OpenLLM-France/wikimedia huggingface dataset:
# LGBT Young Adult Literature/Glossary

Welcome to the Glossary page of the LGBT Young Adult Literature wikibook. This page includes a list of all terms that might need explaining in this wikibook. Please add to it if you feel something is missing.

## Glossary of Terms
### General

* LGBT - stands for Lesbian, Gay, Bisexual, Transgendered, and it is a term used to describe non-heterosexual people. The term LGBT is mostly used in English-speaking countries.
* Queer - is a colloquial term used to describe gay people. It is commonly used in the UK. Historically it was considered derogatory but nowadays has been re-adopted by the community.

### Sexual orientation and gender identity

* Gay/Homosexual - attracted to someone of the same gender. Normally, this term is used for gay males although this term is actually universal.
** Lesbian - a female attracted to other females.
* Bisexual - one who is attracted to two genders, though the attraction can be towards more than two genders.
* Transgender - gender identity that is the opposite of your birth-assigned gender. For example, an AMAB (assigned male at birth) person identifies as a woman.
* Intersex - one who is not born with sex characteristics (e.g. genitalia, chromosomes, etc.) that are defined as male or female.
* Asexual - lacking sexual attraction to anyone.
* Aromantic - lacking romantic attraction to anyone.
* Pansexual - attracted to anyone regardless of gender, i.e. "gender blind" in attraction.

### Media

* Yaoi - A genre of fictional media, including manga, anime, and literature, that focuses on romantic or sexual relationships between male characters. The term primarily targets a female audience and often features themes of romance, intimacy, and emotional connection between male protagonists. While Yaoi content typically portrays male-male relationships, it differs from LGBT representation in that it often caters to fantasies and idealized narratives rather than reflecting real-life experiences or identities.
* Yuri - A genre of fictional media, including manga, anime, and literature, that focuses on romantic or sexual relationships between female characters. Similar to Yaoi, Yuri content is primarily created for and consumed by a female audience. The narratives within Yuri often explore themes of love, intimacy, and emotional connection between female protagonists. While Yuri content may depict same-sex relationships between women, it is distinct from real-life LGBT experiences and identities, often presenting idealized or romanticized portrayals of female-female relationships within fictional contexts.
# LGBT Young Adult Literature/Transsexuality

A transsexual is one who experiences a gender identity that's inconsistent with their assigned sex, and desires to permanently transition to the sex/gender they identify with.
This term is a subset of transgender, although some transsexuals reject the label transgender.
# LGBT Young Adult Literature/Coming Out

Coming out of the closet, also simply called coming out, is a metaphor that describes the act of disclosing your sexual orientation, romantic orientation, and/or gender identity, whether it be publicly or privately (e.g. with family).
This may be seen as a privacy issue as some people tend to out others (tell other people one's sexuality without prior consent), not to mention being a social issue because some employed LGBT people often feel that their job security is compromised. Author Steven Seidman writes that "it is the power of the closet to shape the core of an individual's life that has made homosexuality into a significant personal, social, and political drama in twentieth-century America."

## Advice on coming out

* If you do not feel confident, you will need an LGBT support group or some friends to help you out. Some people think you may lose friends and family support but there are progressive families that tend to support LGBT relatives.
* Try to avoid peer pressure from groups. If you do not wish to come out of the closet, you do not need to.

## References

* ↑ https://archive.org/details/beyondclosettran0000seid_b7u6
# LGBT Young Adult Literature/Parental Interference

If you have come out to your parents as LGBT, there are a few ways they may attempt to make you "normal" (i.e. heterosexual, heteroromantic, and/or cisgender). Remember, it's okay to resist trying to be 'normal' in their eyes.

## Dating

If you identify as gay, bisexual, or a-spec (asexual, aromantic, aroace, etc.), there is a chance that your parents will arrange for you to go on a date with someone of the "opposite gender" as you in order to cover up your true sexuality. For example, a gay boy's parents have arranged a date with a girl, even though this gay boy DOES NOT date females; he only wanted to be with a boy.

## Conversion therapy

Conversion therapy refers to a wide range of pseudoscientific practices that attempt to "repair" one's gender identity, romantic and sexual orientation. It is grounded in the beliefs that any gender identities and sexual/romantic orientations other than straight and cisgender are "abnormal". Torture is the most common form of conversion therapy, as described by victims of these practices.
This harmful practice is often conducted by religious groups with the use of prayer and other religious teachings. LGBT minors (children and teenagers) are often dragged (i.e. forced) into this by their parents. Many medical and psychologist groups condemn this practice as it can lead to depression, anxiety, self-harm, or suicide.
The practice dates back to the 19th century when German psychiatrist Richard von Kraft-Ebing used hypnotic suggestion therapy to turn a gay man straight.

### Jilly Juice

Jilly Juice is a quack pseudo-medicine in the form of a fermented drink. Its creator, Jillian "Mai Tai" Epperly, claims that her "medicine" can cure cancer and autism, regenerate missing limbs, reverse the effects of aging, and even "cure" homosexuality! Ironically, no studies have proven that Jilly Juice works, nor has the Food & Drug Administration (FDA) approved this!

### Bans on conversion therapy

Fortunately, there are some countries that ban these harmful practices.
Asia:
* Cyprus
* Israel
* Singapore
* Taiwan
* Vietnam
Europe:
* Albania
* Belgium
* France
* Germany
* Iceland
* Malta
* Norway
* Portugal
* Spain
North America:
* Canada
* Mexico
Oceania:
* New Zealand
* Samoa
South America
* Argentina
* Brazil
* Chile
* Ecuador
* Paraguay
* Peru
* Uruguay

## References

* ↑ Morris, Meagan (2018-05-24). "Jilly Juice recipe creator claims her concoction can regrow limbs, cure homosexuality – Metro US". www.metro.us. Retrieved 2024-11-26.
# Greek Mythology/Gods/Poseidon

Poseidon (In Greek: Ποσειδῶν, known as Neptune to the Romans) was considered as the god of the ocean. Born of Chronos and Rhea, he had Zeus and Pluto (Hades) for brothers. Poseidon also had three sisters: Hestia, Hera, and Demeter. When Zeus overthrew their father, and the three brothers divided the earth among them, Poseidon became the god of the sea and earthquakes and thunder.

### Origin of the Horse

When Poseidon desired Hestia, she asked Poseidon to create the world's most beautiful animal in an attempt to cool off his advances. As a result, Poseidon created the first horse and also became the God of horses. In some versions of this story, several other types of animals were created before the horse as a result of Poseidon's several unsuccessful attempts. By the time the horse was created, Poseidon's passion for Hestia had longed.

### Symbolism

The symbol Poseidon is associated with is the trident, a weapon which can shake the Earth and shatter any object. Poseidon is often depicted sitting next to his wife, Amphitrite, in his sea chariot pulled by sea creatures.

### Depiction

* Poseidon as depicted on Greek pottery.
* Roman mural of Neptune recovered from Pompeii.
# Chess Opening Theory/1. e4/1...e5/2. Nc3/2...Bc5/3. Qg4

Giraffe Attack/Wade-Stone Variation/Byrd Attack
3. Qg4?!
This move is interesting. Here White aims to take advantage of the Bishop move. Here Black has options:
3...g6 Is by far the most played move. It moves the g-pawn to comfortable square & prepares for f5.
3...Bf8? This is the Ancient Defense. It undevelops the Bishop, & is just a joke alone.
3...Qf6 This is called the Modern Variation. It aims for the Scholar's Mate (with Black) & develops the Queen as well.
3...Nf6?! Is the Old Variation. A bit too aggressive on Master Level.
# Quantum Chemistry/Example 13

Write a question and its solution that shows the specific selection rule for a quantum harmonic oscillator
Calculate the energy for the vibrational transitions $v\rightarrow v+1$ and $v+6\rightarrow v+7$. If they have the same energy gaps comment on why?
$\Delta E_{v\rightarrow v+1}$$=h\nu _{0}{\Bigl(v+1)+{\tfrac {1}{2}}{\Bigr )}-hv_{0}{\Biglv+{\tfrac {1}{2}}{\Bigr )}$
$=h\nu _{0}{\biggl(v+1)+{\tfrac {1}{2}}-{\Biglv+{\tfrac {1}{2}}{\Bigr )}{\biggr )}$
$=h\nu _{0}{\bigglv+1+{\tfrac {1}{2}}-v-{\tfrac {1}{2}}{\biggr )}$
$=hv_{0}$
$\Delta E_{v+1\rightarrow v+2}$$=h\nu _{0}{\Bigl(v+7)+{\tfrac {1}{2}}{\Bigr )}-hv_{0}{\Bigl(v+6)+{\tfrac {1}{2}}{\Bigr )}$
$=h\nu _{0}{\biggl(v+7)+{\tfrac {1}{2}}-{\Bigl(v+6)+{\tfrac {1}{2}}{\Bigr )}{\biggr )}$
$=h\nu _{0}{\bigglv+7+{\tfrac {1}{2}}-v-6-{\tfrac {1}{2}}{\biggr )}$
$=h\nu _{0}$
The energy of the vibrational transitions from from $v\rightarrow v+1$ and $v+6\rightarrow v+7$ have the same energy gap $(h\nu _{0})$ . This is because of the specific selection rule for the quantum harmonic oscillator. The rule states that the molecule is only allowed to move up or down one vibrational energy level for the transition to occur. If the molecule diverges from the rule $\Delta v\pm 1$ then it is considered an overtone, and these are unlikely to occur.
# Quantum Chemistry/Example 26

The square of the angular momentum of a hydrogen atom is measured to be $L^{2}$$=20\hbar ^{2}$. What are the possible values of the z-component of the orbital angular momentum, $L_{z}$, that could be measured for this atom?

#### Solution

The eigenvalues of the square of the angular momentum operator (${\hat {L}}$$^2$) for a quantum mechanical system, such as an electron in a hydrogen atom, are given by:
${\hat {L}}^{2}Y_{l}^{m}(\theta ,\phi )=h^{2}l(l+1)Y_{l}^{m}(\theta ,\phi )$
where $Y_{l}^{m}(\theta ,\phi )$ are the spherical harmonics, which are eigenfunctions of ${\hat {L}}$$^2$, $l$ is the orbital quantum number, and $m$ is the magnetic quantum number. The given value for $L^{2}$ = 20 ħ2 , so we set up the equation:
$\hbar ^{2}l(l+1)=20\hbar ^{2}$
Dividing by $\hbar ^{2}$ and simplifying, we get:
$l(l+1)=20$
$l^{2}+l-20=0$
In this quadratic equation, $l$ can be factored to get:
$(l-4)(l+5)=0$
$l=4$
Since $l$ must be a non-negative integer. The magnetic quantum number $m$ can take on any integer value from $-l$ to $l$, thus for $l=4$, $m$ can be:
$m=-4,-3,-2,-1,0,1,2,3,4$
The z-component of the angular momentum, $L_{z}$, is quantized in units of $\hbar ^{2}$and given by:
$L_{z}=m\hbar$
$L_{z}=-4\hbar ,-3\hbar ,-2\hbar ,-1\hbar ,0\hbar ,1\hbar ,2\hbar ,3\hbar ,4\hbar$.
Therefore, the possible values of the z-component of the orbital angular momentum, $L_{z}$, that could be measured for the atom with a given $L^{2}$$=20\hbar ^{2}$ are $L_{z}=-4\hbar ,-3\hbar ,-2\hbar ,-1\hbar ,0\hbar ,1\hbar ,2\hbar ,3\hbar ,4\hbar$.
# Algebra/Chapter 9

Chapter 9: Quadratic Functions
← Chapter 8 · Chapter 10 →
Introduction

## Sections

* Graphing Quadratic Functions
* Factoring Quadratics
* Completing the Square
* Complex Numbers
* Quadratic Formula
* Quadratic Inequalities
* Quadratic Systems

## Extra

* Review
* Practice Problems
* Solutions to Practice Problems
* Glossary
# Algebra/Chapter 19

Chapter 19: Vectors
← Chapter 18 · Chapter 20 →

## Sections

* Vectors and Their Properties
* Vector Operations
* Vector and Parametric Equations
* Vectors in Three Dimensions
* Dot and Cross Products
* Vectors and Planes
* Determinants
# Quantum Chemistry/Example 19
###### ==Write an example question showing the calculation of the frequency of EM radiation required to excite an H atom from ground to n=2 electronic state.==

The energy of the hydrogen atom is calculated using the following equation:
    '''Energy of Hydrogen Atom''' $E=-\left({\frac {m_{e}e^{4}}{32\pi ^{2}\epsilon _{0}^{2}h^{2}}}\right){\frac {1}{n^{2}}}$ To calculate the change in energy required to go from the ground state to n=2, the following equation is used:
$\delta E=-\left({\frac {m_{e}e^{4}}{32\pi ^{2}\epsilon _{0}^{2}h^{2}}}\right){\frac {1}{n^{2}}}+\left({\frac {m_{e}e^{4}}{32\pi ^{2}\epsilon _{0}^{2}h^{2}}}\right){\frac {1}{n^{2}}}$
$\delta E=-\left({\frac {m_{e}e^{4}}{32\pi ^{2}\epsilon _{0}^{2}h^{2}}}\right){\frac {1}{4}}+\left({\frac {m_{e}e^{4}}{32\pi ^{2}\epsilon _{0}^{2}h^{2}}}\right){\frac {1}{1}}$
$\delta E=\left({\frac {m_{e}e^{4}}{32\pi ^{2}\epsilon _{0}^{2}h^{2}}}\right){\frac {5}{4}}$
$\delta E=\left({\frac {(9.10938x10)^{-31}kg(1.60218x10^{-19}c)^{4}}{32\pi ^{2}8.854x10^{1}2c^{2}S^{2}/kgm^{2})^{2}(1.054x10^{-24}J.s)^{2}}}\right){\frac {5}{4}}$
$\delta E=\left({\frac {(9.10938x10)^{-31}kg(1.60218x10^{-19}c)^{4}}{32\pi ^{2}8.854x10^{1}2c^{2}S^{2}/kgm^{2})^{2}(1.054x10^{-24}J.s)^{2}}}\right){\frac {5}{4}}$
δE =((9.10938x10)−31kg(1.60218x10−19c)432π28.854x1012c2S2/kgm2)2(1.054x10−24J.s)2)54
# Algebra/Chapter 9/Quadratic Systems
## Solving Systems of Simultaneous Equations Involving Equations Of Degree 2

The substitution method should be used for efficiency when solving nonlinear simultaneous equations, unless other methods such as the graphing method provide clear and simple solutions quickly (when they would be faster than substitution).
Example: Solve the system of simultaneous equations.
$y^{2}+(2x+3)^{2}=10\$
$2x+y=1\$
With the second equation, make a given term (here, 2x should be used) the subject.
$2x=1-y\$
Substitute the third equation into the first, and through factorization of the resulting, simplified quadratic with one variable the solutions can be found.
$y^{2}+((1-y)+3)^{2}=10\$
$y^{2}+(4-y)^{2}=10\$
$2y^{2}-8y+16=10\$
$y^{2}-4y+3=0\$
$(y-1)(y-3)=0\$
Hence we know $y=1\$ or $y=3\$
Then, we calculate that the two possibilities are: $y=1\$, $x=0\$ or; $y=3\$, $x=-1\$
# Quantum Chemistry/Integrals in polar coordinates

Integration in spherical polar coordinates is a triple integral in the polar coordinate system. In the Cartesian coordinate system, points in space are defined by their position along the x, y, and z axes. Integration divides the space under a curve into infinitesimal widths along the x-axis. The sum of these strips is the area under the curve.$\int f(x)~{\textrm {d}}x$In spherical polar coordinates, the frame of reference is changed so all points are defined by its position about the origin. The distance from the origin is defined by the radius r and the angle of displacement longitudinally, θ, and latitudinally ϕ. The angular components are limited in their value because of their periodic nature. Their ranges are defined as:$0\leq \phi \leq 2\pi$$0\leq \theta \leq \pi$ Polar coordinates can be converted from Cartesian (r, ϕ, θ)→(x, y, z){\textstyle (r, \phi , \theta )\rightarrow (x, y, z)}:$x=r\cos \phi \cdot \sin \theta$$y=r\sin \phi \cdot \sin \theta$$z=r\cos \theta$For integration in higher dimensions, integration by parts is used. A triple integral is in three dimensions and therefore has three variables. The outer integral relates to the outer variable, and the inner integrals to the inner variable.$\underbrace {\int \underbrace {\int \underbrace {\int f(x,y,z)~{\textrm {d}}x} _{x}~{\textrm {d}}y} _{y}~{\textrm {d}}z} _{z}$In polar coordinates, the strips of infinitesimal widths are converted to wedges and can be converted from Cartesian.${\textrm {d}}x\cdot {\textrm {d}}y\cdot {\textrm {d}}z=r^{2}\sin \theta \cdot {\textrm {d}}r~{\textrm {d}}\theta ~{\textrm {d}}\phi$$\iiint f(x,y,z){\textrm {d}}x~{\textrm {d}}y~{\textrm {d}}z=\int _{0}^{2\pi }\int _{0}^{\pi }\int _{0}^{\infty }r^{2}\cdot f(r\cos \phi \sin \theta ,r\sin \phi \cos \theta ,r\cos \theta )\sin \theta ~{\textrm {d}}r~{\textrm {d}}\theta ~{\textrm {d}}\phi$To solve spherical polar integral, begin at the inner most integral and proceed outwards.

### Example

$\int _{0}^{2\pi }\int _{0}^{\pi }\int _{0}^{\infty }e^{-r}r~{\textrm {d}}r~{\textrm {d}}\theta ~{\textrm {d}}\phi$
integrate the innermost integral, x.
$\int _{0}^{2\pi }\int _{0}^{\pi }{\bigl-r+1{\bigr )}e^{-r}|_{0}^{\infty }~{\textrm {d}}\theta ~{\textrm {d}}\phi$
$\int _{0}^{2\pi }\int _{0}^{\pi }{\bigl-\infty +1{\bigr )}e^{-\infty }-{\bigl-0+1{\bigr )}e^{-0}~{\textrm {d}}\theta ~{\textrm {d}}\phi$
$\int _{0}^{2\pi }\int _{0}^{\pi }0-{\bigr-1{\bigr )}~{\textrm {d}}\theta ~{\textrm {d}}\phi$
$\int _{0}^{2\pi }\int _{0}^{\pi }~{\textrm {d}}\theta ~{\textrm {d}}\phi$
Integrate the next integral, θ.
$\int _{0}^{2\pi }\theta |_{0}^{\pi }~{\textrm {d}}\phi$
$\int _{0}^{2\pi }(\pi -0)~{\textrm {d}}\phi$
Finally, integrate ϕ.
$\pi \phi |_{0}^{2\pi }$
$\pi ({2\pi }-0)$
${2\pi }^{2}$
# Quantum Chemistry/Example 27
## Question

Find the wavenumber (in cm$^{-1}$) of the photon required to excite a CO molecule from its ground vibrational state to its first excited vibrational state, given that the force constant for the CO bond is k=1860 N.m$^{-1}$.

## Solution

The frequency of the photon required for this transition corresponds to the fundamental vibrational frequency of the molecule, $v_{\text{0}}$. To determine the wavenumber of this transition, the first step is to calculate the reduced mass of the molecule. The reduced mass (${\ce {\mu}}$) for a diatomic molecule is given by the following equation:
$\mu ={\frac {m_{1}\cdot m_{2}}{m_{1}+m_{2}}}$
The obtained atomic masses of Carbon and Oxygen are as below:
C= 12.011u, O =15.999u
Plugging these values into the reduced mass equation gives:
$\mu ={\frac {12.011{\text{u}}\cdot 15.999{\text{u}}}{12.011{\text{u}}+15.999{\text{u}}}}$
$\mu =6.861{\text{u}}$
The unit conversion from atomic mass units (u) to kilograms (kg) is as follows:
$1u=1.66054\times 10^{-27}{\text{kg}}$
$\mu =1.139\times 10^{-26}{\text{kg}}$
Now, we can calculate the frequency of the photon using the following equation:
$v_{0}={\frac {1}{2\pi }}\cdot {\biggl{\frac {k}{\mu }}{\biggr )}^{\frac {1}{2}}\qquad {\text{where}}\ \mu =1.139\times 10^{-26}{\text{kg}}$
Plugging in the values for the reduced mass and force constant gives:
$v_{0}={\frac {1}{2\pi }}\cdot {\biggl{\frac {1860{\text{N/m}}}{1.139\times 10^{-26}{\text{kg}}}}{\biggr )}^{\frac {1}{2}}\qquad$
$v_{0}=6.43\times 10^{13}{\text{s}}^{-1}$
Finally, we need to covert the frequency of the photon ($v_{\text{0}}$) in s$^{-1}$ to the wavenumber (ṽ) in cm$^{-1}$:
$v={\frac {v_{0}}{c}}\qquad {\text{where}}\ c=2.998\times 10^{10}{\frac {\text{cm}}{\text{s}}}$
By substituting the known values, the wavenumber of the photon can be determined as follows::
$v={\frac {6.43\times 10^{13}{\text{s}}^{-1}}{2.998\times 10^{10}{\frac {\text{cm}}{\text{s}}}}}\qquad \qquad \therefore v=2.14\times 10^{3}{\text{cm}}^{-1}$
# Quantum Chemistry/Integration by parts
### Example Problem:

Evaluate the triple integral shown below in spherical polar coordinates using the given boundary conditions: $0\leq {r}\leq 1,$ $0\leq {\theta }\leq \pi ,$ $0\leq \phi \leq 2\pi$
$\int ^\int ^{\ }\int ^{\ }r^{2}\sin \theta ~{\textrm {d}}r~{\textrm {d}}\theta ~{\textrm {d}}\phi$
Solution:
The triple integral can be rewritten as an iterated integral under the boundary conditions as follows:
$\int ^\int ^{\ }\int ^{\ }r^{2}\sin \theta ~{\textrm {d}}r~{\textrm {d}}\theta ~{\textrm {d}}\phi$ = $\int _{0}^{\pi }\int _{0}^{1}\int _{0}^{2\pi }r^{2}\sin \theta ~{\textrm {d}}r~{\textrm {d}}\theta ~{\textrm {d}}\phi$
The integral can then be evaluated using Fubini's theorem to separate the multivariable integrals into single variable integrals that can be solved accordingly as follows:
$\int _{0}^{\pi }\int _{0}^{1}\int _{0}^{2\pi }r^{2}\sin \theta ~{\textrm {d}}r~{\textrm {d}}\theta ~{\textrm {d}}\phi$ = $\int _{0}^{\pi }\sin \theta ~{\textrm {d}}\theta ~$ $\int _{0}^{1}\ r^{2}{\textrm {d}}r$ $\int _{0}^{2\pi }{\textrm {d}}\phi$
= $(-cos\theta \mid _{0}^{\pi })$ $\left({\frac {r^{3}}{3}}\mid _{0}^{1}\right)$ $(\phi \mid _{0}^{2\pi })$ = $(2)$ $\left({\frac {1}{3}}\right)$ $(2\pi )$
Multiplying these results together then gives the final solution:
(2) $\left({\frac {1}{3}}\right)$ $(2\pi )$ = $\left({\frac {4\pi }{3}}\right)$
Therefore,
$\int _{0}^{\pi }\int _{0}^{1}\int _{0}^{2\pi }r^{2}\sin \theta ~{\textrm {d}}r~{\textrm {d}}\theta ~{\textrm {d}}\phi$ = $\left({\frac {4\pi }{3}}\right)$
# Embedded Systems/Super Loop Architecture

When programming an embedded system, it is important to meet the time deadlines of the system, and to perform all the tasks of the system in a reasonable amount of time, but also in a good order. This page will talk about a common program architecture called the Super-Loop Architecture, that is very useful in meeting these requirements

## Definition

A super loop is a program structure comprised of an infinite loop, with all the tasks of the system contained in that loop. Here is a general pseudocode for a superloop implementation:
We perform the initialization routines before we enter the super loop, because we only want to initialize the system once. Once the infinite loop begins, we don't want to reset the values, because we need to maintain persistent state in the embedded system.
The loop is in fact a variant of the classic "batch processing" control flow: Read input, calculate some values, write out values. Do it until you run out of input data "cards". So, embedded systems software is not the only type of software which uses this kind of architecture. For example, computer games often use a similar loop. There the loop is called (tight) (main) game loop.

## Power-Save Super Loop

Let's say we have an embedded system which has an average loop time of 1ms, and needs only to check a certain input once per second. It seems a waste to continue looping the program, especially when we don't need to do anything most of the time. In this situation, the program will loop 1000 times before it needs to read the input, and the other 999 loops of the program will just be a countdown to the next read. In this case, it is sorely inefficient to have the processor chugging away at 100% capacity all the time. We will now implement an expanded superloop to build in a delay:
Notice how we added a delay at the end of the super loop? If we build this delay to delay for 999ms, we don't need to loop 1000 times, we can read the input on every loop.
Also, it is important to note that many microcontrollers have power-save modes, where they will require less electrical power, which can be especially good if the system is running off a battery.

## Power Use Calculations

Let's say that we have a microcontroller that uses 20mW of power in "normal mode", but only needs 5mW of power in "Low-Power Mode". Let's also say that we are using the example superloop above, which is in "Low-Power Mode" 99.9% of the time (1ms of calculations every second), and is only in normal mode 0.1% of the time:
$\mathrm {Power} ={\frac {(99.9\%\times 5\ \mathrm {mW} )+(0.1\%\times 20\ \mathrm {mW} )}{100\%}}=5.015\ \mathrm {mW} \quad \mathrm {Average}$
Notice how we can cut down our power consumption by adding in a substantial delay? This is especially important because few embedded applications will require 100% of processor resources. Most embedded systems are able to just sit and wait in a low-power state until needed.