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A cushion filled with stuffing. In geometry, the paper bag problem or teabag problem is to calculate the maximum possible inflated volume of an initially flat sealed rectangular bag which has the same shape as a cushion or pillow, made out of two pieces of material which can bend but not stretch.
Using this term, one can calculate many things in the same way as for a round tube. When the cross-section is uniform along the tube or channel length, it is defined as [1] [2] =, where A is the cross-sectional area of the flow, P is the wetted perimeter of the cross-section.
Precipitators use a rapping system to release the dirt. The crumbling dust falls into the hopper. Once the material in the hopper reaches capacity, it is released through an opening in the bottom with a diameter of about 8–12 inches (20–30 cm). Hoppers are rectangular or circular in cross section but have sides that slope at about a 60° angle.
Packing circles in a square - closely related to spreading points in a unit square with the objective of finding the greatest minimal separation, d n, between points. To convert between these two formulations of the problem, the square side for unit circles will be L = 2 + 2 / d n {\displaystyle L=2+2/d_{n}} .
Rectangle packing is a packing problem where the objective is to determine whether a given set of small rectangles can be placed inside a given large polygon, such that no two small rectangles overlap. Several variants of this problem have been studied.
As the local density of a packing in an infinite space can vary depending on the volume over which it is measured, the problem is usually to maximise the average or asymptotic density, measured over a large enough volume. For equal spheres in three dimensions, the densest packing uses approximately 74% of the volume.
The area required to calculate the volumetric flow rate is real or imaginary, flat or curved, either as a cross-sectional area or a surface. The vector area is a combination of the magnitude of the area through which the volume passes through, A , and a unit vector normal to the area, n ^ {\displaystyle {\hat {\mathbf {n} }}} .
The surface-area-to-volume ratio has physical dimension inverse length (L −1) and is therefore expressed in units of inverse metre (m −1) or its prefixed unit multiples and submultiples. As an example, a cube with sides of length 1 cm will have a surface area of 6 cm 2 and a volume of 1 cm 3. The surface to volume ratio for this cube is thus