Mark Watney begins his day 15 km West and 25 km North of his Mars Habitat. a. Set up a co-ordinate system (draw labeled axis and
tickmarks showing the scale) and draw a vector representing the initial position of Mark Watney. b. Mark spends his day driving his Mars Rover towards Schiaparelli Crater and manages to make it an additional 20 km West but has to go around a hill so ended up 5 km South of his starting point for the day. Draw a vector representing Mark's total position change for the day. c. Using vector addition find Mark's position relative to the Mars Habitat. Give both the numerical description of the vector and show it on your plot.
1 answer:
Answer:
a, b) part a and b on diagram attached
c) sf = -35 i + 20 j
35 km West and 20 km North
Explanation:
For part a and b refer to the attached co-ordinate system:
Note: unit vector i is in West/East direction and unit vector j is in North/South direction.
si = -15 i + 25 j
sf-si = -20 i - 5 j
Hence,
Mark relative position from habitat sf = si + sf/i
sf = ( -15 i + 25 j ) + ( -20 i - 5 j )
sf = -35 i + 20 j
35 km West and 20 km North
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Answer:
3) Transmitted intensity of light if unpolarized light passes through a single polarizing filter = 40 W/m²
- Transmitted intensity of light if an additional polarizer is added perpendicular to the first polarizer in the setup described = 7.5 W/m²
Explanation:
Complete Question
3) What is the transmitted intensity of light if unpolarized light passes through a single polarizing filter and the initial intensity is 80 W/m²?
- What is the transmitted intensity of light if an additional polarizer is added perpendicular to the first polarizer in the setup described in Question 3 (the setup)? Show all work in your answer.
The image of this setup attached to this question as obtained from online is attached to this solution.
Solution
3) When unpolarized light passes through a single polarizer, the intensity of the light is cut in half.
Hence, if the initial intensity of unpolarized light is I₀ = 80 W/m²
The intensity of the light rays thay pass through the first single polarizer = I₁ = (I₀/2) = (80/2) = 40 W/m²
- According to Malus' law, the intensity of transmitted light through a polarizer is related to the intensity of the incident light and the angle at which the polarizer is placed with respect to the major axis of the polarizer before the current polarizer of concern.
I₂ = I₁ cos² θ
where
I₂ = intensity of light that passes through the second polarizer = ?
I₁ = Intensity of light from the first polarizer that is incident upon the second polarizer = 40 W/m²
θ = angle between the major axis of the first and second polarizer = 30°
I₂ = 40 (cos² 30°) = 40 (0.8660)² = 30 W/m²
In the same vein, the intensity of light that passes through the third/additional polarizer is related to the intensity of light that passes through the second polarizer and is incident upon this third/additional polarizer through
I₃ = I₂ cos² θ
I₃ = intensity of light that passes through the third/additional polarizer = ?
I₂ = Intensity of light from the second polarizer that is incident upon the third/additional polarizer = 30 W/m²
θ = angle between the major axis of the second and third/additional polarizer = 60° (although, it is 90° with respect to the first polarizer, it is the angle it makes with the major axis of the second polarizer, 60°, that matters)
I₃ = 30 (cos² 60°) = 30 (0.5)² = 7.5 W/m²
Hope this Helps!!!
