0.5m-1
Perhaps if you read the question properly, you would not have to ask the question!
The ratio is 1/2 square meter per cubic meter.
It is not possible to have a sphere with a surface are of 300 metres squared and a volume of 500 metres cubed. A surface area of 300 sq metres would imply a volume of 488.6 cubic metres or a shape that is non-spherical!
This is pretty easy, just divide 432 by 864 and you get a 1:2 ratio.
Volume= surface area (length x width) x depth re arrange to surface area= depth= Volume/Area Area= Volume/Depth
0.6 m-1 is the ratio of surface area to volume for a sphere.
0.4 m-1 is the ration of surface area 588m2 to volume 1372m3 for a sphere.
Perhaps if you read the question properly, you would not have to ask the question!
The ratio is 1/2 square meter per cubic meter.
It is not possible to have a sphere with a surface are of 300 metres squared and a volume of 500 metres cubed. A surface area of 300 sq metres would imply a volume of 488.6 cubic metres or a shape that is non-spherical!
This is pretty easy, just divide 432 by 864 and you get a 1:2 ratio.
The answer depends on what measure - volume, surface area - equals 375.
1
Volume= surface area (length x width) x depth re arrange to surface area= depth= Volume/Area Area= Volume/Depth
The ratio of surface area to volume for a sphere is constant and equal to 3/r, where r is the radius. Given the measurements, you can calculate the radius of the sphere using the formula for volume of a sphere (V = 4/3 * π * r^3) and then find the ratio as 3/r.
The formula for the surface area of a sphere is 4πr² and the formula for the volume is (4/3)πr³, where r is the radius of the sphere. Setting 4πr² equal to 588 and (4/3)πr³ equal to 1372, you can solve for the radius by equating the two expressions and taking the cube root of the result. Once you have the radius, you can calculate the surface area using the formula and divide it by the volume to find the ratio.
There is no direct relationship between the volume (length*breadth*height) and weight. A given volume of air and the same volume of lead will have ver different weights.