foot. 105 cf then would weigh some 10,500 lbs or about 5 tons. If the soil is fill soil, i.e. sand, clay or loam (combination of sand and clay or sand and silt [silt particles are larger than clay particles, but smaller than sand]) then figure some 120 lbs per cubic ft. That would weigh 105 X 120 = 12,600 lbs or 6 1/2 tons.
Because clay particles are extremely small, clay has a very high surface area. That is to say: If one cubic meter of dry sand and one cubic meter of dry clay were spread out on the ground (so that you could not see the surface) you could expect the clay to cover an area at least ten times that of what the sand would cover. Because of the smaller particle size the air/ pore spaces between the particles of clay are much smaller (than that of sand). For this reason water is held more tightly against the clay particles through cohesive forces and less drains away through gravity. If water were poured on dry sand because of the large pore spaces and lower cohesive forces more water would drain through due to gravity (although some would remain).
He's overweight.
light heavy weight AT 175lbs.
One way, if the clay is malleable, is to reshape it into a cuboid and measure its length (L), width (W) and height (H). Then volume = L*W*H cubic units.Displacement methods will often not work either because the clay will absorb the water (or other solvent), or it might disintegrate.