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Anna71 [15]
3 years ago
12

Outside controlled airspace, the minimum flight visibility requirement for vfr flight above 1,200 feet agl and below 10,000 feet

msl during daylight hours is
Physics
1 answer:
nalin [4]3 years ago
3 0
The least possible flight visibility prerequisite for VFR flight above 1200 feet AFL and below 10000 feet MSL for the duration of daylight hours is 1 mile in the external controlled airspace. In aeronautics, visual meteorological conditions or as called as VMC is an aviation flight group in which visual flight rules or VFR is allowed that is in circumstance in which pilot have adequate visibility to fly the aircraft sustaining visual leave-taking from terrain and other aircraft. They are the contradictory of instrument meteorological conditions. The border standards among instrument meteorological conditions and visual meteorological circumstances are recognized as the visual meteorological conditions minima and are well-defined by visibility, cloud ceilings for take offs and landing and cloud allowances.  
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An astronaut is planning a trip to a newly-discovered planet. According to 2 points
rodikova [14]

Answer:

The new planet has more gravity

Explanation:

gravity force =  G m1 m2 / r^2

   decresing r  or increasing the planet mass will increase the force

3 0
3 years ago
What is a stream’s discharge rate if it has a width of 10 meters, a depth of 2 meters, and a velocity of 2 meters per second?
julia-pushkina [17]

A stream’s discharge rate if it has a width of 10 meters, a depth of 2 meters, and a velocity of 2 meters per second will be 40 m^{3}/sec .

Discharge rate = velocity * area

                 = velocity * depth * width

                 = 2 * 2 * 10 = 40 m^{3}/sec

A stream’s discharge rate if it has a width of 10 meters, a depth of 2 meters, and a velocity of 2 meters per second will be 40 m^{3}/sec .

learn more about discharge rate

brainly.com/question/20709500?referrer=searchResults

#SPJ4

4 0
3 years ago
A small object is attached to a horizontal spring and set in simple harmonic motion with amplitude A and period T. How long does
Lunna [17]

Answer:

1.5T

Explanation:

Since, the amplitude of SHM is A. So, in one time period T the object will travel travel a distance of 4A.

6A-4A= 2A.

Now, this 2A distance must be traveled in T/2 time period.

So, the total time taken to travel a distance of 6A is T+T/2 = 3T/2 = 1.5T

6 0
3 years ago
Two blocks of masses 6 kg and 5.5 kg are
sp2606 [1]

When you squish the spring, you put some energy into it, and after the cord
burns and they go boing in opposite directions, that energy that you stored
in the spring is what gives the blocks their kinetic energy.

But linear momentum still has to be conserved.  It was zero while they were
tied together and nothing was moving, so it has to be zero after they both
take off.

Momentum = (mass) x (velocity)

After the launch, the 5.5-kg moves to the right at 6.8 m/s,
so its momentum is
                               (5.5 x 6.8) = 37.4 kg-m/s to the right.

In order for the total momentum to be zero, the other block has to
carry the same amount of momentum in the opposite direction.

               M x V = (6 x speed) = 37.4 kg-m/s to the left.

Divide each side by  6 :      Speed = 37.4 / 6 =<em>  6.2333... m/s left</em>

(That number is  (6 and 7/30) m/s .)
5 0
3 years ago
An iron wire has a cross-sectional area equal to 5.00×10⁻⁶ m² . Carry out the following steps to determine the drift speed of th
Doss [256]
  1. In mass, there are 55.85 × 10⁻³ kg/mol in in 1 mole of iron.
  2. The molar density of iron is equal to 1.41 × 10⁵ mol/m³.
  3. The density of iron atoms is equal to 8.49 × 10²⁸ atoms/m³.
  4. The number density of conduction electrons is equal to 1.70 × 10²⁹ conduction electrons/m³.
  5. The drift speed of conduction electrons is equal to 2.21 × 10⁻⁴ m/s.

<h3>How to calculate the drift speed of the conduction electrons?</h3>

Mathematically, the drift speed of the conduction electrons can be calculated by using this formula:

V = (m × σ × V)/ρ × e × f × l)

V = I/(n × A × Q)

Where:

  • U represents the drift speed of the conduction electrons, in m/s.
  • m represents the molecular mass of the metal, in kg.
  • e represents the elementary charge, in C.
  • f represents the number of free electrons per atom.
  • σ represents the electric conductivity of the medium at a particular temperature in S/m.
  • ρ represents the density of the conductor, in kg/m³.
  • ℓ represents the length of the conductor, in m.
  • ΔV represents the voltage applied or potential difference across the conductor in V.

<h3>How many kilograms are there in 1 mole of iron? </h3>

Molar mass of iron = 55.85 g/mol.

In Kilograms, we have:

Mass = 55.85 × 1/1000

Mass = 55.85 × 10⁻³ kg/mol.

For the molar density of iron, we have:

Molar density = density/molar mass

Molar density = 7874/0.056

Molar density = 1.41 × 10⁵ mol/m³.

For the density of iron atoms, we have:

Density of iron atoms = Avogadro's constant × molar density

Density of iron atoms = 6.023 × 10²³ × 1.406 × 10⁵

Density of iron atoms = 8.49 × 10²⁸ atoms/m³.

For the number density of conduction electrons, we have:

Fe ---> Fe²⁺ + 2e⁻

Number density of conduction electrons = 2 conduction electrons/1 atom of iron

Number density of conduction electrons = 2 × 8.49 × 10²⁸

Number density of conduction electrons = 1.70 × 10²⁹ conduction electrons/m³.

For the drift speed of conduction electrons, we have:

V = I/(n × A × Q)

V = 30/(1.70 × 10²⁹ × 1.602 × 10⁻¹⁹ × 5 × 10⁻⁶)

Drift speed, V = 2.21 × 10⁻⁴ m/s.

Read more on drift speed here: brainly.com/question/15219891

#SPJ4

Complete Question:

An iron wire has a cross-sectional area of 5.00 x 10-6 m2. Carry out steps (a) through (e) to compute the drift speed of the conduction electrons in the wire.

(a) How many kilograms are there in 1 mole of iron?

(b) Starting with the density of iron and the result of part (a), compute the molar density of iron (the number of moles of iron per cubic meter).

(c) Calculate the number density of iron atoms using Avogadro’s number.

(d) Obtain the number density of conduction electrons given that there are two conduction electrons per iron atom.

(e) If the wire carries a current of 30.0 A, calculate the drift speed of conduction electrons.

4 0
2 years ago
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