Question

Sabendo que duas partículas foram eletrizadas e possuem as cargas elétricas Q1 = 4,0 . 10-6 C e Q2 = 6,0 . 10-6 C e estão separadas, no vácuo, por uma distância de 2,0 . 10-2 m. Sendo K0 = 9 . 109 N.m2/ C2, podemos dizer que a intensidade da força de interação entre elas e a natureza da interação, em newtons, é: ​

Answer 1

Explicação:

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

Temos:

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

• $\sf Q_1=4\cdot10^{-6}~C$

• $\sf Q_2=6\cdot10^{-6}~C$

• $\sf d=2\cdot10^{-2}~m$

Assim:

$\sf F_{el}=\dfrac{9\cdot10^9\cdot4\cdot10^{-6}\cdot6\cdot10^{-6}}{(2\cdot10^{-2})^2}$

$\sf F_{el}=\dfrac{9\cdot4\cdot6\cdot10^9\cdot10^{-6}\cdot10^{-6}}{4\cdot10^{-4}}$

$\sf F_{el}=\dfrac{216\cdot10^{9-6-6}}{4\cdot10^{-4}}$

$\sf F_{el}=\dfrac{216\cdot10^{-3}}{4\cdot10^{-4}}$

$\sf F_{el}=54\cdot10^{-3-(-4)}$

$\sf F_{el}=54\cdot10^{-3+4}$

$\sf F_{el}=54\cdot10$

$\sf \red{F_{el}=540~N}$

Answer 2

Oiieee Td bom?!!?

Temos>>

$• \sf k_0=9\cdot10^{9}~N\cdot m^2/C^2k0=9⋅109 N⋅m2/C2</p><p></p><p>• \sf Q_1=4\cdot10^{-6}~CQ1=4⋅10−6 C</p><p></p><p>• \sf Q_2=6\cdot10^{-6}~CQ2=6⋅10−6 C</p><p></p><p>• \sf d=2\cdot10^{-2}~md=2⋅10−2 m</p><p></p><p>Assim:</p><p></p><p>\sf F_{el}=\dfrac{9\cdot10^9\cdot4\cdot10^{-6}\cdot6\cdot10^{-6}}{(2\cdot10^{-2})^2}Fel=(2⋅10−2)29⋅109⋅4⋅10−6⋅6⋅10−6</p><p></p><p>\sf F_{el}=\dfrac{9\cdot4\cdot6\cdot10^9\cdot10^{-6}\cdot10^{-6}}{4\cdot10^{-4}}Fel=4⋅10−49⋅4⋅6⋅109⋅10−6⋅10−6</p><p></p><p>\sf F_{el}=\dfrac{216\cdot10^{9-6-6}}{4\cdot10^{-4}}Fel=4⋅10−4216⋅109−6−6</p><p></p><p>\sf F_{el}=\dfrac{216\cdot10^{-3}}{4\cdot10^{-4}}Fel=4⋅10−4216⋅10−3</p><p></p><p>\sf F_{el}=54\cdot10^{-3-(-4)}Fel=54⋅10−3−(−4)</p><p></p><p>\sf F_{el}=54\cdot10^{-3+4}Fel=54⋅10−3+4</p><p></p><p>\sf F_{el}=54\cdot10Fel=54⋅10</p><p></p><p>\sf \red{F_{el}=540~N}Fel=540 N</p><p></p><p>$

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