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Os técnicos de aviação devem conhecer os vários tipos de combustíveis e seus cuidados, de forma que o combustível adequado seja usado.

","status","pending_classification","hasRelatedVideo","relatedVideos",[97],{"_98":99,"_117":118,"_52":133,"_152":153,"_170":171},"content",{"_44":100,"_58":101,"_102":-5,"_103":104,"_105":106,"_107":-5,"_46":108,"_111":112,"_113":112,"_114":-5,"_115":116},"162087be-594f-460e-bbdd-a339ac2f10ad","Calor Latente - Parte 1","description","content_type_id","1051eee7-c62e-4b2c-9900-77a29fcec080","resource_id","f2b62fc2-7079-40e7-a5b5-c4a9769b205d","user_id",{"_109":110},"external_id","3659","created_at","2025-08-20T12:27:41.608438+00:00","updated_at","content_group_id","slug","calor-latente-parte-1","prometheusData",{"_119":120,"_126":127,"_128":129,"_130":131,"_28":132},"provider",{"_44":121,"_122":123,"_124":125},"06f120be-2194-4548-8336-7ba44ffbae14","key","grupo_q_proteus","kind","proteus","reference","bf8aacba-3de9-4b10-961c-47ada80dfe8c","state","completed","thumbnail_url","https://proteus.grupoq.io/videos/bf8aacba-3de9-4b10-961c-47ada80dfe8c/thumbnailsthumb.0000000.jpg","https://proteus.grupoq.io/videos/bf8aacba-3de9-4b10-961c-47ada80dfe8c/levels.m3u8",[134],{"_44":135,"_58":136,"_60":137,"_138":139,"_140":141,"_111":142,"_113":142,"_46":143,"_115":137,"_150":151},"96193fa0-9495-45b2-81cc-b547604354af","Física Térmica - Termologia","fisica-termica-termologia","tag_group_id","c1ce2cf2-b217-4cc7-8182-81c56bd3522d","parent_id","c8ab2a95-fd84-4d6f-ade0-835d34ba00f9","2025-04-29T03:44:27.560682+00:00",{"_144":145,"_146":147,"_148":149},"childrenCount",10,"content_count",0,"question_count",8575,"children",[],"video",{"_44":106,"_13":101,"_102":-5,"_154":155,"_156":157,"_28":-5,"_158":-5,"_119":157,"_159":160,"_130":-5,"_161":-5,"_162":-5,"_163":-5,"_164":-5,"_46":165,"_166":147,"_167":-5,"_92":168,"_111":169,"_113":169},"duration",17,"source_type","prometheus","iframe_url","provider_video_id","748a4adb-ad8f-4c5f-88a9-bc522d3b4246","format","quality","file_path","file_size",{"_109":110},"views_count","last_played_at","draft","2025-08-19T11:56:22.11908+00:00","tag",{"_44":135,"_58":136,"_60":137,"_138":139,"_140":141,"_111":142,"_113":142,"_46":172,"_115":137,"_150":173},{"_144":145,"_146":147,"_148":149},[],"similarQuestions",[176,209,229],{"_40":177,"_42":43,"_44":178,"_78":179,"_89":197,"_90":198,"_52":199},"112831da-be9c-458e-94cb-e8292d0dba12","3c0b9441-b878-4896-8bac-fa4ca61e3f13",[180,183,186,190,194],{"_44":181,"_81":82,"_83":182},"265bd71a-6749-41f5-b291-67efd646ca64","y'' + 2y' - 3y = 0",{"_44":184,"_81":87,"_83":185},"6d3c3a1d-8e80-4ff3-b97a-d5b5bca91940","6y'' - y' - y = 0",{"_44":187,"_81":188,"_83":189},"4168ac67-64fd-46ed-a69c-a01dbfcde7ea","C","y'' - 2y' + y = 0",{"_44":191,"_81":192,"_83":193},"cc7bc8df-6995-4264-ac5a-32ea9b4ed7db","D","9y'' + 6y' + y = 0",{"_44":177,"_81":195,"_83":196},"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

",[200],{"_44":201,"_58":202,"_60":203,"_138":139,"_140":204,"_111":205,"_113":205,"_46":206,"_115":203},"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",{"_144":207,"_146":147,"_148":208},4,12302,{"_40":210,"_42":43,"_44":211,"_78":212,"_89":224,"_90":225,"_52":226},"6831a9b9-c9a4-4a76-bd7a-2ce98bfea875","93a32dd4-5fc6-4ccb-a76f-4724bf113742",[213,216,219,221],{"_44":214,"_81":82,"_83":215},"13e3f476-a0a5-421f-b47e-0969e8e9a7e0","

Verdadeiro, verdadeiro, falso

",{"_44":217,"_81":87,"_83":218},"c9a20855-d84b-4860-8b35-9784818282cf","

Falso, verdadeiro, verdadeiro

",{"_44":210,"_81":188,"_83":220},"

Verdadeiro, falso, verdadeiro

",{"_44":222,"_81":192,"_83":223},"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