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4 years ago

What are the minimum requirements for airplane operations under special VFR in Class D airspace at night?

Physics

1 answer:

sdas [7]4 years ago

5 0

Answer:

The minimum requirements for airplane operations under special VFR in Class D airspace at night are;

The pilot must be instrument rated, and

The airplane must be IFR equipped.

Pilots flying into Class D airspace must maintain at least three statute miles visibility.

In addition, pilots must remain at least 500 feet below clouds, 1,000 feet above clouds, and stay 2,000 feet away from horizontal clouds while in Class D airspace.

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State the section of the graph (A, B, C, or D) where the car is moving the slowest.

anastassius [24]

Answer:

D 15m/s

Explanation:

1)you find the speed covered in all the points e.g on speed=distance covered

time taken

a)300m\15s=20m/s

b)rest no distance covered

c)(1000-300)=700m time=(60-35)=25. 700m/25s=28m/s

d)(1200-1000)=300m time=(80-60)=20s

200m/20s=10m/s

thus the lowest speed was 10m/s at D

Average speed=average distance

average time

=1200m/80s. =15m/s

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3 years ago

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jolli1 [7]

Answer:

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Explanation:

6 0

3 years ago

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a football is kicked toward a goal keeper with an initial speed of 20m/s at an angle of 45 degrees with the horizontal .at the m

algol [13]

Answer:

3.18 m/s

Explanation:

Given that

Initial speed of the ball, u = 20 m/s

Angle of inclination, θ = 45°

Distance from the ball, h = 50 m

Using equations of projectile to solve this, we have

We start by finding the time of flight, T

T = 2Usinθ/g

T = (2 * 20 * sin45)/9.8

T = (40 * 0.7071) / 9.8

T = 28.284/9.8

T = 2.89 s

Next we find the Range, R

R = u²sin2θ/g

R = (20² * sin 90) / 9.8

R = (400 * 1) / 9.8

R = 400/9.8 = 40.82 m

Distance the gk must cover

40.82 - 50 m

-9.18 m or 9.18 m in the opposite direction.

Speed of the GK = d/t

9.18 / 2.89 = 3.18 m/s

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4 years ago

Find the magnitude of the torque that acts on the molecule when it is immersed in a uniform electric field of 6.19×105 N/C with

Ivan

Answer:

$\tau=5.81\times 10^5p\ N-m$

Explanation:

We have,

Electric field, $E=6.19\times 10^5\ N/C$

The electric dipole vector at an angle of 69.9 degrees from the direction of the field.

The torque acting on a molecule is given by :

$\tau=p\times E\\\\\tau=pE\sin\theta$

p is electric dipole moment

$\tau=p\times 6.19\times 10^{5}\times \sin (69.9)\\\\\tau=5.81\times 10^5p\ N-m$

So, the magnitude of the torque acting on the molecule is $5.81\times 10^5p\ N-m$ .

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4 years ago

The atomic number of an atom is _____

GrogVix [38]

The proton number? number of protons

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4 years ago

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