Question

Dado o esquema a seguir, qual o módulo da força resultante em q2? Considere q1 = q2= q3 = 1 10-6 C, d = 2 cm, k=9.10^9 N∙m2/C2, teta = 60º e cos(120)=-1/2

Alguem me ajuda?? é pra hj urgente

Answer 1

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⠀⠀⠀☞ O módulo da força resultante em Q₂ vale 22,5 [N] (opção c). ✅

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⠀⠀⠀⭐⠀Para realizar este exercício vamos utilizar a lei de Coulomb e a lei dos cossenos.⠀⭐⠀

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                           $\LARGE\red{\boxed{\pink{\boxed{\begin{array}{lcr}\green{\star}&&\green{\star}\\&\!\!\orange{\bf |F_e| = K_e \cdot \dfrac{|Q_1 \cdot Q_2|}{d^2}}\!\!&\\\green{\star}&&\green{\star}\\\end{array}}}}}$

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$\text{\Large\orange{ \diamond~~\bf |F_e| }~\pink{ \Longrightarrow }~}$ Módulo da força elétrica de interação entre as cargas Q₁ e Q₂ em [N];

$\text{\Large\orange{ \diamond~~\bf K_e }~\pink{ \Longrightarrow }~}$ Constante de Coulomb (≈9 × 10⁹ [N·m²/C²]);

$\text{\Large\orange{ \diamond~~\bf |Q_i| }~\pink{ \Longrightarrow }~}$ Módulo da magnitude da carga i em [C];

$\text{\Large\orange{ \diamond~~\bf d }~\pink{ \Longrightarrow }~}$ Distância linear entre as cargas em [m].

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⠀⠀⠀➡️⠀Sendo assim temos para a interação Q₁Q₂:

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$\Large\blue{\text{ \sf |F_e| = 9 \cdot 10^9 \cdot \dfrac{10^{(-6)} \cdot 10^{(-6)}}{0,02^2} }}$

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$\LARGE\blue{\text{ \sf |F_e| = 9 \cdot 10^9 \cdot \dfrac{10^{(-12})}{0,0004} }}$

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$\LARGE\blue{\text{ \sf |F_e| = \dfrac{9 \cdot 10^{(9-12)}}{0,0004} }}$

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$\LARGE\blue{\sf |F_e| = 22.500 \cdot 10^{(-3)}}$

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$\LARGE\blue{\sf |F_e| = 22,5~[N]}$

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⠀⠀⠀➡️⠀Para a interação Q₃Q₂ temos o mesmo módulo de interação, ou seja, também 22,5 [N]. Agora vamos analisar graficamente a soma vetorial que faremos:

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                                   $\setlength{\unitlength}{0.95cm}\begin{picture}(6,5)\thicklines\bezier(0.7,0.9)(1.3,0.4)(0.7,-0.3)\put(0,0){\Huge \circ }\put(0.4,0.3){\vector(1,2){1}}\put(0.4,0.3){\vector(1,-2){1}}\put(1.3,0.1){\large 120^{\circ} }\put(-0.8,0.1){\Large \bf Q_2 }\put(0.65,1.2){\large \sf F_{q_1\rightarrow q_2} }\put(0.5,-1){\large \sf F_{q_3\rightarrow q_2} }\put(3,0){\Huge \circ }\put(3.4,0.3){\vector(1,2){1}}\put(3.4,0.3){\vector(1,-2){1}}\bezier(3.65,0.8)(4,0.6)(3.9,0.3)\put(3.9,0.7){\small 60^{\circ} }\bezier(5.1,0.8)(4.9,0.6)(4.9,0.3)\put(4.6,0.7){\small 60^{\circ} }\bezier(4.1,1.7)(4.4,1.4)(4.7,1.7)\put(4.2,1.2){\small 60^{\circ} }\put(4.4,2.3){\vector(1,-2){1}}\put(4.4,-1.7){\vector(1,2){1}}\bezier{20}(3.4,0.3)(4.4,0.3)(5.4,0.3)\put(3.6,0){\scriptsize \sf F_{q_1q_2}+F_{q_3q_2} }\end{picture}$

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                            $\Large\red{\boxed{\begin{array}{rcl}&\green{\underline{\footnotesize\sf Esta~imagem~n\tilde{a}o~\acute{e}~visualiz\acute{a}vel~pelo~App~Brainly.}}&\\&\green{\footnotesize\sf \bullet~Experimente~compartilhar\rightarrow copiar~e~acessar}&\\&\green{\footnotesize\sf o~link~copiado~pelo~seu~navegador~ou~Browser.}&\\\end{array}}}$

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⠀⠀⠀➡️⠀Observe que por termos um triângulo equilátero então sabemos que o módulo da força resultante será igual as outras duas forças (22,5 [N], mas vamos praticar um pouco a lei dos cossenos:

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                           $\large\red{\boxed{\pink{\boxed{\begin{array}{lcr}\green{\star}&&\green{\star}\\&\!\!\orange{\bf c^2 = a^2 + b^2 - 2 \cdot a \cdot b \cdot cos(\Theta)}\!\!&\\\green{\star}&&\green{\star}\\\end{array}}}}}$

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⌚ Hora da matemágica✍

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                           $\Large\gray{\boxed{\sf~~\begin{array}{c}~\\\pink{\underline{\underline{\text{\huge \bf\quad \subset\emptyset\,\eta\,\pi\,\alpha\,\ \quad }}}}\\\\\\\blue{\begin{cases}\sf~a = 22,5~[N]\\ \sf~b = 22,5~[N]\\ \sf~\Theta = 60^{\circ} \\\sf~c =~?~[N]\end{cases}}\\\\\!\!\!\!\!\!\!\!^{\underline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad}}\!\!\!\!\!\!\!\!\\\orange{\large\bf c = \pm \sqrt{a^2 + b^2 - 2 \cdot a \cdot b \cdot cos(\Theta)}}\\\!\!\!\!\!\!\!\!^{\underline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad}}\!\!\!\!\!\!\!\!\\\blue{\large\sf c = \pm \sqrt{2 \cdot 22,5^2 - 2 \cdot 22,5^2 \cdot 0,5}}\\\!\!\!\!\!\!\!\!^{\underline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad}}\!\!\!\!\!\!\!\!\\\blue{\sf c = \pm \sqrt{1.012,25 - 506,25}}\\\!\!\!\!\!\!\!\!^{\underline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad}}\!\!\!\!\!\!\!\!\\\\\blue{\sf c = \pm \sqrt{506,25}}\\\!\!\!\!\!\!\!\!^{\underline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad}}\!\!\!\!\!\!\!\!\\\\\green{\boxed{\blue{\sf~|c| = 22,5~[N]}}~\checkmark}\\\\\end{array}~~}}$

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⠀⠀⠀⭐ O que nos leva à opção c). ✌

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                                       $\huge\green{\boxed{\rm~~~\red{c)}~\blue{ 22,5~[N] }~~~}}$ ✅

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                             _______________________________☁

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⠀⠀☀️ L͎̙͖͉̥̳͖̭̟͊̀̏͒͑̓͊͗̋̈́ͅeia mais sobre lei de Coulomb:

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                                     https://brainly.com.br/tarefa/48078017 ✈  

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                                     $\huge\blue{\text{\bf\quad Bons~estudos.}}$ ☕

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                                          $\quad\qquad(\orange{D\acute{u}vidas\ nos\ coment\acute{a}rios})$ ☄

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                             $\bf\large\red{\underline{\qquad \qquad \qquad \qquad \qquad \qquad \quad }\LaTeX}$✍

                                $\sf(\purple{+}~\red{cores}~\blue{com}~\pink{o}~\orange{App}~\green{Brainly}$ ☘☀❄☃☂☻)

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                                                          $\Huge\green{\text{ \underline{\red{\mathbb{S}}\blue{\mathfrak{oli}}~}~\underline{\red{\mathbb{D}}\blue{\mathfrak{eo}}~}~\underline{\red{\mathbb{G}}\blue{\mathfrak{loria}}~} }}$ ✞

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