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2 years ago

8

The angular width of localizers varies between and degrees in order to provide a signal width of approximately 700

feet at the runway thresholds.

1 answer:

natali 33 [55]2 years ago

5 0

Answer: 10 and 35 degrees

Explanation: Localizers width below 10 degree and 35 degree signal arc is unreliable and considered unusable for navigation and as a result, aircrafts may loose alignment

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Alan wrote the following examples of changes in substance.

Yuliya22 [10]

A campfire being lighted and plants converting carbon-dioxide and water into glucose and oxygen are both forms of chemical change.

Therefore, the answer is:

B. Both are examples of chemical change.

5 0

2 years ago

Read 2 more answers

A shot putter releases the shot some distance above the level ground with a velocity of 12.0 m/s, 51.0 ∘above the horizontal. Th

alina1380 [7]

A) Zero

The motion of the shot is a projectile's motion: this means that there is only one force acting on the projectile, which is gravity. However, gravity only acts in the vertical direction: so, there are no forces acting in the horizontal direction. Therefore, the x-component of the acceleration is zero.

B) -9.8 m/s^2

The vertical acceleration is given by the only force acting in the vertical direction, which is gravity:

$F=mg$

where m is the projectile's mass and g is the gravitational acceleration. Therefore, the y-component of the shot's acceleration is equal to the acceleration due to gravity:

$a_y = g = -9.8 m/s^2$

where the negative sign means it points downward.

C) 7.6 m/s

The x-component of the shot's velocity is given by:

$v_x = v_0 cos \theta$

where

$v_0 = 12.0 m/s$ is the initial velocity

$\theta=51.0^{\circ}$ is the angle of the shot

Substituting into the equation, we find

$v_x = (12.0 m/s)(cos 51^{\circ})=7.6 m/s$

D) 9.3 m/s

The y-component of the shot's velocity is given by:

$v_y = v_0 sin \theta$

where

$v_0 = 12.0 m/s$ is the initial velocity

$\theta=51.0^{\circ}$ is the angle of the shot

Substituting into the equation, we find

$v_y = (12.0 m/s)(sin 51^{\circ})=9.3 m/s$

E) 7.6 m/s

We said at point A) that the acceleration along the x-direction is zero: therefore, the velocity along the x-direction does not change, so the x-component of the velocity at the end of the trajectory is equal to the x-velocity at the beginning:

$v_x = 7.6 m/s$

F) -11.1 m/s

The y-component of the velocity at time t is given by:

$v_y(t) = v_y + at$

where

$v_y = 9.3 m/s$ is the initial y-velocity

a = g = -9.8 m/s^2 is the vertical acceleration

t is the time

Since the total time of the motion is t=2.08 s, we can substitute this value into the equation, and we find:

$v_y(2.08 s)=9.3 m/s + (-9.8 m/s^2)(2.08 s)=-11.1 m/s$

where the negative sign means the vertical velocity is now downward.

3 0

2 years ago

A 65 kg students is walking on a slackline, a length of webbing stretched between two trees. the line stretches and sags so that

polet [3.4K]

Answer : Tension in the line = 936.7 N

Explanation :

It is given that,

Mass of student, m = 65 kg

The angle between slackline and horizontal, $\theta=20^0$

The two forces that acts are :

(i) Tension

(ii) Weight

So, from the figure it is clear that :

$mg=2T\ sin\ \theta$

$T=\dfrac{mg}{2\ sin\theta}$

$T=\dfrac{65\ kg\times 9.8\ m/s^2}{2(sin\ 20)}$

$T=936.7\ N$

Hence, this is the required solution.

5 0

2 years ago

Read 2 more answers

A 4 cm diameter "bobber" with a mass of 3 grams floats on a pond. A thin, light fishing line is tied to the bottom of the bobber

Tasya [4]

Answer:

Explanation:

Calculate the volume of the lead

$V=\frac{m}{d}\\\\=\frac{10g}{11.3g'cm^3}$

Now calculate the bouyant force acting on the lead

$F_L = Vpg$

$F_L=(\frac{10g}{11.3g/cm^3} )(1g/cm^3)(9.8m/s^2)\\\\=8.673\times 10^{-3}N$

This force will act in upward direction

Gravitational force on the lead due to its mass will act in downward direction

Hence the difference of this two force

$T=mg-F_L\\\\=(10\times10^{-3}kg(9.8m/s^2)-8.673\times 10^{-3}\\\\=8.933\times10^{-3}N$

If V is the volume submerged in the water then bouyant force on the bobber is

$F_B=V'pg$

Equate bouyant force with the tension and gravitational force

$F_B=T_mg\\\\V'pg=\frac{(8.933\times10^{-2}N)+mg}{pg} \\\\V'=\frac{(8.933\times10^{-2}N)+mg}{pg}$

Now Total volume of bobble is

$\frac{V'}{V^B} =\frac{\frac{(8.933\times10^{-2})+Mg}{pg} }{\frac{4}{3} \pi R^3 }\times100\\\\=\frac{\frac{(8.933\times10^{-2})+(3)(9.8)}{(1000)(9.8)} }{\frac{4}{3} \pi (4.0\times10^{-2})^3 }\times100\\\\$

= $\large\boxed{4.52 \%}$

7 0

2 years ago

If the coefficient of static friction between a table and a uniform massive rope is μs, what fraction of the rope can hang over

astra-53 [7]

Let me give you the procedure like this:
Lets say that F is the fraction of the rope hanging over the table
If its like that then we have to take into account that the <span>friction force keeping on table is given by the following formula:</span>
<span>Ff = u*(1-f)*m*g </span>
and we need to know aso that <span>gravity force pulling off the table Fg is given by this other formula:</span>
<span>Fg = f*m*g </span>
What you need to do is <span>Equate the two and solve for f: </span>

<span>f*m*g = u*(1-f)*m*g </span>
<span>=> f = u*(1-f) = u - uf </span>
<span>=> f + uf = u </span>
=> f = u/(1+u) = fraction of rope
With that you can find the answer

8 0

2 years ago

Read 2 more answers