Question

3 questões de física...
preciso delas com cálculo pra agr de manhã pfvvv
Determinar força elétrica, carga e distância.
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Answer 1

Resposta:

Explicação:

$F = k\frac{Q_{1}.Q_{2}}{d^2}$

1)

$F = 9.10^9\frac{6,5.10^{-4}*3,2.10^{-3}}{2^2} \\\\F = \frac{187,2.10^{9+(-4)+(-3)}}{4} \\\\F = \frac{187,2.10^2}{4} \\\\F = 46,8.10^2\\\\F = 4,68.10^3\:N$

2)

$F = k\frac{Q_{1}.Q_{2}}{d^2} \\\\2.10^2=9.10^9\frac{3,5.10^{-6}*Q_{2}}{1,5^2} \\\\2.10^2=\frac{31,5.10^{9+(-6)}*Q_{2}}{2,25} \\\\2.10^2=\frac{31,5.10^{3}*Q_{2}}{2,25} \\\\\frac{2.10^2*2,25}{31,5.10^3} =Q_{2}\\\\Q_{2}=\frac{4,5.10^2}{31,5.10^3} \\\\Q_{2}=0,142.10^{2-3}\\\\Q_{2}=0,142.10^{-1}\\\\Q_{2}=1,42.10^{-2}\:C$

3)

$F = k\frac{Q_{1}.Q_{2}}{d^2} \\\\1,2.10^2=9.10^9\frac{2,5.10^{-4}*6.10^{-4}}{d^2} \\\\1,2.10^2=\frac{135.10^{9+(-4)+(-4)}}{d^2} \\\\1,2.10^2=\frac{135.10^{1}}{d^2} \\\\1,2.10^2*d^2=135.10^1\\\\d^2=\frac{135.10^1}{1,2.10^2} \\\\d^2=112,5.10^{1-2}\\\\d^2=112,5.10^{-1}\\\\d^2=11,25\\\\d=\sqrt{11,25} \\\\d=3,35\:m$