An area of #"1 hectare"# was filled to a depth of #0.1*m#, and then filled with soil whose density was #1*g*cm^-3#. What mass of soil was used?

1 Answer
Mar 7, 2017

Answer:

#1,000*kg#

Explanation:

We must measure the volume of the plot of the land, and then multiply by the density:

#"1 hectare"# #=# #100*mxx100*m=10000*m^2#

We multiply this value by the depth, to give a volume:

#10000*m^2xx0.1*m=1000*m^3#

And we multiply this volume by the density to get the mass:

#1000*m^3xx1*g*cm^-3=1000*m^3xx1*g*(10^-2*m)^-3=#

#1000*m^3xx1*g*10^6*m^-3=#

#cancel(1000)*cancel(m^3)xx1*cancelgxxcancel(10^-3)*kg*cancel(g^-1)xx10^(cancel(6)3)*cancel(m^-3)=#

#10^3*kg#

I think this is right (all care taken, but no responsibility admitted!). It is always too easy to make an error in these sorts of calculations. We required a mass, and got a mass. Anyway, please check the calculation. Again, I suspect that the actual density of soil would be much more than #1*g*cm^-3#.