Remove eSign Form Now

To now remove sign from your documents, simply upload the document to airSlate SignNow, select the signature you wish to remove, and follow the prompts to delete it. Our intuitive interface makes it easy to manage your eSignatures efficiently.

No, there are no additional fees to now remove sign from your documents on airSlate SignNow. You can manage your signatures and documents as part of your subscription plan, which offers a cost-effective solution for your eSigning needs.

airSlate SignNow provides a range of features that allow you to now remove sign, add new signatures, and customize your documents. With options for bulk sending, templates, and real-time tracking, you can easily streamline your document management process.

Absolutely! airSlate SignNow offers seamless integrations with popular applications like Google Drive, Salesforce, and more. This allows you to efficiently manage your documents and easily now remove sign across platforms.

Using airSlate SignNow provides numerous benefits, including enhanced security, ease of use, and the ability to now remove sign whenever necessary. It empowers businesses to send and eSign documents quickly, helping to improve workflow and save time.

airSlate SignNow employs advanced security measures, including encryption and secure cloud storage, to ensure that your documents are protected. You can confidently now remove sign and manage your eSignatures knowing your data is safe.

You can eSign a variety of document types in airSlate SignNow, including contracts, agreements, and forms. Additionally, if needed, you can easily now remove sign from any document to keep it updated and accurate.

Note-This is a long answer but believe me, it is worth reading.This answer targets every competitive exam in which the English section plays a major role.I am taking an example of English section of bank exams because it is the hardest nut to crack.I am not promoting any apps, websites. I am preparing for this exam also. I have watched a lot of videos of the English strategy. They just beat around the bush and say only two things-Do reading and Improve your vocab. But they don't tell the exact sources that we should follow.That’s why i am writing this answer.English is the toughest section in the bank exams. Most of the people think that I am from English medium background, therefore, cracking this section is a piece a cake for me. This is the reason why the most of the people are not able to clear even sectional cutoff of this section. Just see the cutoff-IBPS PO PRE 2015 CUTOFF-(Out of 30)IBPS PO MAINS 2015 CUTOFF-IBPS PO PRE 2016 CUTOFF-IBPS PO MAINS 2016 CUTOFF-IBPS PO PRE 2017 CUTOFF-IBPS PO MAINS 2017 CUTOFF-(Out of 40)Did you notice something?The cutoff for this section is in single digit out of 30/40 marks. So you can guess the level of this section. That's why I showed screenshot to know the importance of this section.Now how to prepare for this section?There are 3 major parts in this section-1.Reading Comprehension.2.Vocab.3.Grammar.Now let's discuss one by one1.Reading Comprehension-To master this section do at least one- hour editorial reading(Business/Economy related articles) of The Hindu, Indian Express, Economic Times etc. There is no need to buy any newspaper. Just install this app.Daily editorial-The advantage of this app is that when you click on the word the meaning of the word will pop up. So there is no need to use the dictionary again and again.And please don't waste your precious time in jotting down current affairs and vocab. You have two choices either waste your time in jotting down things or follow my advice(i did this thing and believe me making notes is an utter waste of time).For speed reading, I suggest these two apps-1.2.For RC practice-Ideal Test Simulator For The RC Section Found On All Standard Tests.Book to follow for speed reading-How to read better and faster by Norman Lewis.2. Vocab- I have already discussed the importance of these two apps. I will provide the link to the answers at the end of this answer.1.VOLT-2.VOCAB MEME-Follow these lectures for vocabulary-How to build your English Vocabulary - YouTube3. Grammar- This is the section where most the banking aspirants waste their precious time. There are thousands of rules and cramming these rules is a difficult task. Do reading and your half of the work is done for this section. But if you are preparing for SSC then just follow these two books-Plinth to paramount by Neetu Singh.Mirror of common errors by Dr Ashok Kumar Singh.For grammar follow these videos by MERITSHINE-1.GRAMMAR BY SAMIR SIR-Learn the nuances of English Grammar - YouTube2. GRAMMAR BY RATNABHA SHAH SIR-English Grammar for Competitive Exams - YouTubeSome other useful pieces of stuff-Which dictionary should we follow?It depends on person to person but I follow Merriam Webster.What are some important lectures?1.THE EDITORIAL TIMES - YouTube2. Meritshine videos on English.Where to practice?1.Sanctum English2. Daily quizzes of adaa, gradeup, ibpsguide, aspirant zone, olive board.Links of other answers that may be useful for you-1.Nitish Arya's answer to What are good YouTube channels to learn English language?2.Nitish Arya's answer to Which is the best app for learning English vocabulary?3.Nitish Arya's answer to What is the best youtube channel to prepare for bank exam?4.Nitish Arya's answer to How does one read the Hindu newspaper for bank exams?5.Nitish Arya's answer to Which is the best book for reasoning and aptitude?Request-If you know better resources/strategy please tell me in the comment section. Please don't be narrow-minded people who think that if we share our strategy then it will increase the competition. Just remember this quote-Aap jeevan me tab tak safal nhi ho skate jab tak aap dusro ke safal hone m unki madad nhi krte.To get more guidance about bank exams, you can search my channel name on youtube.Channel name-BeNextBankerI will not provide the link. If you are serious banking aspirant then you will definitely search. If i will provide the link then non banking aspirant will click on the link just to see my channel only. They will watch the video only for 1 minute. So why to waste their precious time./-

Yes there has been a signNow change in the working of Indian Embassies around the world especially in the crisis hit nations as the Embassies are now faster in response to the need of the Indian Nationals in that country unlike the earlier years.

As someone who writes mostly technology-related answers, I see the following question so much it makes me want to tear my hair out:“Can iCloud Activation Lock be Bypassed?”For those who don’t know, Apple devices that have an iCloud account active on them with Find My iPhone enabled will lock the device to that Apple ID even if it is restored to factory defaults. This is designed to prevent thievery, since stolen devices (typically iPhones) are useless without the Apple ID password they are locked with to unlock it. It is incredibly common for people to sell devices without removing the lock beforehand (likely because they don’t know it exists, or how to remove it) or because it is stolen. Either way, the lock cannot be bypassed without that password… but that doesn’t stop everyone and their mother from asking if it can be done as if the rules somehow don’t apply to them.Instead of viewing the answers on an existing question, or even asking new people to answer that existing question, they make a new one. Every. Single. Time. Quora is absolutely flooded with these questions, and I get A2A requests for them more than anything else.

The number π is a mathematical constant, the ratio of a circle's circumference to its diameter, commonly approximated as 3.14159.Any circle in naturally occuring Elucidean space, would have a value of π when its circumference is divided by its diameter.Being an irrational number, π cannot be expressed exactly as a fraction (equivalently, its decimal representation never ends and never settles into a permanent repeating pattern). Still, fractions such as [math]\dfrac{22}{7}[/math] and other rational numbers are commonly used to approximate π. The digits appear to be randomly distributed. In particular, the digit sequ...

Trading account is the one which is opened with your trading member. It is this account that you use to place your buy and sell orders. A demat account is simply like a savings account in your bank. IT simply holds your stocks and other dematerialized securities. THe demat account is opened with one of the depositories viz. NSDL or CDSL. When you want to buy a share of a company, as a retail investor, you will need to place it through your stock broker. An investor is not allowed to place orders directly with the stock exchanges. The broker then transmits this order to the exchanges, typically on a real time basis. Inside the exchange, the order goes into the trading system. Depending on the order category, the order is sent for execution (market orders) or to the order book (limit order). [There are some more order categories, but I will not go into those details for the sake of simplicity].A market order is executed against the first opposite order that comes along i.e your buy order is executed against the first sell order available. All orders in the order book are prioritised by price-time priority. So a limit order goes and sits in the order book accordingly and will be executed only when the price of your order is signNowed. Till then it remains in the order book as open order. If the trading session ends before your order is executed, your order is automatically cancelled by the exchange. Once an order is executed, the process of clearing and settlement begins. Here the money is taken from the buyer i.e. you and given to the seller of the security. Simultaneously, the share is taken from the demat account of the seller and transferred to the demat account of the buyer. All these processes are done transparently to both the buyer and the seller and are done jointly by the exchanges, the trading members, the clearing members, the custodians and the banks. To get the complete detailed process of clearing and settlement, you can refer to the NSE website here http://www.nseindia.com/products...The broad outline of the clearing and settlement process is as followsStep 1: Trade details from exchange to NSCCL (clearing house of NSE)Step 2: NSCCL notifies the trade details to Clearing members/ CustodiansStep 3: Download of obligation/ pay-in advice of funds/ securitiesStep 4: Instructions to clearing banks to arrange funds by pay-in timeStep 5: Instruction to depositories for sameStep 6: Pay-in of securitiesStep 7: Pay-in of fundsStep 8: Pay-out of securitiesStep 9: Pay-out of fundsStep 10: Depository informs custodians/ Clearing members through DPStep 11: Clearing banks inform custodians/ Clearing members.It is at the end of step 8 that you as a buyer get your stocks in your demat account and at the end of step 9, the seller of that stock gets the money in his linked bank account. Please note that the pay-in of funds/securities in steps 6 and 7 refer to the obligations of the trading members. As a retail investor, you will typically have to transfer the money/securities (to your stock broker)  along with the order placement. This is an elementary level-1 risk management process followed by the members to ensure that the investors buying  or selling actually have the money to buy or securities to sell.

Before answering this, let’s start with a brief primer on what “half-life” means.Each atoms of a radioactive material will spontaneously split at a random time. And for simplicity, we’ll assume that once it splits, it stops being radioactive and will never split again. So as more and more atoms split and stop being radioactive, the number of remaining radioactive atoms falls exponentially. And because of that, the rate of decay (i.e. the numbers of splits per unit time) also falls exponentially. If a material has a half life of [math]\tau[/math], it means that both the number of radioactive atoms and the rate of decay both falls to half their original level in this time period. We can describe this exponential decay quantitatively with the following formula:[math]\boxed{N(t) = N_0 \cdot 2^{-t/\tau}}[/math]Where [math]N_0[/math] is the starting number of radioactive atoms, and [math]N(t)[/math] is the current number of remaining radioactive atoms at time [math]t[/math].It’s often convenient rewriting this formula so that it uses the base [math]e[/math], as follows:[math]N(t) = N_0 \cdot e^{-t\ln(2)/\tau}[/math]Now, if we want to know who many atoms decay in a comparably short span of time, [math]dt[/math], we simply differentiate this formula:[math]dN(t) = -\dfrac{N_0 \ln(2)}{\tau} \cdot e^{-t\ln(2)/\tau} dt[/math]Why is there a minus sign? Because what we’ve actually calculated here is the reduction of remaining radioactive atoms. The number of splitting events therefore has the minus sign removed:[math]dN(t) = \dfrac{N_0 \ln(2)}{\tau} \cdot e^{-t\ln(2)/\tau} dt[/math]Now we’re finally ready to apply this to the OP’s question.As its name implies, Bismuth-209 has an atomic weight of approximately 209 atomic mass units, therefore [math]6.02214×10^{23}[/math] atoms (i.e. an Avogadro constant worth) have a mass very nearly equal to 209 grams. In other words, a 1 kilo brick of bismuth (see photo below) would have [math]N_0 = 2.88141×10^{24}[/math] atoms.The half life of bismuth is [math]\tau \approx 2.01×10^{19}\,years[/math]So setting [math]t=0[/math] (i.e. now) into our formula, in one year we’d expect approximately:[math]dN(0) = \dfrac{N_0 \ln(2)}{\tau} \cdot e^{0} \approx \dfrac{2.88141×10^{24} \cdot \ln(2)}{2.01×10^{19}} \approx 99,400[/math] decaysIn other words, one decay every 5.3 minutes, or so.So even though bismuth’s half life is very long, with so many atoms the number of splitting events it’s quite detectable. The challenge is actually measuring these decays, since most of the atomic splits will occur deep inside the brick, and their effects will be quickly absorbed by neighboring atoms.And yes, as others have noted, bismuth does oxidize and crystallized to form a variety of cool shapes and colors. However, that has no bearing on this question.

Let’s begin with the one dimensional Schrodinger equation with a potential energy [math]V(x)[/math]. It looks something like this:[math]\dfrac {\hbar^2}{2m}\dfrac {d^2 \psi}{dx^2} = V(x)\psi- E\psi\tag* {}[/math]where [math]V(x)[/math] is not constant and varies from place to place. Let’s re-write it into something equivalent by manipulating it a little bit.[math]\dfrac {d^2 \psi}{dx^2} = \dfrac {2m}{\hbar^2}\left ( V(x) - E\right )\psi\tag* {}[/math]Now let’s factorize and do a little bit of dimensional analysis on the factor [math]2m\left ( V(x) - E\right )[/math].It has units of [math][M][E] = [M][M][D^2][T^{-2}][/math] which is equivalent to [math]\frac {kg^2\cdot m^2}{s^2}[/math] which is the same units with momentum squared i.e [math]p^2(x)[/math]. So let’s make a substitution where [math]p^2(x) \equiv 2m\left ( V(x) - E\right )[/math] and of course [math]p(x) = \sqrt {2m(V(x) - E)}[/math].Now, the Schrodinger equation becomes:[math]\dfrac {d^2 \psi}{dx^2} = \dfrac {p^2}{\hbar^2}\psi\tag {1}[/math](Actually, I will switch the position of the [math]V(x)[/math] and [math]E[/math] inside the square root, and counter that change by putting a minus sign in equation [math](1)[/math]. Nothing is really changed, I just prefer it that way and I am too lazy to redo the LaTeX. Anyways, we now end up with [math]p(x) = \sqrt {2m(E-V(x))}[/math]).Now, let’s assume that [math]E \gt V(x)[/math] so that we can be in the “classical” region — keepin’ it real (pun intended :P).Great. Now that we have the ground prepared, we can begin by making our assumptions about the wavefunction.Typically the wavefunction [math]\psi[/math] is going to be a complex function, which we can express as a composite function of amplitude [math]A[/math] and phase [math]\phi[/math] — both being functions of position. So we can write:[math]\psi = \psi (x) = A(x)e^{i\phi (x)}\tag* {}[/math]Differentiate with respect to x, to obtain:[math]\psi'=(A'+iA\phi')e^{i\phi}\tag* {}[/math]Now differentiate again, to obtain:[math]\psi’’ = \left ( A’’ + 2iA’\phi’ +iA\phi’’ - A(\phi’)^2\right )\tag* {}[/math]Now let’s plug this into equation [math](1)[/math]:[math]\left ( A’’ + 2iA’\phi’ +iA\phi’’ - A(\phi’)^2\right ) = -\dfrac {p^2}{\hbar^2}A\tag* {}[/math]If you separate the real and the imaginary parts, you are left with two equations that look like this:[math]A’’ - A(\phi’)^2 = -\dfrac {p^2}{\hbar^2}A\tag {a}[/math][math]2A’\phi’ + A\phi’’ = 0\tag {b}[/math]The first equation [math](a)[/math] in general cannot be solved. And here’s where the approximation comes into play. We mentioned earlier that the potential [math]V(x)[/math] changes from place to place. If we assume that it changes slowly, then the term [math]A’’[/math] will become negligible. Therefore we can remove it form equation [math](a)[/math] and then solve for the [math]A(x)[/math] and [math]\phi (x)[/math] and substitute them back into the wavefunction and we end up with:[math]\boxed {\psi (x) \approx \dfrac {C}{\sqrt {p(x)}}e^{\pm\frac {i}{\hbar}\int p(x)dx}}\tag* {}[/math]Where of course the most general solution would be a linear combination of the above.You don’t really have to use the Schrodinger equation, I imagine the WKB approximation can be used with several other differential equations and in more dimensions as well. But in any case, that’s pretty much it. Any undergrad textbook in QM will have a chapter and a few examples on this, if you want a deeper understanding!

I don’t think that efficiency is a big issue as many have mentioned, not at least for non-racing uses.The belt drive for bicycles has its best application in city and touring bikes, where it provides a clean alternative to the open air chain, and it is much less bulky than a completely enclosed chain.It is true that belts are not compatible with derailleur gear shifters, but hub gear shifters are already quite commonplace among touring and city bikes so that would not be an issue.One neglected issue of belt drive is that the belt is made in a continuous loop that cannot be opened like a cha...