[{"_1":2,"_332":-5,"_333":-5},"loaderData",{"_3":4,"_9":10},"root",{"_5":-5,"_6":7,"_8":-5},"registerModalConfig","url","https://prouniversitario.com/exercicios/ce418c76-9236-425b-8230-0cf315c0bd17","user","pages/questions/show/index",{"_11":12,"_13":14,"_15":14,"_16":17,"_86":87,"_163":164,"_29":234,"_304":305,"_328":329,"_330":331},"comments",[],"hasVideoHash",false,"isAiGenerated","question",{"_18":19,"_20":19,"_21":22,"_23":24,"_56":57,"_80":19,"_81":82,"_83":84,"_85":44},"correctOptionId","","difficulty","id","ce418c76-9236-425b-8230-0cf315c0bd17","metadata",{"_25":26,"_52":53,"_54":55},"pd",{"_27":28,"_29":30,"_31":32,"_39":40,"_41":42,"_43":44,"_45":46,"_47":44,"_48":49,"_50":51},"area","Outros","tags",[],"subjects",[33],{"_21":34,"_35":36,"_37":38},669639,"name","Termodinâmica","alias","termodinamica","materialId","138833567","answerCount",1,"isIndexable",true,"mainSubject",{"_21":34,"_35":36,"_37":38},"isIndexableSEO","materialFormat","Text","studentDegreeLevel","Graduate","source","PD","externalId","165445959","options",[58,64,68,72,76],{"_21":59,"_60":61,"_62":63},"e41e90f5-dbec-49d5-8075-ec88215e804d","label","A","text","0,38 m",{"_21":65,"_60":66,"_62":67},"7b4010a7-7461-4cf3-a04c-d347186f638f","B","0,37 m",{"_21":69,"_60":70,"_62":71},"7d1bb57d-5039-4158-a9de-4f82eab33f02","C","0,36 m",{"_21":73,"_60":74,"_62":75},"dfc2a9e7-547a-4e6d-b36d-627bc1eda712","D","0,34 m",{"_21":77,"_60":78,"_62":79},"e7519eb8-ffcd-4492-806a-a329297de431","E","0,35 m","solution_explanation","statement","

Usando o método de igual perda de carga através da tabela de redução de área, qual será a dimensão de uma tubulação sendo que o ramal principal tem vazão de 1000 \\, L/s com dimensão de 0,25 \\, m \\times 0,40 \\, m e que o ramal a ser dimensionado precisa manter a altura de 0,25 \\, m e tem uma vazão de 800 \\, L/s:

","status","pending_classification","hasRelatedVideo","relatedVideos",[88],{"_89":90,"_108":109,"_29":124,"_143":144,"_162":125},"content",{"_21":91,"_35":92,"_93":-5,"_94":95,"_96":97,"_98":-5,"_23":99,"_102":103,"_104":103,"_105":-5,"_106":107},"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",{"_100":101},"external_id","3659","created_at","2025-08-20T12:27:41.608438+00:00","updated_at","content_group_id","slug","calor-latente-parte-1","prometheusData",{"_110":111,"_117":118,"_119":120,"_121":122,"_6":123},"provider",{"_21":112,"_113":114,"_115":116},"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",[125],{"_21":126,"_35":127,"_37":128,"_129":130,"_131":132,"_102":133,"_104":133,"_23":134,"_106":128,"_141":142},"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",{"_135":136,"_137":138,"_139":140},"childrenCount",10,"content_count",0,"question_count",8575,"children",[],"video",{"_21":97,"_145":92,"_93":-5,"_146":147,"_148":149,"_6":-5,"_150":-5,"_110":149,"_151":152,"_121":-5,"_153":-5,"_154":-5,"_155":-5,"_156":-5,"_23":157,"_158":138,"_159":-5,"_83":160,"_102":161,"_104":161},"title","duration",17,"source_type","prometheus","iframe_url","provider_video_id","748a4adb-ad8f-4c5f-88a9-bc522d3b4246","format","quality","file_path","file_size",{"_100":101},"views_count","last_played_at","draft","2025-08-19T11:56:22.11908+00:00","tag","similarQuestions",[165,195,213],{"_18":166,"_20":19,"_21":167,"_56":168,"_80":183,"_81":184,"_29":185},"112831da-be9c-458e-94cb-e8292d0dba12","3c0b9441-b878-4896-8bac-fa4ca61e3f13",[169,172,175,178,181],{"_21":170,"_60":61,"_62":171},"265bd71a-6749-41f5-b291-67efd646ca64","y'' + 2y' - 3y = 0",{"_21":173,"_60":66,"_62":174},"6d3c3a1d-8e80-4ff3-b97a-d5b5bca91940","6y'' - y' - y = 0",{"_21":176,"_60":70,"_62":177},"4168ac67-64fd-46ed-a69c-a01dbfcde7ea","y'' - 2y' + y = 0",{"_21":179,"_60":74,"_62":180},"cc7bc8df-6995-4264-ac5a-32ea9b4ed7db","9y'' + 6y' + y = 0",{"_21":166,"_60":78,"_62":182},"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

",[186],{"_21":187,"_35":188,"_37":189,"_129":130,"_131":190,"_102":191,"_104":191,"_23":192,"_106":189},"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",{"_135":193,"_137":138,"_139":194},4,12302,{"_18":196,"_20":19,"_21":197,"_56":198,"_80":210,"_81":211,"_29":212},"6831a9b9-c9a4-4a76-bd7a-2ce98bfea875","93a32dd4-5fc6-4ccb-a76f-4724bf113742",[199,202,205,207],{"_21":200,"_60":61,"_62":201},"13e3f476-a0a5-421f-b47e-0969e8e9a7e0","

Verdadeiro, verdadeiro, falso

",{"_21":203,"_60":66,"_62":204},"c9a20855-d84b-4860-8b35-9784818282cf","

Falso, verdadeiro, verdadeiro

",{"_21":196,"_60":70,"_62":206},"

Verdadeiro, falso, verdadeiro

",{"_21":208,"_60":74,"_62":209},"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?

",[186],{"_18":214,"_20":19,"_21":215,"_56":216,"_80":231,"_81":232,"_29":233},"3ce109b1-6cdb-4b4d-8816-a6af4d74bed0","c478a46c-46a8-4942-ae14-2e79def3dda2",[217,220,223,225,228],{"_21":218,"_60":61,"_62":219},"a5a16a42-3e58-4c76-80f2-c291aa2d8e49","Sim",{"_21":221,"_60":66,"_62":222},"5e6d0254-696c-4da0-b572-82e3a19f433e","Não",{"_21":214,"_60":70,"_62":224},"Indeterminado",{"_21":226,"_60":74,"_62":227},"35403f68-a880-429d-97f4-db507c58de38","Todos",{"_21":229,"_60":78,"_62":230},"2e06d948-d37e-4350-afc7-ced04941bcaf","Nenhum","Os polinômios e números fornecidos estão corrompidos (símbolos “????”), impossibilitando aplicar Ruffini e concluir se são raízes. Sem dados completos, o problema é indeterminado.","Emplea la regla de Ruffini para dictaminar si los siguientes números son o no raíces de los polinomios citados:
a) ????3 de ????4????5
b) ????2 de ????2????????2
c) ????1 de 2????????1
d) ????1 de 2????2????",[186],[186,235,242,252,259,266,267,275,281,286,292,299],{"_21":236,"_35":36,"_37":38,"_129":130,"_131":237,"_102":133,"_104":133,"_23":238,"_106":241},"6b6faf95-0e3a-4f58-abc8-e1444c48041d","92339f4f-ed20-4f4c-ad85-0c3de7e04ae0",{"_135":239,"_137":138,"_139":240},2,11846,"termodinamica-3",{"_21":243,"_35":244,"_37":245,"_129":130,"_131":246,"_102":247,"_104":247,"_23":248,"_106":251},"861ad829-750d-4a44-9569-2f18b430b4df","Engenharia Mecatrônica","engenharia-mecatronica","eb9315c5-bd04-4116-aa46-02c4fde320c0","2025-04-29T02:19:12.828432+00:00",{"_135":249,"_137":138,"_139":250},5,30614,"engenharia-mecatronica-1",{"_21":253,"_35":254,"_37":255,"_129":130,"_131":243,"_102":191,"_104":191,"_23":256,"_106":258},"88fe96b3-2a17-4198-90d5-ddda3d6f5d45","Mecânica","mecanica",{"_135":249,"_137":138,"_139":257},25741,"mecanica-3",{"_21":237,"_35":260,"_37":261,"_129":130,"_131":246,"_102":247,"_104":247,"_23":262,"_106":265},"Engenharia Mecânica","engenharia-mecanica",{"_135":263,"_137":138,"_139":264},11,36318,"engenharia-mecanica-2",{"_21":126,"_35":127,"_37":128,"_129":130,"_131":132,"_102":133,"_104":133,"_23":134,"_106":128},{"_21":268,"_35":269,"_37":270,"_129":130,"_131":-5,"_102":271,"_104":271,"_23":272,"_106":270},"9772e958-72c6-4499-ad47-877cf94d670d","Exatas","exatas","2025-04-29T01:56:45.284172+00:00",{"_135":273,"_137":138,"_139":274},18,135489,{"_21":276,"_35":36,"_37":38,"_129":130,"_131":187,"_102":277,"_104":277,"_23":278,"_106":280},"b21374de-aae5-4a4a-9bf3-0420e731797e","2025-04-29T04:00:14.680867+00:00",{"_135":138,"_137":138,"_139":279},8597,"termodinamica-5",{"_21":190,"_35":282,"_37":283,"_129":130,"_131":246,"_102":247,"_104":247,"_23":284,"_106":285},"Engenharia de Petróleo","engenharia-de-petroleo",{"_135":249,"_137":138,"_139":194},"engenharia-de-petroleo-1",{"_21":132,"_35":287,"_37":288,"_129":130,"_131":268,"_102":247,"_104":247,"_23":289,"_106":291},"Física","fisica",{"_135":263,"_137":138,"_139":290},13777,"fisica-4",{"_21":246,"_35":293,"_37":294,"_129":130,"_131":-5,"_102":295,"_104":295,"_23":296,"_106":294},"Engenharia","engenharia","2025-04-29T01:55:38.635246+00:00",{"_135":297,"_137":138,"_139":298},29,351586,{"_21":300,"_35":36,"_37":38,"_129":130,"_131":253,"_102":301,"_104":301,"_23":302,"_106":303},"f5967cbe-d751-4b40-81f2-04eb5dead2a9","2025-04-29T04:00:13.215003+00:00",{"_135":138,"_137":138,"_139":279},"termodinamica-4","tagsTree",[306,310],{"_21":268,"_35":269,"_37":270,"_129":130,"_131":-5,"_102":271,"_104":271,"_23":272,"_106":270,"_141":307},[308],{"_21":132,"_35":287,"_37":288,"_129":130,"_131":268,"_102":247,"_104":247,"_23":289,"_106":291,"_141":309},[125],{"_21":246,"_35":293,"_37":294,"_129":130,"_131":-5,"_102":295,"_104":295,"_23":296,"_106":294,"_141":311},[312,318,322],{"_21":243,"_35":244,"_37":245,"_129":130,"_131":246,"_102":247,"_104":247,"_23":248,"_106":251,"_141":313},[314],{"_21":253,"_35":254,"_37":255,"_129":130,"_131":243,"_102":191,"_104":191,"_23":256,"_106":258,"_141":315},[316],{"_21":300,"_35":36,"_37":38,"_129":130,"_131":253,"_102":301,"_104":301,"_23":302,"_106":303,"_141":317},[],{"_21":237,"_35":260,"_37":261,"_129":130,"_131":246,"_102":247,"_104":247,"_23":262,"_106":265,"_141":319},[320],{"_21":236,"_35":36,"_37":38,"_129":130,"_131":237,"_102":133,"_104":133,"_23":238,"_106":241,"_141":321},[],{"_21":190,"_35":282,"_37":283,"_129":130,"_131":246,"_102":247,"_104":247,"_23":284,"_106":285,"_141":323},[324],{"_21":187,"_35":188,"_37":189,"_129":130,"_131":190,"_102":191,"_104":191,"_23":192,"_106":189,"_141":325},[326],{"_21":276,"_35":36,"_37":38,"_129":130,"_131":187,"_102":277,"_104":277,"_23":278,"_106":280,"_141":327},[],"videoDislikes",{},"videoLikes",{},"actionData","errors"]

Comentários