{"id":279,"date":"2014-12-13T20:26:12","date_gmt":"2014-12-13T20:26:12","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=279"},"modified":"2024-09-02T16:40:45","modified_gmt":"2024-09-02T16:40:45","slug":"resolucao-comentada-da-cederj-2015","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2015\/exercicios-da-cederj-2015\/resolucao-comentada-da-cederj-2015\/","title":{"rendered":"Resolu\u00e7\u00e3o comentada da CEDERJ 2015"},"content":{"rendered":"

RESOLU\u00c7\u00d5ES<\/b><\/span><\/span><\/span><\/p>\n

01- <\/span><\/span><\/span>Se a <\/b><\/span><\/span><\/span>velocidade angular W<\/b><\/span><\/span><\/span>, pelo enunciado \u00e9 <\/b><\/span><\/span><\/span>constante<\/b><\/span><\/span><\/span> (\u201cvarre o mesmo \u00e2ngulo no mesmo tempo\u201d), a <\/b><\/span><\/span><\/span>velocidade V tamb\u00e9m ser\u00e1 constante<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Velocidade escalar (V) de um MCU<\/b><\/span><\/span><\/span><\/p>\n

Para <\/b><\/span><\/span>qualquer m\u00f3vel em <\/b><\/span><\/span><\/span>MCU<\/b><\/span><\/span><\/span>, o espa\u00e7o percorrido <\/b><\/span><\/span><\/span>(\u0394S)<\/b><\/span><\/span><\/span> durante um <\/b><\/span><\/span><\/span>per\u00edodo (\u0394t=T <\/b><\/span><\/span><\/span>\u2013 tempo que demora para efetuar uma volta completa) ser\u00e1 <\/b><\/span><\/span><\/span>\u0394S=2\u03c0R, onde R \u00e9 o raio da circunfer\u00eancia.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u0394S=2\u03c0R e \u0394t=T<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 <\/span><\/span>V= \u0394S\/\u0394t\u00a0 —\u00a0 V=2\u03c0R\/T<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Observe na <\/b><\/span><\/span><\/span>express\u00e3o acima<\/b><\/span><\/span><\/span> que, para o mesmo intervalo de tempo <\/b><\/span><\/span><\/span>T,<\/b><\/span><\/span><\/span> a <\/b><\/span><\/span><\/span>velocidade V<\/b><\/span><\/span><\/span> (escalar, linear ou simplesmente velocidade) \u00e9 <\/b><\/span><\/span><\/span>diretamente proporcional ao raio R (\u00e0 medida que V varia, R variar\u00e1 proporcionalmente)<\/b><\/span><\/span><\/span> e o gr\u00e1fico VxR ser\u00e1 uma <\/b><\/span><\/span><\/span>reta inclinada.<\/b><\/span><\/span><\/span><\/p>\n

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

02- <\/span><\/span><\/span>Veja abaixo a express\u00e3o matem\u00e1tica da segunda lei de Ohm:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

A resist\u00eancia el\u00e9trica <\/b><\/span><\/span><\/span>R <\/b><\/span><\/span><\/span>\u00e9 <\/b><\/span><\/span><\/span>diretamente proporcional ao comprimento l do fio e \u00e0 resistividade \u03c1<\/b><\/span><\/span><\/span> (caracter\u00edstica do material que constitui o fio) e <\/b><\/span><\/span><\/span>inversamente proporcional \u00e0 \u00e1rea se se\u00e7\u00e3o reta transversal (S<\/b><\/span><\/span><\/span>) \u201cgrossura do fio\u201d.<\/b><\/span><\/span><\/span><\/p>\n

Assim, se voc\u00ea <\/b><\/span><\/span><\/span>dobrar o comprimento do fio estar\u00e1 dobrando o valor da resist\u00eancia el\u00e9trica R<\/b><\/span><\/span><\/span> do mesmo.<\/b><\/span><\/span><\/span><\/p>\n

R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = R e R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = 2R<\/b><\/span><\/span><\/span><\/p>\n

Corrente i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> no fio de <\/b><\/span><\/span><\/span>comprimento L<\/b><\/span><\/span><\/span>submetido \u00e0 voltagem <\/b><\/span><\/span><\/span>U<\/b><\/span><\/span><\/span>— R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = U\/i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> — <\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = U\/R. (I)<\/b><\/span><\/span><\/span><\/p>\n

Corrente i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> no fio de <\/b><\/span><\/span><\/span>comprimento 2L<\/b><\/span><\/span><\/span>submetido \u00e0 voltagem <\/b><\/span><\/span><\/span>U<\/b><\/span><\/span><\/span>— R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = U\/i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> — <\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = U\/2R. (II)<\/b><\/span><\/span><\/span><\/p>\n

Comparando (I) com (II)— <\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> =2i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

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

03- <\/span><\/span><\/span>N\u00e3o existindo atrito o sistema \u00e9 conservativo e, aplicando o teorema da conserva\u00e7\u00e3o da energia mec\u00e2nica nos dois planos inclinados:<\/b><\/span><\/span><\/p>\n

Bloco m<\/b><\/span><\/span><\/span>1 <\/b><\/span><\/span><\/sub><\/span>— <\/b><\/span><\/span>no ponto <\/b><\/span><\/span><\/span>mais alto (A)<\/b><\/span><\/span><\/span> — E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span> = m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.g.h= m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.g.h — <\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span> = m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>gh <\/b><\/span><\/span><\/span>—<\/b><\/span><\/span><\/span><\/p>\n

No ponto <\/b><\/span><\/span><\/span>mais baixo B<\/b><\/span><\/span><\/span>— E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span> = m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.g.h= m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.g.0 — <\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span> = m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span> = E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span> — m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.g.h = m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 — <\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = \u221a(2gh).<\/b><\/span><\/span><\/span><\/p>\n

Bloco m<\/b><\/span><\/span><\/span>2 <\/b><\/span><\/span><\/sub><\/span>— <\/b><\/span><\/span>no ponto <\/b><\/span><\/span><\/span>mais alto (A)<\/b><\/span><\/span><\/span> — E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span> = m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.g.h= m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.g.h — <\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span> = m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>gh <\/b><\/span><\/span><\/span>—<\/b><\/span><\/span><\/span><\/p>\n

No ponto <\/b><\/span><\/span><\/span>mais baixo B<\/b><\/span><\/span><\/span>— E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span> = m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.g.h= m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.g.0 — <\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span> = m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span> = E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span> — m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.g.h = m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 — <\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = \u221a(2gh).<\/b><\/span><\/span><\/span><\/p>\n

V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> (chegam \u00e0 base com a mesma velocidade).<\/b><\/span><\/span><\/span><\/p>\n

Sendo as <\/b><\/span><\/span><\/span>velocidades de m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> e m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> as mesmas<\/b><\/span><\/span><\/span>, observe nos planos inclinados das figuras que <\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> <\/b><\/p>\n

\"\"<\/p>\n

percorre uma dist\u00e2ncia maior que m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> e, consequentemente <\/b><\/span><\/span><\/span>demora mais tempo para chegar \u00e0 base.<\/b><\/span><\/span><\/span><\/p>\n

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

04- <\/span><\/span><\/span>O <\/b><\/span><\/span><\/span><\/span>\u00edndice de refra\u00e7\u00e3o <\/b><\/span><\/span><\/span><\/span>de um meio representa a <\/b><\/span><\/span><\/span><\/span>raz\u00e3o (compara\u00e7\u00e3o) entre a velocidade da luz no v\u00e1cuo e \u00a0a velocidade da luz no meio, ou seja:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Importante: Quanto <\/b><\/span><\/span><\/span><\/span>maior for o \u00edndice de refra\u00e7\u00e3o de um meio<\/b><\/span><\/span><\/span><\/span>, menor ser\u00e1 a velocidade da luz nele e <\/b><\/span><\/span><\/span><\/span>maior ser\u00e1 o desvio sofrido.<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

05- <\/span><\/span><\/span>Pelo enunciado, o <\/b><\/span><\/span><\/span>bloco B desloca-se para a direita<\/b><\/span><\/span><\/span> em rela\u00e7\u00e3o \u00e0 mesa ent\u00e3o ele deve se deslocar<\/b><\/span><\/span><\/span> <\/b>para a <\/b><\/span><\/span><\/span>direita tamb\u00e9m em rela\u00e7\u00e3o ao bloco A<\/b><\/span><\/span><\/span>, <\/b><\/span><\/span>que deve se mover para a <\/b><\/span><\/span><\/span>esquerda e, nele a for\u00e7a de atrito F<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> deve ser para a direita.<\/b><\/span><\/span><\/span><\/p>\n

Sendo as for\u00e7as de atrito contr\u00e1rias ao movimento (ou \u00e0 sua tend\u00eancia), <\/b><\/span><\/span>ambas, sobre o bloco B indicam para a esquerda.<\/b><\/span><\/span><\/span><\/p>\n

Sobre a <\/b><\/span><\/span><\/span>mesa <\/b><\/span><\/span><\/span>a for\u00e7a de atrito <\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span> deve ser para a direita.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Rela\u00e7\u00e3o entre F<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> e F<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>:<\/b><\/span><\/span><\/span><\/p>\n

Blocos id\u00eanticos possuem o mesmo peso P e os coeficientes de atrito tamb\u00e9m s\u00e3o os mesmos \u03bc.<\/b><\/span><\/span><\/span><\/p>\n

F<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=\u03bcN<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=\u03bcP<\/b><\/span><\/span><\/span> — <\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> = \u03bcN<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=\u03bc2P<\/b><\/span><\/span><\/span> — <\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> = 2 F<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

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

06- <\/span><\/span><\/span>a) <\/b><\/span><\/span><\/span>Peso <\/b><\/span><\/span><\/span>do saco de cimento — <\/b><\/span><\/span><\/span>P=<\/b><\/span><\/span><\/span>mg=50.10=<\/b><\/span><\/span><\/span>500N.<\/b><\/span><\/span><\/span><\/p>\n

Deste peso <\/b><\/span><\/span><\/span>P=500N apenas F=250N<\/b><\/span><\/span><\/span> s\u00e3o <\/b><\/span><\/span><\/span>transmitidos ao pedreiro<\/b><\/span><\/span><\/span> atrav\u00e9s do sistema de roldanas (veja figura):<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Como o saco de cimento sobe com <\/b><\/span><\/span><\/span>velocidade constante (equil\u00edbrio din\u00e2mico, for\u00e7a resultante nula)<\/b><\/span><\/span><\/span> a <\/b><\/span><\/span><\/span>for\u00e7a<\/b><\/span><\/span><\/span> que o pedreiro troca com a corda \u00e9 de <\/b><\/span><\/span><\/span>F=250N.<\/b><\/span><\/span><\/span><\/p>\n

b) O <\/b><\/span><\/span><\/span>trabalho realizado<\/b><\/span><\/span><\/span> pelo pedreiro para <\/b><\/span><\/span><\/span>elevar o saco de cimento de peso P=500N a uma altura de<\/b><\/span><\/span><\/span> <\/b>3m<\/b><\/span><\/span><\/span> vale — <\/b><\/span><\/span><\/span>W=F.d<\/b><\/span><\/span><\/span>=P.d=500.3 — <\/b><\/span><\/span><\/span>W<\/b><\/span><\/span><\/span> <\/b>= 1500 J.<\/b><\/span><\/span><\/span><\/p>\n

07- <\/span><\/span><\/span>a) <\/b><\/span><\/span><\/span>Trecho 1<\/b><\/span><\/span><\/span> — expans\u00e3o isob\u00e1rica <\/b><\/span><\/span><\/span>(press\u00e3o constante) P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = P<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span> =P<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

P<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>\/T<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span> = P<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\/T\u2019<\/b><\/span><\/span><\/span> — P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/T<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span> =P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\/T\u2019 — V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/T<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span> =3V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/T\u2019 — <\/b><\/span><\/span><\/span>T\u2019= 3T<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Trecho 2<\/b><\/span><\/span><\/span> — isovolum\u00e9trica a <\/b><\/span><\/span><\/span>volume constante 3V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> <\/b>— <\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/T\u2019 = P<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\/T<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span> — P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.3V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/3T<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span> = (P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/3).3V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/T<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span> <\/b><\/p>\n

— 3V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/3T<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span> = V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/T<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span> — 1\/T<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span> = 1\/T<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span> — <\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span> = T<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

b) Em <\/b><\/span><\/span><\/span>cada trecho o trabalho \u00e9 numericamente igual \u00e0 \u00e1rea:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

08- <\/b><\/span><\/span><\/span><\/span>a) a<\/b><\/span><\/span><\/span><\/span>I<\/b><\/span><\/span><\/span><\/sub><\/span> = (V \u2013 V<\/b><\/span><\/span><\/span><\/span>0<\/b><\/span><\/span><\/span><\/sub><\/span>)\/(t \u2013 t<\/b><\/span><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span><\/span>=(20 \u2013 20)\/(15 \u2013 0)=0\/15 — <\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/span><\/span>I<\/b><\/span><\/span><\/span><\/sub><\/span> = 0.<\/b><\/span><\/span><\/span><\/span><\/p>\n

a<\/b><\/span><\/span><\/span><\/span>II<\/b><\/span><\/span><\/span><\/sub><\/span> = (V \u2013 V<\/b><\/span><\/span><\/span><\/span>0<\/b><\/span><\/span><\/span><\/sub><\/span>)\/(t \u2013 t<\/b><\/span><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span><\/span>=(60 \u2013 0)\/(15 \u2013 0)=60\/15 — <\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/span><\/span>II<\/b><\/span><\/span><\/span><\/sub><\/span> = 4m\/s<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

b) S<\/b><\/span><\/span><\/span>I<\/b><\/span><\/span><\/sub><\/span> = S<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> + V<\/b><\/span><\/span><\/span>I<\/b><\/span><\/span><\/sub><\/span>.t<\/b><\/span><\/span><\/span> — S<\/b><\/span><\/span>I<\/b><\/span><\/span><\/sub> = 0 + 20.t — <\/b><\/span><\/span>S<\/b><\/span><\/span><\/span>I<\/b><\/span><\/span><\/sub><\/span> = 20t.<\/b><\/span><\/span><\/span><\/p>\n

S<\/b><\/span><\/span><\/span><\/span>II<\/b><\/span><\/span><\/span><\/sub><\/span> = S<\/b><\/span><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub><\/span> + V<\/b><\/span><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub><\/span>t + at<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\/2<\/b><\/span><\/span><\/span><\/span> — S<\/b><\/span><\/span><\/span>II<\/b><\/span><\/span><\/span><\/sub> = 0 + 0.t + 4t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup>\/2 — <\/b><\/span><\/span><\/span>S<\/b><\/span><\/span><\/span><\/span>II<\/b><\/span><\/span><\/span><\/sub><\/span> = 2t<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/span> <\/b><\/p>\n

No encontro, <\/b><\/span><\/span>S<\/b><\/span><\/span><\/span>I<\/b><\/span><\/span><\/sub><\/span> = S<\/b><\/span><\/span><\/span>II<\/b><\/span><\/span><\/sub><\/span> — 20t = 2t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> — t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup> \u2013 10t = 0 — t.(t \u2013 10)=0 — <\/b><\/span><\/span>t=10s.<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

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

RESOLU\u00c7\u00d5ES 01- Se a velocidade angular W, pelo enunciado \u00e9 constante (\u201cvarre o mesmo \u00e2ngulo no mesmo tempo\u201d), a velocidade V tamb\u00e9m ser\u00e1 constante. Velocidade escalar (V) de um MCU Para qualquer m\u00f3vel em MCU, o espa\u00e7o percorrido (\u0394S) durante um per\u00edodo (\u0394t=T \u2013 tempo que demora para efetuar uma volta completa) ser\u00e1 \u0394S=2\u03c0R, onde R \u00e9 o raio da<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":277,"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-279","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/279","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=279"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/279\/revisions"}],"predecessor-version":[{"id":281,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/279\/revisions\/281"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/277"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=279"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}