Asked by Anonymous
A spherical buoy of radius 2 meters floats in a calm lake, 40 cm of the buoy
are submerged. Place a coordinate system with the origin at the center of the sphere. Find an equation of the circle formed at the waterline of the buoy.
(A) x^2 + y^2 = 2.04 (B) x^2 + y^2 = 1.75 (C) x^2 + y^2 = 1.44
(D) x^2 + y^2 = 1.11 (E) x^2 + y^2 = 2.31
are submerged. Place a coordinate system with the origin at the center of the sphere. Find an equation of the circle formed at the waterline of the buoy.
(A) x^2 + y^2 = 2.04 (B) x^2 + y^2 = 1.75 (C) x^2 + y^2 = 1.44
(D) x^2 + y^2 = 1.11 (E) x^2 + y^2 = 2.31
Answers
Answered by
mathhelper
Make a sketch of the cross-section of the sphere.
It will have a radius of 2 m
Since .40 m will be submerged, the centre will be 1.60 m above
the water line. So we have to find the radius of the circle at
the water line.
let the radius of that circle be r
r^2 + 1.6^2 = 2^2
r^2 = 4 - 2.56
r^2 = 1.44
the equation is x^2 + y^2 = 1.44
It will have a radius of 2 m
Since .40 m will be submerged, the centre will be 1.60 m above
the water line. So we have to find the radius of the circle at
the water line.
let the radius of that circle be r
r^2 + 1.6^2 = 2^2
r^2 = 4 - 2.56
r^2 = 1.44
the equation is x^2 + y^2 = 1.44
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