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\\tau_{min} = 66 \\text{ MPa}

",{"_43":85,"_80":86,"_82":87},"df3debb2-ea39-47a4-abc8-c078b7c2edd9","B","

\\tau_{min} = 77 \\text{ MPa}

",{"_43":89,"_80":90,"_82":91},"7f9e47b3-7a7e-496c-b8b1-c190b124be0a","C","

\\tau_{min} = 88 \\text{ MPa}

",{"_43":93,"_80":94,"_82":95},"58641256-4997-4293-9136-b9ec46c62708","D","

\\tau_{min} = 99 \\text{ MPa}

",{"_43":97,"_80":98,"_82":99},"a053ffc3-fa14-4a53-b137-1990a47ff149","E","

\\tau_{min} = 110 \\text{ MPa}

","solution_explanation","statement","

Sabendo que a tensão de cisalhamento não deve exceder 110 MPa, qual é o valor mínimo correspondente da tensão de cisalhamento no eixo circular?

","status","pending_classification","hasRelatedVideo","relatedVideos",[108],{"_109":110,"_128":129,"_50":144,"_179":180,"_197":198},"content",{"_43":111,"_56":112,"_113":-5,"_114":115,"_116":117,"_118":-5,"_45":119,"_122":123,"_124":123,"_125":-5,"_126":127},"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",{"_120":121},"external_id","4963","created_at","2025-08-20T12:27:41.608438+00:00","updated_at","content_group_id","slug","optica-geometrica-parte-1","prometheusData",{"_130":131,"_137":138,"_139":140,"_141":142,"_28":143},"provider",{"_43":132,"_133":134,"_135":136},"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",[145],{"_43":146,"_56":147,"_58":148,"_149":150,"_151":-5,"_122":152,"_124":152,"_45":153,"_126":148,"_160":161},"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",{"_154":155,"_156":157,"_158":159},"childrenCount",18,"content_count",0,"question_count",135489,"children",[162],{"_43":163,"_56":164,"_58":165,"_149":150,"_151":146,"_122":166,"_124":166,"_45":167,"_126":170,"_160":171},"ca6fee0e-79c6-4f1c-b500-ee081a071949","Mecânica","mecanica","2025-04-29T02:19:12.828432+00:00",{"_154":168,"_156":157,"_158":169},10,19948,"mecanica-2",[172],{"_43":173,"_56":57,"_58":59,"_149":150,"_151":163,"_122":174,"_124":174,"_45":175,"_126":177,"_160":178},"ac55bdba-154d-42e7-bc25-72dde822663d","2025-04-29T03:44:29.139762+00:00",{"_154":157,"_156":157,"_158":176},3376,"resistencia-dos-materiais-4",[],"video",{"_43":117,"_13":112,"_113":-5,"_181":182,"_183":184,"_28":-5,"_185":-5,"_130":184,"_186":187,"_141":-5,"_188":-5,"_189":-5,"_190":-5,"_191":-5,"_45":192,"_193":157,"_194":-5,"_103":195,"_122":196,"_124":196},"duration",20,"source_type","prometheus","iframe_url","provider_video_id","1cd3d633-d690-4f4b-9995-9b0e44994e35","format","quality","file_path","file_size",{"_120":121},"views_count","last_played_at","draft","2025-08-19T11:56:20.758718+00:00","tag",{"_43":146,"_56":147,"_58":148,"_149":150,"_151":-5,"_122":152,"_124":152,"_45":199,"_126":148,"_160":200},{"_154":155,"_156":157,"_158":159},[201],{"_43":163,"_56":164,"_58":165,"_149":150,"_151":146,"_122":166,"_124":166,"_45":202,"_126":170,"_160":203},{"_154":168,"_156":157,"_158":169},[204],{"_43":173,"_56":57,"_58":59,"_149":150,"_151":163,"_122":174,"_124":174,"_45":205,"_126":177,"_160":206},{"_154":157,"_156":157,"_158":176},[],"similarQuestions",[209,239,259],{"_40":210,"_42":41,"_43":211,"_76":212,"_100":227,"_101":228,"_50":229},"112831da-be9c-458e-94cb-e8292d0dba12","3c0b9441-b878-4896-8bac-fa4ca61e3f13",[213,216,219,222,225],{"_43":214,"_80":81,"_82":215},"265bd71a-6749-41f5-b291-67efd646ca64","y'' + 2y' - 3y = 0",{"_43":217,"_80":86,"_82":218},"6d3c3a1d-8e80-4ff3-b97a-d5b5bca91940","6y'' - y' - y = 0",{"_43":220,"_80":90,"_82":221},"4168ac67-64fd-46ed-a69c-a01dbfcde7ea","y'' - 2y' + y = 0",{"_43":223,"_80":94,"_82":224},"cc7bc8df-6995-4264-ac5a-32ea9b4ed7db","9y'' + 6y' + y = 0",{"_43":210,"_80":98,"_82":226},"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

",[230],{"_43":231,"_56":232,"_58":233,"_149":150,"_151":234,"_122":235,"_124":235,"_45":236,"_126":233},"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",{"_154":237,"_156":157,"_158":238},4,12302,{"_40":240,"_42":41,"_43":241,"_76":242,"_100":254,"_101":255,"_50":256},"6831a9b9-c9a4-4a76-bd7a-2ce98bfea875","93a32dd4-5fc6-4ccb-a76f-4724bf113742",[243,246,249,251],{"_43":244,"_80":81,"_82":245},"13e3f476-a0a5-421f-b47e-0969e8e9a7e0","

Verdadeiro, verdadeiro, falso

",{"_43":247,"_80":86,"_82":248},"c9a20855-d84b-4860-8b35-9784818282cf","

Falso, verdadeiro, verdadeiro

",{"_43":240,"_80":90,"_82":250},"

Verdadeiro, falso, verdadeiro

",{"_43":252,"_80":94,"_82":253},"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