Question

Suppose that the average U.S. household uses 15600 kWh (kilowatt‑hours) of energy in a year. If the average rate of energy consumed by the house was instead diverted to lift a 1.80 × 10 3 kg car 16.8 m into the air, how long would it take? car: s Using the same rate of energy consumption, how long would it take to lift a loaded Boeing 747 airplane, with a mass of 4.05 × 10 5 kg , to a cruising altitude of 9.67 km ? airplane:

Answer 1

Answer:

a) 166.4 s

b) (2.155 × 10⁷) s

Explanation:

15600 KWh for a year,

1 year consists of 365 × 24 hours = 8760 hours.

So, the power consumed in a year for an average household = (Energy/time)

= (15600/8760) = 1.781 KW = 1781 W

a) If the average rate of energy consumed by the house was instead diverted to lift a 1.80 × 10 3 kg car 16.8 m into the air, how long would it take

The power required for this lifting = (mgh/t)

m = 1800 kg

g = 9.8 m/s²

h = 16.8 m

t = ?

P = 1781 W

1781 = (1800×9.8×16.8)/t

t = (296,352/1781)

t = 166.4 s

b) how long would it take to lift a loaded Boeing 747 airplane, with a mass of 4.05 × 10 5 kg , to a cruising altitude of 9.67 km

The power required for this lifting = (mgh/t)

m = 405000 kg

g = 9.8 m/s²

h = 9.67 km = 9670 m

t = ?

P = 1781 W

1781 = (405000×9.8×9670)/t

t = (38,380,230,000/1781)

t = 21,549,820 s = (2.155 × 10⁷) s

Hope this Helps!!!