Find the total area (in terms of k) of the prism. place k last in your formula. t.a. =
- Find the total area (in terms of k) of the prism. place k last in your formula. t.a. =
- Area of a trapezium : how to calculate the area easily
- Triangular prism – volume, surface area, base and lateral
- Find the area of regular polygons
- How to find the area of a circle’s sector
- Surface area of a hexagonal prism – volume & lateral area
Area of a trapezium : how to calculate the area easily
The Sun is the central star of our Solar System. It is a nearly ideal sphere of hot plasma, heated to incandescence in its core by nuclear fusion reactions and radiating the energy predominantly as visible light and infrared radiation. It is, without a doubt, the most vital source of energy for life on Earth. It has a diameter of 1.39 million kilometers (864,000 miles), which is 109 times that of the Earth. It has a mass of around 330,000 times that of Earth and makes up 99.86 percent of the Solar System’s total mass.  a
Hydrogen makes up about three-quarters of the Sun’s mass (73%) and helium (25%) of the remainder, with far smaller amounts of heavy elements such as oxygen, carbon, neon, and iron.
Based on its spectral class, the Sun is a G-type main-sequence star (G2V). As a consequence, it is referred to as a yellow dwarf informally and inaccurately (its light is closer to white than yellow). It originated from the gravitational collapse of matter inside an area of a large molecular cloud around 4.6 billion[a] years ago. The majority of this matter gathered in the core, while the remainder flattened out into the Solar System’s orbiting disk. The central mass became so hot and dense that nuclear fusion started to occur in its core. This mechanism is thought to be responsible for the creation of almost all stars.
Triangular prism – volume, surface area, base and lateral
We’ll use some common geometry formulas in this section. We’ll change our problem-solving approach to accommodate geometry problems. The variables will be named and the equation to solve will be given by the geometry formula. Furthermore, since all of these applications will require shapes of some kind, most people find it helpful to draw a figure and mark it with the information provided. This will be part of the problem-solving strategy for geometry applications’ first phase.
We’ll start with triangle properties to learn about geometry applications. Let’s go over some simple triangle stuff. Triangles have three interior angles and three sides. Each side is usually labeled with a lowercase letter that corresponds to the opposite vertex’s uppercase letter.
To measure the area of a triangle, we must first decide its base and height. The height is a line that creates an angle with the base and links the base to the opposite vertex. We’ll redraw, this time including the height, h. Look into it (Figure).
Find the area of regular polygons
9th of January, 2015
How to find the area of a circle’s sector
Q) I’m trying to figure out what Gaussian beam propagation is, which is defined as the paraxial solution to the Helmholtz equation: http://en.wikipedia.org/wiki/Helmholtz equation#Paraxial approximation; http://en.wikipedia.org/wiki/Helmholtz equation#Paraxial approximation; http://en.wikipedia.org The first few lines of the Wikipedia article state that a given solution to the wave equation must be assumed to have different spatial and time dependences: U(r,t) = A(r)T (t). Why are we able to distinguish this solution’s time and spatial dependence? A) This is a method for solving partial differential equations in general (such as the Helmholtz equation). Separate the variables first, then solve for the equation’s separable “eigen-functions.” Finally, a superposition of these eigen-function solutions yields some general solution.
25th of October, 2014 Q) Why do you say the time-averaged rate of the Poynting vector in Eq.(52) is (1/4)Z0(Jso2) instead of (1/8)Z0(Jso2) on page 95 of your book? Why don’t we take the (cos(w0t))2> definition into account in Eq.(52)? A) I “did” account for the fact that cos2 has a time-average of 12. The power, on the other hand, propagates on “both sides” of the current sheet, that is, on both the +y- and –y-sides. As a result, after time averaging, the result must be multiplied by 2 to get the “total” rate of energy flow out of the layer. The factor 1/8 is multiplied by 2 in this way, giving us 14.
Surface area of a hexagonal prism – volume & lateral area
Tien, Ulrich, and Martin recently addressed a prism–film coupler as a system for efficiently coupling a laser beam through thin-film dielectric light guides. This coupler also allows for precise measurement of the spectrum of propagating modes, which can be used to calculate the refractive index and thickness of the film. A prism–film coupler theory is presented here. A system that incorporates wave and ray optics is used to explain the physical concepts involved. The modes in thin-film light guides are studied, as well as how coupling affects them. We also assess the field distributions in the prism and film, as well as the power transfer between the prism and the film, to arrive at an optimal operating state. In one scenario, 81 percent of the laser power can be guided into any desired mode of film propagation.
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