SOLUTION: A cylinderical can has a radius of 3.4 cm.and when a sphere is placed in it , the water covers it. Now if the sphere is fit in it then find the,
A) total surface area of can.
B)
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Question 1018215: A cylinderical can has a radius of 3.4 cm.and when a sphere is placed in it , the water covers it. Now if the sphere is fit in it then find the,
A) total surface area of can.
B) depth of water in can before the sphere.
Found 3 solutions by Alan3354, ankor@dixie-net.com, ikleyn:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
A cylinderical can has a radius of 3.4 cm.and when a sphere is placed in it , the water covers it. Now if the sphere is fit in it then find the,
A) total surface area of can.
B) depth of water in can before the sphere.
----------
Not enough info.
radius of sphere?
height of cylinder?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A cylindrical can has a radius of 3.4 cm.and when a sphere is placed in it, the water covers it.
To make this problem possible, we have to assume the sphere just fit in the cylinder.
Three sides of sphere touched 3 sides of the cylinder
The top of the sphere was even with the top of the cylinder
Therefore the height of the cylinder and the diameter of the sphere was 6.8 cm
Twice the radius
Now if the sphere is fit in it then find the:
:
A) total surface area of can. (including the bottom)
S.A. = + =
S.A. = 145.267 + 36.317
S.A. = 181.584 sq/cm
:
B) depth of water in can before the sphere.
Find the vol of the cylinder
V =
V = 246.954 cu/cm
Find the vol of the sphere
V =
V = 164.636
Find the vol of the water when the sphere is placed in the cylinder
246.954 - 164.636 = 82.318 cu/cm of water
Find how high (h) the water will be when the cylinder is removed
= 82.318
36.317h = 82.318
h =
h = 2.267 cm is the height of the water in the cylinder when sphere is removed
Answer by ikleyn(52778) (Show Source): You can put this solution on YOUR website!
.
It is well known fact that the volume of the cylinder of the radius R and of the height R minus the volume of the hemi-sphere
of the radius R equals the volume of cone of the radius R and the height R.
I.e. equals one-third of the volume of the cylinder.
Well known after Archimedes.
See the picture in
http://mathcentral.uregina.ca/qq/database/QQ.09.01/rahul1.html
So the answer to question B) is: of the height of the cylinder, which (height) is 2R.
Without any calculations. Simply refer to Archimedes.
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