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nydimaria [60]
1 year ago
10

What is the minimum runway length that will serve? hint: you can solve this problem using ratios without having any additional i

nformation. express your answer in miles.
Physics
1 answer:
ehidna [41]1 year ago
6 0

There are many factors that determine if an aircraft can operate from a given airport. Of course the availability of certain services, such as fuel, access to air stairs and maintenance are all necessary. But before considering anything else, one must determine if the plane can physically land at an airport, and equally as important, take off.

What is the minimum runway length that will serve?

Looking at aerial views of runways can lead some to the assumption that they are all uniform, big and appropriate for any plane to land. This couldn’t be further from the truth.

A given aircraft type has its own individual set of requirements in regards to these dimensions. The classic 150’ wide runway that can handle a wide-body plane for a large group charter flight isn’t a guarantee at every airport. Knowing the width of available runways is important for a variety of reasons including runway illusion and crosswind condition.

Runways also have different approach categories based on width, and have universal threshold markings that indicate the actual width.

To learn more about runway

brainly.com/question/11553726

#SPJ4

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Two long, parallel wires carry currents of different magnitudes. If the amount of current in one of the wires is doubled, what h
Ulleksa [173]

Answer:

The magnitude of the force that each wire exerts on the other will increase by a factor of two.

Explanation:

force on parallel current carrying wire, F = BILsinθ

where;

B is the strength of the magnetic field

L is the length of the wire

I is the magnitude of current on the wire

θ is the angle of inclination of the wire

Assuming B, L and θ is constant, then F ∝ I

F = kI

\frac{F_1}{I_1} = \frac{F_2}{I_2}

When the amount of current is doubled in one of the wires, lets say the second wire;

\frac{F_1}{I_1} = \frac{F_2}{2I_1} \\\\F_2 = \frac{2F_1I_1}{I_1} \\\\F_2 =2F_1

Also, if will double the amount of current on the first wire, then

F₁ = 2F₂

Therefore, the magnitude of the force that each wire exerts on the other will increase by a factor of two.

3 0
3 years ago
a ball is thrown straight up into the air with a speed of 13 m/s. if the ball has a mass of 0.25 kg, how high does the ball go?
evablogger [386]
<h2>Hello!</h2>

The answer is: 8.62m

<h2>Why?</h2>

There are involved two types of mechanical energy: kinetic energy and potential energy, in two different moments.

<h2>First moment:</h2>

Before the ball is thrown, where the potential energy is 0.

<h2>Second moment: </h2>

After the ball is thrown, at its maximum height, the Kinetic Energy turns to 0 (since at maximum height,the speed is equal to 0) and the PE turns to its max value.

Therefore,

E=PE+KE

Where:

PE=m.g.h

KE=\frac{1*m*v^{2}}{2}

<em>E</em> is the total energy

<em>PE</em> is the potential energy

<em>KE</em> is the kinetic energy

<em>m</em> is the mass of the object

<em>g</em> is the gravitational acceleration

<em>h </em>is the reached height of the object

<em>v</em> is the velocity of the object

Since the total energy is always constant, according to the Law of Conservation of Energy, we can write the following equation:

KE_{1}+PE_{1}=KE_{2}+PE_{2}

Remember, at the first moment the PE is equal to 0 since there is not height, and at the second moment, the KE is equal to 0 since the velocity at maximum height is 0.

\frac{1*m*v^{2}}{2}+m.g.(0)=\frac{1*m*0^{2}}{2}+m.g.h\\\frac{1*m*v_{1} ^{2}}{2}=m*g*h_{2}

So,

h_{2}=\frac{1*m*v_{1} ^{2}}{2*m*g}\\h_{2}=\frac{1*v_{1} ^{2}}{2g}=\frac{(\frac{13m}{s})^{2} }{2*\frac{9.8m}{s^{2}}}\\h_{2}=8.62m}

Hence,

The height at the second moment (maximum height) is 8.62m

Have a nice day!

5 0
3 years ago
What is the mass of a bullet moving at 970m/s if the bullet’s KE is 3.9x 10^3J
motikmotik

Answer:

0.08kg

Explanation:

K.E = 1/2 mv^2

v = 970m/s

K.E = 3.9x 10^3J= 3900J

K.E = 1/2 mv^2

3900 = 1/2 m x 970x 970

3900 = 1/2 ×940900m

3900 = 470450m

m = 3900/470450 = 0.00828993516 = 0.008kg

4 0
3 years ago
Read 2 more answers
If mass of the planet is 10% less than that of
Leokris [45]

Answer:

25kg

Explanation:

The formula for acceleration due to gravity is:

g=\dfrac{GM}{r^2}

Here, G represents a constant and M represents the mass of the object. We don't need to know the actual values to solve this, then: We can just multiply by the amount they are multiplied relative to earth.

g=\dfrac{G(0.9M)}{(1.2r)^2}=0.625g

Multiplying this by the weight of the person on Earth of 40kg, you get 0.625*40=25kg.

Hope this helps!

4 0
3 years ago
If the electron has a speed equal to 9.10 x 10^6 m/s, what is its wavelength?
Fudgin [204]

Answer:

\lambda=8.006\times 10^{-11}\ m

Explanation:

Given that,

The speed of an electron, v=9.1\times 10^6\ m/s

We need to find the wavelength of this electron. It can be calculated using De -broglie wavelength concept as :

\lambda=\dfrac{h}{mv}

h is the Planck's constant

\lambda=\dfrac{6.63\times 10^{-34}}{9.1\times 10^{-31}\times 9.1\times 10^6}

\lambda=8.006\times 10^{-11}\ m

So, the wavelength of the electron is 8.006\times 10^{-11}\ m. Hence, this is the required solution.

5 0
3 years ago
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