Tangential velocity 2. Parabolic pathway 3. Projectile 4. Centripetal acceleration 5. Centripetal force a. acceleration towards
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Answer:
(a) 1317.44 W/m²
(b) 1.74×10¹⁵ litres of heating oil
(c) -20.63°C
(d) Energy storage in the Earth and the air
Explanation:
The parameters given are;
Luminosity of the Sun = 3.828 × 10²⁶ Watt
Distance of the Earth from the Sun, d = 152.06 × 10⁶ km
The radius of the Sun = 696 × 10³ km
The radius of the Earth, = 6730 km
The surface area of the Sun = 12000 × Surface area of the Earth
The surface area of the Sun = 6.09 × 10¹² km²
Cross sectional area of the Earth = 1.27 × 10⁴ m² = 0.0127 km²
By the inverse square law, we have;
Where:
R = Solar radiation reaching the Earth
Therefore;
Hence, the energy, E, reaching the Earth in a day is given as follows;
E = R × 4×πײ×60×60×24 = 1317.44 × 4 × π × 6730000² × 60×60×24
E = 6.479×10²² Joules
(b) The number of litres of heating oil is therefore;
6.479×10²² J ÷ (37.3×10⁶ J/litre) = 1.74×10¹⁵ litres of heating oil
(c) 30% of the Energy is reflected, therefore;
0.7 × 6.479×10²² Joules = 4.54×10²² Joules reaches the Earths surface
From Stefan-Boltzmann law, we have;
Where:
α = 0.3
σ = Stefan-Boltzmann constant = 5.6704×10⁻⁸ W/(m²·K⁴)
Therefore;
(d) The heat storage in the Earth and the air