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Tn - general term M - sign for sigma Sn - sum of n termsTn = (M n^2)/nTn = ((n+1)*(2n+1))/6Tn = (2n^2 + 3n +1)/6Sn = M TnSn = 2*(n*(n+1)*(2n+1))/36 + 3n*(n+1)/12 + n/6
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Sn = (n*(n+1)*(2n+1)/18 + n*(n+1)/4 + n/6
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You have to find the coefficient of x^8 in (x^2 + x + 2)/x^6.
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It depends on the sum of sine of the angle subtended by the end points of the wire at the point were the electric field is to be found. For the derivation of the required expression go here :- http://cnx.org/content/m31103/latest/
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That itself is simple. Post the step which is not clear or should I put it in a simpler form. Procedure would remain mainly the same. It is very correct. He has just used the basics.
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Multiply all the given terms. you will get (abcd)^2 = (2*4*8*16) = 1024. Take square root. You get abcd = 32.
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Yes. Very very important . Many times you will have a word reaction having common names of the compounds.
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Can you post a diagram for it.
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Let the point be at xm from the 20kg mass. We have to balance the two torques about this point for getting a horizontal position of the rod.20x=30(5-x) 5x=15 x=3 So, the point is at a distance of 3m from the 20kg mass or at 2m from the 30kg mass.
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If you are thinking that it is because of the negative charge then stop. It is because an electron has a very low mass as compared to that of a proton (mp/me = 1,836.15267245). Now, for a charged particle moving in a magnetic field the expression for time period is T = 2*pi*m/Bq (symbols have their usual meaning). Now the mass of electron is too less. This gives us a very small value for the Time period. For changing the polarity at regular intervals we use an instrument called HFO (High frequency Oscillator). It's almost as big as a big generator. This device has to change the polarity at soon as the electron/proton comes out of the dee's (magnetic field). Even the best of HFOs can't achieve the values even closer to the time period obtained when using an electron. But in the case of a proton the value of time period is achievable. So, an electron can't be accelerated by this machine. But a heavier negatively charged particle can be accelerated it.
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Torque (T) = Moment of Inertia (I) * angular acceleration (@) @ = {w(final) - w(initial)}/t(change of time) @ = (10-2)*2*pi/4 @ = 4*pi T = 112.5*4*pi T = 1413 Nm.
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If You Like Please Take Time To Share! Comments are welcome!
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The general term is (2k-1)^3. k is a Natural number(M is the symbol for sigma) Summing this from k=1 to k=n.You get M 8*k^3 - M 12*k^2 + M 6*k - M 1On solving this you get 2*n^4 - n^2 n^2(2*n^2 - 1)--------------------------------XXXXXXXXX---------------------------
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No. The correct answer is x E nπ ; nEZ (E is symbol for belongs to). You have open the 3x and 2x terms and simply reduce them. I reduced them till here (tan(x))^5 = -(tan(x))^3 Which is possible only when each term is zero. So x E nπ ; nEZ (Solve the equation, you will get the same result.)
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Let v= dx/dt. The d2x/dt2= dv/dt= (dv/dx)(dx/dt)= vdv/dx
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Sorry, I did not know that H3PO3 is phosphorus acid ( I knew it by its other name phosphonic acid). Let me share something (Everyone does not know this) - H3PO2; Hypophosphorous acid or Phosphinic acid, H2PO(OH), monobasic acid, contains P in oxidation state +1
- H3PO3; Phosphorous acid or Phosphonic acid,HPO(OH)2, dibasic acid, contains P in oxidation state +3
- H3PO4; Phosphoric acid, PO(OH)3, tribasic acid, contains P in oxidation state +5
- H3PO5; Peroxomonophosphoric acid, OP(OH)2OOH, tribasic acid, contains P in oxidation state +5
- H4P2O6; Hypophosphoric acid, (OH)2P(O)−P(O)(OH)2, tetrabasic acid, contains P in formal oxidation state +4
- H4P2O7; pyrophosphoric acid, (OH)2(O)P-O-P(O)(OH)2, tetrabasic acid, contains P in formal oxidation state +5
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I have seen a number of my friends in a situation much similar to yours.Follow what the lady above told. There's one thing that I believe in : HAATHI CHALE BAZAAR TO KUTE BHAUNKE HAZAAR. Ignore them.
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From the work energy theorem, 1/2mv^2=Pt (Energy) v@t^1/2 ; as the Mass of the body and Power applied are constant dx/dt@t^1/2 dx@t^1/2dt Integrating, x@t^3/2 @ = Symbol for directly proportional
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