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Determine o momento fletor máximo (em módulo) desenvolvido na viga. Para isso, construa o diagrama de momento fletor. Considere P = 4 \text{ kN}, a = 1,5 \text{ m} e L = 3,6 \text{ m}. (conteúdo da Aula 4 tema 3 ou 4)

","status","published","hasRelatedVideo","relatedVideos",[92],{"_93":94,"_112":113,"_30":128,"_163":164,"_182":183},"content",{"_23":95,"_34":96,"_97":-5,"_98":99,"_100":101,"_102":-5,"_25":103,"_106":107,"_108":107,"_109":-5,"_110":111},"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",{"_104":105},"external_id","4963","created_at","2025-08-20T12:27:41.608438+00:00","updated_at","content_group_id","slug","optica-geometrica-parte-1","prometheusData",{"_114":115,"_121":122,"_123":124,"_125":126,"_6":127},"provider",{"_23":116,"_117":118,"_119":120},"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",[129],{"_23":130,"_34":131,"_36":132,"_133":134,"_135":-5,"_106":136,"_108":136,"_25":137,"_110":132,"_144":145},"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",{"_138":139,"_140":141,"_142":143},"childrenCount",18,"content_count",0,"question_count",135489,"children",[146],{"_23":147,"_34":148,"_36":149,"_133":134,"_135":130,"_106":150,"_108":150,"_25":151,"_110":154,"_144":155},"ca6fee0e-79c6-4f1c-b500-ee081a071949","Mecânica","mecanica","2025-04-29T02:19:12.828432+00:00",{"_138":152,"_140":141,"_142":153},10,19948,"mecanica-2",[156],{"_23":157,"_34":46,"_36":47,"_133":134,"_135":147,"_106":158,"_108":158,"_25":159,"_110":161,"_144":162},"ac55bdba-154d-42e7-bc25-72dde822663d","2025-04-29T03:44:29.139762+00:00",{"_138":141,"_140":141,"_142":160},3376,"resistencia-dos-materiais-4",[],"video",{"_23":101,"_165":96,"_97":-5,"_166":167,"_168":169,"_6":-5,"_170":-5,"_114":169,"_171":172,"_125":-5,"_173":-5,"_174":-5,"_175":-5,"_176":-5,"_25":177,"_178":141,"_179":-5,"_87":180,"_106":181,"_108":181},"title","duration",20,"source_type","prometheus","iframe_url","provider_video_id","1cd3d633-d690-4f4b-9995-9b0e44994e35","format","quality","file_path","file_size",{"_104":105},"views_count","last_played_at","draft","2025-08-19T11:56:20.758718+00:00","tag",{"_23":130,"_34":131,"_36":132,"_133":134,"_135":-5,"_106":136,"_108":136,"_25":184,"_110":132,"_144":185},{"_138":139,"_140":141,"_142":143},[186],{"_23":147,"_34":148,"_36":149,"_133":134,"_135":130,"_106":150,"_108":150,"_25":187,"_110":154,"_144":188},{"_138":152,"_140":141,"_142":153},[189],{"_23":157,"_34":46,"_36":47,"_133":134,"_135":147,"_106":158,"_108":158,"_25":190,"_110":161,"_144":191},{"_138":141,"_140":141,"_142":160},[],"similarQuestions",[194,225,245],{"_19":195,"_21":22,"_23":196,"_64":197,"_83":213,"_85":214,"_30":215},"112831da-be9c-458e-94cb-e8292d0dba12","3c0b9441-b878-4896-8bac-fa4ca61e3f13",[198,201,204,207,210],{"_23":199,"_68":69,"_70":200},"265bd71a-6749-41f5-b291-67efd646ca64","y'' + 2y' - 3y = 0",{"_23":202,"_68":74,"_70":203},"6d3c3a1d-8e80-4ff3-b97a-d5b5bca91940","6y'' - y' - y = 0",{"_23":205,"_68":78,"_70":206},"4168ac67-64fd-46ed-a69c-a01dbfcde7ea","y'' - 2y' + y = 0",{"_23":208,"_68":81,"_70":209},"cc7bc8df-6995-4264-ac5a-32ea9b4ed7db","9y'' + 6y' + y = 0",{"_23":195,"_68":211,"_70":212},"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

",[216],{"_23":217,"_34":218,"_36":219,"_133":134,"_135":220,"_106":221,"_108":221,"_25":222,"_110":219},"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",{"_138":223,"_140":141,"_142":224},4,12302,{"_19":226,"_21":22,"_23":227,"_64":228,"_83":240,"_85":241,"_30":242},"6831a9b9-c9a4-4a76-bd7a-2ce98bfea875","93a32dd4-5fc6-4ccb-a76f-4724bf113742",[229,232,235,237],{"_23":230,"_68":69,"_70":231},"13e3f476-a0a5-421f-b47e-0969e8e9a7e0","

Verdadeiro, verdadeiro, falso

",{"_23":233,"_68":74,"_70":234},"c9a20855-d84b-4860-8b35-9784818282cf","

Falso, verdadeiro, verdadeiro

",{"_23":226,"_68":78,"_70":236},"

Verdadeiro, falso, verdadeiro

",{"_23":238,"_68":81,"_70":239},"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