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!!!
Answer:
Mass of the planet = 6.0 × 
Explanation:
Time period = 2π (R + h) / v
Orbital speed (v) = √GM / (R + h)
T² = 4π² (R + h)² / (GM/ (R + h))
= 4π² (R + h)³ / GM
making m the subject of the formula
m = 4π² (R + h)³ / GT²
= 4π² ( 6.38 × 
+ 230 × 10³ )³ / ( 6.67 × 
) × (89 × 60)²
= 4π² ( 6610000)³ / ( 6.67 × ) × (89 × 60)²
= 5.99 ×
= 6.0 ×
Answer:
Light travels fast in air but i dont know about water
