surface area of sphere = (4/3) pi r^2
surface area of hemisphere = (2/3)pi r^2
here r = 10 m
surface area = (200/3) pi meters^2
there are 10^4 cm^2 per meter^2
surface area = (200/3) pi*10^4 cm^2
volume = surface area * depth
= (200/3) pi *10^4 * 5 cm^3
grams = volume*density
= (200/3) pi *10^4 * 5 (1/19.3)
cost per gram = 43.395
cost = (200/3) pi *10^4 * 5 (1/19.3)*43.395
The volume of the gold would be
V = (pi/4)D^2*(thickness) = 15.7 m^3
= 15.7*10^6 cm^3
The mass would be (density)*V = 303.2 *10^6 g = 303.2*10^3 kg
Multiply that by the cost per kg.
I get about 13 billion dollars
I treated it as a 5 cm layer over a cylinder. Damon treated it as a hemisphere. He is correct.
thank you sooo much damon :)
GOLD DUST FOR SALE
Dear Sir/ Madam,
We are small scale Gold miners here on the sub-region of West Africa. We have some huge
quantity of alluvial Gold Dust for sale at a considerable price of $30,500USD per kilo
which is below world market price, 22 carat and 92.05% purity. If you are interested, do
not hesitate to get back to us as soon as possible for us to give you our full co-operate
MR BENSON IMO
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