{"id":3645,"date":"2018-02-13T13:35:30","date_gmt":"2018-02-13T13:35:30","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=3645"},"modified":"2024-11-14T13:48:34","modified_gmt":"2024-11-14T13:48:34","slug":"resolucao-comentada-trabalho-como-variacao-de-energia","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/enem\/trabalho-como-variacao-de-energia\/resolucao-comentada-trabalho-como-variacao-de-energia\/","title":{"rendered":"Trabalho como varia\u00e7\u00e3o de Energia – EN"},"content":{"rendered":"

Commented Resolution \u00a0<\/span><\/b><\/p>\n

Work as a variation of Energy<\/span><\/b><\/p>\n

01-\u00a0<\/span><\/b>Data \u2014 E = 94,684,781 MWh \u2248 9.5.\u00a0<\/b>10 7<\/b><\/sup>\u00a0\u00a0MWh = 9.5 .\u00a0<\/b>10 13<\/b><\/sup>\u00a0\u00a0Wh \u2014 L = 0.45.\u00a0<\/b>10 8<\/b><\/sup>\u00a0\u00a0J\/kg \u2014 transforming this amount of energy in joules \u2014 E = 9.5.\u00a0<\/b>10 13<\/b><\/sup>\u00a0\u00a0Wh = (9.5.\u00a0<\/b>10 13<\/b><\/sup>\u00a0\u00a0W).(3.6.\u00a0<\/b>10 3<\/b><\/sup>\u00a0\u00a0s) = 3.4.\u00a0<\/b>10 17<\/b><\/sup>\u00a0\u00a0J \u2014 for oil \u2014 E = m L \u2014 m = E\/L \u2014 m = 3.4.\u00a0<\/b>10 17<\/b><\/sup>\u00a0\/0.45\u00a0<\/b>. 10 8<\/b><\/sup>\u00a0\u00a0 = 7.5.\u00a0<\/b>10 9<\/b><\/sup>\u00a0\u00a0kg = 7.5. 10\u00a0<\/b>6<\/b><\/sup>\u00a0\u00a0tons \u2014 m = 7.5 million tons \u2014\u00a0\u00a0<\/b>R- D<\/b><\/span>\u00a0<\/b><\/b><\/b><\/b><\/b><\/b><\/b><\/b><\/b><\/b><\/b><\/b><\/b><\/b><\/b><\/b><\/b><\/b><\/b><\/b><\/b><\/b><\/b><\/b>\u00a0<\/b><\/b><\/p>\n

02-\u00a0<\/span><\/b>The table shows the energy spent in each of the activities, within the time specified in each of the propositions \u2014 note that the one that comes closest to the expenditure of 304 kilocalories corresponds to alternative A.<\/b><\/span>\u00a0<\/b><\/b><\/p>\n

\"\"<\/p>\n

\u00a0\u00a0A- A<\/span><\/b><\/p>\n

03-\u00a0<\/span><\/b>Data \u2014 F = 30 N \u2014 \u2206S = 800 m \u2014 the work (W) of a constant force\u00a0\u00a0<\/b>\u00a0is given by the expression \u2014 W\u00a0<\/b>F<\/b><\/sub>\u00a0= Fdcos\u03b1 \u2014 the force being parallel to the displacement \u2014 \u03b1 = 0\u00b0, cos \u03b1 = 1 \u2014 W\u00a0<\/b>F<\/b><\/sub>\u00a0= 30.(800) = 24,000 J = 24 kJ \u2014\u00a0\u00a0\u00a0<\/b>R- D<\/b><\/span>\u00a0<\/b>\"\"<\/b><\/b><\/b><\/b><\/b><\/b><\/p>\n

04-\u00a0<\/span><\/b>Two forces act on the fish \u2014 pulling force (which the fisherman applies to lift the fish, vertically and upwards) and the weight of the fish (vertically and downwards) \u2014 as the displacement is upwards and the weight of the fish is downwards, they have opposite directions and the work of the weight force is resistant (opposite to the displacement) \u2014\u00a0\u00a0\u00a0<\/b>A- A<\/b><\/span>\u00a0<\/b><\/b><\/b><\/p>\n

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