[{"_1":2,"_223":-5,"_224":-5},"loaderData",{"_3":4,"_9":10},"root",{"_5":-5,"_6":7,"_8":-5},"registerModalConfig","url","https://prouniversitario.com/exercicios/fc8420f8-4fc5-4c31-a435-e51c2dc5f6bd","user","pages/questions/show/index",{"_11":12,"_13":14,"_15":16,"_17":18,"_101":102,"_103":104,"_31":184,"_208":209,"_219":220,"_221":222},"comments",[],"hasVideoHash",false,"isAiGenerated",true,"question",{"_19":20,"_21":22,"_23":24,"_25":26,"_71":72,"_94":95,"_96":97,"_98":99,"_100":14},"correctOptionId","670e9fd6-832e-4eb0-9687-aaa796c9275b","difficulty","","id","fc8420f8-4fc5-4c31-a435-e51c2dc5f6bd","metadata",{"_27":28,"_53":54,"_67":68,"_69":70},"pd",{"_29":30,"_31":32,"_33":34,"_41":42,"_43":44,"_45":16,"_46":47,"_48":14,"_49":50,"_51":52},"area","Outros","tags",[],"subjects",[35],{"_23":36,"_37":38,"_39":40},780225,"name","Dinâmica dos Fluidos","alias","dinamica-dos-fluidos","materialId","83168130","answerCount",1,"isIndexable","mainSubject",{"_23":36,"_37":38,"_39":40},"isIndexableSEO","materialFormat","Text","studentDegreeLevel","Graduate","runic",{"_55":56,"_58":44,"_59":60},"issues",[57],"needs_latex_improvements","issuesCount","issuesDetectionTokenUsage",{"_61":62,"_63":64,"_65":66},"total",604,"prompt",594,"completion",10,"source","PD","externalId","167060257","options",[73,79,82,86,90],{"_23":74,"_75":76,"_77":78},"0a9ed121-e15d-471a-a23c-a0ed55dd683c","label","A","text","3,88x10-6",{"_23":20,"_75":80,"_77":81},"B","5,88x10-6",{"_23":83,"_75":84,"_77":85},"0b9aaf86-a1cc-4e2d-bb0a-f903b0203e1a","C","5,88x10-8",{"_23":87,"_75":88,"_77":89},"1c588813-05e6-41ef-a079-19513be1c295","D","7,55x10-6",{"_23":91,"_75":92,"_77":93},"5550fe30-63e2-47dd-85de-57ca630149a8","E","2,53x10-6","solution_explanation","ν=μ/ρ. Converta ρ: 0,85 kg/dm³ =0,85×1000=850 kg/m³. Logo ν=5×10⁻³ / 850 ≈5,88×10⁻⁶ m²/s.","statement","

Um fluido tem uma viscosidade dinâmica de 5 \\times 10^{-3} \\text{ N.s/m}^2 e uma massa específica de 0,85 \\text{ kg/dm}^3. Determinar a sua viscosidade cinemática em m^2/s.

","status","published","hasRelatedVideo","relatedVideos",[],"similarQuestions",[105,145,163],{"_19":106,"_21":22,"_23":107,"_71":108,"_94":123,"_96":124,"_31":125},"112831da-be9c-458e-94cb-e8292d0dba12","3c0b9441-b878-4896-8bac-fa4ca61e3f13",[109,112,115,118,121],{"_23":110,"_75":76,"_77":111},"265bd71a-6749-41f5-b291-67efd646ca64","y'' + 2y' - 3y = 0",{"_23":113,"_75":80,"_77":114},"6d3c3a1d-8e80-4ff3-b97a-d5b5bca91940","6y'' - y' - y = 0",{"_23":116,"_75":84,"_77":117},"4168ac67-64fd-46ed-a69c-a01dbfcde7ea","y'' - 2y' + y = 0",{"_23":119,"_75":88,"_77":120},"cc7bc8df-6995-4264-ac5a-32ea9b4ed7db","9y'' + 6y' + y = 0",{"_23":106,"_75":92,"_77":122},"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

",[126],{"_23":127,"_37":128,"_39":129,"_130":131,"_132":133,"_134":135,"_136":135,"_25":137,"_144":129},"6516bbbd-f2a3-4710-8382-702240b17836","Mecânica na Engenharia de Petróleo","mecanica-na-engenharia-de-petroleo","tag_group_id","c1ce2cf2-b217-4cc7-8182-81c56bd3522d","parent_id","b787e438-0ab3-454c-ab43-5d3d14c2720d","created_at","2025-04-29T03:44:35.328997+00:00","updated_at",{"_138":139,"_140":141,"_142":143},"childrenCount",4,"content_count",0,"question_count",12302,"slug",{"_19":146,"_21":22,"_23":147,"_71":148,"_94":160,"_96":161,"_31":162},"6831a9b9-c9a4-4a76-bd7a-2ce98bfea875","93a32dd4-5fc6-4ccb-a76f-4724bf113742",[149,152,155,157],{"_23":150,"_75":76,"_77":151},"13e3f476-a0a5-421f-b47e-0969e8e9a7e0","

Verdadeiro, verdadeiro, falso

",{"_23":153,"_75":80,"_77":154},"c9a20855-d84b-4860-8b35-9784818282cf","

Falso, verdadeiro, verdadeiro

",{"_23":146,"_75":84,"_77":156},"

Verdadeiro, falso, verdadeiro

",{"_23":158,"_75":88,"_77":159},"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?

",[126],{"_19":164,"_21":22,"_23":165,"_71":166,"_94":181,"_96":182,"_31":183},"3ce109b1-6cdb-4b4d-8816-a6af4d74bed0","c478a46c-46a8-4942-ae14-2e79def3dda2",[167,170,173,175,178],{"_23":168,"_75":76,"_77":169},"a5a16a42-3e58-4c76-80f2-c291aa2d8e49","Sim",{"_23":171,"_75":80,"_77":172},"5e6d0254-696c-4da0-b572-82e3a19f433e","Não",{"_23":164,"_75":84,"_77":174},"Indeterminado",{"_23":176,"_75":88,"_77":177},"35403f68-a880-429d-97f4-db507c58de38","Todos",{"_23":179,"_75":92,"_77":180},"2e06d948-d37e-4350-afc7-ced04941bcaf","Nenhum","Os polinômios e números fornecidos estão corrompidos (símbolos “????”), impossibilitando aplicar Ruffini e concluir se são raízes. Sem dados completos, o problema é indeterminado.","Emplea la regla de Ruffini para dictaminar si los siguientes números son o no raíces de los polinomios citados:
a) ????3 de ????4????5
b) ????2 de ????2????????2
c) ????1 de 2????????1
d) ????1 de 2????2????",[126],[126,185,193,201],{"_23":186,"_37":187,"_39":188,"_130":131,"_132":127,"_134":189,"_136":189,"_25":190,"_144":192},"8c444991-e4be-4bd9-bc90-7d0f27eb9a2f","Mecânica dos Fluidos na Engenharia de Petróleo","mecanica-dos-fluidos-na-engenharia-de-petroleo","2025-04-29T04:00:14.680867+00:00",{"_138":141,"_140":141,"_142":191},329,"mecanica-dos-fluidos-na-engenharia-de-petroleo-1",{"_23":133,"_37":194,"_39":195,"_130":131,"_132":196,"_134":197,"_136":197,"_25":198,"_144":200},"Engenharia de Petróleo","engenharia-de-petroleo","eb9315c5-bd04-4116-aa46-02c4fde320c0","2025-04-29T02:19:12.828432+00:00",{"_138":199,"_140":141,"_142":143},5,"engenharia-de-petroleo-1",{"_23":196,"_37":202,"_39":203,"_130":131,"_132":-5,"_134":204,"_136":204,"_25":205,"_144":203},"Engenharia","engenharia","2025-04-29T01:55:38.635246+00:00",{"_138":206,"_140":141,"_142":207},29,351586,"tagsTree",[210],{"_23":196,"_37":202,"_39":203,"_130":131,"_132":-5,"_134":204,"_136":204,"_25":205,"_144":203,"_211":212},"children",[213],{"_23":133,"_37":194,"_39":195,"_130":131,"_132":196,"_134":197,"_136":197,"_25":198,"_144":200,"_211":214},[215],{"_23":127,"_37":128,"_39":129,"_130":131,"_132":133,"_134":135,"_136":135,"_25":137,"_144":129,"_211":216},[217],{"_23":186,"_37":187,"_39":188,"_130":131,"_132":127,"_134":189,"_136":189,"_25":190,"_144":192,"_211":218},[],"videoDislikes",{},"videoLikes",{},"actionData","errors"]

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