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Neste caso, a barra está apoiada em apenas uma extremidade, com um apoio do tipo engastado. Trata-se da barra:

","status","pending_classification","hasRelatedVideo","relatedVideos",[109],{"_110":111,"_129":130,"_52":145,"_180":181,"_198":199},"content",{"_44":112,"_58":113,"_114":-5,"_115":116,"_117":118,"_119":-5,"_46":120,"_123":124,"_125":124,"_126":-5,"_127":128},"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",{"_121":122},"external_id","4963","created_at","2025-08-20T12:27:41.608438+00:00","updated_at","content_group_id","slug","optica-geometrica-parte-1","prometheusData",{"_131":132,"_138":139,"_140":141,"_142":143,"_28":144},"provider",{"_44":133,"_134":135,"_136":137},"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",[146],{"_44":147,"_58":148,"_60":149,"_150":151,"_152":-5,"_123":153,"_125":153,"_46":154,"_127":149,"_161":162},"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",{"_155":156,"_157":158,"_159":160},"childrenCount",18,"content_count",0,"question_count",135489,"children",[163],{"_44":164,"_58":165,"_60":166,"_150":151,"_152":147,"_123":167,"_125":167,"_46":168,"_127":171,"_161":172},"ca6fee0e-79c6-4f1c-b500-ee081a071949","Mecânica","mecanica","2025-04-29T02:19:12.828432+00:00",{"_155":169,"_157":158,"_159":170},10,19948,"mecanica-2",[173],{"_44":174,"_58":59,"_60":61,"_150":151,"_152":164,"_123":175,"_125":175,"_46":176,"_127":178,"_161":179},"ac55bdba-154d-42e7-bc25-72dde822663d","2025-04-29T03:44:29.139762+00:00",{"_155":158,"_157":158,"_159":177},3376,"resistencia-dos-materiais-4",[],"video",{"_44":118,"_13":113,"_114":-5,"_182":183,"_184":185,"_28":-5,"_186":-5,"_131":185,"_187":188,"_142":-5,"_189":-5,"_190":-5,"_191":-5,"_192":-5,"_46":193,"_194":158,"_195":-5,"_104":196,"_123":197,"_125":197},"duration",20,"source_type","prometheus","iframe_url","provider_video_id","1cd3d633-d690-4f4b-9995-9b0e44994e35","format","quality","file_path","file_size",{"_121":122},"views_count","last_played_at","draft","2025-08-19T11:56:20.758718+00:00","tag",{"_44":147,"_58":148,"_60":149,"_150":151,"_152":-5,"_123":153,"_125":153,"_46":200,"_127":149,"_161":201},{"_155":156,"_157":158,"_159":160},[202],{"_44":164,"_58":165,"_60":166,"_150":151,"_152":147,"_123":167,"_125":167,"_46":203,"_127":171,"_161":204},{"_155":169,"_157":158,"_159":170},[205],{"_44":174,"_58":59,"_60":61,"_150":151,"_152":164,"_123":175,"_125":175,"_46":206,"_127":178,"_161":207},{"_155":158,"_157":158,"_159":177},[],"similarQuestions",[210,240,260],{"_40":211,"_42":43,"_44":212,"_78":213,"_101":228,"_102":229,"_52":230},"112831da-be9c-458e-94cb-e8292d0dba12","3c0b9441-b878-4896-8bac-fa4ca61e3f13",[214,217,220,223,226],{"_44":215,"_81":82,"_83":216},"265bd71a-6749-41f5-b291-67efd646ca64","y'' + 2y' - 3y = 0",{"_44":218,"_81":87,"_83":219},"6d3c3a1d-8e80-4ff3-b97a-d5b5bca91940","6y'' - y' - y = 0",{"_44":221,"_81":91,"_83":222},"4168ac67-64fd-46ed-a69c-a01dbfcde7ea","y'' - 2y' + y = 0",{"_44":224,"_81":95,"_83":225},"cc7bc8df-6995-4264-ac5a-32ea9b4ed7db","9y'' + 6y' + y = 0",{"_44":211,"_81":99,"_83":227},"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

",[231],{"_44":232,"_58":233,"_60":234,"_150":151,"_152":235,"_123":236,"_125":236,"_46":237,"_127":234},"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",{"_155":238,"_157":158,"_159":239},4,12302,{"_40":241,"_42":43,"_44":242,"_78":243,"_101":255,"_102":256,"_52":257},"6831a9b9-c9a4-4a76-bd7a-2ce98bfea875","93a32dd4-5fc6-4ccb-a76f-4724bf113742",[244,247,250,252],{"_44":245,"_81":82,"_83":246},"13e3f476-a0a5-421f-b47e-0969e8e9a7e0","

Verdadeiro, verdadeiro, falso

",{"_44":248,"_81":87,"_83":249},"c9a20855-d84b-4860-8b35-9784818282cf","

Falso, verdadeiro, verdadeiro

",{"_44":241,"_81":91,"_83":251},"

Verdadeiro, falso, verdadeiro

",{"_44":253,"_81":95,"_83":254},"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