There is a glass in the shape of a truncated cone with radii r1 = 2cm, r2 = 4cm and height h = 10 cm, like the one shown in the following figure, if it is filled with soda to a height of 9cm, where it is observed a radius of 19/5 = 3.8cm and add a 3 cm edge (side) ice cube, using the total differential to estimate the volume increase dV a) Does the liquid overflow or not? b) Explain why c) How many cubes does it take at least for the liquid to overflow? Suggestion: The volume of a truncated cone is given by V = 1 / 3πh ((r1) 2 + r1r2 + (r2) 2), in this case with r1 = 2cm, remaining fixed throughout the process it reduces to V = 1 / 3πh (4 + 2r + (r) 2),
There is a glass in the shape of a truncated cone with radii r1 = 2cm, r2 = 4cm and height h = 10 cm, like the one shown in the following figure, if it is filled with soda to a height of 9cm, where it is observed a radius of 19/5 = 3.8cm and add a 3 cm edge (side) ice cube, using the total differential to estimate the volume increase dV
a) Does the liquid overflow or not?
b) Explain why
c) How many cubes does it take at least for the liquid to overflow?
Suggestion: The volume of a truncated cone is given by V = 1 / 3πh ((r1) 2 + r1r2 + (r2) 2), in this case with r1 = 2cm, remaining fixed throughout the process it reduces to V = 1 / 3πh (4 + 2r + (r) 2), where r (the radius) and h (the height) are variables when the glass is filling.
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