P1 2011 Solucao

André de Freitas Smaira\n1) r = (c / t) * x^2\n\t\nd = (1 / (4y^5)) * g\n\t\tx = (x' + v' * t) * g\n\n\tr_s = c^2 * (p' * v')^2 / y^2\n\t\tm^2 * (x')^2 v^2 \n\t\ty' = e^(c * p' / y \n\n\t\tg = p * (x')^2 * g\n\t\ty^2 * p' * v' * \n\t\ty_s = c^2 * (g' * y_s^2 / p^2 * e^(c * y))\n\n\t=> r = (cd) ** 2 - y^2 * (cd) * x^2\n\t=> A grandeza r é invariável do referencial\n\n2) y' = (y' * v' * g) ** 2\n\t\ty^2 = d^2 * l / (x^2 * y')\n\n3) Considerando x' = 0 → x_0 = lo * (1 - x)\n\tpara y = 0\n\t\t t = [3y_e / g] + √(1 - v^2 / c^2)\n\n\tAssumindo a diferença a partir do Lt → t = t_0 + t_s =>\n\tΔt = t_s + 1 / c * (v / c^2) y^2\n\n\tΔt = (g / (1 - v/c) / v^2) 3)\nK = energia cinética\n\tm * c^2 + m * (a_c) - K + P_{NF}\n\ts = p_a * cos(theta) + p_a * sen(theta)\n\n\to = p_a * sen(theta) - p_{NF}\n\n\tP_{sen}θ = p_{sen} * p_{n}\n\ttg θ_A = −lg on → θ_A = θ_N → |t| = |on|\n\n\tP_a * P_N = 0 → P_{s} = p_a - p_{N}\n\n\tE^2 = K = mp^2 a + k * L_0 \n\to = K_{n} = K_a K_{momentum} \n\n\t= 0 ⇒ mi = m^2 + 2m * m_a c^2 - 2mKNC^2\nm^2 + w_2^2 + 2(m_x^2 + K)\n\nc^2 * (v^2 - 2mv^2)\n= K = (m - m_a)^2 / 2m^4 4)\n\ny_3 \n\n\ny_4 \n\n\ns = (Ee.sig.m / ϵ_0)\n\tS: E = ϵ σ / θ_e\n\tB = 0\n\nS: j = dQ / dt → Q = ∫dF → I = ∫(σV / n)\n\t\t = q/s \n\t\th = L / (cг)\n\t\n∫(Bdl) = μ_0 I\n\t11B = (μ_0 / 11) * q * Qrm → B = (k * q * v_m) / r^2\n\n\tϵ = (Q * q / ϵ_0) / (r^2 * ϵ_0)\n\tϵ = (Qm / p_2 * ϵ_0) E = m0c^2, E = mec^2, K = (m - m0)c^2\n\nE^2 = m0^2c^4 + p^2c^2\n\nE^2 = m0^2c^4 + m^2v^2c^2\n\nE^2 = m0^2c^4 + m^2(v^2/c^2)\n\nE^2 = p^2c^2 + m0^2c^4\n\nNos casos em que m é o mesmo para a luz e para o neutrino, E = pc, ou seja, a energia passa a depender somente do momento da partícula.