Posted by CJ on Thursday, January 27, 2011 at 7:17am.
PLEASE HELP ME! i have this homework due tomorrow and i only have 1 question left and am sooo desperate for an answer:
The diameter of the Dome of the Rock in Jerusalem is 20m. If it is plated with gold that is 5cm thick, find the cost of plating the dome in gold given that 1kg of gold costs $43,395 and 1g of gold is equal to 19.3cm>3
p.s cm>3 means cm cube
- geometry - Damon, Thursday, January 27, 2011 at 7:43am
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
- geometry - drwls, Thursday, January 27, 2011 at 7:52am
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
- geometry - drwls, Thursday, January 27, 2011 at 7:59am
I treated it as a 5 cm layer over a cylinder. Damon treated it as a hemisphere. He is correct.
- geometry - CJ, Thursday, January 27, 2011 at 9:55am
thank you sooo much damon :)
- GOLD DUST FOR SALE - BENSON IMO, Tuesday, January 17, 2012 at 4:47am
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
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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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