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Qual é a classificação dos agregados de acordo com o tamanho das partículas?

","status","pending_classification","hasRelatedVideo","relatedVideos",[257],{"_258":259,"_277":278,"_51":293,"_328":329,"_346":347},"content",{"_43":260,"_55":261,"_262":-5,"_263":264,"_265":266,"_267":-5,"_45":268,"_271":272,"_273":272,"_274":-5,"_275":276},"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",{"_269":270},"external_id","4963","created_at","2025-08-20T12:27:41.608438+00:00","updated_at","content_group_id","slug","optica-geometrica-parte-1","prometheusData",{"_279":280,"_286":287,"_288":289,"_290":291,"_28":292},"provider",{"_43":281,"_282":283,"_284":285},"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",[294],{"_43":295,"_55":296,"_57":297,"_298":299,"_300":-5,"_271":301,"_273":301,"_45":302,"_275":297,"_309":310},"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",{"_303":304,"_305":306,"_307":308},"childrenCount",18,"content_count",0,"question_count",135489,"children",[311],{"_43":312,"_55":313,"_57":314,"_298":299,"_300":295,"_271":315,"_273":315,"_45":316,"_275":319,"_309":320},"ca6fee0e-79c6-4f1c-b500-ee081a071949","Mecânica","mecanica","2025-04-29T02:19:12.828432+00:00",{"_303":317,"_305":306,"_307":318},10,19948,"mecanica-2",[321],{"_43":322,"_55":204,"_57":205,"_298":299,"_300":312,"_271":323,"_273":323,"_45":324,"_275":326,"_309":327},"ac55bdba-154d-42e7-bc25-72dde822663d","2025-04-29T03:44:29.139762+00:00",{"_303":306,"_305":306,"_307":325},3376,"resistencia-dos-materiais-4",[],"video",{"_43":266,"_13":261,"_262":-5,"_330":331,"_332":333,"_28":-5,"_334":-5,"_279":333,"_335":336,"_290":-5,"_337":-5,"_338":-5,"_339":-5,"_340":-5,"_45":341,"_342":306,"_343":-5,"_252":344,"_271":345,"_273":345},"duration",20,"source_type","prometheus","iframe_url","provider_video_id","1cd3d633-d690-4f4b-9995-9b0e44994e35","format","quality","file_path","file_size",{"_269":270},"views_count","last_played_at","draft","2025-08-19T11:56:20.758718+00:00","tag",{"_43":295,"_55":296,"_57":297,"_298":299,"_300":-5,"_271":301,"_273":301,"_45":348,"_275":297,"_309":349},{"_303":304,"_305":306,"_307":308},[350],{"_43":312,"_55":313,"_57":314,"_298":299,"_300":295,"_271":315,"_273":315,"_45":351,"_275":319,"_309":352},{"_303":317,"_305":306,"_307":318},[353],{"_43":322,"_55":204,"_57":205,"_298":299,"_300":312,"_271":323,"_273":323,"_45":354,"_275":326,"_309":355},{"_303":306,"_305":306,"_307":325},[],"similarQuestions",[358,388,408],{"_40":359,"_42":41,"_43":360,"_225":361,"_249":376,"_250":377,"_51":378},"112831da-be9c-458e-94cb-e8292d0dba12","3c0b9441-b878-4896-8bac-fa4ca61e3f13",[362,365,368,371,374],{"_43":363,"_229":230,"_231":364},"265bd71a-6749-41f5-b291-67efd646ca64","y'' + 2y' - 3y = 0",{"_43":366,"_229":235,"_231":367},"6d3c3a1d-8e80-4ff3-b97a-d5b5bca91940","6y'' - y' - y = 0",{"_43":369,"_229":239,"_231":370},"4168ac67-64fd-46ed-a69c-a01dbfcde7ea","y'' - 2y' + y = 0",{"_43":372,"_229":243,"_231":373},"cc7bc8df-6995-4264-ac5a-32ea9b4ed7db","9y'' + 6y' + y = 0",{"_43":359,"_229":247,"_231":375},"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

",[379],{"_43":380,"_55":381,"_57":382,"_298":299,"_300":383,"_271":384,"_273":384,"_45":385,"_275":382},"6516bbbd-f2a3-4710-8382-702240b17836","Mecânica na Engenharia de Petróleo","mecanica-na-engenharia-de-petroleo","b787e438-0ab3-454c-ab43-5d3d14c2720d","2025-04-29T03:44:35.328997+00:00",{"_303":386,"_305":306,"_307":387},4,12302,{"_40":389,"_42":41,"_43":390,"_225":391,"_249":403,"_250":404,"_51":405},"6831a9b9-c9a4-4a76-bd7a-2ce98bfea875","93a32dd4-5fc6-4ccb-a76f-4724bf113742",[392,395,398,400],{"_43":393,"_229":230,"_231":394},"13e3f476-a0a5-421f-b47e-0969e8e9a7e0","

Verdadeiro, verdadeiro, falso

",{"_43":396,"_229":235,"_231":397},"c9a20855-d84b-4860-8b35-9784818282cf","

Falso, verdadeiro, verdadeiro

",{"_43":389,"_229":239,"_231":399},"

Verdadeiro, falso, verdadeiro

",{"_43":401,"_229":243,"_231":402},"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
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Comentários