<span>The Badminton World Federation</span>
Answer:
-6.44 m/s²
Explanation:
Given:
Δx = 60 m
v₀ = 27.8 m/s
v = 0 m/s
Find: a
v² = v₀² + 2aΔx
(0 m/s)² = (27.8 m/s)² + 2a (60 m)
a = -6.44 m/s²
Could it be "Scientific research"?
The ratio of the intensity between light intensity that emerges from the last filter and unpolarized light of intensity, I₀ is It/I₀ = 0.2925
To answer the question, we need to know what polarization of light is.
<h3>What is polarization of light?</h3>
This is when the electric field vector of light is oscillating in one plane.
- Now for light of intensity I' which is initially unpolarized, its intensity after polarization is I = 1/2I'.
- Also, for light initially polarized, its intensity after polarization is I"' = I"cos²Ф where Ф is the angle between the initial direction and the direction of polarization.
<h3>Intensity of light through each polarized filter</h3>
Given that we have 7 polarizing filters, each rotated 17° cw with respect to the previous filter.
So, since the light is initially unpolarized,
- The intensity through the first polarizing filter is I₁ = 1/2I₀ where I₀ is the initial intensity.
- The intensity through the second polarizing filter is I₂ = I₁cos²17°= 1/2I₀cos²17°
- The intensity through the third polarizing filter is I₃ = I₂cos²17° = 1/2I₀cos⁴17°
- The intensity through the fourth polarizing filter is I₄ = I₃cos²17° = 1/2I₀cos⁶17°
- The intensity through the fifth polarizing filter is I₅ = I₄cos²17° = 1/2I₀cos⁸17°
- The intensity through the sixth polarizing filter is I₆ = I₅cos²17° = 1/2I₀cos¹⁰17°
- The intensity through the seventh polarizing filter is I₇ = I₆cos²17° = 1/2I₀cos¹²17°.
<h3>The ratio of the intensity between light intensity that emerges from the last filter and unpolarized light of intensity</h3>
Since I₇ is the last intensity I₇ = It = 1/2I₀cos¹²17°.
So, It/I₀ = 1/2cos¹²17°
= 1/2(0.9563)¹²
= 1/2 × 0.5850
= 0.2925
So, the ratio of the intensity between light intensity that emerges from the last filter and unpolarized light of intensity, I₀ is It/I₀ = 0.2925
Learn more about intensity of polarized light here:
brainly.com/question/25402491
I have a strange hunch that there's some more material or previous work
that goes along with this question, which you haven't included here.
I can't easily find the dates of Mercury's extremes, but here's some of the
other data you're looking for:
Distance at Aphelion (point in it's orbit that's farthest from the sun):
<span><span><span><span><span>69,816,900 km
0. 466 697 AU</span>
</span>
</span>
</span>
<span>
Distance at Perihelion
(</span></span><span>point in it's orbit that's closest to the sun):</span>
<span><span><span><span>46,001,200 km
0.307 499 AU</span> </span>
Perihelion and aphelion are always directly opposite each other in
the orbit, so the time between them is 1/2 of the orbital period.
</span><span>Mercury's Orbital period = <span><span>87.9691 Earth days</span></span></span></span>
1/2 (50%) of that is 43.9845 Earth days
The average of the aphelion and perihelion distances is
1/2 ( 69,816,900 + 46,001,200 ) = 57,909,050 km
or
1/2 ( 0.466697 + 0.307499) = 0.387 098 AU
This also happens to be 1/2 of the major axis of the elliptical orbit.
