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The heat coming from the sun warms the land more quickly than the sea. As a result of these, the air near the land warm up and rises and the cooler air from the sea moves in to replace the risen air. The correct answer is option A
There will be heat transfer from a region of higher temperature to the region of lower temperature. But in the case of land and sea breeze, the transfer of heat are the result of convectional current in nature. Because the land is a better absorber of heat and also has a lower specific heat capacity compare to sea, during the day, the heat coming from the sun warms the land more quickly than the sea. As a result of these, the air near the land warm up and rises.
The cooler air from the sea moves in to replace the risen air.
Why do ocean winds or sea breezes blow toward shore during the day ? It is because air over the beach heats up, rises and is replaced by ocean air.
Therefore, option A is correct
Learn more here : brainly.com/question/1114842
Answer:
B) 18,000 feet MSL
Explanation:
There are three-dimensional parts in the navigation airspace in the world. The class E airspace is mostly used in the regions with coastal areas that are relatively populated. If we consider certain forms of exceptions, the class E airspace can move in the upward direction to few feet (i.e. 1200 ft). However, this doesn't include 18,000 feet MSL.
Not that simple but there are ways. Just make sure your hands can handle it
<span>Answer:
So this involves right triangles. The height is always 100. Let the horizontal be x and the length of string be z.
So we have x2 + 1002 = z2. Now take its derivative in terms of time to get
2x(dx/dt) = 2z(dz/dt)
So at your specific moment z = 200, x = 100âš3 and dx/dt = +8
substituting, that makes dz/dt = 800âš3 / 200 or 4âš3.
Part 2
sin a = 100/z = 100 z-1 . Now take the derivative in terms of t to get
cos a (da./dt) = -100/ z2 (dz/dt)
So we know z = 200, which makes this a 30-60-90 triangle, therefore a=30 degrees or π/6 radians.
Substitute to get
cos (Ď€/6)(da/dt) = (-100/ 40000)(4âš3)
âš3 / 2 (da/dt) = -âš3 / 100
da/dt = -1/50 radians</span>
