{"id":4864,"date":"2020-01-10T03:06:21","date_gmt":"2020-01-10T03:06:21","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=4864"},"modified":"2024-09-04T14:26:43","modified_gmt":"2024-09-04T14:26:43","slug":"resolucoes-das-questoes-de-vestibulares-por-assunto-de-2018-2019","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/vestibulares-recentes\/mecanica\/cinematica-2018-2019\/resolucoes-das-questoes-de-vestibulares-por-assunto-de-2018-2019\/","title":{"rendered":"Resolu\u00e7\u00f5es das quest\u00f5es de vestibulares por assunto de 2018\/2019"},"content":{"rendered":"

Resolu\u00e7\u00f5es das quest\u00f5es de vestibulares por assunto de 2018\/2019<\/b><\/span><\/span><\/span><\/p>\n

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

<\/p>\n

R- A<\/p>\n

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

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

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

<\/p>\n

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

<\/p>\n

<\/p>\n

Se\u00a0<\/b><\/span><\/span>at\u00e9 o encontro A se moveu durante 40 s\u00a0<\/b><\/span><\/span>e\u00a0<\/b><\/span><\/span>B durante 50s,\u00a0<\/b><\/span><\/span>ent\u00e3o\u00a0<\/b><\/span><\/span>B partiu 10 s depois de A.<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

Ou,\u00a0<\/b><\/span><\/span>simples racioc\u00ednio:\u00a0<\/b><\/span><\/span>Se, quando eles se\u00a0<\/b><\/span><\/span>encontram A se moveu durante 40 s,\u00a0<\/b><\/span><\/span>ent\u00e3o com a\u00a0<\/b><\/span><\/span>mesma velocidade, B se moveu durante\u00a0<\/b><\/span><\/span>(90 \u2013 40 =\u00a0<\/b><\/span><\/span>50 s<\/b><\/span><\/span>) o que significa que\u00a0<\/b><\/span><\/span>B partiu 10 segundos depois de A<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

Basta\u00a0<\/b><\/span><\/span>prestar aten\u00e7\u00e3o no enunciado\u00a0<\/b><\/span><\/span>que afirma que o\u00a0<\/b><\/span><\/span>n\u00famero de voltas efetuadas pela roda traseira a cada pedalada \u00e9 calculado dividindo-se a quantidade de dentes da coroa pela quantidade de dentes da catraca\u00a0<\/b><\/span><\/span>e, fazendo isso voc\u00ea\u00a0<\/b><\/span><\/span>obt\u00e9m a tabela abaixo:<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

Voc\u00ea\u00a0<\/b><\/span><\/span>deve escolher a menor raz\u00e3o<\/b><\/span><\/span>\u00a0que \u00e9 aquela em que\u00a0<\/b><\/span><\/span>a roda traseira percorre a menor dist\u00e2ncia para fazer o percurso o mais devagar poss\u00edvel, fornecida pela alternativa IV.<\/b><\/span><\/span><\/span><\/span><\/p>\n

R \u2013 D<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

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

<\/p>\n

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

<\/p>\n

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

Veja as figuras abaixo:<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

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

09-<\/span><\/span><\/span><\/p>\n

Veja abaixo,<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a\u00a0<\/b><\/span><\/span><\/span>figura ilustrativa da situa\u00e7\u00e3o apresentada:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

10-<\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

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

<\/p>\n

Bola 2<\/b><\/span><\/span><\/span><\/p>\n

O mesmo \u00e9 v\u00e1lido para a bola 2 s\u00f3 que agora Epa\u00a0= 0,95%Epd<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>m2gH = 0,95m2.gh2\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>h2\u00a0= 0,95H.<\/b><\/span><\/span><\/span><\/p>\n

Como \u00e9\u00a0<\/b><\/span><\/span><\/span>pedida a diferen\u00e7a h2\u00a0\u2013 h1\u00a0voc\u00ea ter\u00e1\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>h2\u00a0\u2013 h1<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>= 0,95H \u2013 0,75H\u00a0<\/b><\/span><\/span><\/span>= 0,20H<\/b><\/span><\/span><\/span><\/p>\n

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

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

<\/p>\n

Em m\/s<\/b><\/span><\/span><\/span>\u00a0\u00a060\/3,<\/b><\/span><\/span><\/span>6 = 16,66 m\/s\u00a0<\/b><\/span><\/span><\/span>\u00a080\/3,6\u00a0<\/b><\/span><\/span><\/span>= 22,22 m\/s\u00a0<\/b><\/span><\/span><\/span>\u00a0veja na alternativa A que\u00a0<\/b><\/span><\/span><\/span>15 m\/s<\/b><\/span><\/span><\/span>x3,6\u00a0<\/b><\/span><\/span><\/span>= 54 km\/h<\/b><\/span><\/span><\/span>\u00a0e que\u00a0<\/b><\/span><\/span><\/span>25 m\/s<\/b><\/span><\/span><\/span>x3,6\u00a0<\/b><\/span><\/span><\/span>= 90 km\/h,\u00a0<\/b><\/span><\/span><\/span>ou seja,\u00a0<\/b><\/span><\/span><\/span>60 km\/h e 90 km\/h est\u00e3o compreendidos entre 54 km\/h e 90 km\/h.<\/b><\/span><\/span><\/span><\/p>\n

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

13-<\/span><\/span><\/span><\/p>\n

Resumo dos gr\u00e1ficos do espa\u00e7o, velocidade e acelera\u00e7\u00e3o do MUV, com a > 0 e com<\/b><\/span><\/span><\/span><\/p>\n

a < 0.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

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

14-<\/b><\/span><\/span><\/span>
\nColocando a origem das alturas no ponto Q\u00a0<\/b><\/span><\/span><\/span>e calculando a\u00a0<\/b><\/span><\/span><\/span>energia mec\u00e2nica do corpo de massa m nos pontos P e Q:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

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

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

<\/p>\n

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

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

Localizando os m\u00f3veis no plano cartesiano:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

17-<\/span><\/span><\/span><\/p>\n

<\/p>\n

No caso do exerc\u00edcio<\/b><\/span><\/span><\/span>:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

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

18-<\/span><\/span><\/span><\/p>\n

Carro A\u00a0\u00a0VA\u00a0= dA\/tA<\/b><\/span><\/span><\/span>\u00a0\u00a0V = L\/tA\u00a0\u00a0<\/b><\/span><\/span><\/span>tA\u00a0= L\/V.<\/b><\/span><\/span><\/span><\/p>\n

Carro B\u00a0\u00a0VB\u00a0= dB\/tB<\/b><\/span><\/span><\/span>\u00a0\u00a02V = 4L\/tB\u00a0\u00a0tB\u00a0= 4L\/2V\u00a0\u00a0<\/b><\/span><\/span><\/span>tB\u00a0= 2(L\/V) = 2 tA.<\/b><\/span><\/span><\/span><\/p>\n

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

19-<\/span><\/span><\/span><\/p>\n

<\/p>\n

20-<\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

21-<\/span><\/span><\/span><\/p>\n

<\/p>\n

R \u2013 B<\/b><\/span><\/span><\/span><\/p>\n

22-<\/span><\/span><\/span><\/p>\n

Os\u00a0<\/b><\/span><\/span>dois carros A e B\u00a0da figura mant\u00eam-se\u00a0lado a lado\u00a0numa pista\u00a0circular<\/b><\/span><\/span>,\u00a0<\/b><\/span><\/span>no mesmo intervalo de tempo\u00a0\u0394t.<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

Seus\u00a0raios\u00a0nesse\u00a0MCU\u00a0s\u00e3o, respectivamente,\u00a0R<\/b><\/span><\/span>A<\/b><\/span><\/span><\/span>\u00a0e R<\/b><\/span><\/span>B<\/b><\/span><\/span><\/span>, com R<\/b><\/span><\/span>B<\/b><\/span><\/span><\/span>\u00a0> R<\/b><\/span><\/span>A<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

O\u00a0<\/b><\/span><\/span>carro B<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>ter\u00e1\u00a0maior velocidade (escalar, linear) V=\u0394S\/\u0394t, pois dever\u00e1\u00a0percorrer\u00a0maior dist\u00e2ncia \u0394S\u00a0para, no mesmo intervalo de tempo, poder acompanhar o carro A, ou seja, V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/span>\u00a0> V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/span>.\u00a0\u00a0<\/b><\/span><\/span>Mas, como\u00a0<\/b><\/span><\/span>\u201c<\/b><\/span><\/span>varrem\u201d o mesmo \u00e2ngulo (\u0394\u03c6) no mesmo intervalo de tempo<\/b><\/span><\/span>,\u00a0<\/b><\/span><\/span>suas\u00a0<\/b><\/span><\/span>velocidades angulares (W= \u0394\u03c6\/\u0394t) ser\u00e3o iguais, ou seja,\u00a0W<\/b><\/span><\/span>A\u00a0<\/b><\/span><\/span><\/span>= W<\/b><\/span><\/span>B<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

A\u00a0<\/b><\/span><\/span>velocidade angular W \u00e9 a mesma para os dois carros\u00a0<\/b><\/span><\/span>(veja teoria acima) e como a\u00a0<\/b><\/span><\/span>intensidade da for\u00e7a centr\u00edpeta sobre cada um \u00e9 fornecida por F<\/b><\/span><\/span>c<\/b><\/span><\/span><\/span>\u00a0= m.W<\/b><\/span><\/span>2<\/b><\/span><\/span><\/span>.R,<\/b><\/span><\/span>\u00a0essa express\u00e3o indica\u00a0<\/b><\/span><\/span>que F<\/b><\/span><\/span>c<\/b><\/span><\/span><\/span>\u00a0\u00e9 diretamente proporcional ao raio R.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Ent\u00e3o, se voc\u00ea\u00a0<\/b><\/span><\/span>dobrar R estar\u00e1 dobrando F<\/b><\/span><\/span>c<\/b><\/span><\/span><\/span>\u00a0(desde que as massas sejam as mesmas).<\/b><\/span><\/span><\/span><\/span><\/p>\n

R \u2013 C<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

23-<\/span><\/span><\/span><\/p>\n

S\u00e3o dados:<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/span>\u00a0=\u00a0<\/b><\/span><\/span>10 m\/s;\u00a0<\/b><\/span><\/span>V =\u00a0<\/b><\/span><\/span>20 m\/s e\u00a0<\/b><\/span><\/span>t =\u00a0<\/b><\/span><\/span>20 s.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Equa\u00e7\u00e3o da velocidade do MUV\u00a0V = V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/span>\u00a0+ a.t\u00a0<\/b><\/span><\/span>20 = 10 + a.20 a = 10\/20 a = 0,5 m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/span>. (acelera\u00e7\u00e3o do movimento que \u00e9 constante)<\/b><\/span><\/span><\/span><\/span><\/p>\n

Equa\u00e7\u00e3o hor\u00e1ria do espa\u00e7o do MUV<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u2206S = V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/span>.t + a.t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/span>\/2\u00a0<\/b><\/span><\/span>= 10.20 + 0,5.20<\/b><\/span><\/span>2<\/b><\/span><\/span><\/span>\/2 = 200 + 0,5.200<\/b><\/span><\/span><\/span><\/span><\/p>\n

\u2206<\/span>S = 300 m<\/b><\/span><\/span><\/span><\/span><\/p>\n

R \u2013 B<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

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

Corpo abandonado de certa altura h do solo<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

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

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

Velocidade angular (W) de um MCU<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

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

 <\/p>\n

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

<\/p>\n

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

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

Veja o formul\u00e1rio abaixo<\/b><\/span><\/span><\/span><\/p>\n

Corpo abandonado de certa altura h do solo<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

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

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

<\/p>\n

R \u2013 A<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

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

<\/p>\n

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

 <\/p>\n

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

A\u00a0<\/b><\/span><\/span><\/span>incorreta \u00e9 a E, pois no instante t = 10s\u00a0<\/b><\/span><\/span><\/span>eles\u00a0<\/b><\/span><\/span><\/span>apresentam a mesma velocidade (V = 4 m\/s)<\/b><\/span><\/span><\/span>\u00a0e\u00a0<\/b><\/span><\/span><\/span>n\u00e3o a mesma posi\u00e7\u00e3o (localiza\u00e7\u00e3o).<\/b><\/span><\/span><\/span><\/p>\n

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

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

<\/p>\n

<\/p>\n

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

 <\/p>\n

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

<\/p>\n

<\/p>\n

<\/p>\n

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

 <\/p>\n

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

Observe nas equa\u00e7\u00f5es (S = So\u00a0+ V.t) de cada movimento que VA\u00a0= 40 m\/s e VB\u00a0= 50 m\/s.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Colocando\u00a0<\/b><\/span><\/span><\/span>a\u00a0<\/b><\/span><\/span><\/span>origem em B (cB\u00a0= 0), quando t = 0 voc\u00ea ter\u00e1 SA\u00a0\u2013 SB\u00a0= SoA\u00a0\u2013 SoB\u00a0= 50 SoA\u00a0\u2013 0 = 50 SoA\u00a0= 50 m (quando B partiu de SoB\u00a0= 0, A j\u00e1 estava na posi\u00e7ao SoA\u00a0= cA\u00a0= 50 m.<\/b><\/span><\/span><\/span><\/p>\n

Equa\u00e7\u00e3o de A\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>SA\u00a0= 50 + VAt\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>SA\u00a0= 50 + 40t.<\/b><\/span><\/span><\/span><\/p>\n

Equa\u00e7\u00e3o de B\u00a0<\/b><\/span><\/span><\/span>\u00a0SB\u00a0= 0 + 50t\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>SB\u00a0= 50t.<\/b><\/span><\/span><\/span><\/p>\n

No encontro\u00a0\u00a0SA\u00a0= SB<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>50 + 40t = 50t\u00a0\u00a010t = 50\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>t = 5 s.<\/b><\/span><\/span><\/span><\/p>\n

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

 <\/p>\n

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

A\u00a0<\/b><\/span><\/span><\/span>defini\u00e7\u00e3o de repouso ou de movimento<\/b><\/span><\/span><\/span>\u00a0\u00e9 a seguinte: \u201c<\/b><\/span><\/span><\/span>Um corpo est\u00e1 em repouso ou em movimento em rela\u00e7\u00e3o a outro corpo quando a dist\u00e2ncia entre ele variar no decorrer do tempo. Caso contr\u00e1rio estar\u00e1 em repouso\u201d.<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

A\u00a0<\/b><\/span><\/span><\/span>\u00fanica afirmativa que satisfaz a defini\u00e7\u00e3o acima \u00e9 a D.<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

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

 <\/p>\n

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

Ultrapassagem<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

Trem (caminh\u00e3o) de comprimento (x) atravessando um t\u00fanel (ponte) de comprimento ( c )\u00a0<\/b><\/span><\/span><\/span>\u00a0observe que\u00a0<\/b><\/span><\/span><\/span>para atravessar completamente o t\u00fanel (ponte), um ponto P fixo em qualquer parte do<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

trem (caminh\u00e3o) deve percorrer uma dist\u00e2ncia\u00a0<\/b><\/span><\/span><\/span>\u0394S = x + c,<\/b><\/span><\/span><\/span>\u00a0com\u00a0<\/b><\/span><\/span><\/span>velocidade V num intervalo de tempo \u0394t\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>V = .<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

Ponte<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

T\u00fanel<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

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

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

<\/p>\n

<\/p>\n

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

 <\/p>\n

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

Lan\u00e7amento horizontal<\/b><\/span><\/span><\/span><\/p>\n

Decomp\u00f5e-se o movimento em\u00a0<\/b><\/span><\/span><\/span>duas parcelas:<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span>Segundo o eixo X<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00a0 trata-se de um\u00a0<\/b><\/span><\/span><\/span>movimento horizontal uniforme com velocidade constante de<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>intensidade\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span>, que \u00e9 a\u00a0<\/b><\/span><\/span><\/span>velocidade de lan\u00e7amento<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>S = S<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span>\u00a0+ V.t<\/b><\/span><\/span><\/span>\u00a0\u00a0X= 0 + V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span>.t\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span>X =V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span>.t<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span>Segundo o eixo\u00a0<\/b><\/span><\/span><\/span>Y<\/b><\/span><\/span><\/span>\u00a0\u00a0 trata-se de um\u00a0<\/b><\/span><\/span><\/span>movimento uniformemente variado com velocidade inicial<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>o\u00a0<\/b><\/span><\/span>= 0,<\/b><\/span><\/span><\/span>\u00a0ou seja, \u00e9 uma\u00a0<\/b><\/span><\/span><\/span>queda livre com o corpo abandonado da origem<\/b><\/span><\/span><\/span>,\u00a0sujeito\u00a0<\/b><\/span><\/span><\/span>apenas \u00e0 acelera\u00e7\u00e3o da<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>gravidade, de intensidade\u00a0g, dire\u00e7\u00e3o vertical e sentido para baixo<\/b><\/span><\/span><\/span>.\u00a0<\/b><\/span><\/span><\/span><\/p>\n

Equa\u00e7\u00f5es:<\/b><\/span><\/span><\/span><\/span><\/p>\n

S = S<\/b><\/span><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span>\u00a0+ V<\/b><\/span><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span>.t + at<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span>\/2<\/b><\/span><\/span><\/span><\/span>\u00a0\u00a0 Y= 0 + 0.t + gt<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span>\/2\u00a0\u00a0<\/b><\/span><\/span><\/span><\/span>Y = g.t<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span>\/2<\/b><\/span><\/span><\/span><\/span><\/p>\n

V<\/b><\/span><\/span><\/span><\/span>y<\/b><\/span><\/span><\/span>\u00a0= V<\/b><\/span><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/span>\u00a0+ a.t<\/b><\/span><\/span><\/span><\/span>\u00a0\u00a0 V<\/b><\/span><\/span><\/span><\/span>y<\/b><\/span><\/span><\/span>\u00a0= 0 + g.t \u00a0\u00a0<\/b><\/span><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span><\/span>y<\/b><\/span><\/span><\/span>\u00a0= g.t<\/b><\/span><\/span><\/span><\/span><\/p>\n

V<\/b><\/span><\/span><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/span>= V<\/b><\/span><\/span><\/span><\/span>o2<\/b><\/span><\/span><\/span>\u00a0+ 2.a.\u0394S<\/b><\/span><\/span><\/span><\/span>\u00a0\u00a0 V<\/b><\/span><\/span><\/span><\/span>y2<\/b><\/span><\/span><\/span>\u00a0= V<\/b><\/span><\/span><\/span><\/span>oy2<\/b><\/span><\/span><\/span>\u00a0+ 2.g.\u0394h \u00a0\u00a0 V<\/b><\/span><\/span><\/span><\/span>y2<\/b><\/span><\/span><\/span>\u00a0= 0<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span>\u00a0+ 2.g.\u0394h\u00a0\u00a0<\/b><\/span><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span><\/span>y2<\/b><\/span><\/span><\/span>\u00a0= 2.g.\u0394h<\/b><\/span><\/span><\/span><\/span><\/p>\n

No caso do exerc\u00edcio:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

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

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

Para\u00a0<\/b><\/span><\/span><\/span>cada volta completa da roda<\/b><\/span><\/span><\/span>\u00a0a\u00a0<\/b><\/span><\/span><\/span>dist\u00e2ncia percorrida DIST\u2019 \u00e9 fornecida pelo comprimento da circunfer\u00eancia da roda fornecida<\/b><\/span><\/span><\/span>\u00a0por\u00a0<\/b><\/span><\/span><\/span>DIST\u2019 = 2\u03c0R<\/b><\/span><\/span><\/span>\u00a0(uma volta completa) onde\u00a0<\/b><\/span><\/span><\/span>R \u00e9 o raio da roda.<\/b><\/span><\/span><\/span>\u00a0Se a\u00a0<\/b><\/span><\/span><\/span>roda efetua n voltas a dist\u00e2ncia percorrida ser\u00e1 DIST = n.2\u03c0R.<\/b><\/span><\/span><\/span><\/p>\n

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

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

<\/p>\n

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

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

<\/p>\n

 <\/p>\n

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

Resolu\u00e7\u00f5es das quest\u00f5es de vestibulares por assunto de 2018\/2019 01- R- A 02- 03- 04- Se\u00a0at\u00e9 o encontro A se moveu durante 40 s\u00a0e\u00a0B durante 50s,\u00a0ent\u00e3o\u00a0B partiu 10 s depois de A. R- B Ou,\u00a0simples racioc\u00ednio:\u00a0Se, quando eles se\u00a0encontram A se moveu durante 40 s,\u00a0ent\u00e3o com a\u00a0mesma velocidade, B se moveu durante\u00a0(90 \u2013 40 =\u00a050 s) o que significa que\u00a0B<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":4862,"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-4864","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4864","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=4864"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4864\/revisions"}],"predecessor-version":[{"id":4866,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4864\/revisions\/4866"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4862"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=4864"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}