{"id":826,"date":"2015-05-14T17:13:24","date_gmt":"2015-05-14T17:13:24","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=826"},"modified":"2024-09-03T12:44:48","modified_gmt":"2024-09-03T12:44:48","slug":"resolucao-comentada-da-ufrgs-2015","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2015\/exercicios-resolvidos-da-ufrgs-2015\/resolucao-comentada-da-ufrgs-2015\/","title":{"rendered":"Resolu\u00e7\u00e3o Comentada da UFRGS 2015"},"content":{"rendered":"

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

01- <\/span><\/span><\/span>1\u00b0 caso:<\/b><\/span><\/span><\/span> se os dois m\u00f3veis estiverem andando no <\/b><\/span><\/span><\/span>mesmo sentido<\/b><\/span><\/span><\/span>, o valor da <\/b><\/span><\/span><\/span>velocidade relativa<\/b><\/span><\/span><\/span> \u00e9 dado pelo m\u00f3dulo <\/b><\/span><\/span><\/span>da diferen\u00e7a entre os m\u00f3dulos das duas velocidades escalares<\/b><\/span><\/span><\/span>. Dessa forma, temos:<\/b><\/span><\/span><\/span><\/p>\n

|v<\/b><\/span><\/span><\/span><\/em>rel<\/b><\/span><\/span><\/sub><\/span><\/em>\u00a0<\/b><\/i><\/span><\/span><\/span>|= |v<\/b><\/span><\/span><\/span><\/em>A<\/b><\/span><\/span><\/sub><\/span><\/em>\u00a0<\/b><\/i><\/span><\/span><\/span>|- |v<\/b><\/span><\/span><\/span><\/em>B<\/b><\/span><\/span><\/sub><\/span><\/em>\u00a0<\/b><\/i><\/span><\/span><\/span>|<\/b><\/span><\/span><\/span><\/em><\/p>\n

2\u00b0 caso:<\/b><\/span><\/span><\/span> se os dois m\u00f3veis estiverem andando em <\/b><\/span><\/span><\/span>sentidos contr\u00e1rios,<\/b><\/span><\/span><\/span> o <\/b><\/span><\/span><\/span>valor absoluto da<\/b><\/span><\/span><\/span> <\/b>velocidade relativa<\/b><\/span><\/span><\/span> \u00e9 dado pela <\/b><\/span><\/span><\/span>soma dos m\u00f3dulos das duas velocidades escalares<\/b><\/span><\/span><\/span>. Assim, temos:<\/b><\/span><\/span><\/span><\/p>\n

|v<\/b><\/span><\/span><\/span><\/em>rel<\/b><\/span><\/span><\/sub><\/span><\/em>\u00a0<\/b><\/i><\/span><\/span><\/span>|= |v<\/b><\/span><\/span><\/span><\/em>A<\/b><\/span><\/span><\/sub><\/span><\/em>\u00a0<\/b><\/i><\/span><\/span><\/span>|+ |v<\/b><\/span><\/span><\/span><\/em>B<\/b><\/span><\/span><\/sub><\/span><\/em>\u00a0<\/b><\/i><\/span><\/span><\/span>|<\/b><\/span><\/span><\/span><\/em><\/p>\n

No caso do exerc\u00edcio eles se movem em sentidos contr\u00e1rios \"\" <\/b><\/span><\/span><\/span>|v<\/b><\/span><\/span><\/span><\/em>rel<\/b><\/span><\/span><\/sub><\/span><\/em>\u00a0<\/b><\/i><\/span><\/span><\/span>|= |v<\/b><\/span><\/span><\/span><\/em>A<\/b><\/span><\/span><\/sub><\/span><\/em>\u00a0<\/b><\/i><\/span><\/span><\/span>|+ |v<\/b><\/span><\/span><\/span><\/em>B<\/b><\/span><\/span><\/sub><\/span><\/em>\u00a0<\/b><\/i><\/span><\/span><\/span>| = |360<\/b><\/span><\/span><\/span><\/em>\u00a0<\/b><\/i><\/span><\/span><\/span>|+ |360<\/b><\/span><\/span><\/span><\/em>\u00a0<\/b><\/i><\/span><\/span><\/span>|\"\"<\/b><\/span><\/span><\/span><\/em><\/p>\n

|v<\/b><\/span><\/span><\/em>rel<\/b><\/span><\/sub><\/span><\/em>\u00a0<\/b><\/i><\/span><\/span>| = 720 km\/h\/3,6 = 200 m\/s.<\/b><\/span><\/span><\/em><\/span><\/span><\/span><\/p>\n

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

02- <\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = 0 \"\" V = 540\/3,6 = 150m\/s \"\" \u2206t = 2,5minx60 = 150s \"\" a = \u2206V\/\u2206t = <\/b><\/span><\/span><\/span>(150 \u2013 0)\/150<\/b><\/span><\/span><\/span> \"\" a = 1m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

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

03- <\/span><\/span><\/span>Trata-se de um lan\u00e7amento obl\u00edquo onde,na vertical (Y) voc\u00ea tem as fun\u00e7\u00f5es(equa\u00e7\u00f5es), com os respectivos gr\u00e1ficos<\/b><\/span><\/span><\/span><\/p>\n

Equa\u00e7\u00e3o da posi\u00e7\u00e3o (dos espa\u00e7os) ou da altura Y<\/b><\/span><\/span><\/span><\/p>\n

S=S<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.t + a.t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 Y=Y<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>.t \u2013 gt<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 Y=0 + V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>.t \u2013 gt<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 — \u00a0Y= V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>.t \u2013 gt<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2, com V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.sen\u03b1\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Equa\u00e7\u00e3o da velocidade V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span><\/p>\n

V=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ a.t\u00a0 —\u00a0\u00a0 V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 g.t, com V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.sen\u03b1\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Movimento parcial na dire\u00e7\u00e3o horizontal (eixo X)<\/b><\/span><\/span><\/span><\/p>\n

Sendo o movimento apenas na horizontal, a <\/b><\/span><\/span><\/span>proje\u00e7\u00e3o do vetor acelera\u00e7\u00e3o da gravidade (\"\") no eixo X<\/b><\/span><\/span><\/span> \u00e9 <\/b><\/span><\/span><\/span>nula<\/b><\/span><\/span><\/span> e, nesse caso o movimento \u00e9 uniforme (MU) com<\/b><\/span><\/span><\/span> velocidade constante V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>, <\/b><\/span><\/span><\/span>de equa\u00e7\u00e3o<\/b><\/span><\/span><\/span> X = V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>t e cujo gr\u00e1fico da velocidade V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span> , constante, \u00e9 uma reta paralela ao eixo dos tempos:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

04- <\/span><\/span><\/span>I. Correta<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>Segunda lei de Kepler (lei das \u00e1reas)<\/b><\/span><\/span><\/span><\/p>\n

\u201c O <\/b><\/span><\/span><\/span>segmento de reta<\/b><\/span><\/span><\/span> imagin\u00e1rio que une o centro do Sol ao centro do planeta <\/b><\/span><\/span><\/span>descreve \u00e1reas<\/b><\/span><\/span><\/span> <\/b><\/p>\n

\"\"<\/p>\n

proporcionais\u00a0 aos tempos gastos para percorr\u00ea-las\u201d<\/b><\/span><\/span><\/span><\/p>\n

O <\/b><\/span><\/span><\/span>enunciado afirma que o<\/b><\/span><\/span><\/span> planeta percorre esses intervalos de posi\u00e7\u00f5es em tempos iguais.<\/b><\/span><\/span><\/span><\/p>\n

II. Falsa<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>Terceira lei de Kepler (lei dos per\u00edodos)<\/b><\/span><\/span><\/span><\/p>\n

\u201c Os <\/b><\/span><\/span><\/span>quadrados dos per\u00edodos<\/b><\/span><\/span><\/span>\u00a0T\u00a0de revolu\u00e7\u00e3o dos planetas (tempo que demora para efetuar uma volta completa em torno do Sol) <\/b><\/span><\/span><\/span>s\u00e3o proporcionais aos cubos das suas dist\u00e2ncias m\u00e9dias R\u00a0ao Sol<\/b><\/span><\/span><\/span>\u201d<\/b><\/span><\/span><\/span><\/p>\n

T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>=constante=K.<\/b><\/span><\/span><\/span><\/p>\n

No caso do exerc\u00edcio <\/b><\/span><\/span><\/span>R=P<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> = kP<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span> \"\" o correto seria: o <\/b><\/span><\/span><\/span>quadrado <\/b><\/span><\/span><\/span>do per\u00edodo da \u00f3rbita \u00e9 proporcional a <\/b><\/span><\/span><\/span>p<\/b><\/i><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span> .<\/b><\/span><\/span><\/span><\/p>\n

III. Correta<\/b><\/span><\/span><\/span> \"\" Pela lei das \u00e1reas: \u201c O <\/b><\/span><\/span><\/span>segmento de reta<\/b><\/span><\/span><\/span> imagin\u00e1rio que une o centro do Sol ao centro do planeta <\/b><\/span><\/span><\/span>descreve \u00e1reas<\/b><\/span><\/span><\/span> <\/b>proporcionais\u00a0 aos tempos gastos para percorr\u00ea-las\u201d<\/b><\/span><\/span><\/span><\/p>\n

Ent\u00e3o:<\/b><\/span><\/span><\/span><\/p>\n

A<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/\u2206t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>~A<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/\u2206t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=constante=K<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

Observe na <\/b><\/span><\/span><\/span>express\u00e3o acima<\/b><\/span><\/span><\/span> que quando <\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=A<\/b><\/span><\/span><\/span>2 \"\" <\/b><\/span><\/span><\/sub><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>\u2206t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=\u00a0\u2206t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>,<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>ou seja, para o arco maior <\/b><\/span><\/span><\/span>3,4<\/b><\/span><\/span><\/span>, ser percorrido no <\/b><\/span><\/span><\/span>mesmo intervalo de tempo<\/b><\/span><\/span><\/span> que o arco menor <\/b><\/span><\/span><\/span>1,2,<\/b><\/span><\/span><\/span> a <\/b><\/span><\/span><\/span>velocidade em 3,4<\/b><\/span><\/span><\/span> (mais perto do Sol –\u00a0<\/b><\/span><\/span><\/span>peri\u00e9lio<\/b><\/span><\/span><\/span>) deve ser <\/b><\/span><\/span><\/span>maior que a velocidade em 1,2<\/b><\/span><\/span><\/span> (mais afastado do Sol \u2013\u00a0<\/b><\/span><\/span><\/span>af\u00e9lio<\/b><\/span><\/span><\/span>).<\/b><\/span><\/span><\/span><\/p>\n

Portanto\u00a0<\/b><\/span><\/span><\/span>os planetas <\/b><\/span><\/span><\/span>aceleram do af\u00e9lio para o peri\u00e9lio e retardam do peri\u00e9lio para o af\u00e9lio<\/b><\/span><\/span><\/span>. <\/b><\/span><\/span><\/span><\/p>\n

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

05- <\/span><\/span><\/span>Em todo gr\u00e1fico <\/b><\/span><\/span><\/span>Fxt <\/b><\/span><\/span><\/span>o <\/b><\/span><\/span><\/span>impulso<\/b><\/span><\/span><\/span> recebido pela aplica\u00e7\u00e3o da for\u00e7a \u00e9 <\/b><\/span><\/span><\/span>numericamente igual \u00e0 \u00e1rea sob a<\/b><\/span><\/span><\/span> <\/b>curva<\/b><\/span><\/span><\/span> \"\"\"\" <\/b><\/span><\/span><\/span>I = \u00e1rea = (B + b).h\/2<\/b><\/span><\/span><\/span> = (2 + 1).4\/2 = 12\/2 \"\" <\/b><\/span><\/span><\/span>I = 6 N.s.<\/b><\/span><\/span><\/span><\/p>\n

O <\/b><\/span><\/span><\/span>teorema do impulso<\/b><\/span><\/span><\/span> afirma que o <\/b><\/span><\/span><\/span>impulso \u00e9 igual \u00e0 varia\u00e7\u00e3o da quantidade de movimento (momentum linear) \"\" I = \u2206Q = 6 N.s \"\" \u2206Q = m\u2206V = <\/b><\/span><\/span><\/span>m (V \u2013 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>) \"\" 6 = 1.(V \u2013 3)<\/b><\/span><\/span><\/span> \"\" V <\/b><\/span><\/span><\/span>= 6 + 3<\/b><\/span><\/span><\/span> = 9 m\/s.<\/b><\/span><\/span><\/span><\/p>\n

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

06- <\/span><\/span><\/span>Colocando as for\u00e7as que agem sobre cada bloco, na dire\u00e7\u00e3o do movimento:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

I \"\" bloco 1 \"\" F<\/b><\/span><\/span><\/span>R1<\/b><\/span><\/span><\/sub><\/span> = m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.a<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span><\/span>F \u2013 N<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.a<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> (I)<\/b><\/span><\/span><\/span><\/p>\n

I \"\" bloco 2 \"\" F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span> = 3m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.a<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = 3m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.a<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> (II) <\/b><\/span><\/span><\/span><\/p>\n

(I) + (II)<\/b><\/span><\/span><\/span> \"\" F = 4m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>a<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> \"\" a<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = F\/4m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> (III) \"\" (III) em (I) \"\" N<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = 3m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.(F\/4m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>) \"\" <\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = 3F\/4<\/b><\/span><\/span><\/span> <\/b><\/p>\n

II \"\" bloco 1 \"\" F<\/b><\/span><\/span><\/span>R2<\/b><\/span><\/span><\/sub><\/span> = m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.a<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span><\/span> N<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.a<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> (I)<\/b><\/span><\/span><\/span><\/p>\n

II \"\" bloco 2 \"\" F<\/b><\/span><\/span><\/span>R2<\/b><\/span><\/span><\/sub><\/span> = 3m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.a<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> \"\" F \u2013 N<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = 3m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.a<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> (II) <\/b><\/span><\/span><\/span><\/p>\n

(I) + (II)<\/b><\/span><\/span><\/span> \"\" F = 4m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>a<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> \"\" a<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = F\/4m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> (III) \"\" (III) em (I) \"\" N<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.(F\/4m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>) \"\" <\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = F\/4 <\/b><\/span><\/span><\/span><\/p>\n

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

07- <\/span><\/span><\/span>C\u00e1lculo da velocidade <\/b><\/span><\/span><\/span>com que o <\/b><\/span><\/span><\/span>p\u00eandulo X (partindo de P) atinge o p\u00eandulo Y (em Q), <\/b><\/span><\/span><\/span>imediatamente antes da colis\u00e3o<\/b><\/span><\/span><\/span>, pelo teorema da conserva\u00e7\u00e3o da energia mec\u00e2nica:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

E<\/b><\/span><\/span><\/span>mXP<\/b><\/span><\/span><\/sub><\/span> = mV<\/b><\/span><\/span><\/span>X<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + mgh <\/b><\/span><\/span><\/span>= 0,1.0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + 0,1.10.0,2<\/b><\/span><\/span><\/span> <\/b>= 0 + 0,2<\/b><\/span><\/span><\/span> \"\" E<\/b><\/span><\/span><\/span>mXP<\/b><\/span><\/span><\/sub><\/span> = 0,2 J.<\/b><\/span><\/span><\/span><\/p>\n

E<\/b><\/span><\/span><\/span>mXQ<\/b><\/span><\/span><\/sub><\/span> = mV<\/b><\/span><\/span><\/span>Y<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + mgh <\/b><\/span><\/span><\/span>= 0,1.V<\/b><\/span><\/span><\/span>Y<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + 0,1.10.0<\/b><\/span><\/span><\/span> <\/b>= 50V<\/b><\/span><\/span><\/span>Y<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span> + 0<\/b><\/span><\/span><\/span> \"\" E<\/b><\/span><\/span><\/span>mXQ<\/b><\/span><\/span><\/sub><\/span> = 0,05V<\/b><\/span><\/span><\/span>Y<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

E<\/b><\/span><\/span><\/span><\/span>mX<\/b><\/span><\/span><\/span><\/sub><\/span> = E<\/b><\/span><\/span><\/span><\/span>mY<\/b><\/span><\/span><\/span><\/sub><\/span> \"\" 0,2 = 0,05V<\/b><\/span><\/span><\/span><\/span>Y<\/b><\/span><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span> \"\" <\/b><\/span><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span><\/span>XQ<\/b><\/span><\/span><\/span><\/sub><\/span> = V = 2 m\/s.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Utilizando a conserva\u00e7\u00e3o da <\/b><\/span><\/span><\/span>quantidade de movimento antes e depois da colis\u00e3o:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Q<\/b><\/span><\/span><\/span>sa<\/b><\/span><\/span><\/sub><\/span> = m<\/b><\/span><\/span><\/span>X<\/b><\/span><\/span><\/sub><\/span>V + M<\/b><\/span><\/span><\/span>Y<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>Y<\/b><\/span><\/span><\/sub><\/span> = <\/b><\/span><\/span><\/span>100.2 + 100.0<\/b><\/span><\/span><\/span> \"\" Q<\/b><\/span><\/span><\/span>sa<\/b><\/span><\/span><\/sub><\/span> = 200 kgm\/s.<\/b><\/span><\/span><\/span><\/p>\n

Q<\/b><\/span><\/span><\/span>sd<\/b><\/span><\/span><\/sub><\/span> = m<\/b><\/span><\/span><\/span>X<\/b><\/span><\/span><\/sub><\/span>V + M<\/b><\/span><\/span><\/span>Y<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>Y<\/b><\/span><\/span><\/sub><\/span> = <\/b><\/span><\/span><\/span>100.V\u2019 + 100.V\u2019<\/b><\/span><\/span><\/span> \"\" Q<\/b><\/span><\/span><\/span>sa<\/b><\/span><\/span><\/sub><\/span> = 200V\u2019.<\/b><\/span><\/span><\/span><\/p>\n

Q<\/b><\/span><\/span><\/span>sa<\/b><\/span><\/span><\/sub><\/span> = Q<\/b><\/span><\/span><\/span>sd<\/b><\/span><\/span><\/sub><\/span> \"\" 200 = 200V\u2019 \"\" V\u2019 = 1m\/s = V\/2.<\/b><\/span><\/span><\/span><\/p>\n

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

08- <\/span><\/span><\/span>Trabalho como varia\u00e7\u00e3o de energia potencial gravitacional<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>W<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span> =mgh<\/b><\/span><\/span><\/span> = 0,1.10.0,2 <\/b><\/span><\/span><\/span>= 0,2 J.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

09- <\/span><\/span><\/span>Energia mec\u00e2nica do bloco no <\/b><\/span><\/span><\/span>ponto P<\/b><\/span><\/span><\/span> = energia potencial el\u00e1stica <\/b><\/span><\/span><\/span>armazenada pela mola quando ela est\u00e1 <\/b><\/span><\/span><\/span>comprimida de x no <\/b><\/span><\/span><\/span>ponto P<\/b><\/span><\/span><\/span> \"\" E<\/b><\/span><\/span><\/span>mP<\/b><\/span><\/span><\/sub><\/span> =<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>pe<\/b><\/span><\/span><\/sub><\/span> = kx<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Energia mec\u00e2nica do bloco no <\/b><\/span><\/span><\/span>ponto Q, <\/b><\/span><\/span><\/span>onde est\u00e1 em repouso \"\" E<\/b><\/span><\/span><\/span>mQ<\/b><\/span><\/span><\/sub><\/span> = 0.<\/b><\/span><\/span><\/span><\/p>\n

Toda essa <\/b><\/span><\/span><\/span>energia dissipada<\/b><\/span><\/span><\/span> entre os <\/b><\/span><\/span><\/span>pontos P e Q (\u2206E<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>) corresponde ao <\/b><\/span><\/span><\/span>trabalho da for\u00e7a de atrito<\/b><\/span><\/span><\/span> <\/b>(W<\/b><\/span><\/span><\/span>fat<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span> nesse percurso de <\/b><\/span><\/span><\/span>d= 5x<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>\u2206E<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span> = W<\/b><\/span><\/span><\/span>fat<\/b><\/span><\/span><\/sub><\/span> \"\" (0 – kx<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2) = F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>.d.cos\u03b1 \"\" – kx<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2<\/b><\/span><\/span><\/span> <\/b>=\u03bcPdcos180<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span> \"\" <\/b><\/span><\/span><\/span><\/p>\n

– kx<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 = \u03bcmg5x(-1) \"\" kx<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 = 5\u03bcmgx \"\" <\/b><\/span><\/span><\/span>x = 10\u03bcmg\/k.<\/b><\/span><\/span><\/span><\/p>\n

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

10- <\/span><\/span><\/span>Quanto ao <\/b><\/span><\/span><\/span>peso<\/b><\/span><\/span><\/span> \"\" como s\u00e3o feitos de <\/b><\/span><\/span><\/span>mesmo material e R \u00e9 maci\u00e7o e S oco<\/b><\/span><\/span><\/span>, a <\/b><\/span><\/span><\/span>massa de R<\/b><\/span><\/span><\/span> \u00e9 <\/b><\/span><\/span><\/span>maior que a massa de S e, sendo P = mg (com g constante), P<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span> > P<\/b><\/span><\/span><\/span>S<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Quanto ao <\/b><\/span><\/span><\/span>empuxo<\/b><\/span><\/span><\/span> que \u00e9 fornecido por <\/b><\/span><\/span><\/span>E = d<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>\u00e1gua deslocada<\/b><\/span><\/span><\/sub><\/span>.g<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/span>\u00e1gua <\/b><\/span><\/span><\/sub><\/span>\u00e9 a mesma<\/b><\/span><\/span><\/span>, como eles possuem o <\/b><\/span><\/span><\/span>mesmo volume<\/b><\/span><\/span><\/span> (independente do formato) o <\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>\u00e1gua deslocada<\/b><\/span><\/span><\/sub><\/span> \u00e9 o mesmo<\/b><\/span><\/span><\/span> e a <\/b><\/span><\/span><\/span>acelera\u00e7\u00e3o da gravidade g \u00e9 a mesma. Portanto E<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span> = E<\/b><\/span><\/span><\/span>S<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

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

11- <\/span><\/span><\/span>Aplicando a <\/b><\/span><\/span><\/span>equa\u00e7\u00e3o geral dos gases perfeitos<\/b><\/span><\/span><\/span> simultaneamente aos tr\u00eas estados:<\/b><\/span><\/span><\/span><\/p>\n

P<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>\/T<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span> = P<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span>\/T<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span> = P<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>\/T<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span> \"\" 6.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>.0,1\/T<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span> = 4.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>.0,2\/T<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span> = 2.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>.0,3\/T<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span> \"\" 0,6\/T<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span> = 0,8\/T<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span> = 0,6\/T<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span> \"\"<\/b><\/span><\/span><\/span><\/p>\n

T<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>\/3 = T<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>\/3 = T<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span>\/4<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span> = T<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span> < T<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

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

12- <\/span><\/span><\/span>Situa\u00e7\u00e3o I<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span>X est\u00e1 eletrizada com carga positiva e Y, neutra<\/b><\/span><\/span><\/span>. Quando s\u00e3o aproximadas, <\/b><\/span><\/span>X e Y s\u00e3o <\/b><\/span><\/span><\/span>atra\u00eddas <\/b><\/span><\/span><\/span>por indu\u00e7\u00e3o eletrost\u00e1tica.<\/b><\/span><\/span><\/span><\/p>\n

Se voc\u00ea <\/b><\/span><\/span>n\u00e3o domina a teoria, ela est\u00e1 a seguir:<\/b><\/span><\/span><\/span><\/p>\n

Aproxima-se um condutor eletrizado, por exemplo, negativamente de um condutor neutro, sem encostar, mas bem pr\u00f3ximos (figura 1).<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Os el\u00e9trons livres do indutor ser\u00e3o repelidos, ficando o lado direito do induzido com excesso de el\u00e9trons e o lado esquerdo com falta de el\u00e9trons, fen\u00f4meno que recebe o nome de\u00a0indu\u00e7\u00e3o el\u00e9trica (figura 2). Observe na figura 2 que a\u00a0for\u00e7a de atra\u00e7\u00e3o\u00a0entre as cargas negativas e positivas \u00e9\u00a0maior\u00a0que afor\u00e7a de repuls\u00e3o\u00a0entre as cargas negativas\u00a0porque a dist\u00e2ncia entre as cargas\u00a0que se atraem\u00a0\u00e9 menor\u00a0e, quanto\u00a0menor a dist\u00e2ncia\u00a0entre cargas de mesmo m\u00f3dulo,\u00a0maior a for\u00e7a\u00a0entre elas.<\/b><\/span><\/span><\/span><\/p>\n

Portanto,\u00a0um corpo eletrizado pode atrair um corpo neutro,\u00a0figura 3.<\/b><\/span><\/span><\/span><\/p>\n

Situa\u00e7\u00e3o II<\/b><\/span><\/span><\/span> \"\" Trata-se de <\/b><\/span><\/span><\/span>eletriza\u00e7\u00e3o por indu\u00e7\u00e3o. Veja teoria abaixo:<\/b><\/span><\/span><\/span><\/p>\n

Aproxima-se um condutor eletrizado, por exemplo, negativamente. (indutor) de um condutor neutro que se deseja eletrizar (induzido), sem encostar, mas bem pr\u00f3ximos (figura 1).<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Os el\u00e9trons livres do indutor ser\u00e3o repelidos, ficando o lado direito do induzido com excesso de el\u00e9trons e o lado esquerdo com falta de el\u00e9trons, fen\u00f4meno que recebe o nome de\u00a0indu\u00e7\u00e3o el\u00e9trica\u00a0(figura 2). As cargas el\u00e9tricas que se concentram nas duas extremidades opostas (denominadas\u00a0cargas induzidas) s\u00e3o de mesmo m\u00f3dulo, mas de sinais opostos e, por esse motivo o\u00a0induzido continua neutro.\u00a0Em seguida, ainda na presen\u00e7a do indutor voc\u00ea liga o induzido \u00e0 terra (fio terra ou com seu dedo) e observa que el\u00e9trons se deslocam do induzido para a terra (figura 3). Afastando agora o indutor para bem longe, as cargas<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

el\u00e9tricas se repelem e espalham pela superf\u00edcie externa do induzido, que <\/b><\/span><\/span><\/span>fica eletrizado com cargas de sinal oposto ao do indutor (figura 4)<\/b><\/span><\/span><\/span><\/p>\n

Se o indutor tivesse <\/b><\/span><\/span><\/span>cargas positivas<\/b><\/span><\/span><\/span>, ao final de todo o processo o induzido ficaria eletrizado com <\/b><\/span><\/span><\/span>cargas negativas.<\/b><\/span><\/span><\/span><\/p>\n

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

13- <\/span><\/span><\/span>Segunda lei de Ohm \"\" R = \u03c1L\/A.<\/b><\/span><\/span><\/span><\/p>\n

Resistor 1 \"\" R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = \u03c12L\/A \"\" resistor 2 R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = \u03c1L\/2A.<\/b><\/span><\/span><\/span><\/p>\n

Resistor equivalente R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span> = R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> + R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = \u03c12L\/A + \u03c1L\/2A = 5\u03c1L\/2A.<\/b><\/span><\/span><\/span><\/p>\n

R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span> = V\/i i = V\/R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span> = V\/5\u03c1L\/2A i = 2AV\/5\u03c1L.<\/b><\/span><\/span><\/span><\/p>\n

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

14- <\/span><\/span><\/span>Radia\u00e7\u00e3o alfa<\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>(\u03b1)<\/b><\/span><\/span><\/span>: tamb\u00e9m chamada de <\/b><\/span><\/span><\/span>part\u00edculas alfa ou raios alfa<\/b><\/span><\/span><\/span>, s\u00e3o part\u00edculas carregadas por dois pr\u00f3tons e dois n\u00eautrons, sendo, portanto, <\/b><\/span><\/span><\/span>n\u00facleos de h\u00e9lio<\/b><\/span><\/span><\/span>. Apresentam <\/b><\/span><\/span><\/span>carga positiva<\/b><\/span><\/span><\/span> +2 e n\u00famero de massa 4.<\/b><\/span><\/span><\/span>
\n<\/b><\/span><\/span><\/span>Radia\u00e7\u00e3o beta<\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>(\u03b2):<\/b><\/span><\/span><\/span> <\/b>raios beta ou part\u00edculas beta<\/b><\/span><\/span><\/span>, s\u00e3o <\/b><\/span><\/span><\/span>el\u00e9trons, part\u00edculas negativas<\/b><\/span><\/span><\/span> com carga\u00a0 \u2013 1 e n\u00famero de massa 0.<\/b><\/span><\/span><\/span>
\n<\/b><\/span><\/span><\/span>Radia\u00e7\u00e3o Gama<\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>(\u03b3): ou raios gama<\/b><\/span><\/span><\/span>. O comprimento de onda desta radia\u00e7\u00e3o varia de 0,5\"\" a 0,00\"\" (unidade de medida: angstron). As <\/b><\/span><\/span><\/span>radia\u00e7\u00f5es gama s\u00e3o ondas eletromagn\u00e9ticas<\/b><\/span><\/span><\/span>, e possuem <\/b><\/span><\/span><\/span>carga<\/b><\/span><\/span><\/span> e <\/b><\/span><\/span><\/span>massa nulas<\/b><\/span><\/span><\/span>, emitem continuamente calor e t\u00eam a capacidade de ionizar o ar e torn\u00e1-lo condutor de corrente el\u00e9trica<\/b><\/span><\/span><\/span><\/p>\n

Quando as <\/b><\/span><\/span>part\u00edculas penetram no campo magn\u00e9tico<\/b><\/span><\/span><\/span>, todas tem <\/b><\/span><\/span>velocidade vertical e para cima<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 o <\/b><\/span><\/span>campo magn\u00e9tico B penetra no papel<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 pela regra da <\/b><\/span><\/span>m\u00e3o esquerda a for\u00e7a<\/b><\/span><\/span><\/span> que age sobre a <\/b><\/span><\/span>carga positiva \u00e9 horizontal e para a esquerda (\u00e9 a \u03b1)<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 portanto a <\/b><\/span><\/span>carga positiva \u00e9 desviada para a<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

esquerda<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 na <\/b><\/span><\/span>carga negativa o sentido dessa for\u00e7a \u00e9 invertido<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 ela sofre <\/b><\/span><\/span>desvio para a direita<\/b><\/span><\/span><\/span> <\/b>(\u00e9 a \u03b2)\u00a0 —<\/b><\/span><\/span><\/span>\u00a0 a <\/b><\/span><\/span>neutra n\u00e3o sofre desvio\u00a0 —\u00a0 a neutra n\u00e3o sofre desvio (\u00e9 a \u03b3<\/b><\/span><\/span><\/span>)\u00a0 —\u00a0 observe que o desvio da \u03b1 \u00e9 menor, ou seja, ela possui maior in\u00e9rcia e consequentemente maior quantidade de movimento\u00a0e maior massa. \u00a0<\/b><\/span><\/span><\/p>\n

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

15- <\/span><\/span><\/span>Caso particular da refra\u00e7\u00e3o da luz<\/b><\/span><\/span><\/span> \"\" todo raio de luz que incide <\/b><\/span><\/span><\/span>perpendicularmente (radialmente)<\/b><\/span><\/span><\/span> sobre a <\/b><\/span><\/span><\/span>superf\u00edcie de separa\u00e7\u00e3o<\/b><\/span><\/span><\/span> de dois meios homog\u00eaneos e transparentes (no caso, ar-vidro) n\u00e3o sofre desvio e passa pelo ponto O (centro da placa semicircular).<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Ao <\/b><\/span><\/span><\/span>sair para o ar<\/b><\/span><\/span><\/span> passando de um meio <\/b><\/span><\/span><\/span>mais refringente para outro menos refringente<\/b><\/span><\/span><\/span>, o raio de luz <\/b><\/span><\/span><\/span>aumenta a velocidade<\/b><\/span><\/span><\/span> se <\/b><\/span><\/span><\/span>afastando da normal.<\/b><\/span><\/span><\/span> <\/b><\/p>\n

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

16- <\/span><\/span><\/span>Essa <\/b><\/span><\/span><\/span>for\u00e7a eletromotriz de 1V<\/b><\/span><\/span><\/span> gerada pela <\/b><\/span><\/span><\/span>varia\u00e7\u00e3o de fluxo magn\u00e9tico<\/b><\/span><\/span><\/span> no interior do circuito provoca uma <\/b><\/span><\/span><\/span>diminui\u00e7\u00e3o de 1V na ddp da bateria<\/b><\/span><\/span><\/span>, pois se op\u00f5e a ela (lei de Lenz), fazendo com que o <\/b><\/span><\/span><\/span>circuito fique sujeito a uma ddp de U <\/b><\/span><\/span><\/span>= 3 \u2013 1 <\/b><\/span><\/span><\/span>= 2V.<\/b><\/span><\/span><\/span><\/p>\n

R = U\/i<\/b><\/span><\/span><\/span> \"\" 2 = 2\/i \"\" i = 1 A.<\/b><\/span><\/span><\/span><\/p>\n

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

17- <\/span><\/span><\/span>I – incorreta<\/b><\/span><\/span><\/span>. As <\/b><\/span><\/span>ondas P e Q<\/b><\/span><\/span><\/span> possuem a <\/b><\/span><\/span>mesma amplitude A<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span> = A<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/sub><\/span> (veja figura) <\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

II \u2013 correta (veja figura). <\/b><\/span><\/span><\/span><\/p>\n

III \u2013 incorreta \"\" pelo enunciado as velocidades de propaga\u00e7\u00e3o s\u00e3o iguais \"\" V<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span> = V<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/sub><\/span> = V \"\" <\/b><\/span><\/span><\/span><\/p>\n

V<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span> =\u03bb<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>.f<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span> <\/b>= 2\u03bb<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/sub><\/span> \u03bb<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/sub><\/span> \"\"<\/b><\/span><\/span><\/span> <\/b>f<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span> <\/b>= V\/2\u03bb<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/sub><\/span> <\/b>(I)<\/b><\/span><\/span><\/span> \"\" V<\/b><\/span><\/span>Q<\/b><\/span><\/span><\/sub> = <\/b><\/span><\/span>\u03bb<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/sub><\/span>.f<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span>f<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/sub><\/span> = V\/\u03bb<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/sub><\/span> <\/b>(II).<\/b><\/span><\/span><\/span><\/p>\n

(I)\/(II) \"\" f<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>\/f<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/sub><\/span> = V\/2\u03bb<\/b><\/span><\/span><\/span>Q <\/b><\/span><\/span><\/sub><\/span>X \u03bb<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/sub><\/span>\/V \"\"f<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/sub><\/span> = 2f<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

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

18- <\/span><\/span><\/span>Efeito fotoel\u00e9trico<\/b><\/span><\/span><\/span>\u00a0– \u00a0\u00a0Heinrich Rudolf Hertz, em 1887, detectou e gerou em laborat\u00f3rio, em uma <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

s\u00e9rie de experi\u00eancias que,\u00a0<\/b><\/span><\/span><\/span>quando a luz<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>(onda eletromagn\u00e9tica, radia\u00e7\u00e3o eletromagn\u00e9tica)<\/b><\/span><\/span><\/span> de freq\u00fc\u00eancia suficientemente alta incide sobre a <\/b><\/span><\/span><\/span>superf\u00edcie de um metal,<\/b><\/span><\/span><\/span> ela pode <\/b><\/span><\/span><\/span>retirar el\u00e9trons do mesmo<\/b><\/span><\/span><\/span>. Esse fen\u00f4meno ficou conhecido como<\/b><\/span><\/span><\/span>\u00a0efeito fotoel\u00e9trico.<\/b><\/span><\/span><\/span><\/p>\n

A luz tem <\/b><\/span><\/span><\/span>car\u00e1ter dual<\/b><\/span><\/span><\/span>: os fen\u00f4menos de reflex\u00e3o, refra\u00e7\u00e3o, interfer\u00eancia, difra\u00e7\u00e3o e polariza\u00e7\u00e3o da <\/b><\/span><\/span><\/span>luz podem ser explicados pela teoria ondulat\u00f3ria<\/b><\/span><\/span><\/span> e os de <\/b><\/span><\/span><\/span>emiss\u00e3o e absor\u00e7\u00e3o<\/b><\/span><\/span><\/span> podem ser explicados pela <\/b><\/span><\/span><\/span>teoria corpuscular<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

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

19- <\/span><\/span><\/span>O f\u00edsico franc\u00eas <\/b><\/span><\/span><\/span>Louis de Broglie<\/b><\/span><\/span><\/span> apresentou, em 1924, a seguinte hip\u00f3tese que, depois, em 1927 <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

foi comprovada experimentalmente: <\/b><\/span><\/span><\/span>\u201cpart\u00edculas tamb\u00e9m possuem propriedades ondulat\u00f3rias\u201d<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 das equa\u00e7\u00f5es <\/b><\/span><\/span><\/span>E=m.c<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 c \u2013 velocidade da luz no v\u00e1cuo\u00a0 —\u00a0 <\/b><\/span><\/span><\/span>E=h.f<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 c=\u03bbf\u00a0 —\u00a0 f=c\/\u03bb\u00a0—\u00a0 <\/b><\/span><\/span><\/span>mc<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=hf<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 mc<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=hc\/\u03bb\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u03bb=h\/mc<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00a0—\u00a0 mc=Q \u2013 quantidade de movimento\u00a0 —\u00a0 a <\/b><\/span><\/span><\/span>quantidade de movimento Q=m.v evidencia o car\u00e1ter corpuscular da luz enquanto que o comprimento de onda \u03bb evidencia seu car\u00e1ter ondulat\u00f3rio.<\/b><\/span><\/span><\/span><\/p>\n

As naturezas ondulat\u00f3ria e corpuscular da luz s\u00e3o complementares e n\u00e3o antag\u00f4nicas<\/b><\/span><\/span><\/span><\/p>\n

No caso do exerc\u00edcio \"\" <\/b><\/span><\/span><\/span>\u03bb=h\/mc<\/b><\/span><\/span><\/span>\u00a0\"\" c=v \"\" \u03bb = 6,6.10<\/b><\/span><\/span><\/span>-34<\/b><\/span><\/span><\/sup><\/span>\/9.10<\/b><\/span><\/span><\/span>-31<\/b><\/span><\/span><\/sup><\/span>.2,2.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span> = 3,3.10<\/b><\/span><\/span><\/span>-10<\/b><\/span><\/span><\/sup><\/span> m.<\/b><\/span><\/span><\/span><\/p>\n

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

<\/h2>\n

Voltar para os exerc\u00edcios.<\/span><\/a><\/h2>\n","protected":false},"excerpt":{"rendered":"

RESOLU\u00c7\u00d5ES 01- 1\u00b0 caso: se os dois m\u00f3veis estiverem andando no mesmo sentido, o valor da velocidade relativa \u00e9 dado pelo m\u00f3dulo da diferen\u00e7a entre os m\u00f3dulos das duas velocidades escalares. Dessa forma, temos: |vrel\u00a0|= |vA\u00a0|- |vB\u00a0| 2\u00b0 caso: se os dois m\u00f3veis estiverem andando em sentidos contr\u00e1rios, o valor absoluto da velocidade relativa \u00e9 dado pela soma dos m\u00f3dulos<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":824,"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-826","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/826","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=826"}],"version-history":[{"count":1,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/826\/revisions"}],"predecessor-version":[{"id":827,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/826\/revisions\/827"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/824"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=826"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}