Física Básica Vol 2 resolução 4a Ed

H. Moysé s Nussenzveig\nFluidos\nOscilações e Ondas\nCalor\nCURSO DE\nFÍSICA BÁSICA\nedição revista 1)\n\\[ z = x^2 + y^3 + 0 \cdot g \cdot e \cdot h \Longrightarrow \text{los valors:}\ n = 3.14 \ m \ \Rightarrow h = 3.14 \ m \ s \ \text{donde} \ n = 3.94 \ m \ \]\n\n\n\nPREGUNTA 7\n\n\n\\[ TEORÍAS: \ E_{1} - E_{2} \Rightarrow Mq \Longrightarrow mg = \\frac{0}{h^{2}} \\frac{h}{H} \Rightarrow \\frac{A_{1}}{A_{2}} = \\frac{y(v^{2} - h^{2})}{D^{2}} \]\n\n\nD)\n\\[ m = \rho V = \frac{\\rho V}{4} \] \n\n\n\\[ S = \\epsilon Lz \\times \\frac{mdz}{\\pi} \\Rightarrow H = \\frac{\\rho}{D^{2}} \]\n\n2)\n\\[ \\Delta P = F = \\frac{\\pi d^{2}}{A} \\ \]\n\n3)\n\\[ F = \\pi \\Rightarrow (0.33)^{2} 0.9 \\cdot 0.13 \\cdot 10.5 = 9.8921 \\text{ para } 1000 kg \Rightarrow M = 1000 N \]\n\n4) \n\\[ \\Rightarrow p_0 = p_0.cL \\Rightarrow c = 12.50 \cdots 13.80 \mid 1.05 \]\n\n5) \n\\[ P_{\text{total}} = \\textrm{la misma longitud, la misma fuerza desde el borde, los líquidos son}\\text{son defendidos} \]\n\n\nTEORÍAS \n\\[\\rightarrow \\rho_{0}\\Rightarrow 0.72 = 0.897 = 87\\% \]\n\n\\rightarrow f = \\frac{0.92}{D_{1}} = t_{\\text{avg}} \n\n\\rightarrow p_0 / P_{h0} 1) PREGUNTA DE CUESTION DIFÍCIL \n\\[ T_{\text{no}}: E_{1} / A_{1} = \\frac{Mg}{A_{2}} \]\n\n2) \\[ Z = \\frac{m - 0}{m} = \\frac{d}{2h} \\rightarrow Z = 4 \]\n\n3) \\[ m = \\sum_{tio - h_{c}}^{m} E \cdot P + h^{2} = 9.3333N = (D_{1})^{2}\over2D^{2} \]\n\n\\[ F_{E} = \\frac{F D^{1}}{F_{1}} = E (h - h_{E}) \\rightarrow \\Rightarrow 9999 \Rightarrow E = l \]\n\nPueda que el flujo o la sección transversal sea \n\\[ \\Rightarrow F = p_{\\rho} g \cdot V_{s} \\Rightarrow V \\cdots \\rightarrow R = \\] \n\\[ \\Rightarrow \\text{ RES} . 89.7/ \] \n\n\\[ \\text{RES} = \mid 4\\\\ 0\\backslash \\Frame = \\] \n\n\n 12. E = E + F = P \n\\Rightarrow \\rho \\downarrow f_{g} = H_d P \n\\Rightarrow mg = 0.975 \cdot 9.2 = 0.156 \\cdot 10^{-4} \cdots \Rightarrow 3.15 = 0.03 \cdot (0.3 \cdots ) \\Rightarrow 10.500Kg/m^{3}\\ \n\n\\text{estra}\\ \\\ [ R = \\text{onda} F = \\sum \\rightarrow \left( gm \\cdots \\Rightarrow f_{P} = \\cdots p \\\ \] \n\n\\Rightarrow \\pi = \\frac{m - 0}{y} \rightarrow \\text{sumas)} \\cdots\\backslash L_{P} - L_{R} + L\\)\n\n\\Rightarrow \\frac{\\rho}{g^{m}}\cdots p\\\right( \cdots \\cdots ) = \\] \\Rightarrow \\rightarrow { 4.125Kg \rightarrow }~ \\text{RES:} 0.95 Kg \text{ a l}+ \\\ 950 kg/m^{3}\\ [=\\sqrt{g + \\cdots = \\cdots \\cdots } \] 11.10.07\n\nb) T = P3 - E1 - E = -6.203. 4/3 π (0.13)3.9E - h0.2 2/3 π(0.13)2.9E((7/3) r^2(1.93)2)\n\nT = 2 π (0.13)2.9E (6.2932.10-10.92) = 410.6N\n\n\n2. Divisão\n\n\nE:\nFaz = E - P\n\nP = P0 = 4.29. π. 3/p. 3.9E = P. E = 5\n\nFaz = 12.5.T.98 = 4.29-0. 6(1.58N = 629Kg\n\n\nRES: 628kgf\n\n\n\n\n12. dp2 = -g dp = 0 => -ch2 => dz = -g dt => dz3 = 0 => ch3 =>\n\ndz3\n\n3\n\nd3\n\np = 1\n\n\n\n\n<p0 - p0g cosθ\n\n0\n\n...\n\np0 + h + d0\n\n+cal2\n\n\n\n\n\n20. M. Venção) Volume do chumbo: M = 10 - 8.7.10-4 m3\n\nChumbo = 11400\n\n\n=> Volume do alumínio? M = 10 - 3.7.10-3 m3\n\nAlumínio 2700\n\n\n\nparêntese\n\nM => 11400.97.10-4.98= 1200.37.10-4.9 *s = 86.49N\n\nparêntese\n\nM => 2300.37.10-3.9= 1200. 7.4.10-3.95 = 54.12N\n\n\n Alumínio. Pesca média: m = 3.58Kg Lição:\n\ndx/(po g)d(z)\n\n1/2 A(....)\n\ndx = (p0 g)(dz)\n\n\n dz = (p0 g)(dz)\n\n\n\nd 2.\n\nGo: E1 = dz A = dz= 0 =>\n\n3 A\n\nF = 3, 4\n\nF0 =0=>\n\nF = 0 =>\n\n...\n\n...\n\n17.\n\nA = p0 g(z)\n\nB\n\n\nRho = p0 g(sez)\n\nP0V0 = P1 V1 = 0 =>\n\nV0 = P1V1/P0\n\nPa x (h0 = Pz - x1) +Pg [g(h1 + dx)]\n\nPa(x) = (p1)(x)(54.4x 2)\n\nh0 = 255.12 - 7.03x + 49.2x = 0 => o.j.27x2-203x + 19R\n\nx - 0,31 = -2.43(p)(3.9) - 4.9; 235.7 = 207,7 + 184.3 = 1499 m\n\n\n\n40% de V0. INSTITUTO FEDERAL DE EDUCAÇÃO, CIÊNCIA E TECNOLOGIA\n\nMaranhão\n\n\n\n H\n\nh\n\nr\n\n x(t) = x 0 + [P] · cos(ωt + φ) \n\nx(0) = 0; a 0 = \n\n\na(i) = N = 3 = p o = 0 \n\n\nQuando (n) \n\nx[0] = \n\n√ \nh\n\nf(x 1 / 5) (k)\n\n= 1/2O\n(x)\n\n\na - g \n\n... T \(kg \) \n\nbe 1(k) \n\nE = k\n\ngx e\n\n= 2 K \n\nxE = (I/2) (1*{0.6e3 * 5 + 1} +\n\nx\n= (m g) + x/y2 2)( (2.1...\n\n+ √4.10² \ns = 50 μes\n H. Quando a conta é eliminada de um ângulo O, teremos\n7. -mg sen θ\n\n8. I - m g sen θ\n\n\na[i] (mg) [...] \n\nPortanto, θ! θ \n\na[I](m g/d)ãi[0](g) \n\na Equação (g) corres\n\n- \n\nDe b equação do movimento harmônico simples, no \n\na mesma | g ..... [...] \n\nS.\n\nA Equação no movimento será: x(t) = ρ cos(ω t) + mg\n\n\ng \n\nx(0) = 0 \n\nx(t) = N 2g \n\nt \n\nx'(t) = N/2g h - ω ρ sen(ω t)\n\nω; (I) - m g cos l ω + .........\n\n(I)\nm ω²sen(ω t) + ω² sen 2 a \n\nFazendo (2) + (I), ω²ρ(mg) e\n\nk = \n\nx\n\n0 \n\nz\n\n2z h \ng k²\n\n---- \n\n√mg + √kg\n\n I = 1/2 k(θ²) = 1/2 (mg)² / (1 + z k h^(γ)) \n\nk m \n\n1 = 1 / 2 k = m \n\nDou T= K o = T.D \n\nS =→ (MT)\n\nS = T O a = (M) \n\n=\n1 HR²\n\nM = d m 2π²/ I R²\n\n---\n\nLogo: m(g) = \n\n- \n\n(k) H0/(k²)\n\nPd \n\nk = N R²\n\n|| R = mg/K\n\n-------\n\n ANTES\n\nDETALHS\nA\nA\nB\nL\nD\nL\nf\n\nTt\n\nTENSÃO I 2 √\n\nMA\n/ 2²\n⁄M²\n\nH\nH\nˊ\n⟺ W\n\n3\n⟺\n\n 2\n\nL\n12\n/ L × 3 / 2\n\nT =\n\nL0: T² = 4π/I\n\nT1 =\n\n2g\n(20)\n\n\nX\n1Ta\nx\na\n\nP` = Y + 2/4T + 2g(T2) y ANTES\n\nDETALHS\nA\nA\nB\nL\nD\nL\nf\n\nTt\n\nTENSÃO I 2 √\n\nMA\n/ 2²\n⁄M²\n\nH\nH\nˊ\n⟺ W\n\n3\n⟺\n\n 2\n\nL\n12\n/ L × 3 / 2\n\nT =\n\nL0: T² = 4π/I\n\nT1 =\n\n2g\n O momento de inércia I é dado por:\nI = I09 + MP * MS² - MJ * A12HZ²\n12\nI = G = 2π √(I²/12g)\n(11)\nL0\nT² = 4π²/I * 12g\n1/6(12 ² 12²)\n-----------\n√(12gS)\n---------------------\n(12gS)\n\nL0: 24s² * 0² * 12s² = 9 - s²/\n2N²\n\nSUGESTÃO PARA INCLUIR EM TENSORES T = JNA √(12g ²)\n\n16 BOCA DA MINHA, AO PONTE DOAS, MINHAS ALTACOS COMO INDICANDO:\n\nSOMENTE A RESISTÊNCIA FONTE PELS MIIA.\n\nA posição do c = A BOMB SE RECUPTS.\nL1\ng\nh 2x\n√T₀ = √R/4\nT\nT\nT.\n\nTENSÃO I 0 * 8 = H₀ * MA/ (senoθ)\n\n4°\n(mg√R 3) (Ō) \nL0\nT * 2π = 4π\n√(9 / 3g)\n 58\n 22 17.\\[x(t) = P(cos (\\omega t + \\phi))\\quad x_{(t)} = \\mu (t) sin(\\omega t + \\phi)\\rightarrow\\text{ S} \\to k_{v}\\geq k_{1}\\]\n1. A Energia Cinética será dada por \\[K_{f} = \\dfrac{1}{2} m \\mu^{2} \\cdot [\\sin(\\omega t + \\phi) ]\\]\n\\[\\therefore\\left( \\sum_{l} \\,,\\cos(\\phi) \\right)\\rightarrow\\frac{3}{\\sin(\\phi)}\\]\n\n\\[\\text{a. função que ini} K_{\\text{K.}} \\frac{K_{f}}{m}\\]\n\n\\[W_{0} =W(o )\\sqrt{\\dfrac{\\text{A}.\\text{g}}{\\text{T}}}\\]\n\\[K^{2}\cdots T_{1} = m\\left( \\frac{g_{0}}{4} - \\frac{4}{cerque}\right)\\] Usando (1) E (1):\n\\[{Z_{res}} = Z_{m} + Z_{p}\\]\nZ_{p} = \\left( \\frac{\\sqrt{Z_{0}^{2} + m_{f}^{2}}}{4}\\right) = I \\cdot Z_{(ss)}\n\nIM (1) / TORC\n\\[\\left( \\frac{\\left| K P(m|\\right|\\right)\\right)_{0} = 0 \\rightarrow \\Rightarrow \\cdots \\to K^{2}\\cdots m_{f}\\left( m\\right)\\right]}{4}\\]\n\n1. M\' AAC\\[Teórico| M.T.H\\left( 2 \\right)\\]\n\\[\\left( K^{2} \\right)> K_{\\text{res}}\\cdots K_{0}\\right]\\left( 4\\right)\\]\n\\[m^{4}\\cdots K_{m}^{2}\\cdots \\Rightarrow\\based{\\cdots}\\mid\\frac{g^{2}}{4}\\] (NOYÉS)-CAPÍTULO 4\nExercícios Oscilações Forças e Aprofundamentos\n\n1. $\\dfrac{4.2.173}\\leq\\dfrac{K_{f}}{6}$;\\quad$\\dfrac{4.2.207}{x}\\;\dfrac{t^{2}}{m}$\n\n\\[\\left( 41-10\\right) \\left( \\left( \\dfrac{z}{0}\\right) = 0\\cdots \\div |K_f^{2}v| = F_{0}\\div |\\{\\text{Converte, unindo formas: |16'}\\}\\right)\\end{array}\\]\n\n2.0\\\n\na: Raiz do Polinômio: \n\\[2\\text{Por } (\\cdots /vs\\text{interface: Teor}A}\\]\nb. \\[C\\text{ida 2.0evaluation}\n\\] 2. q = dh = W_0 \\cdots x(0) = m + \\cdots \\frac{X(g\\cdots \\sin \\omega t)}{XLL(\\frac{})} + 0\\,\n \\log\\left( x(t) = a\\left( g \\cos(t) - \\cdots \\sin\\cdots w \\cdots t \\right)\\right)\\,\n y \\cdots x(0), s = 0\\,\n \\left(1\\;5\\;t\\right) = \\frac{5}{- \\omega}\\\n \\text{Tema: } = \\frac{d\\cdots}{\\cdots } = A_b\\,\n \\log\\,x(1) = 5*sin(t)\\,\n dS = -4l_{mot}(y)\\,\n 2D = L (d f - 2\\cdots)\\,\n de - 2m\\ / dx^2\\,\n \\left( y(t)\\right) = \\frac{\\text{R}}{0}\\,\n d \\cdots {\\} \\\\\\ 4. Ecuación de movimiento: \\(\\frac{d^2y}{dt^2} + k^2y = 0\\)\n Tema: x(0) = 0.\n \\(x(t) = h_0 e^{\\frac{1}{2} i\\omega t}\\left[x(0)v_0(t-s_0) - v_0(0) s_y(t)\\right]\\)\n \\left|\\left| x(t) \\right|\\right| = V_{0}^{2}\\\n\n a) \\(x(t) = x(1) - 3\\frac{6}{\\rho} \\int_{0}^{y-x} f(t) dt\\)\n P. Plazma o Texto que Francez \\(\\int_0^\\infty e^{-\\frac{t}{T}} dt\\)\n\n H. Luego \\(\\frac{d^2x}{dx} = \\frac{dy}{\\frac{dy}{dt t}}\\)\n\n b) \\(x(0) = x(2) -\\frac{t^2}{2}\\\n \\left | \\left | x(t) \\right | \\right | = \\exp{\\left[\\ln\\left|bh\\frac{}{}=x\\right|\\right]}\\)\n 5. m\\tilde{g} =\\, \\int_0^t\\left[y\\frac{f}{x}\\right] \n Relación experimental: \\(x(z)\\, ver\\, i = sin \\frac{(01) + (03)} dxx g\\)\n Solución (1) \\(f(x) =\\alpha e^{\\beta t} - A g\\)\n Tema: 2\\;.x(1) = A + 9^x - c\\;\n \\left(\\frac{}{}\\right) \n 6. Yo P.R.\n \\(m{x''} - mg \\cdots x = y = \\frac{j}{h}\\)\n Ecuación homogénea\n \\(x''y \\cdots 0 = P(x) = \\alpha\\left[y^3 x^{(\\frac{3}{2})}\\right]\\)\n Solución da Ecuación Homogénea: \\(s^2 = A\\, B e^{-t}\\)\n\n Posible solución particular: K + (u)\n \\(0 + \\cdots u\\)\n \\(y(x)\\,, K\\).\\;\n \\left(\\frac{y^{(0)}}{k} - g\\frac{y^{(1)}}{w}\\right)\\to\\;\n \\left ( i_x \right)\\,\n \\left [\\, x(1) + {g \\neq = 0} \\right]\\, (1,0)\\,\n \\log\\left| x(t) = y_0/x\\, y_0 = 0 \\right| = \\frac{y(u)} + \\frac{x}{x} + \\int ydT\\)\n \\,,\\,\\, \\, dA = \\frac{1}{y} + \\gamma\\;\n \\cdots\\,\\,\\,\n 7. \\( \\ddot{x} + \\gamma \\dot{x} = \\frac{F_0}{m} \\sin(\\omega t) \\)\n\nSistema homogêneo: \\( x'' + \\gamma x' + \\omega_0^2 x = 0 \\)\n\n\\( P(\\alpha z^2, \\beta z^2) = P(\\alpha, \\beta) \\) \n\nSolução do sistema homogêneo: \\( C_1 \\cos(\\omega_0 t) + C_2 \\sin(\\omega_0 t) \\) (1)\n\n\\( \\text{Substituindo:} \\quad N_1 + A_{11} \\ddot{x} + B \\cdot x = \\frac{F_0}{m} \\sin(\\omega t) \\)\n\n\\( \\quad \\quad \\quad N_1 = B \\cdot \\sin(\\omega t) \\)\n\n\\( \\quad \\quad \\quad A(\\omega^2, B) \\to B = \\frac{m F_0}{m \\omega^2 + \\gamma^2} \\)\n\n\\( \\text{Obtemos: } X(\\omega) = 0 = C_1 \\)\n\n\\( \\Rightarrow: \\; x(t) = C_2 \\omega_0 \\sin(\\omega t) + F_0 \\sin(\\omega t) \\frac{1}{m(\\omega^2 - \\omega_0^2)} \\)\n\n\\( \\Rightarrow: C_1 = \\frac{F_0}{m(\\omega^2 - \\omega_0^2)} \\)\n\n\\( \\text{Com os coeficientes obtidos:}\\)\n\n\\( x(t) = C_1 e^{-\\beta t} + \\frac{F_0}{m(\\omega^2 - \\omega_0^2)}e^{-\\beta t} \\) \n \n \n \n \n \n