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2 years ago

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What is the approximate size of the smallest object on the earth that astronauts can resolve by eye when they are orbiting 250 k

m above the earth? assume λ = 500 nm and a pupil diameter of 5.00 mm?

1 answer:

Alex17521 [72]2 years ago

5 0

<span>Skipping OTF and considering a small angle, then 2.38E5 • 1.22(4.56E-7/2.2E-3): 60 meters.</span>

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Interactive Solution 28.5 illustrates one way to model this problem. A 7.11-kg object oscillates back and forth at the end of a

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Explanation:

Given that,

Mass of the object, m = 7.11 kg

Spring constant of the spring, k = 61.6 N/m

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We need to find the time period of oscillation observed by the observed. The time period of oscillation is given by :

$t_o=2\pi \sqrt{\dfrac{m}{k}} \\\\t_o=2\pi \sqrt{\dfrac{7.11}{61.6}} \\\\t_o=2.13\ s$

Time period of oscillation measured by the observer is :

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2 years ago

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3 years ago

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antoniya [11.8K]

Answer:

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Thus, the correct option is "2" When they reach the bottom of the fall

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2 years ago