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60.13N / 25.01E
Joined 19 Sep '11 Moves 56935 An iceberg floats in water so that 90% of its volume is underwater. The iceberg is shaped like a cone, with peak pointing down, base up, and the base rises one foot above the surface of the water. If the iceberg tilts so that base is down and peak up, how high from the water surface will the peak rise?
Omaha, Nebraska, USA
Joined 04 Jul '06 Moves 1114597 R
Removed
Joined 10 Dec '06 Moves 8528 I get the height to be
R
Removed
Joined 10 Dec '06 Moves 8528 Originally posted by joe shmo
I get the height to be
nevermind, thats not correct...looks a bit more complicated at this time of day.
my latest result is
approx
13.45 ft
R
Removed
Joined 10 Dec '06 Moves 8528 Is your silence an indicator that I should try again?
127.0.0.1
Joined 18 Dec '03 Moves 16687 Originally posted by joe shmo
nevermind, thats not correct...looks a bit more complicated at this time of day.
This is what i get.
The total height of the cone is
H = 10 + 3 * 30^(1/3) + 30^(2/3) ~= 28.977
60.13N / 25.01E
Joined 19 Sep '11 Moves 56935 Nice work. =)
SOLUTION
When the peak of the iceberg points down, the underwater
part and the entire iceberg have the same shape.
Let h be the height of the iceberg.
Proportions in objects of the same shape are the same,
except that they are squared for areas and cubed for volumes,
which gives
((h-1) : h)^3 = 9 : 10
from which h = 1 ft / (1 - (9 : 10)^(1/3)) , or about 29 feet.
When the iceberg keels over, the part above water is of
the same shape as the entire iceberg, with volume ratios
1 : 10. Let x be the height of the visible part of the iceberg.
(x : h)^3 = 1 : 10
from which
x = (1 : 10)^(1/3) h
x = 1 ft * (1 : 10)^(1/3) / (1 - (9 : 10)^(1/3))
which is about 13 ft.
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Iceberg
24 Sep '11 23:54
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