Sometime around 2024, astronomers at the European Southern Observatory hope to begin using the E-ELT (European Extremely Large T
elescope), which is planned to have a primary mirror 42 m in diameter. Let us assume that the light it focuses has a wavelength of 550 nm. (a) What is the most distant Jupiter-sized planet the telescope could resolve, assuming its resolution is limited only by diffraction? Express your answer in meters and light years. (b) The nearest known exoplanets (planets beyond the solar system) are around 20 light years away. What would have to be the minimum diameter of an optical telescope to resolve a Jupiter-sized planet at that distance using light of wavelength 550 nm? ( 1 light year = 9.461 × 10 15 m )
1 answer:
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Answer:
Explanation:
a = 4ms⁻², Vf = 180 m/s & Vi = 140m/s
a =
4 =
t = 40/4
t = 10sec
To Measure Distance Use third Equation of Motion:
2aS = Vf²-Vi²
S =
S = 12800/8 = 1600m
Answer:
Distance: 21 yd, displacement: 15 yd, gain in the play: 12 yd
Explanation:
The distance travelled by Sam is just the sum of the length of each part of Sam's motion, regardless of the direction. Initially, Sam run from the 3 yd line to the 15 yd line, so (15-3)=12 yd. Then, he run also 9 yd to the right. Therefore, the total distance is
d = 12 + 9 = 21 yd
The displacement instead is a vector connecting the starting point with the final point of the motion. Sam run first 12 yd straight ahead and then 9 yd to the right; these two motions are perpendicular to each other, so we can find the displacement simply by using Pythagorean's theorem:
Finally, the yards gained by Sam in the play are simply given by the distance covered along the forward-backward direction only. Since Sam only run from the 3 yd line to the 15 yd line along this direction, then the gain in this play was
d = 15 - 3 = 12 yd