C most of the Earths weather occurs in this layer.
Most likely to be b.entering the formal operational stage
Answer:
The diameter of the camera aperture must be greater than or equal to 1.49m
Explanation:
Let the distance separating two objects, x = 6.0 cm = 0.06 m
The distance between the observer and the two objects, d = 160 km = 160000 m
Let ∅ = minimum angular separation between the two objects that the satellite can resolve
tan( ∅) = x/d
Since there is minimum angular separation, tan( ∅) ≈∅
∅ = x/d
∅ = 0.06/160000
∅ = 3.75 * 10⁻⁷rad
For the satellite to be able to resolve the objects,
D ≥ 1.22λ/∅
λ = 560 nm = 560 * 10⁻⁹
D ≥ 1.22 * (560 * 10⁻⁹)/(3.75 * 10⁻⁷)
D ≥ 149.33 * 10⁻² m
D ≥ 1.49 m
Answer:
(a) t = 2.97s
(b) h = 43.3 m
Explanation:
Let t be the time it takes to fall a distance h, then t - 1 (s) is the time it takes to fall a distance of h - 0.56h = 0.44 h
For the ball to fall from rest a distance of h after time t
Also for the ball to fall from rest a distance of 0.44h after time (t-1)
We can substitute the 1st equation into the 2nd one
and divide both sides by g/2
t = 2.97 or t = 0.6
Since t can only be > 1 s we will pick t = 2.97 s
(b) 
What are the "observations above" in this question?
