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A rugosidade superficial dificulta a limpeza, preparação superficial e a remoção do excesso de liquido penetrante. Quais das alternativas abaixo é considerada a solução desse problema?

","status","pending_classification","hasRelatedVideo","relatedVideos",[104],{"_105":106,"_124":125,"_50":140,"_166":167,"_184":185},"content",{"_43":107,"_56":108,"_109":-5,"_110":111,"_112":113,"_114":-5,"_45":115,"_118":119,"_120":119,"_121":-5,"_122":123},"468e3e62-35e0-47be-a15f-68dd43bb32bb","Óptica Geométrica - Parte 1","description","content_type_id","1051eee7-c62e-4b2c-9900-77a29fcec080","resource_id","fa6aaf42-4cc4-4d76-8915-7a67081f7e9a","user_id",{"_116":117},"external_id","4963","created_at","2025-08-20T12:27:41.608438+00:00","updated_at","content_group_id","slug","optica-geometrica-parte-1","prometheusData",{"_126":127,"_133":134,"_135":136,"_137":138,"_28":139},"provider",{"_43":128,"_129":130,"_131":132},"06f120be-2194-4548-8336-7ba44ffbae14","key","grupo_q_proteus","kind","proteus","reference","d80487a0-0f98-42c8-99bb-551165ba3068","state","completed","thumbnail_url","https://proteus.grupoq.io/videos/d80487a0-0f98-42c8-99bb-551165ba3068/thumbnailsthumb.0000000.jpg","https://proteus.grupoq.io/videos/d80487a0-0f98-42c8-99bb-551165ba3068/levels.m3u8",[141],{"_43":142,"_56":143,"_58":144,"_145":146,"_147":-5,"_118":148,"_120":148,"_45":149,"_122":144,"_156":157},"9772e958-72c6-4499-ad47-877cf94d670d","Exatas","exatas","tag_group_id","c1ce2cf2-b217-4cc7-8182-81c56bd3522d","parent_id","2025-04-29T01:56:45.284172+00:00",{"_150":151,"_152":153,"_154":155},"childrenCount",18,"content_count",0,"question_count",135489,"children",[158],{"_43":159,"_56":57,"_58":59,"_145":146,"_147":142,"_118":160,"_120":160,"_45":161,"_122":164,"_156":165},"ca6fee0e-79c6-4f1c-b500-ee081a071949","2025-04-29T02:19:12.828432+00:00",{"_150":162,"_152":153,"_154":163},10,19948,"mecanica-2",[],"video",{"_43":113,"_13":108,"_109":-5,"_168":169,"_170":171,"_28":-5,"_172":-5,"_126":171,"_173":174,"_137":-5,"_175":-5,"_176":-5,"_177":-5,"_178":-5,"_45":179,"_180":153,"_181":-5,"_99":182,"_118":183,"_120":183},"duration",20,"source_type","prometheus","iframe_url","provider_video_id","1cd3d633-d690-4f4b-9995-9b0e44994e35","format","quality","file_path","file_size",{"_116":117},"views_count","last_played_at","draft","2025-08-19T11:56:20.758718+00:00","tag",{"_43":142,"_56":143,"_58":144,"_145":146,"_147":-5,"_118":148,"_120":148,"_45":186,"_122":144,"_156":187},{"_150":151,"_152":153,"_154":155},[188],{"_43":159,"_56":57,"_58":59,"_145":146,"_147":142,"_118":160,"_120":160,"_45":189,"_122":164,"_156":190},{"_150":162,"_152":153,"_154":163},[],"similarQuestions",[193,224,244],{"_40":194,"_42":41,"_43":195,"_76":196,"_96":212,"_97":213,"_50":214},"112831da-be9c-458e-94cb-e8292d0dba12","3c0b9441-b878-4896-8bac-fa4ca61e3f13",[197,200,203,206,209],{"_43":198,"_80":81,"_82":199},"265bd71a-6749-41f5-b291-67efd646ca64","y'' + 2y' - 3y = 0",{"_43":201,"_80":86,"_82":202},"6d3c3a1d-8e80-4ff3-b97a-d5b5bca91940","6y'' - y' - y = 0",{"_43":204,"_80":90,"_82":205},"4168ac67-64fd-46ed-a69c-a01dbfcde7ea","y'' - 2y' + y = 0",{"_43":207,"_80":94,"_82":208},"cc7bc8df-6995-4264-ac5a-32ea9b4ed7db","9y'' + 6y' + y = 0",{"_43":194,"_80":210,"_82":211},"E","y'' - 2y' + 6y = 0","Somente em y''−2y'+6y=0 o polinômio característico r²−2r+6=0 tem Δ<0, gerando raízes complexas conjugadas (1±i√5) e, portanto, solução em termos de exponencial modulando seno e cosseno. Nos demais itens (a–d) o discriminante é ≥0, produzindo raízes reais (distintas ou repetidas).","

Resolva as seguintes EDOs determinando a solução geral e, quando houver condições iniciais do PVI, a solução particular.

(a) y'' + 2y' - 3y = 0

(b) 6y'' - y' - y = 0

(d) 9y'' + 6y' + y = 0

(e) y'' - 2y' + 6y = 0

(f) 4y'' + 9y = 0

(g) 6y'' - 5y' + y = 0; \\, y(0) = 4; \\, y'(0) = 0

(h) y'' + 4y' + 4y = 0; \\, y(-1) = 2; \\, y'(-1) = 1

(i) y'' + y = 0; \\, y\bigg(\\frac{\\pi}{3}\\bigg) = 2; \\, y'\\bigg(\\frac{\\pi}{3}\\bigg) = -4

(j) y'' - 2y' - 3y = 3e^{2t}

(k) y'' + 9y = t^2 e^{3t} + 6

(l) 2y'' + 3y' + y = t^2 + 3 \\sin(t)

(m) y'' + y' - 2y = 2t; \\, y(0) = 0; \\, y'(0) = 1

(n) y'' - 2y' + y = te^t + 4; \\, y(0) = 1; \\, y'(0) = 1

(o) y'' + 4y = 3 \\sin(2t); \\, y(0) = 2; \\, y'(0) = -1

",[215],{"_43":216,"_56":217,"_58":218,"_145":146,"_147":219,"_118":160,"_120":160,"_45":220,"_122":223},"861ad829-750d-4a44-9569-2f18b430b4df","Engenharia Mecatrônica","engenharia-mecatronica","eb9315c5-bd04-4116-aa46-02c4fde320c0",{"_150":221,"_152":153,"_154":222},5,30614,"engenharia-mecatronica-1",{"_40":225,"_42":41,"_43":226,"_76":227,"_96":239,"_97":240,"_50":241},"6831a9b9-c9a4-4a76-bd7a-2ce98bfea875","93a32dd4-5fc6-4ccb-a76f-4724bf113742",[228,231,234,236],{"_43":229,"_80":81,"_82":230},"13e3f476-a0a5-421f-b47e-0969e8e9a7e0","

Verdadeiro, verdadeiro, falso

",{"_43":232,"_80":86,"_82":233},"c9a20855-d84b-4860-8b35-9784818282cf","

Falso, verdadeiro, verdadeiro

",{"_43":225,"_80":90,"_82":235},"

Verdadeiro, falso, verdadeiro

",{"_43":237,"_80":94,"_82":238},"3c98539b-1af9-4d44-aecd-c266754a2075","

Falso, falso, falso

","85,5 % decomposto → 14,5 % restante.\nN/N0 = 0,145 ⇒ t = ln(1/0,145)/(ln2/5730) ≈ 1,6×10⁴ anos.\nItens prováveis: (1) resta 14,5 % do C-14 (V); (2) idade ≈ 1,2×10⁴ anos ou 3 meias-vidas (F); (3) t½ do C-14 = 5730 anos (V). Sequência: V-F-V, correspondendo à alternativa c.","

Qual é a idade da madeira na qual 85,5\\% do C-14 se decompos?

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a) ????3 de ????4????5
b) ????2 de ????2????????2
c) ????1 de 2????????1
d) ????1 de 2????2????",[265],{"_43":216,"_56":217,"_58":218,"_145":146,"_147":219,"_118":160,"_120":160,"_45":266,"_122":223},{"_150":221,"_152":153,"_154":222},[268,270,276,278,280],{"_43":216,"_56":217,"_58":218,"_145":146,"_147":219,"_118":160,"_120":160,"_45":269,"_122":223},{"_150":221,"_152":153,"_154":222},{"_43":271,"_56":57,"_58":59,"_145":146,"_147":216,"_118":272,"_120":272,"_45":273,"_122":275},"88fe96b3-2a17-4198-90d5-ddda3d6f5d45","2025-04-29T03:44:35.328997+00:00",{"_150":221,"_152":153,"_154":274},25741,"mecanica-3",{"_43":142,"_56":143,"_58":144,"_145":146,"_147":-5,"_118":148,"_120":148,"_45":277,"_122":144},{"_150":151,"_152":153,"_154":155},{"_43":159,"_56":57,"_58":59,"_145":146,"_147":142,"_118":160,"_120":160,"_45":279,"_122":164},{"_150":162,"_152":153,"_154":163},{"_43":219,"_56":281,"_58":282,"_145":146,"_147":-5,"_118":283,"_120":283,"_45":284,"_122":282},"Engenharia","engenharia","2025-04-29T01:55:38.635246+00:00",{"_150":285,"_152":153,"_154":286},29,351586,"tagsTree",[289,295],{"_43":142,"_56":143,"_58":144,"_145":146,"_147":-5,"_118":148,"_120":148,"_45":290,"_122":144,"_156":291},{"_150":151,"_152":153,"_154":155},[292],{"_43":159,"_56":57,"_58":59,"_145":146,"_147":142,"_118":160,"_120":160,"_45":293,"_122":164,"_156":294},{"_150":162,"_152":153,"_154":163},[],{"_43":219,"_56":281,"_58":282,"_145":146,"_147":-5,"_118":283,"_120":283,"_45":296,"_122":282,"_156":297},{"_150":285,"_152":153,"_154":286},[298],{"_43":216,"_56":217,"_58":218,"_145":146,"_147":219,"_118":160,"_120":160,"_45":299,"_122":223,"_156":300},{"_150":221,"_152":153,"_154":222},[301],{"_43":271,"_56":57,"_58":59,"_145":146,"_147":216,"_118":272,"_120":272,"_45":302,"_122":275,"_156":303},{"_150":221,"_152":153,"_154":274},[],"videoDislikes",{},"videoLikes",{},"actionData","errors"]

Comentários