The following tutorial uses Photoshop Elements 18.0, but the process is the same for older versions. This tutorial is designed to be simple enough for a beginner but also detailed enough to cover all the subtle nuances and advanced features. User Interface of Photoshop Elements While Photoshop Elements is a more basic graphics editor than Photoshop, it also has a simpler user interface. You’ll find the user interface as shown in Figure 1. Figure 1. Photoshop Elements 18.0 has a simplified user interface. Let’s take a closer look at the elements that are on the main screen. Figure 2. Click on the layers tab to view layers and selections. It’s important to note that you can click on any of the tabs to get a screen similar to what’s shown in Figure 2. The Layers Tab The Layers Tab shows the Layers Panel. It is a panel that shows all the layers that are currently in the document. The other tabs (as shown in Figure 3) are buttons, and not actual screens. Figure 3. All the tabs are buttons, not actual screens. The Background Tab The Background Tab is very similar to the Layers tab. It displays all of the layers in a document’s background. The Levels and Curves tabs are shown in Figure 4, where we can see the Levels tab and the Curves tab. Figure 4. Displaying the Background, Layers, Levels and Curves tabs. The Levels and Curves Tab The Levels and Curves Tab (shown in Figure 5) displays the levels and curves tools that we’ll be working with in this tutorial. This tab is where you control the levels of shadows, highlights, midtones, and colors in your image. Let’s take a closer look at the elements that are on the Levels and Curves Tab. Figure 5. The Levels and Curves tabs. The Tonemapper The Tonemapper (shown in Figure 6) is where you can set the tone of your image. In this tutorial, we are just learning how to quickly correct the tonal range of an image, as well as controlling how we sharpen or soften its edges. Figure 6. The Tonemapper. The Adjustment Layers The Adjustment Layers (shown in Figure 7) has many 05a79cecff

Recommended Links Meerut Jail Complex: the Jails, Rehabilitation, Union reform MIND(A Tribune Research) Published: Aug 22, 2010, 9:13pm IST Two weeks ago, the government told the Supreme Court that it is not right to end the long-waged mass agitation by releasing of the arrested. The agitation in the state started in June,2009 which was accompanied with the process of release of more than 700 of the agitators. Now, the government will hold talks with the leaders of the agitators to end the long-waged agitation. The government has come out with a list of norms for the release of the agitators. To ensure that the ones released after November 11,2011 do not lead again to similar outbreak, the government has promised to set up two jails–one for women and another for men–in the state. The jails will be run by a government-appointed body. Government will provide all help to those who go back home to live,the government will give employment to those who go for rehabilitation and will also provide pension to the prisoners of the rehabilitation jail. The government will provide Rs 25 lakh to each of the parents of the victims and Rs 25 lakh to each of the family members of the victims. The government will also provide Rs 25 lakh to the family members of those who have committed suicide. If the family members do not want to get the sum,the government will get the sum itself and will pay Rs 2 lakh to any of the well-known person for his death. The government will provide Rs 10 lakh to those who want to study after their release. The government will provide Rs 10 lakh to the ex-prisoners who want to study. The government will also give those who want to earn a living while on parole an opportunity to earn a living. The government will also provide cheap price for education for students. The government will also give the students a concession on their fees for a year. The government will also give a concession to the students so that they can buy electronic gadgets and other things. The government will give the concession as a way to build better human relations. The government will also give rebate in electricity bills for all the families living in jail and will give the families Rs 2,500 per month as the family income. The government will also give 6,000 litres of free milk to every family

Q: How do you calculate changing of potential with changing of magnetic field strength? I have read this question, but I am still not convinced. Also, I am confused with how magnetic field strength affects potential energy. I mean, from Maxwell’s Equations, we know that a changing magnetic field $B(x,t)$ produces a force, $\vec F = q (c_{e0} \vec E + c_{m0} \vec B)$. If $E$ and $B$ are changing with time, then shouldn’t there be a change in the potential energy? Of course, that wouldn’t be of much interest when you have a static charged source (like the one I sketched). But what about when the source is moving? When $E$ and $B$ are changing with time, how can potential energy be conserved? What I also don’t understand is that (if my thought is right) how did the Larmor formula come from the Lorentz force equation, which involves $E$ and $B$. A: If you really want to figure out how potential energy changes, the best thing you can do is solve Maxwell’s equations. For example, the simplest way to do it would be to solve Maxwell’s equations for a magnetic field alone, since you’re typically only interested in magnetic fields. You can calculate the electric field using your solution for the magnetic field and your electrical charges. Then you should look at the potential energy of those charges, and use that to figure out what potential energy the charges have when they’re in the magnetic field you found. However, most people would instead solve the PDEs of electromagnetic fields in a region of space and then look for potential solutions (solutions to $\vec{ abla} \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}$ or similar). In that case, your answer is correct, since any potential solution to the PDEs will be an admissible solution to the given problem. Since you want the potential energy of the electric field, you have to solve $\vec{ abla} \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}$. That equation will have a solution for $\vec{E}$ that depends on $\vec{B}$, and if you use that $\vec{E}$ in the Maxwell