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Ex ; Ce ,T) 0 mea Sia 5 0 Space to 6 exe mkaQ saseb t ASSuwe _ 4A €t ea GuaL Oa 3 Rc A k k 4T ale m - 6 not SFow #t 6 = m& AxB Ie) meaSiaLl - JialCs KSBA ...

Question

Ex ; Ce ,T) 0 mea Sia 5 0 Space to 6 exe mkaQ saseb t ASSuwe _ 4A €t ea GuaL Oa 3 Rc A k k 4T ale m - 6 not SFow #t 6 = m& AxB Ie) meaSiaLl - JialCs KSBA V Oea 0 Qe

Ex ; Ce ,T) 0 mea Sia 5 0 Space to 6 exe mkaQ saseb t ASSuwe _ 4A €t ea GuaL Oa 3 Rc A k k 4T ale m - 6 not SFow #t 6 = m& AxB Ie) meaSiaLl - JialCs KSBA V Oea 0 Qe



Answers

As before, $T=e E x$ Now in linear motion, $$ \begin{gathered} \frac{d}{d t} \frac{m_{0} v}{\sqrt{1-v^{2} / c^{2}}}=\frac{m_{0} w}{\sqrt{1-v^{2} / c^{2}}}+\frac{m_{0} w}{\left(1-v^{2} / c^{2}\right)^{3 / 2}} \frac{v}{c^{2}} w \\ =\frac{m_{0}}{\left(1-v^{2} / c^{2}\right)^{3 / 2}} w=\frac{\left(T+m_{0} c^{2}\right)^{3}}{m_{0}^{2} c^{6}} w=e E \\ \text { So, } \quad w=\frac{e E m_{0}^{2} c^{6}}{\left(T+m_{0} c^{2}\right)^{3}}=\frac{e E}{m_{0}}\left(1+\frac{T}{m_{0} c^{2}}\right)^{-3} \end{gathered} $$

We are given a position function. And we have two parts that were asked to find first is defined S and V. When the acceleration is zero. And the second part is to find S. N. A. Position and acceleration When the velocity is zero. So we can tell that we are going to be using the velocity and acceleration functions. The velocity function we could find by taking the derivative position. So that would be three T squared minus 12 T. And the acceleration function we can find by taking the derivative of velocity And that would be 60 minus 12 for part A. In order to find the position and velocity with the acceleration zero, we're going to set the acceleration to equal zero. So we can find the time where that occurs. We'd add the 12 over and we have a T value of two. So the time is too the position function at that time can be found by simply substituting and two for tea. That would give us eight minus 24 Plus one. That's -15. And the acceleration. I'm sorry. The velocity at two seconds Can be found by substituting in two for tea. In the velocity function, That would be three times 2 squared minus 12 Times two. That would be 12 -24 and that's -12. Now for the position acceleration when the velocity is equal to zero, Let's set the velocity function equal to zero. And there's actually two solutions here. We can factor that out by taking a three T out fronts three T, zero and T is zero and t minus 40 when T. S. Four. So there are two different times we're looking at here. The position when the time is zero. If we fill zero and for T. That would give us one And the position at four. We fill four in for tea. That before cubed -6 times four squared plus one, That would be 64 -96 plus one. That's negative 31 for the acceleration at those times The acceleration of zero, if I fill zero in for tea, that will give me 0 -12 or negative 12. And the acceleration at four would give me 24 -12 or

Hi friends. As initially you Raises zero, so total energy is equal to kinetic energy, Kandi Technologies to find us half milk. We went by next week, magnitude squared as we want is perpendicular to be too so kind of technology you will get. Uh huh. Um, I am too over a month plus M2. We went a square Plus B2 sq mm. That's what. Thanks for watching it.

In the question. The given differential question is, have you nowhere left less? Six by t No news from the lash plus four by these were no blue upholstery zero we have, ah, hidden Hoshi. You let we will multiply. But peace Where do they Will it work with peace with the blue Devils slash place six e w nash less four w equals zero Therefore, which is a solution off the form Why equals two p to the power are my Nash equals two r here to the power R minus one on why never lash would be Puerto are R minus one and compared to the power on minus do now, by substituting the value we would get artist where minors are less six are less for equals 20 that is already square bless five are plus four equals +20 Now, using the Madonna spirit mattered our class full and you are plus one would be Crippa zero that is are well equipped to minus four on our big part minus one. So the given general far so be a general form for the given differential equation would be why it was to see one here to the power miners for last Tito paid with the power in my nest. My thank you

In this problem I can write the value of F is equal to divide ET I am not be bye. Underwood. One minor piece square by C square which is equal to m not be by Underwood. One minus B squared by C square plus. I am not B Bye C squared be. Don't be multiplication one by one minus B squared by C square. Or to the power three by two. Now I can write the value up. Ap perpendicular is equal to I'm not Omega by under route 1 -1 beat. A square. Omega is equal to be when omega is but man color to be apple is equal to. Am not Omega by 1-. We tie square to the power three x 2. Omega is equal to P and omega is pedal to be RP omega each. Well ER two p. So this is the final answer.


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