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Determine os diagramas de esforço cortante e de momento fletor para a viga. Qual o momento fletor máximo (em módulo) desenvolvido na viga? Faça P=4 \text{ kN}, a=1,5 \text{ m} e L=3,6 \text{ m}.

","status","pending_classification","hasRelatedVideo","relatedVideos",[105],{"_106":107,"_125":126,"_51":141,"_176":177,"_194":195},"content",{"_43":108,"_57":109,"_110":-5,"_111":112,"_113":114,"_115":-5,"_45":116,"_119":120,"_121":120,"_122":-5,"_123":124},"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",{"_117":118},"external_id","4963","created_at","2025-08-20T12:27:41.608438+00:00","updated_at","content_group_id","slug","optica-geometrica-parte-1","prometheusData",{"_127":128,"_134":135,"_136":137,"_138":139,"_28":140},"provider",{"_43":129,"_130":131,"_132":133},"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",[142],{"_43":143,"_57":144,"_59":145,"_146":147,"_148":-5,"_119":149,"_121":149,"_45":150,"_123":145,"_157":158},"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",{"_151":152,"_153":154,"_155":156},"childrenCount",18,"content_count",0,"question_count",135489,"children",[159],{"_43":160,"_57":161,"_59":162,"_146":147,"_148":143,"_119":163,"_121":163,"_45":164,"_123":167,"_157":168},"ca6fee0e-79c6-4f1c-b500-ee081a071949","Mecânica","mecanica","2025-04-29T02:19:12.828432+00:00",{"_151":165,"_153":154,"_155":166},10,19948,"mecanica-2",[169],{"_43":170,"_57":58,"_59":60,"_146":147,"_148":160,"_119":171,"_121":171,"_45":172,"_123":174,"_157":175},"ac55bdba-154d-42e7-bc25-72dde822663d","2025-04-29T03:44:29.139762+00:00",{"_151":154,"_153":154,"_155":173},3376,"resistencia-dos-materiais-4",[],"video",{"_43":114,"_13":109,"_110":-5,"_178":179,"_180":181,"_28":-5,"_182":-5,"_127":181,"_183":184,"_138":-5,"_185":-5,"_186":-5,"_187":-5,"_188":-5,"_45":189,"_190":154,"_191":-5,"_100":192,"_119":193,"_121":193},"duration",20,"source_type","prometheus","iframe_url","provider_video_id","1cd3d633-d690-4f4b-9995-9b0e44994e35","format","quality","file_path","file_size",{"_117":118},"views_count","last_played_at","draft","2025-08-19T11:56:20.758718+00:00","tag",{"_43":143,"_57":144,"_59":145,"_146":147,"_148":-5,"_119":149,"_121":149,"_45":196,"_123":145,"_157":197},{"_151":152,"_153":154,"_155":156},[198],{"_43":160,"_57":161,"_59":162,"_146":147,"_148":143,"_119":163,"_121":163,"_45":199,"_123":167,"_157":200},{"_151":165,"_153":154,"_155":166},[201],{"_43":170,"_57":58,"_59":60,"_146":147,"_148":160,"_119":171,"_121":171,"_45":202,"_123":174,"_157":203},{"_151":154,"_153":154,"_155":173},[],"similarQuestions",[206,237,257],{"_40":207,"_42":41,"_43":208,"_77":209,"_97":225,"_98":226,"_51":227},"112831da-be9c-458e-94cb-e8292d0dba12","3c0b9441-b878-4896-8bac-fa4ca61e3f13",[210,213,216,219,222],{"_43":211,"_81":82,"_83":212},"265bd71a-6749-41f5-b291-67efd646ca64","y'' + 2y' - 3y = 0",{"_43":214,"_81":87,"_83":215},"6d3c3a1d-8e80-4ff3-b97a-d5b5bca91940","6y'' - y' - y = 0",{"_43":217,"_81":91,"_83":218},"4168ac67-64fd-46ed-a69c-a01dbfcde7ea","y'' - 2y' + y = 0",{"_43":220,"_81":95,"_83":221},"cc7bc8df-6995-4264-ac5a-32ea9b4ed7db","9y'' + 6y' + y = 0",{"_43":207,"_81":223,"_83":224},"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

",[228],{"_43":229,"_57":230,"_59":231,"_146":147,"_148":232,"_119":233,"_121":233,"_45":234,"_123":231},"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",{"_151":235,"_153":154,"_155":236},4,12302,{"_40":238,"_42":41,"_43":239,"_77":240,"_97":252,"_98":253,"_51":254},"6831a9b9-c9a4-4a76-bd7a-2ce98bfea875","93a32dd4-5fc6-4ccb-a76f-4724bf113742",[241,244,247,249],{"_43":242,"_81":82,"_83":243},"13e3f476-a0a5-421f-b47e-0969e8e9a7e0","

Verdadeiro, verdadeiro, falso

",{"_43":245,"_81":87,"_83":246},"c9a20855-d84b-4860-8b35-9784818282cf","

Falso, verdadeiro, verdadeiro

",{"_43":238,"_81":91,"_83":248},"

Verdadeiro, falso, verdadeiro

",{"_43":250,"_81":95,"_83":251},"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