Find the area of a triangle whose vertex is at ... About math :( thanks in advance for the help!?
First find the altitude of the triangle. The altitude will go from the vertex of the triangle to the midpoint of the base which is on the the diagonally opposite edge of the cube. But this is the same length as a diagonal of one of the faces of the cube which measures sqrt(2)*A. The length of the base is A. Since area is one-half altitude times base, the area is (sqrt(2)/2)A^2
Find the area of triangle?
(base*height)/2 Basically draw a rectangle around it and divide it's area in half. A triangle always consumes half it's retangle. same formula for a kite.
What will be the mass moment of inertia of a cube with edges of length "b" unit about an axis passing through an edge?
Thanks for A2A. This is not such a bugging question and answer is straightforward using basic equations and calculus.As you know, moment of inertia[math] I= \int r^2 dm[/math]Let the cube be placed on the coordinates system as shown in the figure belowWe need to find the moment of inertia about an axis through the edge, The vertical axis or z axis, as well as all other axis are passing through the edges in the figure. So we can find the moment of inertia about z axis.If you consider an infinitesimally small mass of size dx , dy and dz then[math]dm= \rho dx dy dz[/math]Where, [math]\rho[/math] is the densityAs per the formule 'r' is the distance from the z-axis to the infinitesimally small mass dm. If the coordinates of the mass 'dm' is (x,y,z) then the distance 'r' is [math] r= \sqrt{ x^2 +y^2}[/math][math] r^2= x^2 +y^2[/math]So the value of x, y and z for the cube ranges from 0 to b as per the length of edges.So substituting in the first equation[math] I= \int r^2 dm[/math][math] =\int_0^b \int_0^b \int_0^b (x^2 +y^2) \rho dx dy dz[/math][math] = \rho \frac{2 b^5}{3} [/math]Mass of the cube [math]m= \rho b^3[/math]Substituting mass in the above equation[math]I= \frac{2 m b ^2}{3}[/math]
A charge Q is placed at the centre of a cube of side A. What is the electric flux through the cube?
According to Gauss’s law,[math]\displaystyle\oint_\,E \cdot ds= Q/ \epsilon[/math]Therefore the total flux through the cube is [math]Q/ \epsilon.[/math]The flux through each side of the cube is [math]Q/ 6\epsilon[/math].
What is the answer if square is inscribed in the circle then the ratio of there area will give what?
Let side of square be 'a'Area of square is a²Diagonal of sqare is √{a²+a²}Or √2aSo diameter of circle is equal to diagonal of sqare.Then diameter=√2aRadius(r)=√2a/2=a/√2Area of circle=πr²=π{a/√2}²=πa²/2Ratio of area of square and area of circle =a²:πa²/2=1:π/2=2:π
What is the electric flux on each side of a cube if +q is placed on any one of the corners of the cube?
The key principle we first need to know is- Gauss’s law for electric flux. It states that the net electric flux through a closed surface is given by:where, Q is the net charge enclosed within the closed surface.Another key thing is, which is obvious, that electric flux, through surfaces which are symmetrical with respect to the geometry of the problem, will be equal. For example,if an electric charge , say +q, is placed at the center of the cube, then the net flux through all the faces should be equal to, as per Gauss’s law- [math]q/episilon0. [/math]And the net flux through each of the faces is- [math]q/(6episilon0)[/math], since the same flux should pass through each face as they are symmetrical to each other.Now, lets approach this problem.The first thing to observe is- faces 1,2 and 3 are symmetric with respect to the position of charge q. So equal amount of flux should pass through each of those faces.What about faces 4,5 and 6? Their planes contain the charge q. So no electric field is perpendicular to them, and hence the flux through faces 4,5 and 6, which is by definition,will be zero, as the angle between E and A is 90 degrees. Below image shows the scenario for a particular face(4,5 or 6).So, now we know that whole of the flux through the cube is passing through faces 1,2 and 3, and in equal amount. So, if F be the net flux through the cube, then through each face, net flux is: F/3.Now, lets find F. Now I am going to create symmetry around the charge q, as shown shown below.I have created 7 other cubes of same dimension as of original cube, the center of which is the position of the charge q. So by symmetry, net flux through through each of the 8 cubes should be the same, which is F.But the net flux through all the cubes will be, by Gauss’s law-[math]q/episilon0[/math]So, through each cube, net flux is: [math]q/8episilon0, [/math]which is equal to F.So, we arrive at the conclusion that, for the original problem geometry, where a charge q was placed at one corner of the cube, the net flux through each of the faces marked by 1,2 and 3, is F/3=q/24episilon0. And through faces marked by 4,5 and 6, the net flux is 0(zero).
How do you find the height of a pyramid with a square base when you are only given the lateral area, which is 624 inches with base edges of 24?
Let LA = lateral area of a square pyramid. Let P = the perimeter of the pyramid’s baseLet L = slant height (height from point C up the face of the triangular face to the top of the pyramid).Let h = the pyramid’s heightCalculationsLA = (1/2)•P•L which becomes[math]624 in^2[/math] = (1/2)( 24 in. •4 ) (L)624 in.^2 = 48 in(L)Dividing by 48 in. givesL = 13 in.Now the interesting part. We use the fact that the distance from the center of a side, point C in Figure 1, to the middle of the pyramid (point M) is half the length of a side or (1/2)•24 in.= 12 in.Note that the distance from M to the top of the pyramid is the height, h.Now we have a right triangle where L = 13 in. is the hypotenuse and 12 in. and h are the leg lengths.[math]h^2 [/math]+ [math]12^2 [/math]= [math]13^2 [/math][math]h^2 [/math]+ 144= 169[math]h^2 [/math]= 25[math]\sqrt(h^2)[/math]= [math]sqrt(25)[/math]h = 5.So, the height of the pyramid is 5 inches.
Charge q is placed at the corner of a cube. What is the flux through one face of a cube?
If charge is at the corner then it will be shared by 8 cubes.Charge for 1 cube = q/8Now in any given cube it is touching 3 9f its faces. So the area vector of that side and the electric field vector will be perpendicular. So flux through those 3 sides will be 0.Equal amount of flux will flow from the other 3 sides.So flux through 1 side = (q/8)/3 = q/24.Hope that helps.
An ant is at a corner of a cubical room of side a. The ant can move with a constant speed u. The minimum time taken to reach the farthest corner of the cube is?
[math]\frac{\sqrt{5}a}{u}[/math]Proof : First of all, note that the optimal path will lie on 2 faces of the cube only. If the path lay on more than two faces, it would have to leave a face and return to it (This is because the cube has only 6 faces, 3 touching the initial vertex and 3 touching the final vertex). Hence we can obtain a shorter path by replacing the trajectory between the two approaches to a face by a straight line.With this information, the proof becomes trivial. The two faces on which the trajectory lies are adjacent, and we can flatten them out to obtain a [math]2a\times a[/math] rectangle. we need to find the shortest path lying on this rectangle between two opposite corners, which is just the length of the diagonal between them.