For a regular six sided object: a cube or rectangular prism (box); just multiply length by width by height. Whatever units you use for the measurements, the answer is in cubic units.
To find the density of an object, it would be D=m/v.So the density would be: 7.3099415204678362573099415204678 kg/L.But in this, there are also significant figures, and so in 6.25, there are 3. Therefore, the answer cannot be more precise than the given equation.So it would round to 7.30.Density = 7.30 kg/L.
Density is just the mass divided by the volume, so 3 grams ÷ 6 mL = 0.5 g/mL
Using Boyle's Law (P1V1 = P2V2), we can rearrange the formula to find the new volume: V2 = (P1 * V1) / P2 V2 = (1 ATM * 2.4 L) / 6 ATM V2 = 0.4 L Therefore, the gas will occupy a volume of 0.4 L at 6 ATM.
6 sided with a square figure
6 kilograms / 8 liters = 6,000 grams / 8,000 cc = 0.75gram per cc. The jug will float in water.
Surface area is the amount of space on the face of a 2D side, like a piece of paper. And to find this we use (for 4 sided object); A = lh where l = length and h = height. This formula changes as the number of sides increase or decrease. Volume is the amount of space inside a 3D object, like a cube. To find this we use(for a 6 sided object); A = lbh where l = length, b = breadth and h = height.
Hexagon
A hexagon
It's a 6 sided shape.
Cube.
A hexagon.
An object with 6 corners is called a hexagon.
'Object' Do you mean 'shape' or 'solid'. a 2-dimensional shape is an hexagon a 3-dimensional solid is a hexahedron.
To find the volume of an irregular object such as a rock, you have to use displacement. If you place the object in a graduated cylinder filled with water, the volume of the object is equal to the amount of water that the object displaces. For example, if a graduated cylinder is filled with 100mL of water, and you place an object such as a rock and the water rises from 100mL to 106mL, then the volume of the rock is 6.
Octahedron.
The volume of a 6 in x 18 in x 6 in object is 648 in3.
They are 1728 mm3 or 1.728 millilitres in volume.