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3 years ago

9

In Dante Controller’s Clock Status tab, you see reports of more than one clock master. The Dante Controller Log also shows devic

es reporting sync loss occasionally. What steps would you take to find the root cause of the problem?

1 answer:

jeka943 years ago

6 0

Answer:

#See solution for details

Explanation:

-Check theswitch configurations for any evidence of the ACLs being configured. Confirm that all PTP traffic can traverse the network in both directions to all switches. Specifically, UDP ports 319 and 320 on 224.0.1.129

-If IGMP snooping is enabled, test the system with no audio and IGMP snooping turned off. If this resolves the problem, you carefully examine the IGMP snooping configuration in each switch and ensure that they follow the guidelines.

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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:

1

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

How to remember physics derivation easier? <br>

polet [3.4K]

Answer:

hmm let's see

Explanation:

ok, ill use the second equation of motion, u see its quite simple

recall that

average velocity= u+v/2

now think of that v like a box you can can call it velocity lol obviously its velocity

now remember the first equation of motion which is

v=u+at now this is inside the box, now you have to replace the v with it

then it becomes u+u+at/2

now u notice there are two initial velocity u know what to do with that

now it becomes

2u + at/2 now since the numerator applies to both then you can simplify it like this

2u/2+at/2

all the same the pont im trying to make is that you can use imaginative ways to mater derivation you can also reverse it if you don't unferstand you can drop your number

7 0

3 years ago

You need to calculate the volume of berm that has a starting cross-sectional area of 118 SF, and an ending cross-sectional area

LekaFEV [45]

6 cubic feet I’m pretty sure that’s the answer

8 0

3 years ago

A 92kg astronaut and a 1200kg satellite are at rest relative to the space shuttle. The astronaut pushes on the satellite, giving

Harman [31]

Answer:

13.7m

Explanation:

Since there's no external force acting on the astronaut or the satellite, the momentum must be conserved before and after the push. Since both are at rest before, momentum is 0.

After the push

$m_av_a + m_sv_s = 0$

Where $m_a = 92kg$ is the mass of the astronaut, $m_s = 1200kg$ is the mass of the satellite, $v_s = 0.14 m/s$ is the speed of the satellite. We can calculate the speed $v_a$ of the astronaut:

$v_a = \frac{-m_sv_s}{m_a} = \frac{-1200*0.14}{92} = -1.83 m/s$

So the astronaut has a opposite direction with the satellite motion, which is further away from the shuttle. Since it takes 7.5 s for the astronaut to make contact with the shuttle, the distance would be

d = vt = 1.83 * 7.5 = 13.7 m

4 0

3 years ago

5. YOU SHOULD USUALLY DRIVE SLOWER

saveliy_v [14]

Answer:

The answer is C.

Explanation:

If there are brake lights then you should drive slower.

3 0

3 years ago

Read 2 more answers