Question

Duas partículas de cargas elétricas Q1 = 4,0 × 10 16 C e Q2 = 6,0 × 10 16 C estão separadas no vácuo por uma distância de 3,0 × 10 -9m. Sendo k = 9,0 × 10 9 N.m2/C2, a intensidade da força de interação entre elas, em
newtons, é de

a) 2,4x10^-5
b) 2,4x10^-3
c) 2,4x10^-6
d) 2,4x10^-4
e) 2,4x10^-2

Answer 1

Explicação:

$\sf F_{el}=\dfrac{k_0\cdot|~Q_1~|\cdot|~Q_2~|}{d^2}$

Temos:

• $\sf k_0=9\cdot10^{9}~N\cdot m^2/C^2$

• $\sf Q_1=4\cdot10^{-16}~C$

• $\sf Q_2=6\cdot10^{-16}~C$

• $\sf d=3\cdot10^{-9}~m$

Assim:

$\sf F_{el}=\dfrac{9\cdot10^9\cdot4\cdot10^{-16}\cdot6\cdot10^{-16}}{(3\cdot10^{-9})^2}$

$\sf F_{el}=\dfrac{9\cdot4\cdot6\cdot10^9\cdot10^{-16}\cdot10^{-16}}{9\cdot10^{-18}}$

$\sf F_{el}=\dfrac{216\cdot10^{9-16-16}}{9\cdot10^{-18}}$

$\sf F_{el}=\dfrac{216\cdot10^{-23}}{9\cdot10^{-18}}$

$\sf F_{el}=24\cdot10^{-23-(-18)}$

$\sf F_{el}=24\cdot10^{-23+18}$

$\sf F_{el}=24\cdot10^{-5}$

$\sf \red{F_{el}=2,4\cdot10^{-4}~N}$

Letra D

Answer 2

°•Oieeee Td bom

$Fel=d2k0⋅∣ Q1 ∣⋅∣ Q2 ∣</p><p></p><p>Temos:</p><p></p><p>• \sf k_0=9\cdot10^{9}~N\cdot m^2/C^2k0=9⋅109 N⋅m2/C2</p><p></p><p>• \sf Q_1=4\cdot10^{-16}~CQ1=4⋅10−16 C</p><p></p><p>• \sf Q_2=6\cdot10^{-16}~CQ2=6⋅10−16 C</p><p></p><p>• \sf d=3\cdot10^{-9}~md=3⋅10−9 m</p><p></p><p>Assim:</p><p></p><p>\sf F_{el}=\dfrac{9\cdot10^9\cdot4\cdot10^{-16}\cdot6\cdot10^{-16}}{(3\cdot10^{-9})^2}Fel=(3⋅10−9)29⋅109⋅4⋅10−16⋅6⋅10−16</p><p></p><p>\sf F_{el}=\dfrac{9\cdot4\cdot6\cdot10^9\cdot10^{-16}\cdot10^{-16}}{9\cdot10^{-18}}Fel=9⋅10−189⋅4⋅6⋅109⋅10−16⋅10−16</p><p></p><p>\sf F_{el}=\dfrac{216\cdot10^{9-16-16}}{9\cdot10^{-18}}Fel=9⋅10−18216⋅109−16−16</p><p></p><p>\sf F_{el}=\dfrac{216\cdot10^{-23}}{9\cdot10^{-18}}Fel=9⋅10−18216⋅10−23</p><p></p><p>\sf F_{el}=24\cdot10^{-23-(-18)}Fel=24⋅10−23−(−18)</p><p></p><p>\sf F_{el}=24\cdot10^{-23+18}Fel=24⋅10−23+18</p><p></p><p>\sf F_{el}=24\cdot10^{-5}Fel=24⋅10−5</p><p></p><p>\sf \red{F_{el}=2,4\cdot10^{-4}~N}Fel=2,4⋅10−4 N</p><p></p><p>$

Assim confirmado que é a letra

D>>Veja em anexo ↗️↗️

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