How far apart are these graduations?

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Homework Help Overview

The discussion revolves around a graduated cylinder with a specified diameter and the calculations related to the spacing of graduations and the rise in water level when a sphere is submerged. The subject area includes geometry and volume calculations in fluid mechanics.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to determine the distance between graduations on a graduated cylinder and the rise in water level due to a submerged sphere. Some participants provide calculations for the height of the cylinder based on volume and radius, while others confirm the correctness of these calculations.

Discussion Status

The discussion includes various calculations and confirmations of the original poster's approach. Some participants express agreement with the calculations presented, indicating a productive exchange of reasoning, though no explicit consensus is reached on the overall problem.

Contextual Notes

The problem involves specific measurements and assumptions about the cylinder and sphere, which may influence the calculations. The context suggests a focus on understanding the relationships between volume, height, and the geometry of the objects involved.

dami
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A certain graduated cylinder has an inside diameter of 4cm.
The graduations on its side are labeled "cc" (cubic centimeters).
1. How far apart are these graduations?
2. The cylinder is partially filled with water, and a solid sphere of radius 1.2 cm is then totally submerged therein. Through what distance does the water level rise when the sphere is so immersed?
 
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Is this correct?
1. Find the height of a cylinder whose radius is 2 cm and whose volume is 1.00 cm^3.

V = pi*r^2*h
h = V/(pi*r^2) = 1/
h = 0.0796 cm

The graduations would be 0.0796 cm apart

2. Find the height of a cylinder of radius 2 which has a volume equal to that of the sphere.

Using R = radius of sphere = 1.2 cm
V(sphere) = (4/3)pi*R^3 = 7.238229474 cm^3

V(sphere) = volume of cylinder = pi*r^2*h
h = V(sphere)/(pi*r^2) = 0.576 cm

So the level of water would rise by 0.576 cm
 


That all looks right to me...

12.566 == area of 4cm diameter circle
0.0795 == height to contain 1cc volume
7.238 == volume of 1.2cm radius sphere
0.576 == height to contain 7.238cc volume
 


Nice bit of homework?
 

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