In five-dimensional geometry, a 5-cube is a name for a five-dimensional hypercube with 32 vertices, 80 edges, 80 square faces, 40 cubic cells, and 10 tesseract 4-faces. It is represented by Schläfli symbol {4,3,3,3} or {4,3 3 }, constructed as 3 tesseracts, {4,3,3}, around each cubic ridge .

If there exists a point charge of 5Q at a radius of less than (it sits within the void inside the conducting shell), and the total charge of the conducting shell is 3Q, what is the magnitude of the charge on the outer surface of the shell?

In five-dimensional geometry, a rectified 5-cube is a convex uniform 5-polytope, being a rectification of the regular 5-cube. There are 5 degrees of rectifications of a 5-polytope, the zeroth here being the 5-cube, and the 4th and last being the 5-orthoplex. Vertices of the rectified 5-cube are located at the edge-centers of the 5-cube.

Chromosome 5q deletion syndrome is an acquired, hematological disorder characterized by loss of part of the long arm (q arm, band 5q33.1) of human chromosome 5 in bone marrow myelocyte cells. This chromosome abnormality is most commonly associated with …

Mar 11, 2023 · In this case, the charge enclosed within the cube is simply 5q, since the point charge is at the center of the cube. Therefore, we have: Φ = E*A = (5q) / ε0. To find the electric field, we can use Coulomb's law, which states that the electric field due to a point charge q at a distance r is given by: E = k*q / r^2, where k is the Coulomb ...

Aug 9, 2021 · What is the total electric flux through the five faces of the cube other than $\mathrm{ABCD}$? My answer comes out to be $Q/\epsilon$. The solution is: Flux due to inner charge only through fives faces $= 5q/6\epsilon$ Net flux due to outer charge $=0$ (Gauss's Law) Flux due to outer charge only through one face $= q/6\epsilon$

Show that the square of any positive integer cannot be of the form 5q + 2 or 5q + 3 for any integer q. Let a be an arbitrary positive integer. ⇒ a 2 = 5 (5m 2 + 2mr) + r 2, where, 0 < r < 5 … (i) Where, q = 5m 2 is an integer. Where, q = (5m 2 + 2m) is an integer. where, q = (5m 2 + 4m) is an integer. Where, q = (5m 2 + 6m + 1) is an integer.

To expand an expression using the distributive property, multiply each term inside a set of parentheses by each term outside the parentheses, and then simplify by combining like terms. …

Nov 25, 2017 · Prove that the square of any positive integer is of the form 5q, 5q + 1, 5q + 4 for some integer q.

Show that the square of any positive integer cannot be of the form 5q + 2 or 5q + 3 for any integer q. Solution: Assume the positive integer to be = a. Using Euclid’s division lemma, a = bm + r. It is given that, b = 5. a = 5m + r. So, r = 0, 1, 2, 3, 4. Case 1: When r = 0, a = 5m. Case 2: When r = 1, a = 5m + 1. Case 3: When r = 2, a = 5m + 2.