The gem amethyst gets its purplish color from trace amounts of iron and aluminum within its crystal lattice structure. Amethyst is a variety of quartz, which is a silicon dioxide mineral. The coloration occurs due to the presence of impurities or color centers that interact with light, absorbing certain wavelengths and transmitting others.
In the case of amethyst, the purple color is primarily attributed to the presence of iron impurities (Fe3+) within the crystal lattice of quartz. The iron impurities absorb certain wavelengths of light in the visible spectrum, particularly in the green and yellow regions, leaving behind the purplish hues to be transmitted to our eyes.
The exact shade of purple can vary in amethyst gemstones, ranging from light lilac to deep violet, depending on the concentration of iron impurities and other factors during their formation. Heat treatment or exposure to radiation can also influence the color of amethyst, but the natural variety gets its beauty from the fascinating interplay of these impurities within the crystal structure.
image distance,di=10 cm
object distance,do=20cm
magnification, m=di/do
=10/20
=0.5
since the image is virtual, magnification is negative.
therefore m=-0.5
Answer:
1. t_reaction = 1.08 s
2. v₀₁ = 16.365 m/s
Explanation:
1. This is a kinematics exercise, let's analyze the situation a bit, we can calculate the braking distance and the rest of the distance we can use to calculate the reaction time.
Braking distance
v² = v₀² + 2 a x
when he finishes braking the speed is v = 0
0 = v₀² + 2 a x
x = -v₀² / 2a
x = - 17²/2 (-7)
x = 20.64 m
the distance for the reaction is
d = x_reaction + x
x_reaction = d - x
x_reaction = 39 - 20.64
x_reaction = 18.36 m
as long as it has not reacted the vehicle speed is constant
v = x_reaction / t_reaction
t_reaction = x_reaction / v
t_reaction = 18.36 / 17
t_reaction = 1.08 s
2. Let's find the distance traveled in the reaction time of t1 = 1.21983 s
as the speed is constant
v = x / t
x₁ = v t₁
the distance traveled during braking is
v² = v₀² + 2a x₂
0 = v₀² + 2 a x₂
x₂ = -v₀² / 2a
v = v₀
the total distance is
x_total = x₁ + x₂
x_total = v₀ t₁ + v₀² / 2a
39 = v₀ 1.21983 + v₀²/14
v₀² + 17.08 vo - 546 =0
we solve the second degree equation
v₀ = [ -17.08 ±√(17.08² + 4 546) ]/2
v₀ = [-17.08 ± 49.81 ]/2
v₀₁ = 16.365 m/s
v₀₂ = - 33.445 m/s
as the acceleration is negative the correct result is v₀₁ = 16.365 m/s