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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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kramer

Answer:

Tiene que patear el balón cerca del nivel del suelo para conseguir el gol.

Explicación:

Leo Messi necesita patear el balón no en el aire, en el nivel del suelo para poder meter el gol para el equipo. Delante del fútbol hay muchos jugadores del equipo contrario dispuestos a detener el tiro libre de Leo Messi. Cuando Leo Messi patea el balón en el aire, los jugadores también saltan en el aire para detener el tiro libre que atraviesa la cabeza y para conseguir el gol Leo Messi tiene que evitar patear el balón en el aire y patear el balón cerca del nivel del suelo. que puede atravesar a los jugadores oponentes cuando lo saltan para detener el balón a través de sus cabezas para que el balón pase por debajo de sus pies y llegue a la portería.

7 0
3 years ago
A particular molecule has a small molecular mass of approximately 1000 daltons. What should be done to make this molecule more a
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Answer:

Bind it to a large protein

Explanation:

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3 years ago
A rocket moves upward, starting from rest with an acceleration of +30.0 m/s2 for 5.00 s. It runs out of fuel at the end of this
olchik [2.2K]

Answer:

The distance covered by the rocket after fuel ran out is 3442.04 m

Explanation:

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The distance it travelled within this time is given by  s=ut+ \frac{1}{2} at^2                                                                                                  =0 \times 5+ \frac{1}{2} (30\times25)=375 m

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v=0+30\times5=150m/s

Given that at this height it runs out of fuel but travels further. Here final velocity v=0(maximum height), initial velocityu=150 m/s  and time to zero velocity t=\frac{v}{g} = \frac{150}{9.8} =15.3 s.

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s= ut+\frac{1}{2} gt^2=150 \times 15.3+1/2 \times9.8 \times 15.3^2=3442.04 m

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