{"id":2461,"date":"2016-09-11T01:10:09","date_gmt":"2016-09-11T01:10:09","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=2461"},"modified":"2024-08-22T17:42:27","modified_gmt":"2024-08-22T17:42:27","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-grandezas-fisicas-sistema-de-unidades-e-analise-dimensional","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/introducao\/sistema-de-unidades-analise-dimensional\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-grandezas-fisicas-sistema-de-unidades-e-analise-dimensional\/","title":{"rendered":"Grandezas f\u00edsicas, Sistema de Unidades e An\u00e1lise Dimensional"},"content":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/p>\n

Grandezas f\u00edsicas, <\/b><\/span><\/span><\/span>Sistema de Unidades e An\u00e1lise Dimensional<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

\u00a0<\/span><\/p>\n

01-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>[V]=K.M<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sup><\/span>.T<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/sup><\/span>.(F\/A)<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/sup><\/span>=K.M<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sup><\/span>.T<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/sup><\/span>.(M.T<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>.L<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 [L]<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>=K.M<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sup><\/span>.T<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/sup><\/span>.M<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/sup><\/span>.T<\/b><\/span><\/span><\/span>-2n<\/b><\/span><\/span><\/sup><\/span>.L<\/b><\/span><\/span><\/span>-1n<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 (L<\/b><\/span><\/span><\/span>3+n<\/b><\/span><\/span><\/sup><\/span>).(T<\/b><\/span><\/span><\/span>2n<\/b><\/span><\/span><\/sup><\/span>) = K.(M<\/b><\/span><\/span><\/span>a + n<\/b><\/span><\/span><\/sup><\/span>).T<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 3 + n = 0\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0n= – 3\u00a0<\/b><\/span><\/span><\/span>\u00a0—\u00a0\u00a0 2n = b\u00a0 —\u00a0 2.(-3)=b\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0b= – 6\u00a0<\/b><\/span><\/span><\/span>\u00a0—\u00a0 a + n=0\u00a0 ===\u00a0 a \u2013 3=0\u00a0 —<\/b><\/span><\/span><\/span>\u00a0 a=3<\/b><\/span><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>G=<\/b><\/span><\/span><\/span>\u03c1<\/b><\/span><\/span><\/span><\/span>.c.A.v.(1\/V).<\/b><\/span><\/span><\/span>\u03b8<\/b><\/span><\/span><\/span><\/span>\u00a0 —\u00a0 [G]=((M.L<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>).(L<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.t<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span>\u03b8<\/b><\/span><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>).(L<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>).(L.t<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>).(L<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>).(<\/b><\/span><\/span><\/span>\u03b8<\/b><\/span><\/span><\/span><\/span>)\u00a0 —\u00a0\u00a0 [G]=M.L<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>.t<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Calorias corresponde \u00e0 unidade de energia, que no SI \u00e9 medida em joules (J)\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/span>a)r<\/b><\/span><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span><\/sub><\/span>=2,1.10<\/b><\/span><\/span><\/span><\/span>5<\/b><\/span><\/span><\/span><\/sup><\/span>.0,30\u00a0 —\u00a0 r<\/b><\/span><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span><\/sub><\/span>=0,63.10<\/b><\/span><\/span><\/span><\/span>5<\/b><\/span><\/span><\/span><\/sup><\/span>m=\u00a0 —\u00a0 b) (1\/r<\/b><\/span><\/span><\/span><\/span>f<\/b><\/span><\/span><\/span><\/sub><\/span>)\/(1\/r<\/b><\/span><\/span><\/span><\/span>s<\/b><\/span><\/span><\/span><\/sub><\/span>)=2,1.10<\/b><\/span><\/span><\/span><\/span>5<\/b><\/span><\/span><\/span><\/sup><\/span>\/0,63.10<\/b><\/span><\/span><\/span><\/span>5<\/b><\/span><\/span><\/span><\/sup><\/span>=3,33<\/b><\/span><\/span><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>O ano-luz \u00e9, ent\u00e3o, uma unidade de\u00a0<\/b><\/span><\/span><\/span>comprimento<\/b><\/span><\/span><\/span><\/span>\u00a0 —\u00a0 ela corresponde ao espa\u00e7o percorrido por um raio de luz em 1 ano, no v\u00e1cuo, com velocidade de 3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>m\/s\u00a0 —\u00a0 \u00e9 uma medida grande demais para nossas aplica\u00e7\u00f5es comuns aqui na Terra\u00a0 —\u00a0 essa unidade se destina a marcar dist\u00e2ncias no espa\u00e7o c\u00f3smico, entre as estrelas de uma mesma gal\u00e1xia ou entre gal\u00e1xias distintas. Ela \u00e9 \u00fatil para os astr\u00f4nomos.<\/b><\/span><\/span><\/span>\"\"<\/p>\n

06-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>d) 10<\/b><\/span><\/span><\/span>-8<\/b><\/span><\/span><\/sup><\/span>\u00a0g\u00a0 —\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span>R-B<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 veja teoria<\/b><\/span><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0erro \u2013 2h e n\u00e3o 2hrs\u00a0 —\u00a0 2<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>erro \u2013 110km e n\u00e3o 110Km (k min\u00fasculo e n\u00e3o K mai\u00fasculo)\u00a0 —\u00a0 3<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0erro \u2013 20min e n\u00e3o 20MIN\u00a0 —\u00a0 2kh e n\u00e3o 2KG\u00a0 —\u00a0 4<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>erro – 4L e n\u00e3o 5ltrs\u00a0 —\u00a0 5<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0erro \u2013 2h e n\u00e3o 2h.(com ponto)\u00a0 —\u00a0 6<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0erro – 80km\/h e n\u00e3o 80KM\/H\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- E<\/b><\/span><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 veja teoria<\/b><\/span><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe que 32% da luminosidade da superf\u00edcie vale\u00a0 —\u00a0 I=(32\/100)l\u00ado\u00a0 —\u00a0 I = l<\/b><\/span><\/span><\/span>0<\/b><\/span><\/span><\/sub><\/span>.0,8<\/b><\/span><\/span><\/span>h\/40<\/b><\/span><\/span><\/sup><\/span>\u00a0\u00a0 —\u00a0 (32\/100)l0 = l0.0,8h<\/b><\/span><\/span><\/span>\u00a0h\/40<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 cancelando I0\u00a0 —\u00a0 (32\/100) = 0,8<\/b><\/span><\/span><\/span>\u00a0h\/40<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 fazendo o logaritmo em ambos os lados da igualdade\u00a0 —\u00a0 log(32\/100) = log[0,8<\/b><\/span><\/span><\/span>\u00a0h\/40<\/b><\/span><\/span><\/sup><\/span>)]\u00a0 —\u00a0 propriedades do log\u00a0 —\u00a0\u00a0 log32 – log100 = (h\/40).log0,8\u00a0 —\u00a0 log2<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\u00a0– log10\u00b2 = (h\/40).log(8\/10)\u00a0 —\u00a0 5log2 – 2log10 = (h\/40).(log8 – log10)\u00a0 — 5.0,3 – 2 = (h\/40).(log2\u00b3 – 1)\u00a0 —\u00a0 1,5 – 2 = (h\/40).(3log2 – 1)\u00a0 —\u00a0 -0,5 = (h\/40).(3.0,3 – 1)\u00a0 —\u00a0 -0,5 = (h\/40).(-0,1)\u00a0 —\u00a0 (-0,5)\/(-0,1) = h\/40\u00a0 —\u00a0 5 = h\/40\u00a0 —\u00a0 h = 5.40\u00a0 —\u00a0 h = 200 cm=2,0m\u00a0 —\u00a0 R- C<\/b><\/span><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>E=X.c\u00a0 —\u00a0 X=E\/c=P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.\u2206t\/v\u00a0 —\u00a0 X=(F.v).\u2206t\/v\u00a0 —\u00a0 X=F.\u2206t=(m.a). \u2206t=m.(\u2206v\/\u2206t). \u2206t\u00a0 —\u00a0 X=m. \u2206t\u00a0 —\u00a0 unidades de X\u00a0 —\u00a0 kg.m\/s\u00a0 —\u00a0 esta unidade corresponde \u00e0 unidade de impulso ou de quantidade de movimento cujas unidades dimensionais s\u00e3o [X]=KG.M\/S=m.l.t<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>\u00a0 —<\/b><\/span><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Pela defini\u00e7\u00e3o, a for\u00e7a eletromotriz (<\/b><\/span><\/span><\/span>\u03b5<\/b><\/span><\/span><\/span><\/span>) \u00e9 o quociente de uma energia por uma carga el\u00e9trica (<\/b><\/span><\/span><\/span>\u03b5<\/b><\/span><\/span><\/span><\/span>=W\/Q)\u00a0 —\u00a0 voc\u00ea j\u00e1 viu na express\u00e3o (U=W\/Q), que a diferen\u00e7a de potencial (U)\u00a0 tamb\u00e9m \u00e9 o quociente de uma energia por uma carga el\u00e9trica\u00a0 —\u00a0 logo, for\u00e7a eletromotriz e diferen\u00e7a de potencial (e consequentemente potencial el\u00e9trico) s\u00e3o grandezas f\u00edsicas da mesma esp\u00e9cie\u00a0 —\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Nas unidades do sistema internacional (SI)\u00a0 —\u00a0 I=P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/S\u00a0 —\u00a0 I = W\/m<\/b><\/span><\/span><\/span>2\u00a0 —\u00a0 Po=W\/\u2206t\u00a0 —\u00a0 I=(J\/s)\/m2<\/b><\/span><\/span><\/sup><\/span>= J\/(s.m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>) = (N.m)\/(s.m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>)\u00a0 —\u00a0I = N\/(s.m) = (kg.m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>)\/(s.m) = kg.s\u00ad<\/b><\/span><\/span><\/span>3\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/sup><\/span>I=kg.s3\u00a0<\/b><\/span><\/span><\/sup><\/span>\u00a0—\u00a0 f=1\/T=1\/s\u00a0 —<\/b><\/span><\/span><\/span>\u00a0f = s\u00ad<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u03c1<\/b><\/span><\/span><\/span><\/span>=m\/v\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u03c1<\/b><\/span><\/span><\/span><\/span>= kg\/m<\/b><\/span><\/span><\/span>3\u00a0<\/b><\/span><\/span><\/sup><\/span>\u00a0—\u00a0\u00a0<\/b><\/span><\/span><\/sup><\/span>v = m\/s<\/b><\/span><\/span><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>a=m<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 substituindo na express\u00e3o dada e levando em conta que o coeficiente 2<\/b><\/span><\/span><\/span>\u03c0<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u00e9 adimensional\u00a0 —\u00a0 kg.s<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>=(s<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sup><\/span>(kg\/m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>).(m\/s).(m)<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 s<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>=s<\/b><\/span><\/span><\/span>-1x<\/b><\/span><\/span><\/sup><\/span>.m<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>.m.s<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>.m<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sup><\/span>\u00a0\u00a0 —\u00a0 s<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>=s<\/b><\/span><\/span><\/span>-x \u2013 1<\/b><\/span><\/span><\/sup><\/span>.m<\/b><\/span><\/span><\/span>y-2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 -x \u2013 1=-3\u00a0 —\u00a0 x=2\u00a0 —\u00a0 y-2=0\u00a0 —\u00a0 y=2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- A<\/b><\/span><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>O Sistema Internacional de Unidades (SI) adota o joule (J) para unidade de energia, em substitui\u00e7\u00e3o \u00e0 antiga unidade a caloria, que se referia mais especificamente ao calor, pois antigamente pensavam que o calor era um fluido que passava de um corpo para outro e n\u00e3o uma forma de energia (teoria do Cal\u00f3rico)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- A<\/b><\/span><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A primeira est\u00e1 errada\u00a0 —\u00a0 a correta \u00e9 a segunda, pois\u00a0 —\u00a0 F=2Ma=2(m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0+ m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>).v\/t\u00a0 —\u00a0 F=2(m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0+ m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>).vt<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 lembre-se de 2 \u00e9 um n\u00famero, portanto grandeza adimensional.<\/b><\/span><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>1kW equivale a 10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>W\u00a0 —\u00a0 1h equivale a 3,6.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>s\u00a0 —\u00a0 rela\u00e7\u00e3o entre Joule e w.s\u00a0 —\u00a0\u00a0 1J = 1w.s\u00a0 —\u00a0 1kWh = 10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>W . 3,6.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>s = 3,6.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>W.s\u00a0 —\u00a0 1kWh = 3,6.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>J\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>01- Correto\u00a0 —\u00a0 p=mg=M.(L\/T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 peso=M.(L.T<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span><\/p>\n

02. Correto\u00a0 —\u00a0 P=<\/b><\/span><\/span><\/span>\u03c1<\/b><\/span><\/span><\/span><\/span>.g.h=(M\/L<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>).(L\/T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>).(L)=M.L<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>.T<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span><\/p>\n

04) Correto\u00a0 —\u00a0 W=P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.t=(Fv)t=mavt\u00a0 —\u00a0 W=(M).(L\/T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>).(L\/T).(T)\u00a0 —\u00a0 W=ML<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>T<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span><\/p>\n

08) Falsa\u00a0 —\u00a0 a correta \u00e9 U=ML<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>T<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>I<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span><\/p>\n

R- (1 + 2 + 4) = 7\u00a0<\/b><\/span><\/span><\/span><\/p>\n

17-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>As unidades representadas s\u00e3o coulomb (C); watt (W); segundo (s); volt (V); tesla (T); metro (m); amp\u00e8re (A) e newton (N)\u00a0 —\u00a0\u00a0 a unidade tesla \u00e9 unidade de campo magn\u00e9tico, que aparece na express\u00e3o da for\u00e7a magn\u00e9tica em um fio associada a uma corrente el\u00e9trica que percorre este fio, ou seja, F = B.i.L, onde L \u00e9 o comprimento deste fio\u00a0 —\u00a0 em unidades SI\u00a0 —\u00a0 F = B.i.L\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\u00a0N = T.A.m\u00a0 —\u00a0 mas a unidade amp\u00e8re, A, equivale a coulomb\/segundo, C\/s\u00a0 —\u00a0 N = T.A.m\u00a0 —\u00a0 N = T.(C\/s).m\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

C = N.s\/(T.m)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span><\/p>\n

\u00a018-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Considere uma part\u00edcula de massa m e velocidade\u00a0\"\"\u00a0\u00a0—\u00a0 por defini\u00e7\u00e3o, a intensidade do momento angular\u00a0\"\"\u00a0em um determinado ponto O \u00e9 fornecida por I=m.r.v.sen<\/b><\/span><\/span><\/span>\u03b8<\/b><\/span><\/span><\/span><\/span>\u00a0 —\u00a0\u00a0\"\"\u00a0\u00e9 o vetor que une O ao produto dos vetores\u00a0\"\"\u00a0e\u00a0\"\"\u00a0—\u00a0 o\u00a0momento angular\u00a0\"\"\u00a0\u00e9 perpendicular ao plano que cont\u00e9m os vetores\u00a0\"\"\u00a0e\u00a0\"\"\u00a0\u00a0—\u00a0 I=m.r.v.sen<\/b><\/span><\/span><\/span>\u03b8<\/b><\/span><\/span><\/span><\/span>\u00a0 —\u00a0 sen<\/b><\/span><\/span><\/span>\u03b8<\/b><\/span><\/span><\/span><\/span>\u00a0\u00e9 uma grandeza adimensional\u00a0 —\u00a0 momento angular=(M).(L).(L\/T)=ML<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\"\"<\/p>\n

T<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/span><\/p>\n

19-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Trata-se da equa\u00e7\u00e3o de Torricelli, V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 2.a.h\u00a0 — com V=0\u00a0 —\u00a0 cuja dedu\u00e7\u00e3o \u00e9 feita a partir de V<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=<\/b><\/span><\/span><\/span>\u0394<\/b><\/span><\/span><\/span><\/span>S\/<\/b><\/span><\/span><\/span>\u0394<\/b><\/span><\/span><\/span><\/span>t\u00a0 —\u00a0 onde\u00a0\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=(V + V<\/b><\/span><\/span><\/span>0<\/b><\/span><\/span><\/sub><\/span>)\/2=h\/t\u00a0 —\u00a0 h=(V + V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)t\/2\u00a0 —\u00a0 h=V<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>.t\/2\u00a0 —\u00a0 observe nessa equa\u00e7\u00e3o que h \u00e9 proporcional a V<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>\u00a0e a t\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/span><\/p>\n

20-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Por defini\u00e7\u00e3o, pot\u00eancia \u00e9 a raz\u00e3o entre o trabalho realizado por uma for\u00e7a e o intervalo de tempo de realiza\u00e7\u00e3o\u00a0 —\u00a0\u00a0 P = W\/\u2206t\u00a0 —\u00a0 o trabalho por sua vez \u00e9 o produto entre a for\u00e7a e o deslocamento\u00a0 —\u00a0 W = F.d\u00a0 —\u00a0 for\u00e7a resultante (lei fundamental da din\u00e2mica)\u00a0 —\u00a0\u00a0 F = m.a\u00a0 —\u00a0 acelera\u00e7\u00e3o\u00a0 —\u00a0\u00a0 a = v\/\u2206t\u00a0 —\u00a0 v = d\/t\u00a0 —\u00a0 P = W\/\u2206t = (F.d)\/\u2206t = (m.a.d)\/\u2206t = [m.( v\/\u2206t).d]\/ \u2206t = (m.v.d)\/( \u2206t)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u00a0 —\u00a0 P=[m.( d\/\u2206t).d]\/(\u2206t)<\/b><\/span><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sup><\/span>\u00a0—\u00a0 P = m.d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/\u2206t<\/b><\/span><\/span><\/span>3\u00a0<\/b><\/span><\/span><\/sup><\/span>\u00a0—\u00a0 nas unidades SI\u00a0 —\u00a0\u00a0 P\u00a0 —\u00a0\u00a0 kg.m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/s<\/b><\/span><\/span><\/span>3\u00a0<\/b><\/span><\/span><\/sup><\/span>\u00a0—\u00a0<\/b><\/span><\/span><\/span>\u00a0R- C\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/p>\n

21-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Considerando S<\/b><\/span><\/span><\/span>mil\u00edmetros<\/b><\/span><\/span><\/sub><\/span>\u00a0a \u00e1rea da esfera maior e S<\/b><\/span><\/span><\/span>micr\u00f4metros<\/b><\/span><\/span><\/sub><\/span>\u00a0a da menor\u00a0 —\u00a0 S<\/b><\/span><\/span><\/span>mil\u00edmetros<\/b><\/span><\/span><\/sub><\/span>\/S<\/b><\/span><\/span><\/span>micr\u00f4metros<\/b><\/span><\/span><\/sub><\/span>=4<\/b><\/span><\/span><\/span>\u03c0<\/b><\/span><\/span><\/span><\/span>R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/4<\/b><\/span><\/span><\/span>\u03c0<\/b><\/span><\/span><\/span><\/span>r<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>= (R\/r)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=(10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>\/10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>\u00a0 – –\u00a0 S<\/b><\/span><\/span><\/span>milimetros<\/b><\/span><\/span><\/sub><\/span>=10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>S<\/b><\/span><\/span><\/span>micr\u00f4metros<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 a esfera maior cont\u00e9m 10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>\u00a0vezes mais cargas que a menor\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>maior<\/b><\/span><\/span><\/sub><\/span>=10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>.Q<\/b><\/span><\/span><\/span>menor<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 10.000 cargas elementares=10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>maior<\/b><\/span><\/span><\/sub><\/span>=10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>\u00a0 — Q<\/b><\/span><\/span><\/span>maior<\/b><\/span><\/span><\/sub><\/span>=10<\/b><\/span><\/span><\/span>10<\/b><\/span><\/span><\/sup><\/span>\u00a0cargas elementares\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/span><\/p>\n

22-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>O\u00a0el\u00e9tron-volt\u00a0\u00e9 uma\u00a0unidade que mede energia\u00a0 —\u00a0 equivale a 1,602.10<\/b><\/span><\/span><\/span>\u221219<\/b><\/span><\/span><\/sup><\/span>\u00a0joules (J)\u00a0 —\u00a0 seu s\u00edmbolo \u00e9\u00a0eV\u00a0e seu plural,<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/span>l\u00e9trons-volt\u00a0 —\u00a0 um el\u00e9tron-volt \u00e9 a quantidade de\u00a0energia\u00a0cin\u00e9tica ganha por um \u00fanico\u00a0el\u00e9tron\u00a0quandoacelerado por uma\u00a0diferen\u00e7a de potencial (ddp)\u00a0de um\u00a0volt (V), no\u00a0v\u00e1cuo\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span>\"\"<\/p>\n

Observa\u00e7\u00e3o\u00a0 —\u00a0 M\u00faltiplos do El\u00e9tron volt:<\/b><\/span><\/span><\/span><\/p>\n

keV<\/b><\/span><\/span><\/span>\u00a0(quilo eV): mil el\u00e9trons-volt =\u00a010<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0el\u00e9trons-volt.<\/b><\/span><\/span><\/span><\/p>\n

MeV<\/b><\/span><\/span><\/span>\u00a0(mega eV): 1 milh\u00e3o de el\u00e9trons-volt =\u00a010<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>\u00a0el\u00e9trons-volt.<\/b><\/span><\/span><\/span><\/p>\n

GeV<\/b><\/span><\/span><\/span>\u00a0(giga eV): 1 bilh\u00e3o (mil milh\u00f5es) de el\u00e9trons-volt =\u00a010<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>\u00a0el\u00e9trons-volt.<\/b><\/span><\/span><\/span><\/p>\n

TeV<\/b><\/span><\/span><\/span>\u00a0(tera eV): 1 trilh\u00e3o (mil bilh\u00f5es) de el\u00e9trons-volt = 10<\/b><\/span><\/span><\/span>12<\/b><\/span><\/span><\/sup><\/span>\u00a0el\u00e9trons-volt.<\/b><\/span><\/span><\/span><\/p>\n

23-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>De acordo com o texto o di\u00e2metro de um fio de cabelo pode medir cerca de 100.000nm=10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>nm=10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>.10<\/b><\/span><\/span><\/span>-9<\/b><\/span><\/span><\/sup><\/span>m =10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>m =10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>m =100.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>m=100mm\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- E<\/b><\/span><\/span><\/span><\/p>\n

24-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Joule \u00e9 a unidade que expressa a energia (calor) trocada pelos corpos em uma mudan\u00e7a de estado f\u00edsico ou em uma varia\u00e7\u00e3o de temperatura\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- A<\/b><\/span><\/span><\/span><\/p>\n

25-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A=F\/x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=MLT<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>\/L<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 A=MT<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>\/L\u00a0 —\u00a0 A=ML<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>T<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/p>\n

26-<\/b><\/span><\/span><\/span> <\/b>R- B\u00a0<\/b><\/span><\/span><\/span>\u00a0—\u00a0 veja teoria<\/b><\/span><\/span><\/span><\/p>\n

27-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>(01) Correta: pascal\u00a0 —\u00a0\u00a0 press\u00e3o =for\u00e7a\/\u00e1rea=m.a\/A=(kg).(m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>).(1\/m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>)=m<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>.kg.s<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

(02) Errada: newton\u00a0 —\u00a0 for\u00e7a = massa.acelera\u00e7\u00e3o =(kg).(m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>)=kg.m.s<\/b><\/span><\/span><\/span>-2\"\"<\/b><\/span><\/span><\/sup><\/span><\/p>\n

(04) Errada: joule\u00a0 —\u00a0 energia (trabalho) = for\u00e7a.deslocamento =m.a.\u2206S=(kg).(m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>).(m)=kg.m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.s<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span><\/p>\n

(08) Correta: watt\u00a0 —\u00a0\u00a0 pot\u00eancia =trabalho (energia)\/tempo=(kgm<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>s<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>).(1\/s)=kg.m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>s<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span><\/p>\n

R- (01 + 08) = 09<\/b><\/span><\/span><\/span><\/span><\/p>\n

28-<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

29-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Cuidado, a posi\u00e7\u00e3o, \u00e9 uma grandeza escalar, o deslocamento, pode ser escalar ou vetorial\u00a0 —\u00a0 no entanto, h\u00e1 o\u00a0<\/b><\/span><\/span><\/span>respectivamente<\/b><\/span><\/span><\/span><\/span>, esclarecendo que a grandeza A \u00e9 vetorial e a grandeza B \u00e9 escalar\u00a0 —\u00a0 as op\u00e7\u00f5es (c) e (d) s\u00e3o prontamente descartadas, pois apresentam grandezas que s\u00e3o ambas vetoriais e ambas escalares, respectivamente\u00a0 —\u00a0 momento de uma for\u00e7a \u00e9 uma grandeza vetorial e trabalho \u00e9 uma grandeza escalar, tendo ambos a mesma unidade, resultando do produto de uma for\u00e7a por um comprimento (N.m), apesar de seus significados f\u00edsicos serem totalmente diferentes\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span><\/p>\n

30-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Dados: B<\/b><\/span><\/span><\/span>CMS<\/b><\/span><\/span><\/sub><\/span>\u00a0= 4 teslas; B<\/b><\/span><\/span><\/span>Terra<\/b><\/span><\/span><\/sub><\/span>\u00a0= 30 micro-teslas = 30<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/sub><\/span>10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>\u00a0tesla.<\/b><\/span><\/span><\/span><\/p>\n

Raz\u00e3o\u00a0 —\u00a0 B<\/b><\/span><\/span><\/span>CMS<\/b><\/span><\/span><\/sub><\/span>\/B<\/b><\/span><\/span><\/span>Terra<\/b><\/span><\/span><\/sub><\/span>=4\/30.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>=133.333\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/span><\/p>\n

31-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Justificando a incorreta:<\/b><\/span><\/span><\/span><\/p>\n

08) A unidade quilograma (kg) pode ser utilizada\u00a0<\/b><\/span><\/span><\/span>apenas<\/b><\/span><\/span><\/span>\u00a0como unidade de massa\u00a0 —\u00a0 a grandeza peso \u00e9 uma for\u00e7a e, no SI, \u00e9 medida em\u00a0<\/b><\/span><\/span><\/span>newton<\/b><\/span><\/span><\/span>.\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/p>\n

R- (01 + 02 + 04) = 07<\/b><\/span><\/span><\/span><\/p>\n

32-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>1. Correta\u00a0 —\u00a0 6 m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= 6<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>(100 cm)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= 6<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/sub><\/span>10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>\u00a0cm<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= 60.000 cm<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

2. Correta\u00a0 —\u00a0 216 km\/h = 216\/3,6\u00a0m\/s = 60 m\/s.<\/b><\/span><\/span><\/span><\/p>\n

3. Errada\u00a0 —\u00a0 3.000 m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0= 3.000<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>(1.000 L) = 3.000.000 L.<\/b><\/span><\/span><\/span><\/p>\n

4. Correta\u00a0 —\u00a0 7.200 s =7.200\/3.600h = 2 h.<\/b><\/span><\/span><\/span><\/p>\n

5. Correta\u00a0 —<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>2,5<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\u00a0g = 2,5.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span> = \u00a02,5<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/sub><\/span>10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0kg = 250 g\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/p>\n

R- B<\/b><\/span><\/span><\/span><\/p>\n

33-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A errada \u00e9 a alternativa D\u00a0 —\u00a0 veja porqu\u00ea\u00a0 —\u00a0 PV=nRT\u00a0 —\u00a0 R=PV\/nT=press\u00e3o (atm).volume (litro)\/n(mols).temperatura (K)\u00a0 —\u00a0 R=atm.litro.mol<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>.K<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

34-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>I. Falsa\u00a0 —\u00a0 peso \u00e9 grandeza vetorial.<\/b><\/span><\/span><\/span><\/p>\n

II. Correta.<\/b><\/span><\/span><\/span><\/p>\n

III. Correta.<\/b><\/span><\/span><\/span><\/p>\n

IV. Falsa\u00a0 —\u00a0 temperatura \u00e9 grandeza\u00a0 escalar.\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span>R- D<\/b><\/span><\/span><\/span><\/span><\/p>\n

35-<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

R- E<\/b><\/span><\/span><\/span><\/p>\n

36-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>21,30 \u2013 9,39=11,61=11 + 0,61=11 + 61\/100\u00a0 —<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

37-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>De acordo com o enunciado, uma carga el\u00e9trica q de valor 1 statcoulomb, quando colocada a 1 cm de outra carga de mesmo valor, trocar\u00e1 for\u00e7a de intensidade de 1 dina\u00a0 —\u00a0 usando os dados e fazendo as convers\u00f5es de unidades para o S.I.\u00a0 —\u00a0 1N = 10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\u00a0dina\u00a0 —\u00a0\u00a0 1 dina = 10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/sup><\/span>\u00a0N\u00a0 —\u00a0 1 cm = 10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>\u00a0m\u00a0 —\u00a0 equa\u00e7\u00e3o da for\u00e7a el\u00e9trica\u00a0 —\u00a0 F=kQq\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

10<\/b><\/span><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/span><\/sup><\/span>=9.10<\/b><\/span><\/span><\/span><\/span>9<\/b><\/span><\/span><\/span><\/sup><\/span>q<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\/(10<\/b><\/span><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 q=1\/3.10<\/b><\/span><\/span><\/span><\/span>-9<\/b><\/span><\/span><\/span><\/sup><\/span>C\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/span><\/span><\/p>\n

38-<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

39-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>U=E.d\u00a0 —\u00a0 U=(F\/q).d\u00a0 —\u00a0 U=m.a.d\/q\u00a0 —\u00a0 i=q\/\u2206t\u00a0 —\u00a0 q=i. \u2206t\u00a0\u00a0—\u00a0 U=m.a.d\/i.\u2206t\u00a0 —\u00a0 U=(kg).(m\/s2).(m)\/A.s\u00a0 —\u00a0 U=kg.m2\/s3.A\u00a0 —\u00a0\u00a0R- B<\/b><\/span><\/span><\/p>\n

 <\/p>\n

40-\u00a0<\/b><\/span><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

41-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Pelo enunciado, voc\u00ea deve escolher (separar, distinguir) da melhor maneira poss\u00edvel as subst\u00e2ncias indicadas no gr\u00e1fico reflet\u00e2ncia x comprimento de onda\u00a0 —\u00a0 observe que, na faixa entre 0,4\u00b5m e 0,5\u00b5m as curvas est\u00e3o muito pr\u00f3ximas e os valores da reflet\u00e2ncia variando em torno de 2% a 8%, sendo muito dif\u00edcil distinguir essas subst\u00e2ncias\u00a0 —\u00a0 \u00e0 medida que o comprimento de onda vai aumentando, a diferen\u00e7a entre as reflet\u00e2ncias dessas subst\u00e2ncias vai aumentando, at\u00e9 que, entre 0,4=8\u00b5m e 0,9\u00b5m elas s\u00e3o as maiores fornecidas, j\u00e1 sendo poss\u00edvel distinguir essas subst\u00e2ncias pela reflet\u00e2ncia\u00a0 —\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span>R- E<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 Observa\u00e7\u00e3o: a \u00e1gua n\u00e3o est\u00e1 nessa faixa, mas, nessa faixa, quando a subst\u00e2ncia indicar reflet\u00e2ncia zero, voc\u00ea pode afirmar que se trata da \u00e1gua.<\/b><\/span><\/span><\/span><\/p>\n

42- <\/b><\/span><\/span><\/span>a) Do texto\u00a0 —\u00a0 \u201cSegundo todos os c\u00e1lculos, as futuras usinas de fus\u00e3o nuclear poder\u00e3o extrair de 1 metro c\u00fabico de \u00e1gua uma quantidade de energia igual \u00e0 de 2 mil barris de petr\u00f3leo\u201d\u00a0 —\u00a0 regra de tr\u00eas\u00a0 —\u00a0 1 m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 2.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0barris\u00a0 —\u00a0 100 m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 n barris\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

n=2.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 n=2.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\u00a0barris de petr\u00f3leo\u00a0 —\u00a0 como cada barril cont\u00e9m 1,5.10<\/b><\/span><\/span><\/span>6\u00a0<\/b><\/span><\/span><\/sup><\/span>kcal, 2.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\u00a0barris conter\u00e3o\u00a0 —\u00a0 W=2.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\u00a0barris x<\/b><\/span><\/span><\/span><\/p>\n

1,5.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>kcal\/barril\u00a0 —\u00a0 W=3,0.10<\/b><\/span><\/span><\/span>11<\/b><\/span><\/span><\/sup><\/span>\u00a0kcal=3,0.10<\/b><\/span><\/span><\/span>14<\/b><\/span><\/span><\/sup><\/span>cal\u00a0 —\u00a0 do enunciado\u00a0 —\u00a0 1 BEP (Barril Equivalente de Petr\u00f3leo), equivale a 1,45.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>\u00a0cal\u00a0 —\u00a0 regra de tr\u00eas\u00a0 —\u00a0 1 BEP \u2013 1,45.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>\u00a0cal\u00a0 —\u00a0 n\u2019 BEP \u2013 3.10<\/b><\/span><\/span><\/span>14<\/b><\/span><\/span><\/sup><\/span>\u00a0cal\u00a0 —\u00a0 n\u2019=3.10<\/b><\/span><\/span><\/span>14<\/b><\/span><\/span><\/sup><\/span>\u00a0cal\/1,45.10<\/b><\/span><\/span><\/span>9\u00a0<\/b><\/span><\/span><\/sup><\/span>cal\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

n\u2019\u22482,07.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\u00a0BEP<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

b) Do texto: \u201cOs centros dos n\u00facleos dos \u00e1tomos de hidrog\u00eanio devem estar a 1.10<\/b><\/span><\/span><\/span>-15<\/b><\/span><\/span><\/sup><\/span>\u00a0metros um do outro para que ocorra a fus\u00e3o\u201d\u00a0 —\u00a0 ainda do texto\u00a0 —\u00a0 \u201cessa fus\u00e3o \u00e9 o processo no qual dois n\u00facleos de \u00e1tomos leves (por exemplo, o hidrog\u00eanio \u2013 cujo n\u00facleo \u00e9 constitu\u00eddo por 1 pr\u00f3ton com carga el\u00e9trica elementar \u00e9 1,6.10<\/b><\/span><\/span><\/span>-19<\/b><\/span><\/span><\/sup><\/span>C) se combinam, ou se fundem, constituindo um elemento mais pesado. Os n\u00facleos, ent\u00e3o, carregados positivamente, devem se aproximar suficientemente um do outro, ou seja, vencer a for\u00e7a de repuls\u00e3o eletrost\u00e1tica entre eles\u201d\u00a0 —\u00a0 portanto, s\u00e3o dados\u00a0 —\u00a0 d=1.10<\/b><\/span><\/span><\/span>-15<\/b><\/span><\/span><\/sup><\/span>m\u00a0 —\u00a0 .|Q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>|= |Q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>|=1,6.10<\/b><\/span><\/span><\/span>-19<\/b><\/span><\/span><\/sup><\/span>C\u00a0 —\u00a0 k=9.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>N.m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/C<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

Lei de Coulomb\u00a0 —\u00a0\u00a0 F=k.|Q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>|. |Q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>|.\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= 9.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>.1,6.10<\/b><\/span><\/span><\/span>-19<\/b><\/span><\/span><\/sup><\/span>.1,6.10<\/b><\/span><\/span><\/span>-19<\/b><\/span><\/span><\/sup><\/span>\/(1.10<\/b><\/span><\/span><\/span>-15<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u00a0 —\u00a0 F=23,04.10<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0F=230,4 N<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

43<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe na express\u00e3o D=K.V2, onde K \u00e9 constante e, nela voc\u00ea observa que a velocidade V \u00e9 diretamente proporcional ao quadrado da velocidade\u00a0 —\u00a0 assim, se a velocidade dobra passando de 80kmh para 160kmh, a dist\u00e2ncia percorrida pelo carro at\u00e9 parar fica 22=4 vezes maior\u00a0 —\u00a0\u00a0R- C<\/b><\/span><\/span><\/p>\n

 <\/p>\n

Voltar para os exerc\u00edcios<\/span><\/a><\/h3>\n","protected":false},"excerpt":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre\u00a0 Grandezas f\u00edsicas, Sistema de Unidades e An\u00e1lise Dimensional \u00a0 \u00a0 01-\u00a0[V]=K.Ma.Tb.(F\/A)n=K.Ma.Tb.(M.T-2.L-1)n\u00a0 —\u00a0 [L]3=K.Ma.Tb.Mn.T-2n.L-1n\u00a0 —\u00a0 (L3+n).(T2n) = K.(Ma + n).Tb\u00a0 —\u00a0 3 + n = 0\u00a0 —\u00a0\u00a0n= – 3\u00a0\u00a0—\u00a0\u00a0 2n = b\u00a0 —\u00a0 2.(-3)=b\u00a0 —\u00a0\u00a0b= – 6\u00a0\u00a0—\u00a0 a + n=0\u00a0 ===\u00a0 a \u2013 3=0\u00a0 —\u00a0 a=3 02-\u00a0G=\u03c1.c.A.v.(1\/V).\u03b8\u00a0 —\u00a0 [G]=((M.L-3).(L2.t-2.\u03b8-1).(L2).(L.t-1).(L-3).(\u03b8)\u00a0 —\u00a0\u00a0 [G]=M.L-1.t-3\u00a0 —\u00a0\u00a0R- C 03-\u00a0Calorias<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":2456,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"om_disable_all_campaigns":false,"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"footnotes":""},"class_list":["post-2461","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2461","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/comments?post=2461"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2461\/revisions"}],"predecessor-version":[{"id":10784,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2461\/revisions\/10784"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2456"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=2461"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}