Solucionario Fisica 3 Examen Parcial 2024-1, Exámenes de Física

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PA UNIVERSIDAD NACIONAL DE INGENIERIA is FACULTAD DE INGENIERIA QUIMICA Y TEXTIL y DEPARTAMENTO ACADEMICO DE CIENCIAS BASICAS Bl SOLUCIONARIO Solución 1 Ser UP: First find the current. The equivalent resistance across the bartery is 30.0 8), s0 the total current 154.00 A Malf o£whuch goes through the bar. Applying Newton 's second Law to the bas grves DF =ma =F, mg —ILB=mg EXECUTE: Solving for the acceleration gives am tLB—mg _ (20AX1:50mX1.60 T) - 300N m (3.00 N/9.80 ns”) -5.£8 0/5 The direction is upward. EVALUATE: Once the bars free of the conducting wires, its acceleration will become 9 8 1019 dowwxard since only gravity will be acting on it. Solución 2 28.63. IDENTIFY: Apply Y F =0 to oue of the wises. The force one wire exerts on the otuer depeads caso Y F =0 gives nvo equations for the nuvo unknowns 7 and. SETUR: The force diagram for one of the wires is given in Figure 28.63. Ton 8 pr T : . Sl "The force one wire exers on the oteris F | LL), E z 247 0 Tsinó where 7 =2(0.040 m)sin 6 =8.362x10” m is the | distance berween the Tio wizes. Figure 28.63 28-20 Chapter28 Extcure: DF =0 gives TcosO=mg and T =mg/cos0 NA -0 gives F =Tsinó =(mg/cos6)sinó =mgtanó And m=AL, so F=2Lgtanó Pur (22) atsuno 2ar 1. fisrens (4/27) l- [(0.0125 kg/mx(9 80 m/s)'(ran 6.00*X8.36210” m) y 2:10 T-vA =232A Solución 3