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A Homeless man approached a 24-year-old woman - 2 minutes later his unexpected question left her speechless
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We tend to look at homeless people as faceless men living on the streets.
But it is further from the truth. Homeless people also have thoughts, feelings and dreams.
And sometimes, the people who have the least are the ones who end up giving the most - something that 24-year-old Caroline Santana has witnessed herself.
Caroline Santana lives in Sao Paulo, Brazil.
One night, she was about to board a bus back home. But when she began searching her purse, she found that her monthly free pass got lost and that she didn't have enough money for the ticket.
Then, a man approached her. It was clear that he was homeless and was begging for money.
"Can you wait a minute? I think I lost my free pass", Caroline wrote on Facebook.
Caroline couldn't imagine the man's reaction.
"How much do you need for a ticket?", the homeless man asked her. "I only have 4 reals (about $1.2) but I can give them to you".
Caroline was very moved by this, but refused to accept his money.
"Are you sure? I can give you the money. I can't let you go home alone. I can help you", he said.
Then Caroline looked at her purse again and found her bus free pass.
She thanked the man for his kindness and beautiful gesture. Then, she started walking towards the bus.
But after a few steps, she turned back. She wanted to take a picture with the homeless man and post it on Facebook so that everyone can see what a big heart he has.
"Write that my name is Cesar. That's my name!" The man asked.
Caroline found 2 more reals in her purse and gave them to the man before their ways separated.
She wrote the story in a Facebook post, that since its original posting date has received more than 650,000 likes and got shared more than 120,000 times.
"Was it a simple gesture? I'm sure others are ignoring and condemning him for his looks - many people probably think he can't behave and that he might be a criminal, but no, he has a big heart.
This story is about character, humility, wisdom, and love! Never judge anyone by appearance. God writes straight in crooked lines."
What a wonderful story. And it just shows you, as Caroline wrote, to never judge a person by appearance.
Homeless man asked a Burger King employee 'What can I buy with 50 cents?' - the boy's answer amazed everyone
Their son got his first paycheck - when his parents saw what he was wasting it on, they couldn't stop crying
She stopped to talk to this homeless person every day, until he gave her a piece of paper that changed his life forever!
Student bought coffee and pastry for a homeless man, an hour later he put a wrinkled note in her hands that left me in tears
For years hospitals use this to help children fall asleep, find the method that can help you too
Here's the new pill that makes your farts smell like chocolate and flowers
Here's why Japanese women don't get fat and their life expectancy is the longest in the world! The secret is just amazing
He bought six IKEA kitchen cabinets. What he did with them dropped my jaw to the floor!
6-year-old autistic boy was banned and not invited to any birthday party, so his Dad did what no one else dared
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Today, I'm going to share some more of the last minute repurposed cabinet door projects that I took to Glendale 2015, an easy welcome sign.
Painting cabinet doors and any other projects are so easy with my Homeright Finish Max and my favorite DIY chalky paint primer. After mixing some plaster of paris with my favorite paint, I sprayed a couple of cabinet doors, then still had some paint in the cup, and grabbed a couple of more.
Yep, I have many cabinet doors of all shapes and sizes to choose from.
The one on the right is already done, and I'm about to do the one on the left.
My favorite color seems to be red these days. One of the black primered doors got a dose of dry brushed Reclaim paint.
This is how the black cabinet door looks after applying a light (dry) coat of red paint with a cheap chip brush.
The red paint dries much deeper than it appears it will. You can see that this is a new chip brush and it was shedding a little. Many people use these and throw them away. I keep them forever. The more I use them, the less they shed. Wanna know why I love these brushes so much? They are incredibly easy to clean and they don't waste a lot of paint (down the drain, so to speak).
After the paint dried, I sanded it with some brown paper, then cut out a WELCOME stencil with my Silhouette Portrait and some contact paper. You can see how I apply the contact paper for stenciling in my post Repurposed Pallet Coffee Cup Rack.
After applying the contact paper stencil, I weeded the letters and readied it for paint. I use Martha Stewart Foam Pouncers for all of my stenciling.
I used my favorite d-rings on the back so the sign can be hung.
Or it can be leaned against a wall in the entryway or on the porch. I love, love, love this color! I thought it would fly out of the booth at Glendale, but it took several hours for just the right person to fall in love with it.
You will love this Welcome Sign at McCall Manor from a new board, made to look old.
I left one of the cabinet doors black, and added WELCOME in white vinyl. Sad, but this is one of the few items that actually came home with me from Glendale this year.
Would you like to see more great repurposed cabinet doors?
I'm excited that 2 of the narrow cabinet doors were leftover when we moved into the little house. They're very pretty maple, but I think a fun sign is in order. My question is where do you find your cabinet doors? I want to find one that wider to use as a shopping list. Haven't decided if I want it to be a chaulk board or a paper roll held down with a strip of metal at the bottom to tear it off.
So, I say make both!
As for cabinet doors, a good friend once gave me about 40 that she bought at an auction. I have junk malls in my area, and there are booths that sell just cabinet stuff and wood. In addition the Habitat ReStore is a good option to find them. Be sure to check online (freecycle and craigslist) and last but not least, if you're on facebook-put a status on there that you're looking for some. Maybe one of your family or friends has re-done their kitchen to open shelving or just removed the doors.
Love your welcome sign! What a great upcycle of the cabinet doors!
Thank you Holly! I love reusing old cabinet doors for easy projects and more!
It appears I LOVE everything you do. I excited at the thought of going to your booth, but I best take my husband….. or I might find myself in big trouble!
There's no end to your awesome signs (and supply of cabinet doors)! Thanks for sharing it at the Vintage Inspiration Party!
I really need to try the contact paper trick! Love these!
Love this look! Would you ever consider making stencils out of contact paper and selling those?
Custom work would cost a little more.
Beautiful, I like both equally! You do such a nice neat job on those doors.
Thank you Julie, I do love, love, love using cabinet doors for projects.
I just love looking at your projects! Seriously! .. Thanks for sharing the link to your D-ring source the other day too! I just received my order and am thrilled with the savings over buying them at our local store!
I go through mine quickly, but most people will not need to replenish very often.


text:
Recovery May Be More Effective In Community Setting, Studies Suggest
By Paul Gaita 09/25/15
Studies have concluded that Oxford Houses have lower rates of substance abuse.
Oxford House in Mississippi. Photo via
Since 1991, Dr. Leonard A. Paul, a professor of psychology at DePaul University, has conducted research into the efficacy of the Oxford House Model of substance abuse treatment, which emphasizes a collaborative house setting with community rules and responsibilities for its residents.
His findings, which have been published in more than 100 articles, suggest that individuals seeking treatment at one of the more than 1,900 Oxford Houses across the United States, had significantly lower substance use rates than those who employed other models of treatment, including outpatient programs and self-help groups.
The Oxford House Model, which was established in Silver Springs, Maryland, in 1975, is a self-supporting, drug-free home that typically hosts between six and 15 residents. They employ no professional staff and do not advocate a single method of recovery; rather, residents must decide for themselves if they will seek help outside of the house through substance abuse treatment or 12-step program.
Residents pay all expenses and house decisions are made through a democratic process; there is also no limit to the amount of time they can stay at the house, barring any substance use or disruptive behavior.
According to the studies, the houses are most effective when they are located in safe neighborhoods with strong community connections, which, according to Jason, creates a "win-win situation" for both parties with neighborhoods benefiting from individuals in strong recovery and the residents receiving support from the community at large.
In the course of their two decades of studies, Paul and his research team examined the recovery rates for more than 2,000 Oxford House residents, and found that after 24 months, residents experienced far fewer incidents of relapse, as well as lower incarceration rates and higher monthly income than other treatment scenarios.
"Oxford House offers residents the freedom to decide which treatment they desire while receiving constant support and guidance within an abstinent communal setting," said Jason. "It is a low-cost, safe and effective way to treat substance abuse in a collaborative housing setting."
Jason and his team have been awarded a five-year grant of nearly $3 million from the National Institutes of Health's National Institute on Alcohol Abuse and Alcoholism to study the causes of relapse, as well as further analysis of the social structure within Oxford House to determine its impact on residents' success rates.
Not all parties are convinced of Oxford House's efficacy. Mississippi Governor Phil Bryant recently asked his state's Department of Health to end financial support to Oxford House, Inc., when residents expressed concern over child safety and plummeting property values after a house opened in Jackson.
outpatient group
Paul Gaita
Paul Gaita lives in Los Angeles. He has contributed to the Los Angeles Times, Variety, LA Weekly, Amazon.com and The Los Angeles Beat, among many other publications and websites.
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Comment Section in a Recovery Magazine—A Love Poem
. . . . . . . . . And that's how I got better. End of piece. End of piece is a pseudonym for...

redpajama_set_name: RedPajamaCommonCrawl


A study carried out by security management vendor FireMon at this month's Infosecurity Europe in London has given brutally honest insight into the immense pressure cyber security professionals are under to carry out their jobs and meet outside regulations. A staggering 28% admitted to compromising their ethics to pass audits, a figure that is up 6% from five years ago when the same question was posed in a similar survey. This is likely due to growing network complexity and all of the disparate technology, security and otherwise, used to keep cyber criminals at bay.
When asked if they felt that they spend most of their day fire-fighting rather than doing meaningful security work, 51% of the IT security professionals surveyed agreed. A further 56% admitted they had added a product purely to meet compliance regulations, even though they knew it offered no other business benefit.
When it comes to demands from the business side, 52% of IT security pros admitted to adding access that they know had decreased their organisation's security posture.
Get Visibility– IT security managers can't manage what they don't know is there. Having detailed visibility into firewall rules and policy effectiveness allows organizations to clean up outdated or redundant rules and close security gaps, lowering overall firewall complexity and level of risk.
Get Intelligence– By taking into account knowledge of the vulnerabilities in the networked environment on well-known threat entry points and combining it with real-time monitoring and vulnerability mapping, the security team has the situational awareness it needs to identify and remediate problematic issues before they evolve.
Integrate– Exchange of information between disparate systems cannot be underestimated. The ability to share security information in real time without restricting it to a single application, system or device can empower managers to make decisions.
Automate– Change workflow automation can help security teams to assess the impact of any new access being provided and restrict or vet it against the corporate security policy to ensure it does not break compliance or introduce unacceptable risk.


Download Instagram For Free: Exactly what is this trendy thing called Instagram that all the awesome youngsters appear to be right into? It's been around for a couple of years, silently picking up traction mostly thanks to everyone's brand-new fascination with mobile digital photography, so do not really feel embarrassed to ask if you have absolutely no hint what it's about.
Instagram is a social networking application made for sharing pictures as well as video clips from a mobile phone.
Just like Twitter Or Facebook, every person who produces an Instagram account has a profile as well as a news feed.
When you publish a picture or video on Instagram, it will be presented on your account. Other customers that follow you will certainly see your posts in their own feed. Furthermore, you'll see blog posts from other users whom you decide to follow.
Instagram is readily available for free on iOS and Android tools.
It can also be accessed on the web from a computer, but individuals could only publish as well as share photos or videos from their gadgets.
Before you can start making use of the app, Instagram will ask you to produce a free account.
You could subscribe via your existing Facebook account or by email. All you require is a username as well as a password.
You might be asked if you wish to follow some pals who get on Instagram in your Facebook network. You can do this right now or skip through the procedure as well as come back to it later on.
It's always a great idea to personalize your account by adding your name, a photo, a brief bio as well as a site web link if you have one when you first get on Instagram.
When you start adhering to people and also trying to find people to follow you back, they'll wish to know who you are as well as just what you're all about.
As previously mentioned, Instagram is all about visual sharing, so everyone's major purpose is to share and also locate just the most effective images as well as videos. Every user account has a "Fans" and "Complying with" count, which represents how many individuals they comply with and the amount of other users are follow them.
Every customer profile has a switch you can touch to follow them. If a user has their account set to personal, they will certainly have to authorize your request initially.
Bear in mind that when your account is produced as well as readied to public, any individual could locate and view your profile, in addition to all your photos and also video clips. Learn ways to set yours to private if you only desire the followers you authorize to be able to see your messages.
Engaging on blog posts is enjoyable as well as very easy. You can double tap any kind of message to "such as" it or include a comment at the bottom. You could also click the arrowhead switch to share it with somebody using straight message.
If you wish to locate or include even more friends or interesting accounts to comply with, use the search tab (noted by the magnifying glass symbol) to browse through customized blog posts recommended to you.
You could additionally make use of the search bar on top to try to find particular users or hashtags.
Instagram has actually come a long way since its early days in regards to posting choices. When it first launched in 2010, users could only post images with the application and add filters without any added modifying functions.
Today, you can upload both straight via the app or from existing photos/videos on your device. You could additionally upload both photos and also video clips as much as one complete minute in length, and also you have a whole bunch of extra filter choices plus the capacity to modify as well as modify.
When you tap the middle Instagram publishing tab, you could choose the electronic camera or video symbol to allow the application recognize whether you want to post an image or a video.
Capture it via the application, or tap the photo/video preview box to bring up a formerly recorded one.
Instagram has up to 23 filters you can opt to apply to both pictures as well as videos. By touching the Edit choice at the bottom of the image editor, you can also use editing results that permit you to modify modifications, illumination, comparison as well as framework. For videos, you could cut them and also choose a cover framework.
If you intend to modify your picture or video within the Instagram application, merely touch the wrench symbol and also select a function from the bottom menu. You can readjust the comparison, heat, saturation, highlights, darkness, vignette, tilt change and sharpness.
After you've used an optional filter and also possibly made some edits, you'll be taken to a tab where you could fill in an inscription, tag other users to it, tag it to a geographical area as well as simultaneously publish it to a few of your other social media networks.
Once it's released, your fans will certainly have the ability to view it and also interact with in their feeds. You could always delete your articles or modify their information after you released them by touching the three dots on top.
You can configure your Instagram account to have actually pictures posted on Facebook, Twitter, Tumblr or Flickr. If these sharing configurations are all highlighted, instead of staying grey and non-active, after that all of your Instagram images will instantly be uploaded to your socials media after you push Share. If you do not desire your picture shared on any type of certain social media, merely touch any one of them so that it's grey and set to Off.
Instagram recently presented its new Stories attribute, which is an additional feed that appears at the really top of your main feed. You could see it noted by little image bubbles of the users you follow.
Tap any kind of among these bubbles to see that customer's story or stories that they released over the last 24-HOUR. If you're familiar with Snapchat, after that you'll possibly see how comparable Instagram's tales feature is to it.
To publish your personal tale, all you have to do is tap your own picture bubble from the main feed or swipe precisely any kind of tab to access the tales camera tab. If you intend to figure out even more concerning Instagram tales, take a look at this malfunction of how it differs from Snapchat.


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Shale gas health review by Public Health England
Michael Fallon responds to the draft Public Health England Review of the 'Potential public health impacts of exposures to chemical and radioactive pollutants as a result of shale gas extraction'.
In response to the release of the draft Public Health England Review of the 'Potential public health impacts of exposures to chemical and radioactive pollutants as a result of shale gas' on Thursday 31 October;
Energy Minister Michael Fallon welcomed the report's findings saying:
"I welcome this report which shows that the potential risk to public health from shale gas production in the UK is low.
"The UK has the most robust regulatory regime in the world for shale gas and companies will only be granted permission to frack for shale if their operations are safe.
"Public safety and health is paramount and Government will continue to work with industry to ensure the stringent safety guidelines are upheld as they explore the great potential for shale."


text:
\section{Introduction}
\label{sec1}
General relativity (GR) is able to predict and describe a number of astrophysical objects \cite{Berti:2015itd}. One of the most important predictions of GR is black holes \cite{Barack:2018yly}. Due to their causal structure, black holes have a surface to which any particle or wave passing through this surface cannot return, this surface is known as the event horizon. Recently, the study of black holes has become more important. Two events that contributed to this were the detection of gravitational waves by the LIGO/VIRGO collaboration \cite{Abbott:2016blz,Abbott:2017vtc,LIGOScientific:2018mvr,TheLIGOScientific:2016pea,TheLIGOScientific:2016wfe,TheLIGOScientific:2016htt,Abbott:2020khf}, and the first image of the environment of a black hole obtained by the Event Horizon Telescope (EHT) collaboration \cite{Akiyama:2019cqa, Akiyama:2019brx, Akiyama:2019sww,Akiyama:2019bqs,Akiyama:2019fyp,Akiyama:2019eap}. Although these experiments represent a major development in black hole research, the presence of a shadow or gravitational ringdown are not definitive proof of this type of solution \cite{Herdeiro:2021lwl,Cardoso:2016rao}.
In addition to black holes, there are a number of solutions that can be obtained through GR. Some of these solutions are for example; wormholes \cite{Morris:1988cz,Morris:1988tu,Visser:1995cc,Lobo:2017oab,Bronnikov:2016xvj,Bronnikov:2016osp,Bronnikov:2018vbs,Blazquez-Salcedo:2020czn,Bolokhov:2021svp,Bronnikov:2021piw}, boson stars \cite{Colpi:1986ye,Liebling:2012fv,Yoshida:1997qf}, black bounce \cite{Simpson:2018tsi,Simpson:2019cer,Lobo:2020kxn,Huang:2019arj,Nascimento:2020ime,Lobo:2020ffi,Franzin:2021vnj} and regular black holes \cite{Zaslavskii,Fan-Wang,Bronnikov:2017tnz,Capozziello:2014bqa,Rodrigues:2015,NED2,NED3,NED4,NED5,Bronnikov:2006fu,Hollenstein:2008hp,NED10,Rodrigues:2016,Rodrigues:2017,Rodrigues:2018,bambi,neves,toshmatov,DYM,ramon,berej,Rodrigues:2019,Silva:2018,Junior:2020} among others. For some of these solutions to exist, coupling with auxiliary fields such as scalar, electromagnetic or Dirac fields are necessary. The black bounce, for example, is a type of solution that may arise from the coupling of the gravitational theory with a scalar field and an electromagnetic field \cite{Huang:2019arj}.

In the context of black holes, the problem of singularities arises, i.e., places in spacetime where the geodesics are interrupted \cite{Bronnikov:2012wsj}. Possibly, the existence of singularities is a failure of the theory because it is a classical theory of gravity. In this sense, a quantum theory of gravity could solve the problem of the existence of singularities \cite{cap2}.

We do not yet have a complete theory of quantum gravity, but there are some alternatives to eliminate the problem of the singularity. One of these alternatives is the regular black hole, so named because there are no singularities. The first regular solution was proposed by James Bardeen \cite{Bardeen}. Since the Bardeen solution did not satisfy the Einstein equations for the vacuum, Beato and Garcia showed that this solution can be obtained if one takes into account the coupling of gravitational theory with nonlinear electrodynamics \cite{Beato1}. In the literature, the study of regular solutions has become quite extensive, with a large number of proposed solutions and with the analysis of the properties \cite{Macedo:2014uga,Macedo:2015qma,Macedo:2016yyo,Paula:2020yfr,Lima:2020wcb}.

Another type of solution that avoids the presence of singularities is the black bounce \cite{Simpson:2018tsi}. Unlike the regular metrics mentioned above, this type of solution has a minimal area that is not zero. Depending on the parameters of this solution, it is possible to obtain the Schwarzschild metric, a kind of regular black hole where there is a bounce instead of a singularity, and a Morris-Thorne wormhole \cite{Lobo:2020ffi}. Another novelty is that, if we consider GR, these solutions are not described only by coupling with nonlinear electrodynamics, which requires an additional matter \cite{Franzin:2021vnj,Huang:2019arj}. Bronnikov and his collaborators also proposed a solution with a structure similar to the black bounce, the black universe \cite{Bronnikov:2016xvj,Bronnikov:2016osp,Bronnikov:2006fu}. Several papers have appeared on black bounces; quasinormal modes, shadows, absorption and others \cite{Tsukamoto:2020bjm,Zhou:2020zys,Cheng:2021hoc,Tsukamoto:2021caq,Yang:2021cvh,Bambhaniya:2021ugr,Lima:2020auu}.

Classically, black holes are objects from whose interior nothing can escape. Thus, one would expect that no radiation can emanate from them \cite{Hawking:1971vc}. By semiclassical analysis, it has been shown that black holes can indeed emit radiation and that they also have a temperature and entropy \cite{Hawking:1975vcx}. In fact, there is a whole thermodynamics associated with the study of black holes. Just as we have the standard laws of thermodynamics, there are four laws for black hole thermodynamics \cite{Bardeen:1973gs}. In particular, the first law provides us with a kind of energy conservation equation, and thus it is possible to relate the entropy of the black hole to its parameters such as mass, charge, and rotation \cite{Wald:1999vt}. Since it is a thermodynamic system, it is also possible to analyze the thermodynamic stability of black holes. For example, the Schwarzschild solution has a negative heat capacity, so it is a thermodynamically unstable solution \cite{Wald:1999vt}.

The entropy of a black hole and the area are related by $S=A/4$ \cite{Bekenstein:1973ur}. However, for some solutions, if the first law of thermodynamics is maintained, the relation between area and entropy is no longer preserved \cite{Kumar:2020cve}. Since entropy can be determined by Wald's formula \cite{Wald:1999vt} and we consider general relativity, the relation between entropy and the area of the black hole is not changed. In this way, the first law of thermodynamics must be modified depending on the type of solution we consider \cite{Ma:2014qma,Zhang:2016ilt}. This type of modification can affect the thermodynamic stability of a solution, since its temperature and heat capacity also change \cite{Maluf:2018lyu}.

The structure of this paper is as follows. In section \ref{sec2} we introduced and analysed some general properties, such as the spacetime conditions of regularity. In section \ref{sec3} we start analysing the thermodynamics of this solution using the temperature, the Smarr formula and the first law of thermodynamics. Section \ref{sec4} is devoted to the study of the heat capacity and its importance for thermodynamic stability. In section \ref{sec5} we obtain the equation of state and the possibility to analyse this solution from the point of view of virtual micromolecules. In section \ref{sec6} we analyse the Helmholtz free energy and the isothermal compressibility to compare them with the heat capacity results. In section \ref{sec7} we see how the thermodynamic functions behave near the critical points to obtain the critical exponents. Our conclusions and perspectives can be found in section \ref{sec8}.

In this paper we consider natural units, where $c=\hbar=G=1$, and the metric signature $(+,-,-,-)$. We adopt the convention that Greek indices run from $0$ to $3$, so that $x^0=t$, $x^1=r$, $x^2=\theta$, and $x^3=\phi$.

\section{Bardeen-Kiselev solution with cosmological constant}
\label{sec2}
We consider a spherically symmetric spacetime
\begin{equation}
ds^2=f(r)dt^2-\frac{1}{f(r)}dr^2-r^2\left(d\theta^2+\sin^2\theta d\phi^2\right).
\end{equation}

The solution is magnetically charged and has cosmological constant. The action that describes this theory is
\begin{equation}
S=\int d^4x\sqrt{-g}\left[R+2\lambda+L_{ NED }\right],
\end{equation}
where $R$ is the curvature scalar, $\lambda$ is the cosmological constant, and $L_{ NED }$ is the nonlinear Lagrangian of electromagnetic theory. The Lagrangian is a nonlinear function of the electromagnetic scalar $\mathcal{F}=F^{\mu\nu}F_{\mu\nu}$, where $F^{\mu\nu}$ is the Maxwell-Faraday tensor.

For a spherically symmetric spacetime that is only magnetically charged, the only non-zero component of $F_{\mu \nu}$ is \cite{NED4} \footnote{In \cite{PlebanskiKrasinski}, the authors explicitly show how the non-zero components are obtained for a spherically symmetric electromagnetic field.}
\begin{equation}
F_{23}=q \sin\theta,
\end{equation}
and the scalar $\mathcal{F}$ is
\begin{equation}
\mathcal{F}=\frac{2q}{r^4},
\end{equation}
where $q$ is the magnetic charge.

Let us consider the Lagrangian
\begin{equation}
L(\mathcal{F})=\frac{24 \sqrt{2} M q^2}{\kappa ^2 \left( \sqrt{\frac{2 q^2}{\mathcal{F}}}+2 q^2\right)^{5/2}}-\frac{6\omega c \left(\frac{2\mathcal{F}}{q^2}\right)^{\frac{3}{4} (\omega +1)}}{\kappa ^2}.
\end{equation}
When $c\rightarrow 0$, the Bardeen solution is recovered. In the limit $\mathcal{F}\rightarrow 0$

\begin{equation}
L(\mathcal{F})\approx -\frac{6 c 2^{\frac{3}{4} (\omega +1)} \omega \mathcal{F}^{\frac{3 (\omega +1)}{4}} }{\kappa ^2\left(q^2\right)^{\frac{3}{4} (\omega +1)}}+\frac{12 \sqrt [4]{2}\mathcal{F} \mathcal{F}^{1/4} m}{\kappa ^2 \sqrt [4]{q^2}}+O\left(\mathcal{F}\mathcal{F}^{3/4}\right).
\end{equation}
When $\omega=1/3$, we find
\begin{equation} L(\mathcal{F})\approx -\frac{4 c \mathcal{F} }{\kappa ^2 q^2}+\frac{12 \sqrt [4]{2}\mathcal{F} \mathcal{F}^{1/4} m}{\kappa ^2 \sqrt [4]{q^2}}+O\left(\mathcal{F}\mathcal{F}^{3/4}\right),
\end{equation}
so that for very small values of $\mathcal{F}$ the linear term dominates,
\begin{equation}
L(\mathcal{F})\approx \mathcal{F}, \mathcal{F}\rightarrow 0.
\end{equation}
This means that electromagnetism in the Maxwell limit behaves approximately like Maxwell theory.

In the presence of the cosmological constant the Einstein equations are
\begin{equation}
R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R+\lambda g_{\mu\nu}=\kappa^2 T_{\mu\nu}.
\end{equation}
Using the stress-energy tensor to nonlinear electrodynamics
\begin{equation}
T_{\mu\nu}=g_{\mu\nu}L(\mathcal{F})-\frac{dL}{d\mathcal{F}}F_{\mu}^{\ \alpha}F_{\nu\alpha},
\end{equation}
and solving the Einstein equations, we obtain the metric function
\begin{equation}
f(r)=1-2 c r^{-3 \omega -1}-\frac{2 M r^2}{\left(q^2+r^2\right)^{3/2}}-\frac{\lambda r^2}{3}.\label{fBK}
\end{equation}
For the limit $c\rightarrow 0$ we find the Bardeen-(anti)de Sitter solution and for $q\rightarrow0$ we find the Kiselev-(anti)de Sitter solution with cosmological constant \cite{Malakolkalami:2015cza}.

To obtain the horizons, we need to solve $f(r)=0$. However, we cannot obtain analytical expressions for the radius of the horizon. The number of horizons depends on the values of the parameters. In Fig. \ref{fig1} we see that the number of horizons can be up to four.

\begin{figure}
	\includegraphics[scale=.6]{f1.eps}
	\includegraphics[scale=.6]{f2.eps}
	\includegraphics[scale=.6]{f3.eps}
	\includegraphics[scale=.6]{f4.eps}
	\caption{graphic representation of $f(r)$ with respect to the radial coordinate with different values of $q$, $\lambda$, $c$ and $\omega$.}
	\label{fig1}
\end{figure}

If we want to check the existence of curvature singularities, since spacetime is spherically symmetric, we only need to check the Kretschmann scalar \cite{Lobo:2020ffi}, $K=R^{\mu\nu\alpha\beta}R_{\mu\nu\alpha\beta}$, which is given by

\begin{eqnarray}
K&=&\frac{4 \left(2 c r^{-3 \omega -1}+\frac{1}{3} r^2 \left(\lambda +\frac{6
		M}{\left(q^2+r^2\right)^{3/2}}\right)\right)^2}{r^4}+\frac{4 \left(2 c (3 \omega +1) r^{-3 \omega -2}+\frac{2 M
		r \left(r^2-2 q^2\right)}{\left(q^2+r^2\right)^{5/2}}-\frac{2 \lambda r}{3}\right)^2}{r^2}\nonumber\\
	&+&\left(2 c (9 \omega
(\omega +1)+2) r^{-3 (\omega +1)}+\frac{2}{3} \left(\lambda +\frac{M \left(-33 q^2 r^2+6 q^4+6
	r^4\right)}{\left(q^2+r^2\right)^{7/2}}\right)\right)^2.
\end{eqnarray}
The spacetime is regular only for some values of $\omega$. When $\omega\leq -1$, the solution is regular in $r\rightarrow 0$ and to $\omega\geq-1$ it is regular in $r\rightarrow \infty$ but is not asymptotically flat. For $\omega=-1$ we have the following limits
\begin{eqnarray}
&&\lim\limits_{r\rightarrow \infty}K(r)=\frac{8}{3} (6 c+\lambda )^2,\\
&&\lim\limits_{r\rightarrow 0}K(r)=\frac{8 \left(12 M \left(q^2\right)^{3/2} (6 c+\lambda )+q^6 (6 c+\lambda )^2+36 M^2\right)}{3 q^6}.
\end{eqnarray}
For this value of $\omega$, the solution is everywhere curvature regular everywhere, since we have only a shift in the cosmological constant.

\section{Thermodynamics}\label{sec3}
To start the study of black hole thermodynamics, we can obtain the temperature of a black hole using the surface gravity by \cite{Wald:1999vt}
\begin{equation}
T_k=\frac{k}{2\pi},
\end{equation}
where
\begin{equation}
k=\left.\frac{f'(r)}{2}\right|_{r=r_+},
\end{equation}
is the surface gravity and $r_+$ is the event horizon radius. Considering Eq. \eqref{fBK}, we obtain
\begin{equation}
T_k=\frac{1}{4 \pi}\left(2 c (3 \omega +1) r_+^{-3 \omega -2}+\frac{2 M r_+ \left(r_+^2-2 q^2\right)}{\left(q^2+r_+^2\right)^{5/2}}-\frac{2
		\lambda r_+}{3}\right).\label{Tk}
\end{equation}

The first law of black hole thermodynamics for a charged and static black hole is given by \cite{Gunasekaran:2012dq}
\begin{equation}
dM=T_HdS + \Phi dq+VdP,\label{flW}
\end{equation}
where $M$ is the total energy of the system, $\Phi$ is the magnetic potential, $P$ is the thermodynamic pressure, written as
\begin{equation}
P=-\frac{\lambda}{8\pi},
\end{equation}
and $S$ is the entropy of the system. The area law states that \cite{Bekenstein:1973ur}
\begin{equation}
S=\frac{A}{4}=\pi r_+^2,\label{Entropy}
\end{equation}
where $A$ is the area of the black hole.

The first law, as given in Eq. \eqref{flW}, is not valid if we have nonlinear electrodynamics \cite{Ma:2014qma,Zhang:2016ilt,Maluf:2018lyu}. The temperature and magnetic potential can be derived from the first law as follows
\begin{equation}
T_H=\frac{\partial M}{\partial S},\ \mbox{and} \ \Phi=\frac{partial M}{\partial q}.\label{Temperaturewrong}
\end{equation}
We can obtain $M$ from the condition $f(r_+)=0$, which leads to
\begin{equation}
M=-\frac{1}{6} \left(q^2+r_{+}^2\right)^{3/2} r_{+}^{-3 \omega -3} \left(6 c+\lambda r_{+}^{3 \omega +3}-3 r_{+}^{3 \omega+1}\right).\label{mass}
\end{equation}
The temperature in \eqref{Temperaturewrong}, with \eqref{mass}, is given by
\begin{equation}
T_H=\frac{sqrt{q^2+r_{+}^2} r_{+}^{-3 \omega -5} \left(6 c \left(q^2 (\omega +1)+r_{+}^2 \omega \right)+r_{+}^{3 \omega +1} \left(-2
	q^2-\lambda r_{+}^4+r_{+}^2\right)\right)}{4 \pi }.
\end{equation}
The temperature $T_k$ is different from $T_H$. To solve this problem, we need to modify the first law. If the first law is written as \cite{Singh:2020xju}
\begin{equation}
d\mathcal{M}=TdS + \Phi dq+PdV,\label{fltr}
\end{equation}
where
\begin{equation}
d\mathcal{M}=dM\left(1+4\pi \int_{r_+}^{\infty}r^2\frac{\partial T^0_{\ 0}}{\partial M}\right)=W(r_+,q)dM,
\end{equation}
we find that the temperature is
\begin{equation}
T=W\left(r_+,q\right)\frac{\partial M}{\partial S}=T_k=\frac{r_{+}^{-3 \omega -2} \left(6 c \left(q^2 (\omega +1)+r_{+}^2 \omega \right)+r_{+}^{3 \omega +1} \left(-2 q^2-\lambda
	r_{+}^4+r_{+}^2\right)\right)}{4 \pi \left(q^2+r_{+}^2\right)},
\end{equation}
where $W(r_+,q)$ is the correction factor, given by
\begin{equation} W(r_+,q)=\frac{r_{+}^3}{\left(q^2+r_{+}^2\right)^{3/2}}.\label{correctionfactor}
\end{equation}
The temperatures for Bardeen-(anti)de Sitter and Kiselev-(anti)de Sitter are given by
\begin{eqnarray}
T_{kB}&=&-\frac{2 q^2+\lambda r_+^4-r_+^2}{4 \pi q^2 r_++4 \pi r_+^3},\\
T_{kK}&=&\frac{6 c \omega r_+^{-3 \omega -1}-\lambda r_+^2+1}{4 \pi r_+ }.
\end{eqnarray}

\begin{figure} \centering \includegraphics[scale=0.6]{Tkc.eps} \includegraphics[scale=0.6]{Tkq.eps} \caption{graphic representation of the temperature associated to the solution \eqref{fBK} with $M^2\lambda=-0.08\pi$ and $\omega=-2/3$.} \label{FigTemperature}
\end{figure}
It is not so obvious what the differences are between this temperature and the temperature of the equation \eqref{Tk}. To better analyze this situation, let us see how the temperature behaves when we change certain parameters in Fig \ref{FigTemperature}. If we assign a value to $r_+$, the temperature decreases as the charge and the parameter $c$ increase. These results show that the solution\eqref{fBK} is colder than the Bardeen- (anti)de Sitter solution and the Kiselev- (anti)de Sitter solution.

To obtain the complete first law, we must first find the Smarr formula, a relation that connects the mass (energy) to the other parameters through a homogeneous function. We can derive the Smarr formula from the properties of a homogeneous function \cite{Homogenea}.

In terms of the entropy, mass can be written as
\begin{equation}
M(S, q, c, \lambda)=\frac{1}{6} \left(q^2+\frac{S}{\pi }\right)^{3/2} S^{-\frac{3}{2} (\omega +1)} \left((3 \pi -\lambda S) S^{\frac{3 \omega }{2}+\frac{1}{2}}-6 c \pi ^{\frac{3 (\omega +1)}{2}}\right).\label{massS}
\end{equation}
To determine the degree of homogeneity, we write
\begin{equation}
M(l^aS, l^bq, l^dc, l^h\lambda)=\frac{1}{6} \left(l ^{2b}q^2+\frac{l ^a S}{\pi }\right)^{3/2} l ^{-\frac{3a}{2} (\omega +1)} S^{-\frac{3}{2} (\omega +1)} \left(\left(3 \pi -l^{h}\lambda l^a S\right) l^{\frac{a}{2} (3\omega +1)}S^{\frac{1}{2} (3\omega +1)}-6 \pi ^{\frac{3 (\omega +1)}{2}} l^{d}c\right).
\end{equation}
To isolate $l$, we assume $h=-a$, $d=a(3 \omega +1)/2$, $b=a/2$, and $a=1$, which leads to
\begin{equation}
M(lS, l^{1/2}q, l^{(3 \omega +1)/2}c, l^{-1}\lambda)=-\frac{1}{6} \sqrt{l} \left(q^2+\frac{S}{\pi }\right)^{3/2} S^{-\frac{3}{2} (\omega +1)} \left(6 \pi^{\frac{3 (\omega +1)}{2}} c+\left(\lambda S-3 \pi \right) S^{\frac{1}{2} (3 \omega +1)}\right),\label{Masshomogeneous}
\end{equation}
such that the mass is a homogeneous function with degree of homogeneity $n=1/2$ \cite{Homogenea}.

Euler's identity states that for a homogeneous function with degree $n$, we have \cite{Homogenea}
\begin{equation}
n.f(x_1,x_2 ... x_{m})= a_{1}x_{1}\frac{\partial f}{\partial x_{1}}+a_{2}x_{2}\frac{\partial f}{\partial x_{2}}+a_{3}x_{3}\frac{\partial f}{\partial x_{3}}+...+a_{m}x_{m}\frac{\partial f}{\partial x_{m}}.
\end{equation}
Thus we obtain
\begin{equation}
\frac{1}{2}M(S, q, c, \lambda)= T_H S+\frac{1}{2}\Phi q+\frac{1}{2}\left(1+3\omega\right)A_c c-A_{\lambda}\lambda,\label{Smarrcorrect}
\end{equation}
where $T_H$ and $\Phi$ are given by \eqref{Temperaturewrong}, and $A_c$ and $A_{\lambda}$ are
\begin{equation}
A_c=\frac{\partial M}{\partial c},\ \mbox{and} \ A_{\lambda}=\frac{\partial M}{\partial \lambda}.\label{Smarrrelations}
\end{equation}
With Eq. \eqref{Temperaturewrong} and \eqref{Smarrrelations} we get that \eqref{Smarrcorrect} and \eqref{massS} are equal.

The first law is written as
\begin{equation}
d\mathcal{M}=T_k dS+ W(S,q)\Phi dq+W(S,q)A_cdc+W(S,q)A_{\lambda}d\lambda.\label{FistLaw}
\end{equation}

For regular black holes, it is normal that the Lagrangian depends explicitly on some parameters, such as charge and mass. Fluctuations in the matter sector are not a problem because they are compensated by fluctuations in the geometry sector by the first law, \eqref{FistLaw}.

\section{Heat Capacity}\label{sec4}
From thermodynamics, we know that the stability of a thermodynamic system requires $C_P\geq C_V\geq0$ and $k_T\geq k_S\geq0$, where $C_P$ and $C_V$ are the heat capacity at constant pressure and volume, respectively, and $k_T$ and $k_S$ are the isothermal compressibility and isentropic compressibility, respectively \cite{Stanley}.

In black thermodynamics, compressibility and heat capacity also give us information about the stability of the system. The isothermal compressibility and heat capacity at constant pressure are given by \cite{Davies:1977bgr}
\begin{eqnarray}
C_P=\left.T_k\frac{\partial S}{\partial T_k}\right|_P,\label{capacity}\\
k_T=-\left.\frac{1}{V}\frac{\partial V}{\partial P}\right|_T.\label{compress}
\end{eqnarray}

We will focus on the heat capacity. We get that the heat capacity at constant pressure is
\begin{eqnarray}
C_P&=&\left\{2 S \left(\pi  q^2+S\right) \left(S^{\frac{3 \omega }{2}+\frac{1}{2}} \left(2 \pi ^2 q^2+\lambda  S^2-\pi  S\right)-6 c \pi ^{\frac{3 (\omega +1)}{2}} \left(\pi  q^2 (\omega +1)+S
\omega \right)\right)\right\}\times\nonumber\\
&\times&\Big\{6 c \pi ^{\frac{3 (\omega +1)}{2}} \left(\pi  q^2 S (\omega  (6 \omega +7)+4)+\pi ^2 q^4 (\omega +1) (3 \omega +2)+S^2 \omega  (3 \omega +2)\right)\nonumber\\
&+&S^{\frac{3	\omega }{2}+\frac{1}{2}} \left(\pi  S^2 \left(3 \lambda  q^2+1\right)-7 \pi ^2 q^2 S-2 \pi ^3 q^4+\lambda  S^3\right)\Big\}^{-1}\label{heat}
\end{eqnarray}
This result is valid only if we derive temperature from the surface gravity. If we use \eqref{Temperaturewrong}, we get
\begin{equation}
\bar{C}_P=\frac{2 S \left(\pi q^2+S\right) \left(S^{\frac{3 \omega }{2}+\frac{1}{2}} \left(2 \pi ^2 q^2+\lambda S^2-\pi S\right)-6 c \pi ^{\frac{3 (\omega +1)}{2}} \left(\pi q^2 (\omega +1)+S
	\omega \right)\right)}{6 c \pi ^{\frac{3 (\omega +1)}{2}} \left(2 \pi q^2 S \left(3 \omega ^2+5 \omega +2\right)+\pi ^2 q^4 \left(3 \omega ^2+8 \omega\hspace{-0.1cm} +\hspace{-0.1cm}5\right)+S^2 \omega (3 \omega
	+2)\right)+S^{\frac{3 \omega }{2}+\frac{1}{2}} \left(\pi S^2\hspace{-0.1cm}-\hspace{-0.1cm}4 \pi ^2 q^2 S\hspace{-0.1cm}-\hspace{-0.1cm}8 \pi ^3 q^4\hspace{-0.1cm}+\hspace{-0.1cm}\lambda S^3\right)}.
\end{equation}
These two heat capacities are connected by
\begin{equation}
C_P=W(S,q)\frac{\partial T_H}{\partial T_k}\bar{C}_P.\label{CCrelation}
\end{equation}

The heat capacity to the Bardeen-(anti)de Sitter and Kiselev-(anti)de Sitter solutions are
\begin{eqnarray}
C_{PB}=\frac{2 S \left(\pi q^2+S\right) \left(2 \pi ^2 q^2+\lambda S^2-\pi S\right)}{\pi S^2 \left(3 \lambda q^2+1\right)-7 \pi ^2 q^2 S-2 \pi ^3 q^4+\lambda S^3},\\
C_{ PK }=\frac{2 S \left(-6 c \pi ^{\frac{3 (\omega +1)}{2}} \omega +\lambda S^{\frac{3 (\omega +1)}{2}}-\pi S^{\frac{3 \omega }{2}+\frac{1}{2}}\right)}{6 c \pi ^{\frac{3 (\omega +1)}{2}} \omega (3 \omega +2)+\lambda S^{\frac{3 (\omega +1)}{2}}+\pi S^{\frac{3 \omega }{2}+\frac{1}{2}}}.
\end{eqnarray}
In Fig. \ref{figC1} we see how behaves the heat capacity to the solution \eqref{fBK}. There are both positive and negative values for $C_P$, so the solution is thermodynamically stable for some values of entropy. This behavior is different from the Schwarzschild solution where the heat capacity is always negative, $C_P=-2S$. We also see that in the range of small values of entropy and negative heat capacity, the temperature is also negative and therefore this range has no physical meaning. In the Bardeen- (anti)de Sitter solution, the first phase transition occurs at a higher temperature value than in the other cases, while the second phase transition occurs at a lower temperature value than in the other cases. In this way, the Bardeen solution has a smaller temperature range for the intermediate phase. From Fig. \ref{figC2} we see that the Kiselev-(anti)de Sitter solution has only one phase transition, while the solution \eqref{fBK} has two phase transitions.
\begin{figure}
	\includegraphics[scale=.6]{Cvc1.eps}
	\includegraphics[scale=.6]{Cvc2.eps}
	\includegraphics[scale=.6]{Cvc3.eps}
	\includegraphics[scale=0.6]{TS1.eps}
	\caption{Heat capacity at constant pressure and temperature associated to the solution \eqref{fBK} with $q=0.2M$, $M^2\lambda=-0.08\pi$ and $\omega=-2/3$.}
	\label{figC1}
\end{figure}

\begin{figure}
	\includegraphics[scale=.6]{Cvq1.eps}
	\includegraphics[scale=.6]{Cvq2.eps}
	\caption{Heat capacity at constant pressure and temperature associated to the solution \eqref{fBK} with $M^2\lambda=-0.08\pi$, $\omega=-2/3$ and $Mc=0.2$.}
	\label{figC2}
\end{figure}

\section{Equation of State}\label{sec5}
From the modified first law \eqref{FistLaw}, the thermodynamic volume is given by
\begin{equation}
V=\frac{\partial \mathcal{M}}{\partial P}=\frac{4\pi r^3_+}{3}.\label{Vol}
\end{equation}
With \eqref{Tk} and \eqref{Vol} we obtain the equation of state $P(T,V)$, i.e.
\begin{equation}
P(T,V)=\frac{4 q^2 \left(6 \pi ^{4/3} T \sqrt [3]{V}+(6 \pi )^{2/3}\right)+6 (6 \pi )^{2/3} T V-3 \sqrt [3]{6}V^{2/3}}{36 \sqrt [3]{\pi } V^{4/3}}-\frac{ 4^{\omega } \pi ^{\omega +\frac{2}{3}}c }{3^{\omega+\frac{2}{3}}V^{\omega +\frac{5}{3}}} \left(2 \sqrt [3]{2} q^2 (\omega
+1)+\left(\frac{3V}{\pi }\right)^{2/3} \omega \right).\label{press}
\end{equation}
The pressure to the Bardeen-(anti)de Sitter and Kiselev-(anti)de Sitter solutions are
\begin{eqnarray}
P_B(T,V)&=&\frac{\sqrt [3]{\frac{\pi }{6}} \left(2 q^2+3 T V\right)}{3 V^{4/3}}+\frac{2 \pi q^2 T}{3 V}-\frac{1}{2\ 6^{2/3} \sqrt [3]{\pi } V^{2/3}},\\
P_K(T,V)&=&-c \left(\frac{4 \pi }{3}\right)^{\omega } \omega V^{-\omega -1}+\frac{\sqrt [3]{\frac{\pi }{6}} T}{\sqrt [3]{V}}-\frac{1}{2\ 6^{2/3} \sqrt [3]{\pi } V^{2/3}}.
\end{eqnarray}
The critical points result from the conditions \cite{Dehyadegari:2017flm,Dayyani:2017fuz,Wei:2020poh}
\begin{equation}
\left(\frac{\partial P}{\partial V}\right)_T=0, \ \mbox{and} \ \left(\frac{\partial^2P}{\partial V^2}\right)_T=0.\label{Pressurecondition}
\end{equation}
Unfortunately, we are not able to solve this equation analytically, but if we fix some values for parameters $q$, $\omega$ and $c$, we get numerical values for $T_c$, $P_c$ and $V_c$. In Fig. \ref{fig3} we show the isotherm considering three cases; $T<T_c$, $T>T_c$ and $T=T_c$, where $T_c$ is the critical temperature. It is clear that the pressure is higher for higher temperature values. At a fixed temperature, the pressure increases as the charge and $c$ increase. For the Kiselev-(anti)de Sitter solution it is not possible to solve the condition \eqref{Pressurecondition} because there is no inflection point in $P(V)$. So we are not able to find the critical points.

\begin{figure}
	\includegraphics[scale=.6]{PV.eps}
	\includegraphics[scale=0.6]{PVc.eps}
	\includegraphics[scale=0.6]{PVq.eps}
	\caption{Graphic representation of the $P-V$ behavior of the solution \eqref{fBK} with $q=0.2M$, $Mc=0.2$ and $\omega=-2/3$. There is also the behavior at a fixed temperature and different values of $q$ and $c$.}
	\label{fig3}
\end{figure}

Since we have the equation of state $P\left(T,V\right)$, we can obtain the compressibility factor $Z$, which is defined as \cite{Stanley}
\begin{equation}
Z=\frac{PV}{T}.\label{ CF }
\end{equation}
For an ideal gas, i.e., a gas in which there is no interaction between the particles that compose it, we have $Z=1$ \cite{Stanley}. For a real gas where there is interaction between the particles that make it up, $Z$ depends on factors such as pressure and temperature \cite{Stanley}. Thus, the compressibility factor basically tells the deviation of the behavior of a real gas from that of an ideal gas. Since $P\left(T,V\right)$ is a function of temperature and volume, $Z$ is also a function of $T$ and $V$ and is written as
\begin{equation}
Z(T,V)=\frac{4 q^2 \left(6 \pi ^{4/3} T \sqrt [3]{V}+(6 \pi )^{2/3}\right)+6 (6 \pi )^{2/3} T V-3 \sqrt [3]{6}V^{2/3}}{36 \sqrt [3]{\pi V} T}-\frac{ c }{T}\left(\frac{4\pi}{3V}\right)^{\omega +\frac{2}{3}} \left( q^2 (\omega
+1)+\left(\frac{3V}{4\pi }\right)^{2/3} \omega \right).\label{ZTV}
\end{equation}
To Bardeen-(anti)de Sitter e Kiselev-(anti)de Sitter we find
\begin{eqnarray}
Z_K(T,V)&=&-\frac{c \left(\frac{4 \pi }{3}\right)^{\omega } \omega V^{-\omega }}{T}-\frac{\sqrt [3]{V}}{2\ 6^{2/3} \sqrt [3]{\pi } T}+\sqrt [3]{\frac{\pi }{6}} V^{2/3},\\
Z_B(T,V)&=&\frac{\sqrt [3]{\frac{\pi }{6}} \left(2 q^2+3 T V\right)}{3 T \sqrt [3]{V}}+\frac{2 \pi q^2}{3}-\frac{\sqrt [3]{V}}{2\ 6^{2/3} \sqrt [3]{\pi } T}.
\end{eqnarray}
In Fig. \ref{FigZ} we see how $Z$ behaves when we increase the pressure for different values of the temperature. The compressibility factor diverges to small values of the pressure. As said, for a real gas $Z\neq 1$, but for small values of pressure some gasses behave like an ideal gas. For a van der Waals fluid, the compressibility factor at the critical point is $Z_c=0.375$ \cite{Dayyani:2017fuz}. For the solution \eqref{fBK} we find $Z_c\approx 0.75$, which is twice as much as in the van der Waals case.

\begin{figure}
	\includegraphics[scale=0.6]{Z.eps}
	\caption{Compressibility factor in terms of the ratio $P/P_c$ to $q=0.2M$, $Mc=0.2$ and $\omega=-2/3$.}
	\label{FigZ}
\end{figure}

In the context of fluids, the compressibility factor, when analyzed microscopically, provides information about the interaction of the molecules that make up the fluid \cite{James}. The most important point is this interpretation of the molecules in the context of black holes. According to Wei and Liu, it is possible to analyze a black hole microscopically through the idea of virtual molecules \cite{Wei:2015iwa}. Thus, it is interesting to use the number density of virtual micromolecules of the black hole, i.e.
\begin{equation}
n=\frac{1}{v}=\frac{1}{2 l_p^2r_+},
\end{equation}
where $v$ is the specific volume of the black hole fluid and $l_p=\sqrt{\hbar G/c^3}$ is the Planck length. Since we are considering natural units, $l_p=1$ and $n=1/v=1/2 r_+$. Note that the specific volume is linear with the radius of the event horizon. In terms of the number density, the temperature is
\begin{eqnarray}
T(n)=\frac{3 c\left(2n\right)^{3 \omega +2}n \left(4 n^2 q^2(\omega +1)+\omega \right)-8 n^4 q^2+2 \pi P+n^2}{2 n\pi\left(4 n^2 q^2+1 \right)}.\label{EqTn}
\end{eqnarray}
For Bardeen-(anti)de Sitter and Kiselev-(anti)de Sitter we find.
\begin{eqnarray}
T_B(n)&=&\frac{-8 n^4 q^2+n^2+2 \pi P}{8 \pi n^3 q^2+2 \pi n},\\
T_K(n)&=&\frac{6 c 2^{3 \omega} \omega n^{3 \omega +2}}{\pi }+\frac{P}{n}+\frac{n}{2 \pi }.
\end{eqnarray}
In Fig. \ref{FigTn} see how the temperature behaves when we change the number density for different values of the pressure. There is a maximum value of $n$ and, depending on the value of $P$, there are other limits in the number density. For example, for small values of pressure, there is a range where the temperature is negative, so these values of $n$ are not allowed. Towards larger values of charge, the number density increases. As $c$ increases, the number density decreases.
\begin{figure}
	\includegraphics[scale=0.6]{Tn.eps}
	\includegraphics[scale=0.6]{Tnc.eps}
	\includegraphics[scale=0.6]{Tnq.eps}
	\caption{behavior of ratio $T_k/T_c$ in terns of number density to $\omega=-2/3$.}
	\label{FigTn}
\end{figure}

\section{Helmholtz free energy and isothermal compressibility}\label{sec6}
To verify the phase transition, we should analyze the Helmholtz free energy \cite{Dehyadegari:2017flm,Dayyani:2017fuz,Wei:2020poh,Liu:2014gvf,Nam:2018tpf,Dehyadegari:2020tau,NaveenaKumara:2020biu}, $F$, given by
\begin{equation}
F=M- TS.
\label{HFE}
\end{equation}
With \eqref{mass}, \eqref{Tk}, and \eqref{Entropy}, we obtain
\begin{equation}
F=\frac{1}{6} \left(\frac{\left(q^2+r_{+}^2\right)^{3/2} \left(\left( 8 \pi P r_{+}^2+3\right) r_{+}^{3 \omega +1}-6 c\right)}{r_{+}^{3 (\omega +1)}}+\frac{3 r_{+}^2 \left(6 c r_{+}^{-3 \omega }-8 \pi P r_{+}^3-3 r_{+}\right)}{2 \left(q^2+r_{+}^2\right)}-\frac{6 c}{r_{+}^{3 \omega }} -3 c (3 \omega +1) r_{+}^{-3 \omega }+3 r_{+}\right).
\label{helm}
\end{equation}
To Bardeen-(anti)de Sitter and Kiselev-(anti)de Sitter we find
\begin{eqnarray}
F_B&=& \frac{r_+}{2}-\frac{r_+^3 \left(8 \pi P r_+^2+3\right)}{4 \left(q^2+r_+^2\right)}+\frac{\left(8 \pi P r_+^2+3\right) \left(q^2+r_+^2\right)^{3/2}}{6r_+^2},\\
F_K&=&-\frac{c}{r_+^{3 \omega}}\left(1+\frac{3\omega}{2}\right) -2 \pi P r_+^3+\frac{r_+}{6} \left(8 \pi P r_+^2+3\right)-\frac{r_+}{4}.
\end{eqnarray}

In Fig. \ref{FigHel} we show the behavior of the Helmholtz free energy. For small values of the temperature, we have only small black holes. As we increase the temperature, there is more than one possible state. There is a temperature value for which there are three different states, but two of them have the same energy. Interestingly, there is a black hole, the intermedite one, for which the heat capacity is negative, that is, it is thermodynamically unstable. When we increase the pressure, the intermediate black hole disappears and the phase transition occurs from a small black hole to a large black hole. The Kiselev solution has only two phases and as the charge increases, the temperature required for the phase transition decreases. In the Bardeen solution, there are three phases, but as $c$ increases, one of the phases disappears at positive temperatures.
\begin{figure}
	\includegraphics[scale=0.6]{Fhelm1}
	\includegraphics[scale=0.6]{Fhelm2}
	\includegraphics[scale=0.6]{Fhelmq.eps}
	\includegraphics[scale=0.6]{Fhelmc.eps}
	\caption{Helmholtz free energy as a function of temperature to $\omega=-2/3$.}
	\label{FigHel}
\end{figure}

Another way to analyze the phase transition is through isothermal compressibility. If we use \eqref{compress} with \eqref{press}, we get
\begin{eqnarray}
k_T&=&24 \pi \left(r_{+}^2+q^2\right)r_{+}^{3\omega +5}\left[r_{+}^{3 \omega+1} \left(q^2 r_{+}^2 \left(24 \pi Pr_{+}^2+7 \right)+8 \pi P r_{+}^6+2q^4-r_{+}^4\right)\right.\nonumber\\
&-&\left.6 c\left(q^2 r_{+}^2 (\omega (6 \omega +7)+4)+q^4 (\omega +1) (3\omega +2)+r_{+}^4 \omega (3 \omega+2)\right)\right]^{-1}.\label{kt}
\end{eqnarray}
To Bardeen-(anti)de Sitter and Kiselev-(anti)de Sitter we get
\begin{eqnarray}
k_{ TB }&=&\frac{24 \pi r^{4} \left(q^2+r^2\right)}{q^2 r^2 \left(24 \pi P
 r^{2}+7\right)+8 \pi P r^6+2q^4 -r^4},\\
k_{ TK }&=&\frac{24 \pi r^{3\omega +3}}{6 c \omega - (18 c \omega^2 +18 c \omega)+\left(8 \pi P r^2-1\right)r^{3\omega +1}}.
\end{eqnarray}
When the isothermal compressibility is positive, the solution is thermodynamically stable \cite{Stanley}. In Fig. \ref{figkt} we see that the solution is stable for a small and large black. As we already know from the heat capacity, the intermediate black hole is thermodynamically unstable.
\begin{figure}
	\includegraphics[scale=.6]{kt.eps}
	\caption{Isothermal compressibility associated to the solution \eqref{fBK} with $q=0.2M$, $M^2P=0.01$, $\omega=-2/3$, and $Mc=0.2$.}
	\label{figkt}
\end{figure}

\section{Critical exponents}\label{sec7}
As we have seen, some physical quantities such as isothermal compressibility and heat capacity diverge during a phase transition. Even though these quantities diverge, we can analyze their behavior near the transition point. For this purpose, we use the critical exponents \cite{Dayyani:2017fuz,Wei:2020poh}. To obtain the critical exponents, we use
\begin{equation}
t=\frac{T-T_c}{T_c}, \ v=\frac{V-V_c}{V_c} \ \mbox{and} \ p=\frac{P}{P_c}.\label{pvt}
\end{equation}

The critical exponents $\alpha$, $\beta$, $\gamma$, and $\delta$ are obtained from \cite{Stanley}
\begin{eqnarray}
C_V=T\left(\frac{\partial S}{\partial T}\right) \propto \left|t\right|^{-\alpha},\\
\eta=V_1-V_2 \propto \left|t\right|^{\beta},\\
k_T=-\frac{1}{V}\frac{\partial V}{\partial P}\propto \left|t\right|^{-\gamma},\\
\left|P-P_c\right|\propto \left|V-V_c\right|^{\delta},
\end{eqnarray}
where $\eta$ is the difference in volume between two phases.

Before proceeding, we have analyzed the phase transition graphically for certain parameter values. Thus, in the next analyzes, although it is not obvious, we consider the same values as before, $\omega=-2/3$, $q=0.2M$ and $M c=0.2$.

If we use \eqref{pvt} and rewrite \eqref{press} and expand it for small values of $t$ and $v$, we get
\begin{equation}
p=1+t (a-b v)-d v^3+O\left(v^4,tv^2\right),\label{paprox}
\end{equation}
where $a$, $b$, and $d$ are just constants that are combinations of $\omega$, $q$, and $c$. If we consider $t$ as constant and derive $p$ to $v$, we get
\begin{equation}
dp=-\left(bt+3dv^2\right)dv.
\end{equation}
This relation is important to apply Maxwell's area law, which states \cite{Dayyani:2017fuz,Wei:2020poh}
\begin{equation}
\int VdP=0.
\end{equation}
Thus, considering Maxwell's area law and the fact that the pressure is constant at the phase transition, we obtain
\begin{eqnarray}
p&=&1+t (a-b v_1)-d v_1^3=1+t (a-b v_2)-d v_2^3,\\
0&=&\int_{v_1}^{v_2}\left(v+1\right)\left(bt+3dv^2\right)dv.
\end{eqnarray}
Solving these equations yields the nontrivial solution as follows
\begin{equation}
v_1=-v_2 \propto \sqrt{t} \ \mbox{and} \ \eta\propto\sqrt{t}.
\end{equation}
This means that one of the critical exponents is $\beta=1/2$.

We can also use \eqref{paprox} to calculate the isothermal compressibility and we find that
\begin{equation}
k_T\propto \frac{1}{t},
\end{equation}
such that $\gamma=1$.

We can rewrite the condition $\left|P-P_c\right|\propto \left|V-V_c\right|^{\delta}$ as $\left|p-1\right|\propto \left|v\right|^{\delta}$. From \eqref{paprox} we get
\begin{equation*}
p-1\propto t (a-b v)-d v^3.
\end{equation*}
However, in the critical isotherm we have $t=0$, so that,
\begin{equation}
p-1\propto v^3.
\end{equation}
With this result, we find the critical exponent $\delta=3$.

To determine $\alpha$, we need to analyze the heat capacity at constant volume. To obtain $C_V$, we need to rewrite the entropy as a function of volume, which is given by
\begin{equation} S(V)=\sqrt [3]{\frac{9V^2\pi }{16}}.
\end{equation}
If the volume is constant, the heat capacity at constant volume is
\begin{equation} C_V=\left.T\left(\frac{\partial S}{\partial T}\right)\right|_V =0.
\end{equation}
This means that the critical exponent is $\alpha=0$.
We see that these exponents satisfy the Griffiths, Rushbrooke and Widom equalities \cite{Wei:2020poh,Stanley,Griffiths};
\begin{eqnarray}
\alpha+\beta\left(1+\delta\right)=2,\ \mbox{Griffiths}\\
\gamma\left(\delta+1\right)=\left(2-\alpha\right)\left(\delta-1\right),\ \mbox{Griffiths}\\
\alpha+2\beta+\gamma=2, \ \mbox{Rushbrooke}\\
\gamma=\beta \left(\delta-1\right),\ \mbox{Widom}
\end{eqnarray}
which says that there are only two independent exponents.

\section{Conclusions}\label{sec8}
In this article, we have obtained and analyzed a magnetically charged solution that mimics the Bardeen regular black hole, magnetically charged, with cosmological constant, and surrounded by quintessence. The Lagrangian of this solution has as a limit the Bardeen case at $c=0$. We find a metric function that can have up to 4 horizons depending on the parameters. By analyzing the Kretschmann scalar, we find that the solution is regular: in the whole spacetime up to $\omega=-1$, at infinity of the radial coordinate when $\omega\leq-1$, in $r=0$ up to $\omega\geq-1$.

It was also possible to analyze the solution from the point of view of black hole thermodynamics. We obtained two different temperatures $T_k$ and $T_H$ reflecting the use of nonlinear electrodynamics, and therefore we modified the first law by the correction factor \eqref{correctionfactor}. Moreover, analyzing \eqref{Masshomogeneous}, we found that the mass is a homogeneous function with a degree of homogeneity $n=1/2$, and thus we found the Smarr formula \eqref{Smarrcorrect}.

We obtain two values for the heat capacity connected by \eqref{CCrelation}. Analyzing Fig. \ref{figC1} and Fig. \ref{figC2}, we conclude that $C_P$ can take both positive and negative values as a function of entropy, i.e., there are regions where the solution is thermodynamically stable $(C_P\geq0)$ and in others it is unstable $(C_P < 0)$. The equation of state \eqref{press} behaves like a Van der Walls fluid for $q=0.2M$, $Mc=0.2$ and $\omega=-2/3$, and at the critical point the compressibility factor is twice as large as in the Van der Walls case. We also construct the temperature as a function of the number density of virtual micromolecules of the black hole, as shown in Fig. \ref{FigTn}. To obtain a positive temperature, $n$ cannot take any arbitrary value, so there must be some bounds on this variable.

By analyzing the Helmholtz free energy and the isothermal compressibility of the system, we show that the solution has two thermodynamically stable states, namely small and large black holes. During the phase transition, a third intermediate phase occurs, but it is thermodynamically unstable. Finally, fixing the values $q=0.2M$, $Mc=0.2$, and $\omega=-2/3$, we calculated the critical exponents $\beta=1/2$, $\gamma=1$, $\delta =3$, and $\alpha=0$, but only two of them are independent since they are connected by the Griffiths, Widom, and Rushbrooke equalities.

In future work, we hope to explore some topics that are missing in this paper, such as geodesic analysis, shadows, causal structure, maximum analytical extension, and the analysis of the interaction with bosonic and fermionic fields.

\vspace{1cm}

{\bf Acknowledgements}: M. E. R.  thanks Conselho Nacional de Desenvolvimento Cient\'ifico e Tecnol\'ogico - CNPq, Brazil, for partial financial support. This study was financed in part by the Coordena\c{c}\~{a}o de Aperfei\c{c}oamento de Pessoal de N\'{i}vel Superior - Brasil (CAPES) - Finance Code 001.

redpajama_set_name: RedPajamaArXiv


Clearfield County, PA – The first ever Veritas Film Festival starts today in Clearfield County and continues for the next week.
Filmmakers from all around the globe will be at the Rowland Theatre in Philipsburg and the Ritz Theater in Clearfield, starting at 9 each morning and finishing at around 5 each afternoon at both locations. An award ceremony will also be held each night in Philipsburg.
Each night has a different theme.
Spencer Folmar is the artistic director of the festival and owner and producer of Hard Faith films, which include Generational Sins and the upcoming film Shooting Heroin.
Tickets are $10 for a day pass or $35 to see films throughout the entire week.
For more information, the full schedule, and links to buy tickets, visit their Facebook page or veritasfilms.com.


I'm into buying a SSD (Samsung Pro) to replace my laptop's HDD. The laptop is a Toshiba Satellite S55-B (Core i7, 8GB RAM, Radeon Graphics).
I was wondering how to transfer all the content and partitions from the HDD to the SSD, because something I've been asking myself if the recovery partition will be transferred to the SSD.
Also, the other question is if it's worth the change to SSD. I mean, like ''Hey, you will really feel the difference in speed".
Use easeus disk copy (it's free), remember if it will copy everything so you will end up with an ssd with mbr and not gpt.
Also cloning the single partition won't help, you will need the boot records and boot files. If they reside on c:\ then you may not have issues swapping out.
However if they reside in the small 100 meg primary partition then you will need to clone that as well. If you do run into issues swapping out drives then you will have no choice but to clone the entire hdd. Unless of course you are using a custom boot loader, as you could simply just clone (partition) and mount. Making sure you set up the boot loader first of course.
The easiest method is to move all your excess data off hdd, clone the entire thing to your ssd and then move what data you want back. This way you have an external back if things go side ways.
Last edited by TechFreak1; 12-02-2015 at 11:08 AM.
Can I get some help getting my SMS to an Android?
My Lumia 520 Suddenly Can't Connect to GSM Service, Why?


text:
Recycling in Billerica, MA
Home / Recycling / Massachusetts / Billerica, MA
How is Casella an industry professional in Recycling Service? We start by providing residents, homeowners and municipalities with environmentally conscious Recycling Service in Billerica, MA, that brings trash management to a different level. Each year Casella renews and processes enough raw waste to fill the capacity of the Empire State building 5 times. Simply contact Casella now to receive a price quote on our recycling in Billerica, MA.
Casella Recycling in Billerica, MA: High-Quality Service, Competitive Pricing
The Casella approach to Recycling Service in Billerica, MA is not only innovative, but plays an important role in renewing the life-cycle of the products and materials. We're not only dedicated to creating a greener tomorrow, but we're dedicated to our service to you and ensuring we meet all your requirements. By investing efforts into future waste renewal technologies, Casella continues to provide Recycling Service in Billerica, MA to residents that have come to rely and trust the Casella name.
For our business professionals, we are prepared and trained to always bring safe and reliable recycling in Billerica, MA that has a low impact on the environment. It's the personal touch that makes Casella stand apart from other companies offering recycling in Billerica, MA.
When you take a look at what Casella offers as compared to other recycling in Billerica, MA, you'll see that there truly is no comparison. We pride ourselves on providing exemplary service over the last 40 years, not only for our customers, but for the environment and the communities in which we live and work. At Casella, we make a sincere effort at being responsible community members that protect our environmental resources, develop and nurture our people and continuously strive to improve where we live and work.


The Gemini AS-12BLU-PK is an all-in-one active Bluetooth party speaker set. Everything you need to enhance the ambiance is included: 12" speaker, heavy duty speaker stand, FM radio option, pro-grade microphone, and a remote control – a complete sound system package that gets any party or gathering started right out of the box. Its packaged speaker stand not only elevates your sound at adjustable heights from 43" – 74" tall for a superior listening experience.


Every morning I take a walk through the Koguro park. When I used to go to work(round trip of 3 hrs…), I did not have time to take a walk. But now work globally from my SOHO, and I enjoy the ultimate combination of the local living and the global business.
Every day I feel the seasons changes and I find thing in the beautiful scenery and flowers. Koguro park is a small park but it is like a big garden to me. I sense in the air that lots of people's life-style will change soon.
Please click here to see "Koguro Park" video that I made!


text:
Questions about specific movies, TV shows and more
These are questions relating to specific titles. General questions for movies and TV shows are here. Members get e-mailed when any of their questions are answered.
Unanswered general questions
Answered general questions
The Man in the Iron Mask (1998)
Question: How did Athos realise that Raoul's death was actually planned by Louis? How could he know that the king wanted his son to die?
Answer: But it is later revealed that Louis deliberately set it up in order to get Raoul killed. Christine becomes Louis' mistress after finding out about Raoul's death.
Answer: Athos knew the king well enough that he was returning Raoul to battle solely so he could pursue Christine, Raoul's fiancée, as his mistress. The king presumed Raoul would be killed by having him placed in a dangerous position at the front. Although Raoul subsequently died, he conceivably could have survived. After becoming Louis' mistress, Christine later wrote to someone in the military who revealed it was the king who issued the order to have Raoul placed at the battlefront. Christine, distraught, considered his actions to be murder; arguably the king mostly wanted Raoul out of the way, though he still was responsible. Louis had no long-term interest in Christine and had a continual succession of mistresses, soon becoming bored with them.
raywest
But it is later revealed that Louis deliberately set it up in order to get Raoul killed. Christine becomes Louis' mistress after finding out about Raoul's death.
Digimon (1999)
Show generally
Question: Why are Digimon called digital monsters? They don't look digital.
Answer: Digimon started as a series of virtual pets, making them digital. In the television series, Digimon live in the "digital world."
Bishop73
Disclosure (1994)
Question: Was Garvin the actual mastermind behind the plan to blame Tom for the CD-ROM problems from the very beginning or was that Meredith's idea? It would be unwise for Garvin to have set up Tom from the beginning since he appeared to have confidence in the merger going through prior to the sexual harassment case. And if Meredith was going to make changes to cut costs, why would she not learn about how the drives work and any potential problems that she may cause?
Movielover1996
Question: When Peter and Gwen hear a boy being carried out of the Oscorp training...what is the kid screaming...is it something about Peter or someone taking his badge, so therefore they wouldn't let him in?
Answer: He says "Look! Listen, tell them Rodrigo Guevara is down here. Please, just tell them, Rodrigo Guevara is down here! I am Rodrigo!" That was the name on the badge Peter took, but we don't hear him say anything about someone taking his badge.
Black Hawk Down (2001)
Question: When the rangers are being deployed, Nelson puts something in his mouth saying the last rope he almost bit his tongue off.
Answer: I'm assuming the question is, what was it? It was a mouth guard to keep him from biting his tongue when he hits the ground at the bottom of the rope.
kayelbe
American Dad (2005)
Question: TV Tropes said there was an episode where Francine talks about having been molested in the girls' locker room shower in high school and Stan seems to be aroused by the story - does anyone know what this episode was called?
Answer: S04e01, "1600 Candles."
X-Men 3 (2006)
Question: The awkward delivery Kitty makes when she says "Who's hiding... d**khead?!" to Juggernaut always struck me as really strange. Is it just poor delivery of a generic insult? Or is she trying to make a (bad) one-liner about the fact his helmet is maybe, kinda-sorta vaguely shaped like a man's glans? The weird pause and emphasis on the word "d**k" in "d**khead" made me think it was the latter, but I'm not sure.
TedStixon
Answer: She was scared, going into battle for the first time and going up against a very powerful mutant. It was the best one-liner she could think of at that moment. She purposely said it to agitate him, to put him in such a blinding rage, he wouldn't notice his powers were gone because of "LEECH."
Swamp Thing (1982)
Question: What did Arcane believe he was going to turn into after drinking the formula?
Answer: He believed he would turn into a being of Supreme Intelligence, the next step in human evolution, like Professor X. An advanced mutant.
The House (2017)
Question: Is there any indication how much of this film was unscripted or how much Amy and Will improvised their lines. I can't tell if it's just the acting style the actors chose, or if it was the way the director wanted it, but a lot (especially the opening scenes) comes across as bad improv. Or am I the only one that gets that vibe?
Answer: A lot of the time with modern comedies, they'll usually do a scene as written, then do a number of improv takes where the actors and director come up with jokes on the fly. Then in editing, they'll pick the (presumably) best bits from both the scripted and improv takes to put into the finished film. And Will Ferrell confirmed that there was indeed a lot of improvising on-set of "The House." Unfortunately, without a copy of the shooting script, it's basically impossible to tell exactly how much of the finished film is made up of the unscripted improv. But it's fair to say that at least a decent chunk of the movie was made up of improv comedy. So chances are it's not just you.
Question: After Frank fell to his death, Jack and Jason stay in the tent for the night. Jason has things dropped from his bag, and he puts something (looks like a tin can maybe?) into the bag right away and Jack notices it and stares at him with some kind of betrayed looks on his face. Does it imply Jason was keeping food or water for himself behind Jack's back? Please explain the scene.
Bunch Son
Chosen answer: It's their last can of Sterno...basically, canned jellied alcohol (or ethanol) that can be lit on fire and used as a localized heat source. It's common amongst campers, as it can be used as a burner or to heat a small tent, which is how they've been using it. The look they exchange is an unspoken acknowledgement that if they don't get more fuel, or find somewhere warm, they will freeze to death soon.
Season 1 generally
Question: Why did Eobard believe that pretending to be crippled would divert any suspicion away from him? He could simply walk around even after the accelerator exploded and people could just say that he was lucky that he didn't get injured, mutilated or even killed.
Harry Potter and the Prisoner of Azkaban (2004)
Question: How come from the whole time from when Hagrid gives Scabbers to Ron, until he bites him and runs away after the execution, Ron keeps holding him instead of putting him in his pocket?
Answer: To start with, Ron recently believed that Scabbers was dead, and is relieved to have him back. So Ron is probably not in the mood to shove him into a pocket again. And why shouldn't anyone be holding their pet?
Poltergeist III (1988)
Question: Whatever happened to Scott, Donna's boyfriend? Donna and everybody else returned after Tangina left with Kane, but did he?
Answer: At the end of the movie, it was a different actress who played Carol Ann as the actress had already died by that point. Rumor has it that the actor who played Scott was so distraught over her death that he couldn't come back to film the scene.
lartaker1975
Question: Darnell stated that he knew a guy that had a car like Christine once and that he killed himself in it. Are the guy he knew and LeBay's brother the same person?
Answer: Yes.
Question: I have two questions. 1) If Christine can fix herself, why was she in such a state when Arnie first bought her? 2) Did Arnie do all the repairs on Christine or did Christine fix herself bit by bit to give the illusion that Arnie was doing it?
Answer: She is influenced by the person owning her. It's love. The love was gone when Arnie bought her, since her last owner died. But Arnie was in love with her and Christine fell in love with him overtime because of his dedication to fixing her. Because of the love for each other, the magic that powers Christine becomes more powerful, until she is eventually able to fix herself. During his time fixing her, she eventually takes over the repairs, but at first he was the one doing the fixing. It's possible her state contributed to her fixing herself, so the more she is repaired, the more she is able to do herself as well.
Home Alone (1990)
Question: I know it's just a movie, but how were Gus Polinski and his wife unable to even realise they left their own son behind at a funeral home, let alone leaving him there all day, as Gus was explaining to Kate near the end of the movie (especially under the assumption that it began the night before during a typical service or even during calling hours)? Surely, any parent with common sense would know not to leave their children unsupervised in an event like that.
zredman
Answer: It can be any number of things. High stress. Forgetfulness. Just being tired. Maybe they got hyper focused and it simply slipped their mind. Etc. (And since it's a funeral home in his case, it could be all of those things, plus more.) And it's really not all that uncommon. One survey I found online showed that over 10% of parents admitted to accidentally forgetting their child was in their car with them and leaving them in the back seat. And in another survey, 9% of parents have admitted to accidentally either losing track of or leaving a child behind somwhere while they were out. It just... happens.
Answer: While Gus doesn't state many of the particulars of the event, I've attended numerous funerals where kids are constantly on the move. Going in and out of family rooms, going outside, etc. It's possible to lose track or think they're with another relative. Gus is used to being on the road and isn't always in "Dad mode." My own parents forgot me at school numerous times.
Fantastic Four: The Animated Series (1994)
Question: Why was the first season of this cartoon received so negatively?
Answer: Gleaned from the Internet: Much of the negative reaction was due to the poor, simplistic, and outdated quality of the animation where the characters moved stiffly and clumsily through scenes. There were also some obvious animation compositing gaffs. The animation colors were also criticized as being overly bright and cheery and out of character to the established story's darker tone. The broad humor of the first season was replaced by a more mature narrative between the characters.
Answer: Gleaned from the Internet: Some critics criticized the premise of so many one-off battles that were introduced and concluded within one or two episodes. Others felt the first-season plot lines were overly simplistic and never progressed or developed through the season. Also, the rivalry between Iron Man and the Mandarin also failed to capture viewer's attention.
Silver Surfer (1998)
Question: Why was this series cancelled after one season?
Answer: From Wikipedia: Eight episodes were written for the second season before production was shut down. According to series creator Larry Brody, the cancellation was the result of a legal dispute between Marvel and Saban Entertainment.
Last Man Standing (2011)
Extrasensory Deception - S8-E20
Question: When Mike and Vanessa are eating at one point, there is a light behind Mike. Anyone know what it is?
Answer: If you're referring to near the end, after Ryan leaves, there's a recess in the wall with shelves that has vases and decorations on it (next to the clock on the wall). There are light on the underside of one of the shelves and the plates on the shelves are reflecting those lights.


Many air-cleaning systems filter out particles. Some destroy germs, and some attack odors. The PureAir™ line combats all three major types of indoor air contaminants. The PureAir™ S Air Purification System is a smart whole-home solution designed to be paired with the iComfort S30 and a Lennox communicating system providing sensor based visibility to the filter life and purifier (UVA) life knowing the exact time for replacement needs. The Healthy Climate PureAir™ is designed to be paired with non-Lennox communicating systems and non-Lennox systems.


I have been madly in love with the recent offerings of hedgehog marginalia from Medieval Animal Data Network. These adorable looking creatures, are often rendered in the rich golden and red hues of fall, and harvesting grapes (despoiling the vineyards) using their spines like cocktail toothpicks. But most wonderful of all...a group of artists got together and created an animated film of the hedgehog in action, based on the art and narrated in Latin. Totally awesome.
De Herinacio. On the Hedgehog from obrazki nunu on Vimeo.
English, about 1250 - 1260; Book of Hours of Charlotte of Savoy created in Paris, France, in the 15th century. New York, The Pierpont Morgan Library, Ms. M. 1004, fol. 82v; Bestiary- Therouanne, ca 1270; Latin bestiary Cambridge, 1320.
I am still in awe of these amazing little Medieval illustrations of the classical Greek and Latin heroines of Ovid writing imaginary epistles. Ovid's work was translated into French by poet Octavien de Saint-Gelais, and it so delighted Louise of Savoy (1476-1531), mother of the future king Francis I of France, that she commissioned artist Robinet Testard to decorate a sumptuous manuscript of Ovid with these beautiful miniature portraits of the heroines writing their own stories. Each one is so different ..caught in the unique drama of her life -- some beautiful, some rather violent. (And really...so many bad husbands and rotten boyfriends...) Here's a sample of some of my favorites ...but do, stop by La Folia Magazine and indulge in all the others. All these heroines writing their own stories provides fuel and words for my own characters who are impatient to write down their (mis)adventures.
Wouldn't it be wonderful if we could publish our novels with such lovely marginalia? There is something so unexpectedly tender about these two snails meeting and kissing in the margin. Marginalia art has a vital life of its own -- existing side by side with text but not necessarily commenting on the text. It's the mind reading and wandering a little bit, hearing the whispered words in the head, but delighting in the visual art that sidetracks and then returns one to the text, refreshed. And all in a matter of moments.
Art: Bibliothèque municipale de Lyon, Ms 6881, detail of f. 72. Book of Hours, use of Chalon. 15th century.


text:
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NSString* const SPAuthURL = @"https://auth.simperium.com/1/";
NSString* const SPWebsocketURL = @"wss://api.simperium.com/sock/1";
NSString* const SPTermsOfServiceURL = @"https://simperium.com/tos/";

NSString* const SPAPIVersion = @"1.1";

#if TARGET_OS_IPHONE
NSString* const SPLibraryID = @"ios";
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// TODO: Update this automatically via a script that looks at current git tag
NSString* const SPLibraryVersion = @"1.9.0";

/// SSL Pinning
///
/// 1. Extract the PEM:
///     > openssl s_client -showcerts -host api.simperium.com -port 443
/// 2. Calculate the Public Key Hash
///     > ./External/TrustKit/get_pin_from_certificate.py certificate.pem
/// 3. Verify PEM Expiration Date
///     > openssl x509 -enddate -noout -in certificate.pem
///
NSString* const SPPinnedDomain = @"api.simperium.com";
NSString* const SPPinnedPublicKeyHash = @"T97MoWLAydq7mqpZDJ4t6zRNZIUAD3alp/vqGkVPUIw=";

redpajama_set_name: RedPajamaGithub


I mean really. Do we have to hear from these guys this many times? These aren't debates. These are moderated discussions. They've cut out Kucinich so they look like they are from the left.
Bill Clinton wasn't the first black president. He was the maybe first modern 'born poor' president.
For good measure, he made sure to oversee the execution of convicted killer Ricky Ray Rector, a brain-damaged black man, in the heat of the primaries.
Obama is fortunate he wasn't busted [for cocaine use] during Bill Clinton's years in office. Clinton left behind a larger, darker prison population than when he took office. Black incarceration rates during the Clinton years surpassed those during Ronald Reagan's eight years.
One Clinton initiative required citizens, mostly black, in public housing to surrender their Fourth Amendment, or privacy, rights.
But I am just as sure that if Obama went after Hillary Clinton to reveal the real record of the period she seems intent on restoring, h e would be savagely attacked for playing the race card by the very same media that is fawning over him now.
At least they're letting Dennis's buddy Ron Paul in the debates still. Like Dennis, he's the only one talking about real issues. At least they'll still be the only Congressmen with any balls the next coule-a years.


text:
Éric Wolfer (born 27 February 1966) is a French volleyball player. He competed in the men's tournament at the 1992 Summer Olympics.

References

1966 births
Living people
French men's volleyball players
Olympic volleyball players of France
Volleyball players at the 1992 Summer Olympics
Sportspeople from Brussels

redpajama_set_name: RedPajamaWikipedia


text:
Black Man Sits Down to Have an 'Uncomfortable Conversation' With a Room Full of Police Officers
Emmanuel Acho
Emmanuel Acho has now had 9 different conversations on his YouTube channel "Uncomfortable Conversations with a Black Man." He started the channel because he wanted to create a "safe place to have the uncomfortable conversations about race."
Especially after George Floyd's death and the Black Lives Matter movement, Acho felt it was important to sit down and have these difficult yet important conversations.
Acho has talked to a variety of different people. His very first conversation was with Matthew McConaughey. He also sat down with Chip and Joanna Gaines along with all of their children.
In Acho's most recent video, he sat down with police officers from the Petaluma Police Department. Before starting the conversation, he talked to the camera and explained that "so much of the dissension in the world, so much of the dissension in America is based around the interaction between police officers, white police officers, and black people."
Then the camera moved out to reveal the police officers that were joining him for the discussion. He mentioned that not only was he joined by the Petaluma Police Department but also a live audience. This is the first conversation that included an audience in the studio.
The audience wasn't made up of random individuals. The audience members also included police officers, one of whom took part in the conversation.
Throughout the conversation, Acho asked difficult questions, like if the officers approach a situation differently when they are responding to a report of a black suspect as opposed to a white suspect.
The officers made the conversation uncomfortable for Acho as well, asking him difficult yet important questions like if he reacts differently when approached by a white police officer as opposed to a black police officer and why.
The conversation is eye-opening no matter your race. As Acho points out, "You can't fix a problem you don't know exists. I'm letting you know it exists." Watch Acho's conversation with the white police officers in the video below.
In the comments, multiple viewers have expressed that they would like to see more conversations like this between Acho and the police. One comment reads, "I feel like this conversation, in particular, needs to be extended into more episodes. The conversation between officers and a black man brought a bit of healing to my mind and to my soul. To see this conversation be had was moving. But what really touched me was to see a little bit of transparency between both parties(Emmanuel and the police). The way you both allowed for this to be an open safe space."
Another viewer wrote, "Please keep having THIS particular conversation."
What did you learn from this "uncomfortable conversation"? Do you think Acho should do more conversations like this with white police officers? What questions would you want him to ask?


This week Royboy lays out his big idea for 2017. An extended road trip with the goal of shooting 48 cars in 48 states for a book project.
These folks are making 48 Cars In 48 States possible! Support them, let them know that you appreciate them supporting this adventure.
My buddy Ryno is handling getting the Galaxie ready for the trip. If you want your car done right, Ryno is one of the few guys that I trust.


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Please be courteous and respectful of others.
Individuals may only use the facility during open hours and when an Aquaplex Lifeguard is on duty.
Individuals under the influence of alcohol or drugs will not be permitted at the facility.
Pool activities vary. Please check the facility schedule, or speak with a staff member to determine what activities are available. Classes, practices, and rentals may restrict the availability of some areas.
Aquaplex staff is responsible for the safety of all facility users; follow their directions at all times. Aquaplex staff members have the final say in policy interpretation and enforcement.
In an emergency, you will hear one long blast of a whistle; stop whatever you are doing and follow the Aquaplex staff's instructions.
Please report instances of personal injury immediately.
Proper swimming attire is required; including bathing suits, board shorts, or rash guards. No jeans, cut-offs, street clothes, or inappropriate (G-string) bathing attire.
"Deck Changing" or changing your clothes on the swimming pool deck is prohibited at all times.
Please do not enter the water if you have open cuts, sores, wounds, infectious disease, or have had diarrhea, stomach flu or been severely sick in the last two weeks (14 days).
Before entering the pool area, a hot soapy shower is highly recommended.
Please put sun screen on at least 15 minutes before entering the pool.
Rough housing and horseplay are not permitted in the facility.
Dangerous pool entries, dives, and jumps are not permitted into pools or the spa.
Please do not run on the pool deck.
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Food and beverages may be brought in plastic containers. No glass containers are allowed. No food or beverages are allowed in pools.
No roller-skates, roller-blades, skateboards, scooters, or bicycles are permitted within the facility.
Participants are encouraged to lock their valuables at all times. Aztec Recreation is not responsible for lost or stolen items.
All guests must exit the water and pool deck ten (10) minutes before the posted closing time. All guests must exit the facility by the designated closing time.
Water basketball and volleyball equipment is available to use on a first-come-first-serve basis. Please check out equipment from the office and return after use.
Please return equipment (kick boards, pull buoys, rafts, etc) to proper storage bins after use.
Children under the age of eighteen (18) may only use the facility during posted Family Swim hours and must be supervised by an adult (over 18 years of age) at all times.
All flotation devices must be US Coast Guard approved. Individuals wearing a USCG approved lifejacket are reserved to the Recreation pool and must be within arm's length of a responsible adult at all times.
Participants are advised to have a physical examination prior to participating in any physical activity. Associated Students/SDSU assumes no responsibility for accidents or injuries that take place on Aztec Recreation premises.
Repeated policy violations or actions that threaten the safety and enjoyment of others may result in the loss of access and membership privileges.
Lap lanes are for lap swimming only. If more than two (2) people are sharing a lane please circle swim by staying to the right.
Do not swim under the bulkhead or through lap lanes.
Starting blocks are not for recreational use. Qualified instructors or coaches are required for supervision during usage.
Only one (1) person on a board at a time.
Be sure that there is no one swimming under the board and the area is clear before diving or jumping off.
Do not run on diving boards.
Only one bounce is permitted on diving boards.
Dive straight off the end of the board; do not attempt to jump off the side of the board.
Swim to the nearest ladder or side of the pool and exit the diving area. Do not swim directly under the diving board.
No unorthodox dives/jumps or dangerous activities or games will be permitted on the diving boards.
The fulcrum must be kept below three (3) during open diving.
You must be eighteen (18) years of age to use the Hydro-therapy Spa.
Elderly persons, pregnant women, and those with health conditions should consult a physician before using the Hydro-therapy Spa.
Do not use the Hydro-therapy Spa while under the influence of medications that can cause drowsiness or that raise or lower blood pressure.
Never use the Hydro-therapy Spa without an Aquaplex Lifeguard on duty.
Observe reasonable time limits of fifteen (15) minutes in the Hydro-therapy Spa. Then leave the water to cool down before returning for another brief stay.
Please do not submerge your head under the water for long periods of time.
When maximum occupancy occurs please limit your time to fifteen (15) minutes.
Children under the age of eighteen (18) must be supervised by an adult at all times.
Children who are not "water safe" must be within an arm's reach of an adult at all times while in the water.
"Water safe" is defined as a child who can safely tread water, control breathing, and demonstrate forward momentum above and below the water surface. Aquaplex Lifeguards have the discretionary authority to require an adult to be in the water with any child.
Children, who are not toilet trained, MUST wear proper swim diapers.
Children under the age of eighteen (18) are not allowed in the Competition Pool unless they have passed the deep water test.
The deep water swim test includes one length (25 yards) demonstrating controlled breathing and constant forward momentum on the surface of the water; 30 seconds treading water, demonstrating ability to keep one's head above water surface.
For safety reasons, when floating lounges are in use there will not be "free-swim" in that section of the pool.
Children under the age of eighteen (18) may not use the floating lounges.
Non-USCG (United States Coast Guard) approved floatation devices are not allowed. Please check with Aquaplex staff to ensure any floatation device you bring in is acceptable prior to using it.
Children wearing US Coast Guard approved flotation devices are not allowed in the Competition Pool and must be accompanied by an adult (in the water) at all times.
Please do not hang on lanes, ladders, ropes, or rails.
Children may bring appropriate swim toys. No hard toys, balls, or squirt guns.
Children are not permitted in the Hydro-therapy Spa.


Alone, the creature is a minimal threat, but Harriers hunt in packs as small as six and as large as 20, making them a powerful predator of even the hardiest Tusk. Most small packs (8 or fewer Harriers) will only go after prey such as deer or elk. Larger packs tend to be bolder and will strike against a solitary humanoid or small group (4 or fewer).
A common tactic is to panic their prey, allowing others spread around it to strike. Often times, a small number (2-4) will run into scrub to flush prey for several others waiting in nearby trees to pick off.


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Reinventing The Steel (20th Anniversary Edition) - PANTERA [Limited Silver Vinyl]
€46.00 €50.00 You Save 8% (€4.00)
Pantera's final opus, Reinventing The Steel, represented a recommitment to everything the band loved about heavy metal. Released in 2000 at the peak of nu-metal's popularity, the album's back-to-basics approach flew in the face of the trend and served as a potent reminder of the enduring power of primal metal. The album received widespread critical acclaim as well as high praise from fans, who voted the album as Album of The Year 2000 as well as voting the single "Revolution is My Name" Single of The Year 2000. The album also ranked No.2 on Guitar World's readers' poll for Top 10 Guitar Albums of 2000. Pantera's 9th and final studio album turns 20 this year and is celebrated here.
Collections: COLOUR VINYL, Vinyl
Category: Colour Vinyl, nonsquare-cover, Release Date 08.01.21, Vinyl


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