Indices Fisicos - Jaque e Sarah

PUC Minas 21/09 Lista de Exercícios - Índice físico RA: 667860, 637009 → N: 5 1a) V = 5,98 x 10⁻⁴ m³ mₛ = 0,918 𝜌ₛ g = 10 m/s² m = 1,010 kg 𝜒 = 26,5 kN/m³ 𝜌 = ? e = Vv/ Vₛ Vₛ = 26,5 x 10³ = 0,918 x 10 Vₛ 26,500 Vₛ = 9,18 Vₛ = 3,46 x 10⁻⁴ Vv = V - Vₛ ë = 2,52 x 10⁻⁴ 𝑒 = 0,728 b) 𝜂 = Vv/V → 2,52 x 10⁻⁴ / 5,98 x 10⁻⁴ x 100 𝜂 = 42,19% c) W = Mw/M𝑠 Mw = M - Mₛ Mw = 92 g W = 92 / 918 x 100 W = 10,02% d) Sr = Vwu/ Vv Vwu = mw" + 92 cm³ 9,2 x 10⁻⁵ m³ / 2,52 x 10⁻⁴ x 100 Sr = 36,51% e) 𝜒 = M/V → 1010 kg x 10 / 5,98 x 10⁻⁴ m³ = 16,89 kN/m³ g = 10 m/s² 2a) 𝜒 = 19,5 kN/m³ 𝜒ₛ = 26,9 kN/m³ W = 29% Xₑd = X/1+u → 19,5 / 1+0,29 = 15,12 kN/m³ b) 𝑒 = 𝜒ₛ(1+u) - 1 X → 26,9 (1+0,29) - 1 / 19,5 26,9 x (1,801) - 1 / 19,5 = 0,779 c) 𝜂 = 𝑒/1+𝑒 → 0,779 / 1+0,779 x 100 = 43,79% d) Sr = w x 𝜒ₛ/𝑒 x 𝜒w → 0,29 x 26,9 x 100 / 0,779 x (1x10) Sr = 100,14% c) η = ? η = e / 1 + e η = 0.767 x 100 / η = 43.41% i + 0.767 d) Se es Vmu x 100 = Vmu = mu = Vmu = 20 g Vv V mu = 20 cm³ V1s = Vs x V Vv = 21.71 100 se = 20 x 100 se = 92.12% 21.71 e) γ = ? γ = m x g γ = 25 x 10 γ = 19 kN/m³ V 50 f) γd = ? γd = 19 γd = 15 x N/m³ 1 + w 1 + 0.267 01 a) Indice de vazios: * Mlw = Mtotal - Mconjuntoseco = 62.929 - 62.011 = 6.918 g * Ms = Mconjuntoseco - Mrecipiente = 62.011 - 35.046 = 26.965 g * Vs = Ms = 26.965 x = 9.46 cm³ ps 2.85 g/cm³ * M = Ms + Mlw = 26.965 + 6.918 = 33.913 g * p = M = 2.85 = 33.913 V = 33.913 = 11.89 cm³ V V 2.85 * Vv = V - Vs = 11.89 - 9.46 = 2.43 cm³ e = 2.43 = (0.25) 9.46 b) teor de umidade: w = Mlw = 6.918 = 0.2569 (0.2337) Ms 26.965 c) porosidade: n = Vv = 2.43 = 0.2044 * 20.44% V 11.89 08. V = 1000 m³ e = 1.20 *etl = Vv = 1.25 x 3000 - 1.0 = Vo = 7856.55 m³ Vs Vs **e = Vv = o,35 = Vv = 13.9372.97 m³ de aterro/ Vs Vv 1 7856.55 09. e = ? η = ? V - D³ Vs2 = π x D³ Vs= 0.524 D³ 6 V1 = V - Vs o³ = 0.524 D³ + Vv o,4760³ et = Vv ¿e = 0,476 D³ e = 0.91 Vs o.524 D³ η = e η = 0.91 x 100 η = 47.64% 1 + e 1 + 0.91 10. r = 5 cm e = 1.0 e' : 0.6 V = 4 π * r³ 3 V = 4 π (10)² = 523.60 cm² 3 3 e = Vv 1 e = Vv. 4488.2 523.60 ←Ⅴo = 261.8 cm³ Vs Vs 261.8 = Vv Vv = 418.28 cm³ e' = Vv ¯> 0.16 = Vv ¯>l 1.0 = 2bl.8 = Vv. = 418.28 cm³ Vo ∴ : O novo volume da esfera é 418.28 cm³ 11) γc=20kN/m^3 r=2,2cm γs=27,5kN/m^3 V=4πx r^3/3 Vs=4πx(0,022)^3/3 V=4,160x10^-5m^3 a) w=7 w=γs-γ/γs w=27,5-20/27,5 x100 w=27,3% b) e= γs (1+w)-1/γ e=27,5(1+0,273)-1 20 e=0,749 c) η= e/1+e η=0,749x100/1+0,749 η=42,8 d) γd=γ/1+w γd=20/1+0,272 γd=15,7kN/m^3