{"id":44,"date":"2023-06-26T21:13:28","date_gmt":"2023-06-26T21:13:28","guid":{"rendered":"https:\/\/physigeek.com\/pt\/terceira-lei-de-newton-ou-principio-de-acao-e-reacao\/"},"modified":"2023-06-26T21:13:28","modified_gmt":"2023-06-26T21:13:28","slug":"terceira-lei-de-newton-ou-principio-de-acao-e-reacao","status":"publish","type":"post","link":"https:\/\/physigeek.com\/pt\/terceira-lei-de-newton-ou-principio-de-acao-e-reacao\/","title":{"rendered":"Terceira lei de newton (ou princ\u00edpio de a\u00e7\u00e3o e rea\u00e7\u00e3o)"},"content":{"rendered":"<p>Este artigo explica o que \u00e9 a terceira lei de Newton, tamb\u00e9m conhecida como princ\u00edpio de a\u00e7\u00e3o e rea\u00e7\u00e3o, e o que ela diz. Voc\u00ea poder\u00e1 ver exemplos da terceira lei de Newton, bem como sua f\u00f3rmula matem\u00e1tica. Al\u00e9m disso, voc\u00ea pode praticar exerc\u00edcios resolvidos passo a passo da terceira lei de Newton. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"%C2%BFCual-es-la-tercera-ley-de-Newton\"><\/span> Qual \u00e9 a terceira lei de Newton?<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Na f\u00edsica, <strong>a declara\u00e7\u00e3o da terceira lei de Newton<\/strong> , tamb\u00e9m chamada de princ\u00edpio de a\u00e7\u00e3o e rea\u00e7\u00e3o, afirma o seguinte: <\/p>\n<div style=\"background:linear-gradient(to bottom, #FFFFFF 0%, #E1F5FE 100%);  padding-right: 27px; padding-left: 30px; padding-top: 23px; padding-bottom: 0.5px; border: 2px dashed #4FC3F7; border-radius:25px;\">\n<p> Se um corpo exercer uma for\u00e7a sobre outro corpo, ele exercer\u00e1 uma for\u00e7a de mesma magnitude e dire\u00e7\u00e3o, mas na dire\u00e7\u00e3o oposta, sobre o primeiro corpo.<\/p>\n<\/div>\n<p> Em outras palavras, se o corpo A exerce uma for\u00e7a horizontal de 10 N para a direita sobre o corpo B, o corpo B exercer\u00e1 uma for\u00e7a horizontal de 10 N para a esquerda sobre o corpo A.<\/p>\n<p> Portanto, as for\u00e7as entre dois corpos ou sistemas s\u00e3o sempre iguais, mas em dire\u00e7\u00f5es opostas.<\/p>\n<p> Logicamente, o princ\u00edpio de a\u00e7\u00e3o e rea\u00e7\u00e3o \u00e9 chamado de terceira lei de Newton em homenagem ao f\u00edsico Isaac Newton, que primeiro formulou esta lei. No total, existem tr\u00eas leis de Newton:<\/p>\n<ul>\n<li> A primeira lei ou princ\u00edpio da in\u00e9rcia de Newton.<\/li>\n<li> Segunda lei de Newton ou princ\u00edpio fundamental da din\u00e2mica.<\/li>\n<li> Terceira lei de Newton ou princ\u00edpio de a\u00e7\u00e3o-rea\u00e7\u00e3o.<\/li>\n<\/ul>\n<p> Voc\u00ea pode conferir o que \u00e9 cada lei de Newton em nosso site, ingenierizado.com. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Formula-de-la-tercera-ley-de-Newton\"><\/span> F\u00f3rmula para a terceira lei de Newton<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> A terceira lei de Newton (ou princ\u00edpio de a\u00e7\u00e3o e rea\u00e7\u00e3o) diz que se um corpo exerce uma for\u00e7a sobre outro corpo, o primeiro corpo recebe uma for\u00e7a aplicada pelo segundo corpo de mesma magnitude, mas na dire\u00e7\u00e3o oposta. <strong>A terceira lei de Newton pode, portanto, ser expressa pela seguinte f\u00f3rmula<\/strong> : <\/p>\n<figure class=\"wp-block-image aligncenter size-full is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/troisieme-loi-de-newton.png\" alt=\"Terceira lei de Newton ou princ\u00edpio de a\u00e7\u00e3o e rea\u00e7\u00e3o\" class=\"wp-image-763\" width=\"244\" height=\"244\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/troisieme-loi-de-newton-300x300.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/troisieme-loi-de-newton-150x150.png 150w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/troisieme-loi-de-newton.png 512w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\"><\/figure>\n<p> Ouro<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-b60e67f5879bf157e9f528efae793758_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F_{12}\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"25\" style=\"vertical-align: -3px;\"><\/p>\n<p> \u00e9 a for\u00e7a que o corpo 1 exerce em rela\u00e7\u00e3o ao corpo 2. E<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-b60e67f5879bf157e9f528efae793758_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F_{12}\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"25\" style=\"vertical-align: -3px;\"><\/p>\n<p> \u00e9 a for\u00e7a que o corpo 2 exerce sobre o corpo 1.<\/p>\n<p> Assim, para que a equa\u00e7\u00e3o da terceira lei de Newton seja cumprida, as duas for\u00e7as devem ter o mesmo m\u00f3dulo mas o seu sinal deve ser oposto, ou por outras palavras, as for\u00e7as devem ser opostas.<\/p>\n<p> A primeira for\u00e7a produzida tamb\u00e9m \u00e9 chamada <strong>de for\u00e7a de a\u00e7\u00e3o<\/strong> . Da mesma forma, a for\u00e7a que resulta de uma rea\u00e7\u00e3o \u00e0 primeira for\u00e7a exercida \u00e9 chamada de <strong>for\u00e7a de rea\u00e7\u00e3o<\/strong> . <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejemplos-de-la-tercera-ley-de-Newton\"><\/span> Exemplos da Terceira Lei de Newton<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Agora que conhecemos a defini\u00e7\u00e3o da terceira lei de Newton, vejamos v\u00e1rios exemplos do mundo real para compreender completamente o conceito.<\/p>\n<ol style=\"color:#4fd12f; font-weight: bold;\">\n<li style=\"margin-bottom:25px\"> <span style=\"color:#101010;font-weight: normal;\">Um exemplo t\u00edpico da terceira lei de Newton \u00e9 uma pessoa exercendo uma for\u00e7a sobre uma parede. Quando voc\u00ea aplica uma for\u00e7a na parede, ela exerce outra for\u00e7a de mesma magnitude sobre a pessoa. Portanto, a pessoa n\u00e3o conseguir\u00e1 mover a parede, mas perceber\u00e1 que est\u00e1 sendo empurrada para tr\u00e1s devido \u00e0 for\u00e7a de rea\u00e7\u00e3o que a parede exerce sobre ela.<\/span><\/li>\n<li style=\"margin-bottom:25px\"> <span style=\"color:#101010;font-weight: normal;\">Outro exemplo do princ\u00edpio de a\u00e7\u00e3o e rea\u00e7\u00e3o \u00e9 a for\u00e7a normal. A Terra exerce uma for\u00e7a gravitacional que nos empurra em dire\u00e7\u00e3o ao centro do planeta, j\u00e1 que a for\u00e7a normal \u00e9 a for\u00e7a de rea\u00e7\u00e3o que o solo exerce sobre as pessoas e contraria essa for\u00e7a. Ent\u00e3o, gra\u00e7as \u00e0 for\u00e7a normal, podemos permanecer na superf\u00edcie da Terra.<\/span><\/li>\n<li> <span style=\"color:#101010;font-weight: normal;\">Finalmente, quando saltamos, estamos na verdade exercendo uma for\u00e7a no solo, e ent\u00e3o o solo reage e exerce uma for\u00e7a da mesma magnitude sobre n\u00f3s, impulsionando-nos para cima. Assim, quanto mais for\u00e7a exercermos no solo, mais for\u00e7a o solo exercer\u00e1 sobre n\u00f3s e, portanto, mais saltaremos.<\/span><\/li>\n<\/ol>\n<p> Observe que a terceira lei de Newton n\u00e3o significa que as duas for\u00e7as se oponham e, portanto, se anulem. Em vez disso, a for\u00e7a de a\u00e7\u00e3o atua sobre um corpo e a for\u00e7a de rea\u00e7\u00e3o atua sobre outro corpo.<\/p>\n<p> Al\u00e9m disso, embora a for\u00e7a de ac\u00e7\u00e3o e a for\u00e7a de reac\u00e7\u00e3o tenham a mesma magnitude, n\u00e3o t\u00eam o mesmo efeito, uma vez que actuam em corpos diferentes. Seguindo o primeiro exemplo explicado acima, quando uma pessoa exerce uma for\u00e7a sobre uma parede, obviamente ela n\u00e3o a move, por\u00e9m, a for\u00e7a de rea\u00e7\u00e3o que a parede exerce sobre a pessoa a move. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejercicios-resueltos-de-la-tercera-ley-de-Newton\"><\/span> Exerc\u00edcios resolvidos da terceira lei de Newton<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h3 class=\"wp-block-heading\"> Exerc\u00edcio 1<\/h3>\n<p> Se uma for\u00e7a vertical descendente de 15 N \u00e9 exercida sobre um objeto com massa de 4 kg, que for\u00e7a o solo deve exercer para que o objeto fique em equil\u00edbrio? <\/p>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__FFF8E1\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#FFF8E1\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>veja a solu\u00e7\u00e3o<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> O objeto estar\u00e1 em equil\u00edbrio se n\u00e3o estiver em movimento e, para que isso aconte\u00e7a, o solo deve exercer uma for\u00e7a que contrarie a for\u00e7a do peso do objeto mais a for\u00e7a aplicada.<\/p>\n<p class=\"has-text-align-left\"> Ent\u00e3o primeiro calculamos o peso do objeto:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-031f3807bdf941c2a1b1b553653e61f6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P=m\\cdot g=4\\cdot 9,81=39,24 \\ N\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"244\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> A soma das duas for\u00e7as que empurram o objeto para baixo \u00e9, portanto:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-1a78bff8432a797873cdd5b7cc6be8d2_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F=15+39,24=54,24\\ N\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"209\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Concluindo, o solo deve exercer uma for\u00e7a vertical ascendente de 54,24 N sobre o objeto para que ele esteja em equil\u00edbrio.<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h3 class=\"wp-block-heading\"> Exerc\u00edcio 2<\/h3>\n<p> Um corpo de 0,3 kg est\u00e1 suspenso por um fio, da mesma forma, outro corpo de 0,1 kg est\u00e1 suspenso do anterior por outro fio, conforme mostra a imagem a seguir. Se uma for\u00e7a de 6 N for exercida para cima, qual ser\u00e1 a acelera\u00e7\u00e3o de todo o sistema? E qual \u00e9 a tens\u00e3o do segundo fio? <\/p>\n<figure class=\"wp-block-image aligncenter size-full is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-de-la-troisieme-loi-de-newton.png\" alt=\"Problema da terceira lei de Newton\" class=\"wp-image-772\" width=\"161\" height=\"270\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-de-la-troisieme-loi-de-newton-179x300.png 179w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-de-la-troisieme-loi-de-newton.png 318w\" sizes=\"auto, (max-width: 179px) 100vw, 179px\"><\/figure>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__FFF8E1\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#FFF8E1\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>veja a solu\u00e7\u00e3o<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Neste caso, precisamos de utilizar a segunda lei de Newton e a terceira lei de Newton para resolver o problema.<\/p>\n<p class=\"has-text-align-left\"> Primeiro, calcularemos a for\u00e7a do peso que atua sobre cada corpo:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-c0cdb663ec9f8fe79fbecd960b50fc39_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P=m\\cdot g\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"75\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-2ff3d69e4cc3ce44f1802620f61d1779_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_1=m_1\\cdot g=0,3\\cdot 9,81=2,94\\N\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"243\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-0ea943ecb0e66b05dfb9ca77df25e2b0_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_2=m_2\\cdot g=0,1\\cdot 9,81=0,98\\N\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"243\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Agora aplicamos a equa\u00e7\u00e3o da segunda lei de Newton a todo o sistema:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-f925629131b789d77252b47fd9709f0a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\sum F=m\\cdot a\" title=\"Rendered by QuickLaTeX.com\" height=\"25\" width=\"103\" style=\"vertical-align: -8px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-6a5cb351f075fbd330f605e188b0d9dd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"6-P_1-P_2=m\\cdot a\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"151\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Substitu\u00edmos os dados e apagamos a acelera\u00e7\u00e3o para encontrar seu valor:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-2a1c49635c13339807137ccbe4a5015e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"6-2.94-0.98=(0.3+0.1)\\cdot a\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"241\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-41a0811c05576ed5597a686bc2971f65_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"2,08=0,4\\cdot a\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"106\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-c36983f6e932ef6e1385b460a4a51208_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"a=\\cfrac{2.08}{0.4}=5.2\\ \\cfrac{m}{s^2}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"138\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Por outro lado, a for\u00e7a que o corpo 1 exerce sobre o corpo 2 ser\u00e1 oposta \u00e0 for\u00e7a que o corpo 2 exerce sobre o corpo 1. Al\u00e9m disso, como conhecemos a acelera\u00e7\u00e3o do corpo 2 e o seu peso, reformulamos a equa\u00e7\u00e3o for\u00e7as mas desta vez apenas no corpo 2: <\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-f925629131b789d77252b47fd9709f0a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\sum F=m\\cdot a\" title=\"Rendered by QuickLaTeX.com\" height=\"25\" width=\"103\" style=\"vertical-align: -8px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-3cbfbfdadaa2a91cab52745cc42f488a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"T-P_2=m_2\\cdot a\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"122\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-5bd1d9368a3cada47b4e4ed131913219_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"T-0,98=0,1\\cdot 5,2\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"156\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-cb6b0f2baf22cc6afeab9828d1c80c0e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"T=0,52+0,98\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"127\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-456eb8a31ddbd5aa8e4f30e2f58d4375_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"T=1,5\\N\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"61\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Em resumo, a acelera\u00e7\u00e3o do sistema \u00e9 de 5,2 m\/s <sup>2<\/sup> e a tens\u00e3o da segunda corda \u00e9 de 1,5 N.<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n","protected":false},"excerpt":{"rendered":"<p>Este artigo explica o que \u00e9 a terceira lei de Newton, tamb\u00e9m conhecida como princ\u00edpio de a\u00e7\u00e3o e rea\u00e7\u00e3o, e o que ela diz. Voc\u00ea poder\u00e1 ver exemplos da terceira lei de Newton, bem como sua f\u00f3rmula matem\u00e1tica. Al\u00e9m disso, voc\u00ea pode praticar exerc\u00edcios resolvidos passo a passo da terceira lei de Newton. Qual \u00e9 &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/physigeek.com\/pt\/terceira-lei-de-newton-ou-principio-de-acao-e-reacao\/\"> <span class=\"screen-reader-text\">Terceira lei de newton (ou princ\u00edpio de a\u00e7\u00e3o e rea\u00e7\u00e3o)<\/span> Leia mais &raquo;<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"","footnotes":""},"categories":[5],"tags":[],"class_list":["post-44","post","type-post","status-publish","format-standard","hentry","category-dinamico"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v21.4 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>\u25b7 Terceira lei de Newton (princ\u00edpio de a\u00e7\u00e3o e rea\u00e7\u00e3o)<\/title>\n<meta name=\"description\" content=\"Explicamos o que \u00e9 a terceira lei de Newton (princ\u00edpio de a\u00e7\u00e3o e rea\u00e7\u00e3o). 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