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Q: Can conic projections result from projecting a spherical surface onto a cone?
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Continue Learning about Math & Arithmetic

What is the greatest number of pieces that can result from slicing a spherical orange with 3 straight cuts?

8


How many one dollar bills wold it take to go from one end to the other end of the earth?

The earth is [approximately] spherical and, as a result, it has no ends!


Why do you use geodesic domes?

A geodesic domeis a spherical or partial-spherical shell structure or lattice shell based on a network of great circles(geodesics) on the surface of a sphere. The geodesics intersect to form triangular elements that have local triangular rigidity and also distribute the stress across the structure. When completed to form a complete sphere, it is a geodesic sphere. A dome is enclosed, unlike open geodesic structures such as playground climbers.Typically a geodesic dome design begins with an icosahedron inscribed in a hypothetical sphere, tiling each triangular face with smaller triangles, then projecting the vertices of each tile to the sphere. The endpoints of the links of the completed sphere are the projected endpoints on the sphere's surface. If this is done exactly, sub-triangle edge lengths take on many different values, requiring links of many sizes. To minimize this, simplifications are made. The result is a compromise of triangles with their vertices lying approximately on the sphere. The edges of the triangles form approximate geodesic paths over the surface of the dome.Geodesic designs can be used to form any curved, enclosed space. Standard designs tend to be used because unusual configurations may require complex, expensive custom design of each strut, vertex and panel.See the related link.


How do you work out the surface area of a triangular piece of cake?

times the base and the height and halve the result. simple!


What is the gravitational potential energy of an object equal to?

It is the product of the mass of the object in Kg, the gravitational acceleration which is 9.81 m/sec2, and the height of the object above earth's surface in meters. Result is in Joules