un conductor de plata (p=1,6x10°-6ohm x m) tiene una seccion de 5x10°-6 m°2 y una longitud de 110 m . calcular la resistencia
1 answer:
Responder:
35,2 ohm.
Explicación:
Dado:
La resistencia específica del conductor es,
La longitud del conductor es,
El área de la sección transversal del conductor es,
Sabemos que la resistencia de un conductor es directamente proporcional a su longitud e inversamente proporcional al área de la sección transversal.
Por lo tanto, la resistencia se puede expresar como:
Ahora, conecte los valores dados y resuelva para 'R'. Esto da,
Por lo tanto, la resistencia del conductor es de 35,2 ohm.
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Answer:
charge C = greatest net force
charge B = the smallest net force
ratio = 9 : 1
Explanation:
we know that in Electrostatic Forces, when 2 charges are at same sign then they repel each other and if they are different signed charges then they attract each other
so as per Coulomb's formula of Electrostatic Forces
F = .....................1
and here k is 9 × N.m²/c² and we consider each charge at distance d
so two charge force at A to B is
F1 =
and force between charges at A to C, at 2d distance
F1 = =
force between charges at A to D, 3d distance
F1 = =
so
Charge a It receives force to the left from b and c and to the right from d
so at a will be
F(a) = -F1 - F2 + F3 ....................2
put here value
F(a) =
solve it
F(a) =
F(a) = = 1.13 F1
and
Charge b It receives force to the right from a and d and to the left from c
F(b) = F1 - F1 + F2 ....................3
F(b) =
F(b) = = 0.25 F1
and
Charge c It receives forces to the right from all charges.
F(c) = F2 + F 1 + F 1 ....................4
F(c) =
F(c) = = 2.25 F1
and
Charge d It receives forces to the left from all charges
F(d) = - F3 - F2 -F 1 ....................5
F(d) =
so
F(d) = = 1.36 F1
and
now we get here ratio of the greatest to the smallest net force that is
ratio =
ratio = 9 : 1