Answer:
1.324 × 10⁷ m
Explanation:
The centripetal acceleration, a at that height above the earth equal the acceleration due to gravity, g' at that height, h.
Let R be the radius of the orbit where R = RE + h, RE = radius of earth = 6.4 × 10⁶ m.
We know a = Rω² and g' = GME/R² where ω = angular speed = 2π/T where T = period of rotation = 1 day = 8.64 × 10⁴s (since the shuttle's period is synchronized with that of the Earth's rotation), G = gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg², ME = mass of earth = 6 × 10²⁴ kg. Since a = g', we have
Rω² = GME/R²
R(2π/T)² = GME/R²
R³ = GME(T/2π)²
R = ∛(GME)(T/2π)²
RE + h = ∛(GMET²/4π²)
h = ∛(GMET²/4π²) - RE
substituting the values of the variables, we have
h = ∛(6.67 × 10⁻¹¹ Nm²/kg² × 6 × 10²⁴ kg × (8.64 × 10⁴s)²/4π²) - 6.4 × 10⁶ m
h = ∛(2,987,477 × 10²⁰/4π² Nm²s²/kg) - 6.4 × 10⁶ m
h = ∛75.67 × 10²⁰ m³ - 6.4 × 10⁶ m
h = ∛(7567 × 10¹⁸ m³) - 6.4 × 10⁶ m
h = 19.64 × 10⁶ m - 6.4 × 10⁶ m
h = 13.24 × 10⁶ m
h = 1.324 × 10⁷ m
Evaporation is more contrasted with the rainfall which is less in the correlation of tropical locale. In tropical locale, protection is very little in light of mists. Alongside temperature, it is the main consideration in adding to changes in the thickness of seawater and consequently sea flow. Saltiness is the way to understanding the worldwide water cycle. 97% of the Earth's free water dwells in the seas. The water cycle is commanded by precipitation and evaporation.
Answer:
685.6 J
Explanation:
The latent heat of vaporization of ammonia is
L = 1371.2 kJ/kg
mass of ammonia, m = 0.0005 Kg
Heat = mass x latent heat of vaporization
H = 0.0005 x 1371.2
H = 0.686 kJ
H = 685.6 J
Thus, the amount of heat required to vaporize the ammonia is 685.6 J.
Hydroelectricity is the best answer.
This is an article by the EIA, but the pie graph is the most helpful: https://www.eia.gov/energyexplained/?page=us_energy_home
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