All categories

4 years ago

Which cruising altitude is appropriate for vfr flight on a magnetic course of 135°

Physics

1 answer:

Nata [24]4 years ago

5 0

Answer:odd thousands plus 500 feet.

Explanation:

On a magnetic course of zero through 179 degrees, select an odd thousand foot cruising altitude plus 500 feet, such as 3,500, 5,500, up to and including 17,500. Even and odd thousands are reserved for those aircraft on an active instrument flight plan. Even thousands plus 500 feet are for aircraft flying between 180 and 359 degrees.

You might be interested in

What are the agency that monitors the tectonic phenomena?

matrenka [14]

Answer:

ASDASDSAD

Explanation:

ASDSADASDSADASD

8 0

3 years ago

if change in blood pressure between the brain and the feet is 1.88×10^4 pa .what will be the volume flow rate from head to feet

klio [65]

Answer: $3765.66 \frac{m^{3}}{s}$

Explanation:

We can solve this problem using the Poiseuille equation:

$Q=\frac{\pi r^{4}\Delta P}{8\eta L}$

Where:

$Q$ is the Volume flow rate

$r=23 cm \frac{1 m}{100 cm}=0.23 m$ is the effective radius

$L=6 ft \frac{0.3048 m}{1 ft}=1.8288 m$ is the length

$\Delta P=1.88(10)^{4} Pa$ is the difference in pressure

$\eta=3(10)^{-3} Pa.s$ is the viscosity of blood

Solving:

$Q=\frac{\pi (0.23 m)^{4}(1.88(10)^{4} Pa)}{8(3(10)^{-3} Pa.s)(1.8288 m)}$

$Q=3765.66 \frac{m^{3}}{s}$

4 0

4 years ago

Given two vectors A⃗ =−2.00i^+ 4.00 j^+ 4.00 k^ and B⃗ = 1.00 i^+ 2.00 j^−3.00k^, do the following.

ra1l [238]

For Part A.
The absolute value of vector A is
|A | = 6
This is simply the square root of the sum of the squares of the coordinates.

For Part B
Do the same thing as in Part A
| B | = √14

Part C
Simply subtract the corresponding elements resulting in
(-3, 2, 7)

Part D
√62

Part E
Yes

5 0

4 years ago

How can the momentum of an object be changed?

pickupchik [31]

C. You’re welcome

Leave a like

3 0

4 years ago

Three ideal polarizing filters are stacked, with the polarizing axis of the second and third filters at 21 degrees and 61 degree

kvv77 [185]

Answer:

When second polarizer is removed the intensity after it passes through the stack is

$I_f_3 = 27.57 W/cm^2$

2 When third polarizer is removed the intensity after it passes through the stack is

$I_f_2 = 102.24 W/cm^2$

Explanation:

From the question we are told that

The angle of the second polarizing to the first is $\theta_2 = 21^o$

The angle of the third polarizing to the first is $\theta_3 = 61^o$

The unpolarized light after it pass through the polarizing stack $I_u = 60 W/cm^2$

Let the initial intensity of the beam of light before polarization be $I_p$

Generally when the unpolarized light passes through the first polarizing filter the intensity of light that emerges is mathematically evaluated as

$I_1 = \frac{I_p}{2}$

Now according to Malus’ law the intensity of light that would emerge from the second polarizing filter is mathematically represented as

$I_2 = I_1 cos^2 \theta_1$

$= \frac{I_p}{2} cos ^2 \theta_1$

The intensity of light that will emerge from the third filter is mathematically represented as

$I_3 = I_2 cos^2(\theta_2 - \theta_1 )$

$I_3= \frac{I_p}{2}(cos^2 \theta_1)[cos^2(\theta_2 - \theta_1)]$

making $I_p$ the subject of the formula

$I_p = \frac{2L_3}{(cos^2 \theta [cos^2 (\theta_2 - \theta_1)])}$

Note that $I_u = I_3$ as $I_3$ is the last emerging intensity of light after it has pass through the polarizing stack

Substituting values

$I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (61-21)])}$

$I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (40)])}$

$=234.622W/cm^2$

When the second is removed the third polarizer becomes the second and final polarizer so the intensity of light would be mathematically evaluated as

$I_f_3 = \frac{I_p}{2} cos ^2 \theta_2$

$I_f_3$ is the intensity of the light emerging from the stack

substituting values

$I_f_3 = \frac{234.622}{2} * cos^2(61)$

$I_f_3 = 27.57 W/cm^2$

When the third polarizer is removed the second polarizer becomes the

the final polarizer and the intensity of light emerging from the stack would be

$I_f_2 = \frac{I_p}{2} cos ^2 \theta_1$

$I_f_2$ is the intensity of the light emerging from the stack

Substituting values

$I_f_2 = \frac{234.622}{2} cos^2 (21)$

$I_f_2 = 102.24 W/cm^2$

7 0

3 years ago