Respuesta :
Answer:
5.71×10¹¹ gallons
4.88×10¹⁷ J
1.17×10² megatons
Step-by-step explanation:
Volume of the storm:
(60 km) (400 km) (12 km) (1000 m / km)³ (100 cm / m)³ = 2.88×10²⁰ cm³
Mass of water:
(2.88×10²⁰ cm³) (7.5×10⁻⁷ g/cm³) = 2.16×10¹⁴ g
Volume of water:
(2.16×10¹⁴ g) (1 gallon / 3785 g) = 5.71×10¹¹ gallons
Total energy:
(2.16×10¹⁴ g) (2260 J/g) = 4.88×10¹⁷ J
(4.88×10¹⁷ J) (1 megaton / 4.18×10¹⁵ J) = 1.17×10² megatons
Answer:
- 5.71×10^10 gallons
- 4.88×10^17 Joules
- 117 megatons
Step-by-step explanation:
a) The volume of the storm is ...
(60 km)(400 km)(12 km) = 2.88×10^5 km^3
= (2.88×10^5 km^3)×((10^5 cm)/(1 km))^3 = 2.88×10^20 cm^3
Then the volume of water is ...
(2.88×10^20 cm^3)×(7.5×10^-7 g/cm^3) = 2.16×10^14 g
In gallons, this is ...
(2.16×10^14 g)×(1 gal)/(3.785×10^3 g) ≈ 5.71×10^10 gal
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b) The energy released is ...
(2.16×10^14 g)(2.260×10^3 J/g) = 4.8816×10^17 J
In megatons, this is ...
(4.8816×10^17 J)/(4.18×10^15 J/megaton) ≈ 117 megatons
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Comment on the storm
The average rainfall over the storm area would be about 0.9 cm, about 0.35 inches.