F.S Corso

Apuntes

F.S Corso al (2) M = 1500 kg F = ? T=125 F = MA A = 761.76 / 2 = 1000 / 125 A = 212.76 / 2 = 212.76 A = 2000 / 2 = 2000 F= 1500 kg (169.76%) = 25000 F= 1500 kg Peso del Hombre = 8000 P= 5000, Heller Corso = ? A = 1/2 M A = 5000 / M M = 1/2 M / 2 = 8000 / 96.76% R= 1/2 M A = 5000 / 96.76 = (8000 / 96.76%) F= Corso (F4) F = MA A = 4800 kg F = MA ... A = 20 kg MA = ? | La demanda con la 2de lugar de Nuestro A o (1/2) 1. Peso: MA = 1/2 → MA = 1/2 MA Justificación: Corso A = 1/2 M = M = constante Ma Aa = Ma Aa ma Ma (5000) = Ma (2000) Ma = 20 kg Ma = 20 kg = 4800 kg Ma = 1/2 F= Corso (F5) Ma = 1.09, Heller F = ? F= Ma Aa = (1.09) (4800 kg) = (4000 kg) F= MA = 2.0 kg F= 4.0 kg al ante F= 5.0 kg al ante T = 0.005, Heller T= (1800) al ante T= (1800) al ante F= MA = Ma Aa Aa = (1.09) + 4.0 kg 0.095 = (4000) 0.005 = (4000) F= Corso (F6) F= MA = 2.0 kg F= MA = 4.0 kg F= MA = 5.0 kg F= MA = 10.0 kg F= Corso (F7) F= MA = 2.0 kg F= MA = 4.0 kg F= MA = 5.0 kg F= Corso (F8) F= MA = 2.0 kg F= MA = 4.0 kg F= MA = 5.0 kg F= Corso (F9) F= MA = 2.0 kg F= MA = 4.0 kg F= MA = 5.0 kg F= Corso (F10) F= MA = 2.0 kg F= MA = 4.0 kg F= MA = 5.0 kg F= Corso (F11) F= MA = 2.0 kg F= MA = 4.0 kg F= ...