The answer from question 2 shows that the rock lost from each three foot by three foot plot of land in happy valley each year weighs a bit less than a small hamburger patty. but, how thick is the layer of rock that is lost each year? we need the density of the rock to calculate the thickness lost. the density of calcite, the main mineral in limestone, is about 2700 kilograms per cubic meter (2700 kg/m3), which is almost three tons for each three foot cube. (rock is there is a tiny bit of space between some of the grains in the rock, so let's use 2600 kilograms per cubic meter (2600 kg/m3) for the density. for simplicity, let's round off the answer from question 2, to obtain one-tenth of a kilogram from each square meter each year, or 0.1 kg/m2/yr (that is 0.22 pounds or 3.5 ounces, which is a bit less than the 4 ounces in a hamburger patty). dividing this yearly rate at which each square meter of the valley is losing kilograms of rock by the density in kilograms per cubic meter yields the rate at which the valley surface is being lowered in meters per year (m/yr). the lowering rate is 0.1 kg/m2/yr divided by 2600 kg/m3 =/yr (note: your calculator probably shows a whole bunch of digits; just choose the answer below that is closest).a. 0.38 (your calculator might also show this as 3.8x10-5 or something similar)b. 0.08 (your calculator might also show this as 8.0x10-2 or something similar)c. 0.38 (your calculator might also show this as 3.8x10-1 or something similar)d. 0.0038 (your calculator might also show this as 3.8x10-3 or something similar)e. 0.18 (your calculator might also show this as 1.8x10-5 or something similar)