Answer:As sunlight passes through the atmosphere, all UVC and most UVB is absorbed by ozone, water vapour, oxygen and carbon dioxide. UVA is not filtered as significantly by the atmosphere. Is there a connection between ozone depletion and UV radiation? Ozone is a particularly effective absorber of UV radiation.
Explanation:
I had old notes
Complete question is;
The energy flow to the earth from sunlight is about 1.4kW/m²
(a) Find the maximum values of the electric and magnetic fields for a sinusoidal wave of this intensity.
(b) The distance from the earth to the sun is about 1.5 × 10^(11) m. Find the total power radiated by the sun.
Answer:
A) E_max ≈ 1026 V/m
B_max = 3.46 × 10^(-6) T
B) P = 3.95 × 10^(26) W
Explanation:
We are given;
Intensity; I = 1.4kW/m² = 1400 W/m²
Formula for maximum value of electric field in relation to intensity is given as;
E_max = √(2I/(ε_o•c))
Where;
ε_o is electric constant = 8.85 × 10^(-12) C²/N.m²
c is speed of light = 3 × 10^(8) m/s
Thus;
E_max = √(2 × 1400)/(8.85 × 10^(-12) × 3 × 10^(8)))
E_max ≈ 1026 V/m
Formula for maximum magnetic field is;
B_max = E_max/c
B_max = 1026/(3 × 10^(8))
B_max = 3.46 × 10^(-6) T
Formula for the total power is;
P = IA
Where;
A is area = 4πr²
We are given;
Radius; r = 1.5 × 10^(11) m
A = 4π × (1.5 × 10^(11))² = 2.82 × 10^(23) m²
P = 1400 × 2.82 × 10^(23)
P = 3.95 × 10^(26) W
Answer:
Vi = 24.14 m/s
Explanation:
If we apply Law of Conservation of Energy or Work-Energy Principle here, we get: (neglecting friction)
Loss in K.E of the Rock = Gain in P.E of the Rock
(1/2)(m)(Vi² - Vf²) = mgh
Vi² - Vf² = 2gh
Vi² = Vf² + 2gh
Vi = √(Vf² + 2gh)
where,
Vi = Rock's Speed as it left the ground = ?
Vf = Final Speed = 17 m/s
g = 9.8 m/s²
h = height of rock = 15 m
Therefore,
Vi = √[(17 m/s)² + 2(9.8 m/s²)(15 m)]
Vi = √583 m²/s²
<u>Vi = 24.14 m/s</u>
Heat transfer in a closed system is the addition of changes in internal energy and the total amount of work done by it. The final energy of the system is 35.5kJ.
<h3>What is heat transfer? </h3>
Heat transfer is the transfer of heat energy due to temperature differences.
The paddle-wheel paintings are quantities of workdone, 500 N.m or 0.5kJ.
The preliminary (initial) power of the device is 10 kJ.
Total warmness transferred in the course of the method is 30 kJ
Total warmness misplaced in the course of the method to the encompassing air is 5 kJ.
The energy of the system is given as:
The energy of the system = Energy in - Energy out
The energy of the system = Initial energy + Energy transferred + Work done - Energy lost
Energy of the system = 10 + 30 + 0.5 - 5 kJ
Energy of the system = 35.5 kJ
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