Question

Ayuda porfa es para mañanaa

Answer 1

SOLUCIÓN :

 2 ) Frq3 =?

       F23 = 9*109 N*m2/C2 * 6*10-6 C * 4*10-6 C /( 0.5 m)²= 0.864 N

       F13 = 9*109 N*m2/C2*2*10-6C *4*10-6C /(0.5m)2 = 0.288 N

     Ley del coseno :

      Frq3 = √ F13²+ F23²- 2*F23*F13 * cos α

      Frq3 = √( 0.864 N) ²+ ( 0.288 N)²- 2* ( 0.864 N *0.288N * cos 120º

       Frq3 = 1.038 N .

  3 ) Frq2 =?

       F12 = 9*109 N*m2/c2 * 3*10-9 C * 2*10-9 C /( 3.5 m )² = 4.408*10-9 N

       F23 = 9*109 N*m2/c2* 2*10-9 C * 1*10-9 C /(3.5m )² = 1.469 *10-9 N

      F42 = 9*109 N*m²/c2 * 5*10-9 C * 2*10-9 C / ( 4.94 m)² = 3.687*10-9 N

      d42 = √( 3.5²+3.5²  = 4.94 m

      Fr1 = √ ( 4.408*10-9N )²+ (1.469*10-9 N)²

      Fr1 = 4.64*10-9 N

    Frq2 = Fr1 - F42 = 4.64*10-9 N - 3.687*10-9 N = 9.53*10-10 N .