Anyone help me please ?

Reception of signals from a radio facility, located off the airway being flown, may be inadequate at the designated mea to ident

lakkis [162]

The altitude ensures acceptable navigational signal coverage only within 22 NM of a VOR.

<h3>What is altitude?</h3>

Altitude or height exists as distance measurement, usually in the vertical or "up" approach, between a reference datum and a point or object. The exact meaning and reference datum change according to the context.

The MOCA exists in the lower published altitude in effect between fixes on VOR airways, off-airway routes, or route segments that satisfy obstacle support conditions for the whole route segment. This altitude also ensures acceptable navigational signal coverage only within 22 NM of a VOR.

The altitude ensures acceptable navigational signal coverage only within 22 NM of a VOR.

Therefore, the correct answer is 22 NM of a VOR.

To learn more about altitudes refer to:

brainly.com/question/1159693

#SPJ4

3 0

1 year ago

Define Electromechanical systems.

dedylja [7]

Answer:

Electromechanical systems or devices are systems or devices that involves the interaction between electrical systems and mechanical systems in which the motion of mechanical parts is converted to electrical energy or made to interact with energy or in which electrical energy is converted to mechanical energy or interacts with a moving mechanical system

Therefore;

Electromechanical systems convert <u>electrical energy</u> input into a <u>mechanical energy</u> output

Explanation:

3 0

2 years ago

My computer has a mass of 0.031080997078386 slug the Earth's surface.

poizon [28]

Answer:

The answer is "0.187 lbm and 1 lbf".

Explanation:

The mass = $0.031080997078386\ slug$

Calculating mass on Mars:

$\to m=m_g\frac{g}{g_e}$

$=0.031080997078386 \times \frac{32.2}{5.35}\\\\=0.187 \ lbm$

$\to W=mg_e$

$=0.187 \times 5.35\\\\=1 \ lbf$

4 0

2 years ago

Find the mass of electron

Natali5045456 [20]

Answer:

9.10938356 × 10^-31 kilograms

Explanation:

i suck at chem lol

6 0

2 years ago

A uniform edge load of 500 lb/in. and 350 lb/in. is applied to the polystyrene specimen. If it is originally square and has dime

aliya0001 [1]

The image of the load applied to the polystyrene is missing, so i have attached it.

Answer:

a_new = 2.00302 in

b_new = 2.00552

Explanation:

From the image attached, we can see that the load of 500 lb/in is applied in the x-direction while the load of 350 lb/in acts in the y-direction.

Now, formula for stress is;

Stress(σ) = Force/Area

We are not given force and area but the load and plate thickness.

Thus, stress = load/thickness

We are given;

Load in x - direction = 500 lb/in.

Load in y - direction = 350 lb/in.

Thickness; t = 0.25 in

Thus;

σ_x = 500/0.25

σ_x = 2000 ksi

σ_y = 350/0.25

σ_y = 1400 ksi

From Hooke's law for 2 dimensions, strain is given by the formula;

ε_x = (1/E)(σ_x - vσ_y)

ε_y = (1/E)(σ_y - vσ_x)

We are given v_p = 0.25 and Ep = 597 × 10³ psi

Thus;

ε_x = (1/(597 × 10^(3)))(2000 - (0.25 × 1400)

ε_x = 0.00276

ε_y = (1/(597 × 10^(3)))(1400 - (0.25 × 2000)

ε_y = 0.00151

From elongation formula, we know that;

Startin is: ε = ΔL/L

Thus; ΔL = Lε

We are given a = 2 and b = 2

Thus;

ΔL_x = 2 × 0.00276

ΔL_x = 0.00552

ΔL_y = 2 × 0.00151

ΔL_y = 0.00302

New dimensions are;

a_new = 2 + 0.00302

a_new = 2.00302 in

b_new = 2 + 0.00552

b_new = 2.00552

3 0

2 years ago