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Sekilde bir damperli kamyon ona ait bir kaldirma mekanizmast gorilmektedir. Mekanizma UZUv boyutlari ve 2 nolu uzvun aCISI sekil izerinde verilmistir: 014 aCISIL Fr...

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Sekilde bir damperli kamyon ona ait bir kaldirma mekanizmast gorilmektedir. Mekanizma UZUv boyutlari ve 2 nolu uzvun aCISI sekil izerinde verilmistir: 014 aCISIL Freudenstein denklemini kullanarak hesaplayiiz. 013 aCISII analitik geometri bilgileri ile bulunuz. Agirhgi P=I0 kN olan kasayi sekilde goriilen pozisyonda dengede tutabilmek icin yatayla $ = 1300lik act yapan hidrolik silindirin uygulamast gereken kuvvetin biyiikliiguni Analitik yintemle hesaplayiniz. Mekanizma UZuv boyutlari:

Sekilde bir damperli kamyon ona ait bir kaldirma mekanizmast gorilmektedir. Mekanizma UZUv boyutlari ve 2 nolu uzvun aCISI sekil izerinde verilmistir: 014 aCISIL Freudenstein denklemini kullanarak hesaplayiiz. 013 aCISII analitik geometri bilgileri ile bulunuz. Agirhgi P=I0 kN olan kasayi sekilde goriilen pozisyonda dengede tutabilmek icin yatayla $ = 1300lik act yapan hidrolik silindirin uygulamast gereken kuvvetin biyiikliiguni Analitik yintemle hesaplayiniz. Mekanizma UZuv boyutlari: 012 758 _ 0.6 m ; @2 0.9 m ; @3 0.46 m; a4 0.64 m; AoE 0.6 m; BoC = C2 0.5 m cudrot 012 &1 Bo_ Ao 7117TA Ehid



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Out of $\mathrm{TiF}_{6}^{2-}, \mathrm{CoF}_{6}^{3-}, \mathrm{Cu}_{2} \mathrm{Cl}_{2}$ and $\mathrm{NiCl}_{4}^{2-}$, the colourless $\begin{array}{ll}\text { species are : } & \text { [C.B.S.E. (P.M.T.) 2009] }\end{array}$ (a) $\mathrm{Cu}_{2} \mathrm{Cl}_{2}$ and $\mathrm{NiCl}_{4}^{2-}$ $\square$ (b) $\mathrm{TiF}_{6}^{2-}$ and $\mathrm{Cu}_{2} \mathrm{Cl}_{2}$ (c) $\mathrm{CoF}_{6}^{3-}$ and $\mathrm{NiCl}_{4}^{2-}$ $\square$ (d) $\mathrm{TiF}_{6}^{2-}$ and $\operatorname{CoF}_{6}^{3-}$

Question Number 45. The position function is given as X equal Do 9000 b squared minus 8000. Thank you. Now for the first part, Brenda Hey is Accardo zero point 0 to 5 Said guns. Then X is it will do 9000 indoor. See? You know point 0 to 5 is squired minus 8000 in new 0.0 25 you This is a photo four point play 75 Meet that Put it up. I'd be let us first point the acceleration function Acceleration is going through the double differentiation off the for Duchenne function. So exploration is equal to 9000 in Toto minus 8000 Sorry, it is 1000 Indo six in tow. Hey, this is the acceleration function now 40 equal to 0.25 The acceleration will become 18,000 minus 48 0000 Indo zero point 0 to 5 Solving this we will get the insulation. 6000 made that parts again Miss Well, this is the isolation. No using the pasta question off motion the velocity while living The end of the battle is we could do you which is zero less ex elation. 6000 in two time 0.0 five So that we lost Key is 1 50 me that but sickening in the Parsi The force at equal to zero is Mass 1.5 Katie In two x elation. We will put the value off people to zero in the X elation function. Then we get 9000 in Toto solving this we will get the force at equal to zero 700 zero, new done and the fourth at PayPal too. 0.25 The force is equal to Ford's equals to mass in through the X elation. We just 6000 solving this. We will get the force 9000 near done.

In this problem, I'm writing the reaction. Just look at it carefully. Any energy for at be your ford four H two will react to form any and at four at people for plus four as to and and ME energy for add bill for will react to form and the beauty plus an STD plus as to So according to the option. In this problem option. See each correct. Was there not that we have any Beauty which reacts with metallic oxide to give colored or to post fails.

So we have a, uh, sulfurous acid solution at 0.1 Moeller and we are given the two k is for the two protons associations. And we've got five choices for what would be true here. Now, uh, that first Proton is not gonna be strong acid. So it is not going to disassociate 2.1. So we're not gonna wind up with, um, Mawr than 0.1. Can't store geometrically wind up with more than 0.1 of the hydrogen sulfide ions. So that one is out. Um, this one, huh? Wonder about that is the hydro Nia, my on 0.1. Moeller. Well, I don't think so, but let's calculated and see if we wrote out the standard equilibrium and made the assumption that X would be small. Then we would wind up with X, which is the Hydro Nehemiah concentration equal to the square root of Casey and the hydro name ion. Concentration would be 0.36 If you did the quadratic formula and actually worked out, you'll find that it's actually 0.30 So yeah, we're not going to get point when Moeller uh hi. Drawing on my own. Ah, and In fact, we're above already 0.13 with the first proton. Looks like about 20 room three now. So we're left with these two choices. Are we going to get another, uh, 0.6 This again? That's 36% association. We calculated that assuming we dropped the minus X on the 360.10 Ah, but we can only do that if we have less than 5%. Association is 36%. So that is not the answer. But let's see why this is theano, sir. Now, when you take account of the first association, it produces again by the quadratic 0.30 of the conjugal base and 0.30 of the hydro Nia, my on. So when we start looking here at the K two equilibrium, those air our initial concentrations and then take the ice table from there and as we have seen before, then once we plug in, uh, we read the K expression from the reaction we plug in the expressions from the Ellen into the K expression well, where X is added or subtracted. It's probably small. We always check that we found here that it wasn't small, but here X is 6.3 times 10 to the minus eight Moeller, and that is the sole fight concentration. And that is the answer to our question. Both the additional age plus as well as the contra get base. Ah, in the second, equilibrium tend to be equal to or do you come out to be equal to the value of the equilibrium constant?

The solution off. The answer is there. That position off the chicken sized objects along the barrel is that is X equal to 9.0, multiplied by 10 to the power 3 m per second squared T square minus 8.0 multiplied by 10 to the power 4 m. But take in Cuba take you her tea at this time equal to 0.25 seconds. Now the length of barrel is AL equal to 9.0, multiplied by 10 to the power 3 m per second squared multiplied by 0.25 second hole square minus 8.0 multiplied by 10 to the power 4 m per second squared. Reply by 0.25 second Whole Cube. Now putting the values that X, we'll apply by 0.25 seconds equal to 9.0, multiplied by 10 to the power 3 m per second square 0.25 2nd square minus 8.0, multiplied by 10 to the powerful meter per cube meter per second cube multiplied by 0.25 second cube here, Alec called a 4.4 am now for me the velocity off object when it leaves the end off barrel that is vehicle to deep by DT off X. Now in this from the question that we to multiply by 9.0 multiplied by 10 to re power 3 m per second squared the minus 38.0 multiplied by gentry Powerful medical second cube T square hair Now solving this we get 300 m per second No part c there is mass equal toe 1.5 kg Now acceleration that is equal to D Y d v by duty had divided the off we is equal toe multiply by nine point zero multiplied by 10 trip our 3 m per second square minus six bracket 8.0 multiplied by 20 Power 4 m per second cube T at equal to zero equal to 1.8 multiplied by tended power 4 m per second squared so required force that is ethical toe Emma Here we substitute the value So the F is equal to 2.7 multiplied by 10 to the power for Newton at equal to 0.25 seconds. Here solving the equation, we get equal toe 6000 m per second squared. So required for that is F A ethical toe. M A is equal to 1.5 kg. Yeah, multiply by 6000 m per second square which is equal to nine multiplied by 10 to the power three Newton. So this is a detail solution step by step. Please go through this.


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