Part 1
When the solar atmosphere accumulates a lot of magnetic energy
to a point that cannot accumulate more, all that magnetic energy is suddenly released,
and with it, a lot of radiation. So much, that in fact it covers all of the
electromagnetic spectrum; from radio waves to gamma rays. That burst of
radiation is called a solar flare. In a single solar flare the amount of
radiation released is millions of times greater than all the nuclear bombs in
the face if the earth exploding together. Lucky for us, most of the high-energy
radiation dissipates before reaching the Earth, and the radiation that do reach
us, is deflected by the Earth’s magnetic field.
Part 2
1. Not all the radiation
of solar flares that reach the Earth is deflected by its magnetic field; some
of them reach us and charges the upper atmosphere with ionized particles. Those
particles react with the gases in the atmosphere and produce a light; that
light is what we call Auroras borealis or southern nights; One the most beautiful
natural spectacles in earth, who thought Auroras begin their lives as deadly
solar flares.
2. Solar flares
contain a lot of high-energy radiation that is extremely dangerous for our
electronic devices; when they reach the Earth, they can damage sensible
electronics like satellites. A very powerful solar flare could even damage all
the electronic devices on the surface of the Earth.
Answer:
a) y= 3.5 10³ m, b) t = 64 s
Explanation:
a) For this exercise we use the vertical launch kinematics equation
Stage 1
y₁ = y₀ + v₀ t + ½ a t²
y₁ = 0 + 0 + ½ a₁ t²
Let's calculate
y₁ = ½ 16 10²
y₁ = 800 m
At the end of this stage it has a speed
v₁ = vo + a₁ t₁
v₁ = 0 + 16 10
v₁ = 160 m / s
Stage 2
y₂ = y₁ + v₁ (t-t₀) + ½ a₂ (t-t₀)²
y₂ = 800 + 150 5 + ½ 11 5²
y₂ = 1092.5 m
Speed is
v₂ = v₁ + a₂ t
v₂ = 160 + 11 5
v₂ = 215 m / s
The rocket continues to follow until the speed reaches zero (v₃ = 0)
v₃² = v₂² - 2 g y₃
0 = v₂² - 2g y₃
y₃ = v₂² / 2g
y₃ = 215²/2 9.8
y₃ = 2358.4 m
The total height is
y = y₃ + y₂
y = 2358.4 + 1092.5
y = 3450.9 m
y= 3.5 10³ m
b) Flight time is the time to go up plus the time to go down
Let's look for the time of stage 3
v₃ = v₂ - g t₃
v₃ = 0
t₃ = v₂ / g
t₃ = 215 / 9.8
t₃ = 21.94 s
The time to climb is
= t₁ + t₂ + t₃
t_{s} = 10+ 5+ 21.94
t_{s} = 36.94 s
The time to descend from the maximum height is
y = v₀ t - ½ g t²
When it starts to slow down it's zero
y = - ½ g t_{b}²
t_{b} = √-2y / g
t_{b} = √(- 2 (-3450.9) /9.8)
t_{b} = 26.54 s
Flight time is the rise time plus the descent date
t = t_{s} + t_{b}
t = 36.94 + 26.54
t =63.84 s
t = 64 s
The solution for this problem is:
For 1st minimum, let m be equal to 1.
d = slit width
D = screen distance.
Θ = arcsin (m * lambda/ (d))
= 0.13934 rad, 7.9836 deg
y = D*tan (Θ)
y = 6.50 * tan (7.9836)
= 0.91161 m is the distance from the central maximum to the first-order minimum
Gravity is the correct answer.
