A cross section is cut from a brass cylinder with walls that are 2 centimeters thick. The radius of the cross section, measured to the outer portion of the ring, is 5.5 centimeters. What is the maximum volume of a ball that could pass through the interior of the brass ring?

Respuesta :

Here, Radius to the inner portion = Radius to outer portion - thickness
r = 5.5 - 2 = 3.5 cm

Now, ball with radius of 3.5 cm can pass through the ring. 
So, volume = 4/3 πr³
v = 4/3 * 3.14 * (3.5)³
v = 4.186 * 42.875
v = 179.50 cm³

In short, Your Answer would be: 179.50 cm³

Hope this helps!

The maximum volume of a ball that could pass through the interior of the brass ring is  179.50 cm³   .                                                                                                            

           

What is cross section?

The cross-sectional area is the area of a two-dimensional shape that is obtained when a three-dimensional object - such as a cylinder - is sliced perpendicular to some specified axis at a point. For example, the cross-section of a cylinder - when sliced parallel to its base - is a circle.  

What is volume of sphere?

The volume of sphere is the capacity it has. It is the space occupied by the sphere. The volume of sphere is measured in cubic units, such as m3, cm3, in3, etc. The shape of the sphere is round and three-dimensional. It has three axes as x-axis, y-axis and z-axis which defines its shape.

Formula of volume of sphere :

volume of sphere = [tex]\frac{4}{3} \pi r^{3}[/tex]

According to the question

Thickness of  cross section is cut from a brass cylinder  = 2cm  

Radius of the cross section = 5.5 cm

Radius to the inner portion = Radius to outer portion - thickness

                                             = 5.5 - 2

                                             = 3.5 cm

Now, The maximum volume of ball that could pass through the interior of the brass ring

as ball is a sphere  

so,

volume of sphere = [tex]\frac{4}{3} \pi r^{3}[/tex]  

volume of ball = [tex]\frac{4}{3} \pi (3.5)^{3}[/tex]  

                         = [tex]\frac{4}{3} * 3.14 * 42.875[/tex]  

                         =  179.50 cm³

Hence, the maximum volume of a ball that could pass through the interior of the brass ring is  179.50 cm³   .                                                                                                            

           

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