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Musya8 [376]
3 years ago
5

With certain exceptions, Class E airspace extends upward from either 700 feet or 1,200 feet AGL to, but does not include,A) 14,5

00 feet MSL.* B) 18,000 feet MSL.C) 10,000 feet MSL.
Physics
1 answer:
DedPeter [7]3 years ago
6 0

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.

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Answer:

Wind the long piece of thin wire around the uniform glass rod multiple times, find the length of the total diameters using the metre ruler, and divide by the number of times you wound it around the rod.

Explanation:

Since the diameter of one long piece of thin wire is too thin to be measured by a metre ruler, you can wind it multiple times and push it side by side to get a length you can measure.

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A man with a mass of 65.0 kg skis down a frictionless hill that is 5.00 m high. At the bottom of the hill the terrain levels out
anzhelika [568]

Answer:

The horizontal distance is 4.823 m

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As per the question:

Mass of man, m = 65.0 kg

Height of the hill, H = 5.00 m

Mass of the backpack, m' = 20.0 kg

Height of ledge, h = 2 m

Now,

To calculate the horizontal distance from the edge of the ledge:

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KE_{initial} + PE_{initial} = KE_{final} + PE_{final}

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PE = Potential energy

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0 + mgH = \frac{1}{2}mv^{2} + 0

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mv = (m + m')v'

65.0\times 9.89 = (65.0 + 20.0)v'

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Now, time taken for the fall:

h = \frac{1}{2}gt^{2}

t = \sqrt{\frac{2h}{g}}

t = \sqrt{\frac{2\times 2}{9.8} = 0.638\ s

Now, the horizontal distance is given by:

x = v't = 7.56\times 0.638 = 4.823\ m

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