A basic geometry question.?

Question

If a sphere with an 8m diameter costs $400 to fill, how much will a sphere with a 16m diameter cost to fill? How do I set it up as well?

Answer 1

The volume of a sphere is
(4/3)(pi)R^3
where R is the radius.

So find the volume of the 8m sphere...
(4/3)(pi)(8^3) = (2048/3)(pi)

Take $$ / Volume to get the cost per cubic meter...

400 / (2048/3 * pi) = 0.18651...

Now find the volume of the 16m sphere...
(4/3)(pi)(16^3) = (16384/3)(pi)

Multiply that by the cost per cubic meter...
(16384/3)(pi)(0.18651) = 3200

The result is $3200

Answer 2

The volume of the sphere will 2^3, according to the volume formula 4/3 pi. r^3 so that the cost will be $400 * 2^3 = $3200

Answer 3

For similar figures (all spheres are similar), ratio of volumes (cost is a function of the volume filled) is the 3rd power of ratio of sides (in this case diameters). So...

x/400 = (16/8)^3 = 2^3 = 8
x = $3200.

Answer 4

The volume of a sphere varies with the cube of the radius. You have doubled the radius, so the volume will increase by a factor of 2^3 = 8.

$400 * 8 = $3200

Answer 5

First, figure out the volume of both spheres.

volume=(4/3)(pi)(r^3)

You know diameter, so you can find r.

Then divide the cost (400) by the volume of the 8m sphere. That'll give you cost per cubic meter.

Then multiply that by the volume of the 16m sphere.

Answer 6

Since you only want to know how much it cost to fill, we will compute this using pure ratio and proportion and the formula of the volume of the sphere.

Formula for volume of sphere = (4*pi*r^3)/3
Diameter = (4*pi*((r/2)^3))/3
=(4*pi*r^3/8)/3
=(pi*r^3/2)/3
=(pi*r^3/6) : 400
=(pi*((r*2)^3)/6 : x

Therefore:

(pi*r^3)/6 : 400 = (8*pi*r^3)/6 : x

We can cancel, pi*r^3/6 from both sides, leaving:

1:400 = 8:x
x = 400*8
x= 3200

Answer 7

Volume for a sphere = (4/3)*pi*r^3
where r = radius of sphere = half the diameter of the sphere

let V1 be the volume of the 8m diameter sphere and r1 the radius of that, which is 4m. similarly for V2 and r2.
V1 = (4/3)*pi*r1^3
V2 = (4/3)*pi*r2^3
V2/V1 = (r1^3)/(r2^3)
= ((2^3)^3) / ((2^2)^3)
= 2^3
= 8
Therefore the cost to fill V2 = 8 x cost to fill V1
= $3200

Answer 8

The volume of a sphere is
V=(4/3)*pi*R^3

The first sphere has radius R1=4
and Volume V1=(4/3)*pi*64

The second has radius R2=8 so
the volume V2=(4/3)*pi*64*8
So the second needs 8 times more
filling ,3200$

Answer 9

The total vol of the 8D sphere is 268. the total for the 16 D is 2144. Then use porportions, for every 268, it takes 400$ to fill, therefore for a 16D sphere, it costs 3200$ to fill.

Answer 10

Volume of sphere is 4/3 x PI x r^3

With 8mm volume = 2144.66058mm^2

With 16mm volume = 17157.284679mm^2

Increase in volume = x8

Cost increases x8
therefore cost = 400x8=$3200