\u00a0Commented Resolution<\/p>\n
Work (Energy) \u2013 Power<\/p>\n
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01-\u00a0The energy stored by\u00a0water<\/span><\/a>\u00a0is the gravitational potential energy of that mass of water at a height of 120m \u2014 W=mgh=m.10.120 \u2014 W=1,200m \u2014 Power=energy\/time \u2014 P\u00a0o<\/sub>\u00a0=512,000,000W \u2014 \u2206t=1s \u2014 P\u00a0o<\/sub>\u00a0=W\/\u2206t \u2014 512,000,000=1,200m\/1 \u2014 m=512,000,000\/1,200 =512.106\/\u00a012.102<\/sup>\u00a0\u2014\u00a0\u00a0m=42.7.104\u00a0=<\/sup>\u00a0427,000kg + (10% of 427,000=42,700)\u00a0\u00a0\u2014<\/sup>\u00a0\u00a0<\/a><\/p>\n m=469.700kg.<\/p>\n Considering the density of water as 1kg\/L, we will have that a mass of 1kg occupies a volume of 1L of water \u2014\u00a0\u00a0R- E<\/p>\n 02-\u00a0Power corresponds to the measure of how quickly energy is transferred \u2014 Po\u00a0=<\/sub>\u00a0W\/\u0394t \u2014 note in this expression that the most efficient (highest power Po\u00a0)<\/sub>\u00a0is the one that provides the greatest amount of energy (W) in the shortest amount of time (\u0394t), that is, it is directly proportional to the amount of energy (W) and inversely proportional to the operating time interval (\u0394t) \u2014\u00a0\u00a0\u00a0R- C<\/p>\n 03-\u00a0m=1 ton = 1,000kg \u2014 rule of three \u2014 1,000kg \u2013 100% \u2014 1.5kg \u2013 p \u2014 1000p=1.5\u00d7100 \u2014 p=150\/1000 \u2014 p=0.15%<\/p>\n R-B<\/p>\n 04-\u00a0Since the tractor is the same, the energy it consumes can be considered constant for any type of\u00a0fuel<\/span><\/a>\u00a0you use \u2014 lower fuel consumption per liter of a given fuel will imply a greater amount of energy released by the same liter, making the tractor run for longer \u2014 therefore, the energy released by a liter of sunflower oil will be greater than the energy released by a liter of diesel oil \u2014\u00a0\u00a0A- E\u00a0\u00a0<\/a><\/p>\n