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4 years ago

12

When an unpressurized aircraft's static pressure system is leak checked to comply with the requirements of Section 91.411, what

aircraft instrument may be used in lieu of a pitot-static system tester

1 answer:

velikii [3]4 years ago

6 0

ANSWER::: The PLUTONIC
will be used in lieu of a pitot-static’s system tester.

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A photon of wavelength 2.78 pm scatters at an angle of 147° from an initially stationary, unbound electron. What is the de Brogl

Elena-2011 [213]

Answer:

2.07 pm

Explanation:

The problem given here is the very well known Compton effect which is expressed as

$\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)$

here, $\lambda$ is the initial photon wavelength, $\lambda^{'}$ is the scattered photon wavelength, h is he Planck's constant, $m_e$ is the free electron mass, c is the velocity of light, $\theta$ is the angle of scattering.

Given that, the scattering angle is, $\theta=147^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.$

Here, the photon's incident wavelength is $\lamda=2.78pm$

Therefore,

$\lambda^{'}=2.78+4.45=7.23 pm$

From the conservation of momentum,

$\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}$

where, $\vec{P_\lambda}$ is the initial photon momentum, $\vec{P_{\lambda^{'}}}$ is the final photon momentum and $\vec{P_e}$ is the scattered electron momentum.

Expanding the vector sum, we get

$P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta$

Now expressing the momentum in terms of De-Broglie wavelength

$P=h/\lambda,$

and putting it in the above equation we get,

$\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}$

Therefore,

$\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm$

This is the de Broglie wavelength of the electron after scattering.

6 0

4 years ago

Which energy transformations BEST illustrates why people can walk? A) Chemical energy is being converted into light energy. B) C

Gekata [30.6K]

The answer is letter c.

The explanation behind this is when human consumes a plant or when a human being eats an animal that ate a plant. The chemical energy kept in the plant (or animal) cells is progressed into the cells of the human's body. All of the body progressions, like ingestion, breathing, pumping blood, are driven by cells changing the kept chemical energy into heat and work, in a procedure called respiration. In the muscle cells of the human (or any animal), the chemical energy is converted into mechanical work and heat. The muscle pacts, the legs thrust, and the body jumps into the air.

5 0

4 years ago

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When two light waves arrive at the same place at the same time they create a?

Montano1993 [528]

Hey there,
<span>They interfere essentially like any other form of wave.
</span>
Hope this helps :))

~Top

3 0

4 years ago

When n liters of fuel were added to a tank that was already 1 3 full, the tank was filled to 7 9 of its capacity. In terms of n,

Zarrin [17]

Answer: 9/4n

Explanation:

Let's take T to represent tank

When n litres of fuel was added to tank that was 1/3 full = n + 1/3T

The tank was 7/9T

n+1/3T = 7/9T

n =7/9T - 1/3T

n = 4/9

If a full tank is taken to be 1 = 9/9

Hence, we will have 1 = x × n

1 = x × 4/9

x = 9/4

Hence 9/4 of n = 9/4n

8 0

3 years ago

The velocity of a 150 kg cart changes from 6.0 m/s to 14.0 m/s. What is the magnitude of the impulse that acted on it?

aleksandrvk [35]

M = 150 kg.

Final velocity, v = 14 m/s

Initial Velocity, u = 6 m/s

Impulse = m(v - u)

= 150*(14 - 6)

= 150*8 = 1200 kgm/s or 1200 Ns

8 0

4 years ago

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