1018

• Run Multiple Choice Generator at h:\shares\physics
• Run Core Problems by giving the program qzz as your 'course'--enter this when asked for your 3-digit course number.  
• Do Precalculus Quiz 1018

If we shave a square in each direction so that its dimensions decrease to 20.0 % of their original dimensions, then the area of the square becomes 20.0 % of 20.0 % of its original area, or .200 * .200 = .200 ^2 = .0400 of the original.

If we shave a cube in each direction so that its dimensions decrease to 20.0 % of their original dimensions, then the volume of the cube becomes 20.0 % of 20.0 % of 20.0 % of its original volume, or .200 * .200 * .200 = .200 ^3 = .00800 of the original.

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The surface area of a 3-dimensional object can be though of as being covered of tiny squares. If we scale up the object so that its linear dimensions change, without changing the proportion of those dimensions, then the surface area of the object will be related to any given linear dimension x by

A = k x^2, where k is a constant number.

Any quantity Q that uniformly covers area will therefore obey a relationship Q = k x^2 with that linear dimension.

For example if the linear dimension we choose is length of the diagonal, then we will have

A = k * diagonal^2.

For instance, then, the amount of paint required to cover the surface will be in the relationship P = k x^2 with the diagonal.

The volume of a 3-dimensional object can be though of as being composed of tiny cubes. If we scale up the 3-dimensional object so that its linear dimensions change, without changing the proportion of those dimensions, then the volume of the object will be related to any given linear dimension x by

V = k x^3, where k is a constant number.

For example if the linear dimension we choose is altitude, then we will have

V = k * altitude^3.

Any quantity Q that uniformly occupies volume will therefore obey a relationship Q = k x^3 with that linear dimension.

For instance, then, the amount of mass which uniformly occupies a volume will be in the relationship M = k x^3 with the altitude.

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If the volume and surface area of a 3-dimensional object with altitude 10.8 are, respectively, 1200.0 and 151. , then we know that
 

• V = k * altitude^3,

where

• k = V / altitude^3 =
• volume / altitude^3 =
• 1200.0 / (10.8 )^3 = .980 .
   

It follows that
 

• V = .980 * altitude^3,

so when altitude = 10.7 we have
 

• V = .980 * 10.7 ^3 = 1170.0 .

We also have the surface area relationship A = k * altitude^2, so that
 

• k = A / altitude^2 = 151. / (10.8 )^2 = 1.31 .
   

Thus

 

• A = 1.31 * altitude^2,

and where altitude is 10.7 we get

• area = A = 1.31 * 10.7 ^2 = 147. .

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