Print Takulator Periodic Table Question 12 of 13 Sapling Learning Suppose that t

Question

Print Takulator Periodic Table Question 12 of 13 Sapling Learning Suppose that the average U.S. household uses 11600 kWh (kilowatt-hours) of energy in a year. If the average rate of energy consumed by the house was instead diverted to lift a 2130 kg car 12.8 m into the air, how long would it take? Number Using the same rate of energy consurmption, how long would it take to lift a loaded 747, with a mass of 415000 kg, to a cruising altitude of 8.92 km? Number Hint Previous @check Answer ) Nextxitr

Explanation / Answer

Energy = 11600 (k J/s) (3600 s)

= 4.176 x 10^10 J

P = Energy / time = (4.176 x 10^10 J) / (365 x 24 x 3600 s)

P = 1324.2 Watt

Energy required = m g h = 2130 x 9.81 x 12.8

= 267459.84 J

t = (267459.84) / 1324.2 = 202 sec ........Ans

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Energy = 415000 x 9.81 x 8.92 x 10^3

= 3.63 x 10^10 J

t = (3.63 x 10^10 J) / (1324.2 W)

= 27.4 x 10^6 sec ..........Ans

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