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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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Check the correctness of formula t=2π √m/k, dimensionally

pantera1 [17]

Hi there! Lets see!

m is mass, and its units are kg
k is the elastic constant measured in newtons per meter (N/m), or kilograms per second squared kg/s²

Therefore:

$\sqrt{\dfrac{m}{k}} =\sqrt{\dfrac{[kg]}{[\dfrac{kg}{s^2}]}} =\sqrt{\dfrac{[kg]}{[kg]}\cdot s^2} = \sqrt{[s]^2} = s$

The period is given in seconds so the formula is dimensionally correct.

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2 years ago

In January 2004, NASA landed exploration vehicles on Mars. Part of the descent consisted of the following stages:

fenix001 [56]

Acceleration is given by:

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time interval

Let's apply the formula to the different parts of the problem:

A) $-20.5 m/s^2$

Let's convert the quantities into SI units first:

$u = 19300 km/h \cdot \frac{1000 m/km}{3600 s/h} = 5361.1 m/s$

$v=1600 km/h \cdot \frac{1000 m/km}{3600 s/h} =444.4 m/s$

t = 4.0 min = 240 s

So the acceleration is

$a=\frac{444.4 m/s-5361.1 m/s}{240 s}=-20.5 m/s^2$

B) $-3.8 m/s^2$

As before, let's convert the quantities into SI units first:

$u = 444.4 m/s$

$v=321 km/h \cdot \frac{1000 m/km}{3600 s/h} =89.2 m/s$

t = 94 s

So the acceleration is

$a=\frac{89.2 m/s - 444.4 m/s}{94 s}=-3.8 m/s^2$

C) $-53.0 m/s^2$

For this part we have to use a different formula:

$v^2 - u^2 = 2ad$

where we have

v = 0 is the final velocity

u = 89.2 m/s is the initial velocity

a is the acceleration

d = 75 m is the distance covered

Solving for a, we find

$a=\frac{v^2-u^2}{2d}=\frac{0^2-(89.2 m/s)^2}{2(75 m)}=-53.0 m/s^2$

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3 years ago

How does the efficiency of the inclined plane vary with the inclination angle?

svlad2 [7]

Friction , as the angle gets huger and higher , this is less and less normal force into the inclined plane .

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3 years ago

The relationship among mass force and acceleration is explained by

Anettt [7]

Newton's second law of motion. F = m a .

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2 years ago

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The force of repulsion between two like–charged table tennis balls is 8.2 × 10-7 newtons. If the charge on the two objects is 6.

Allisa [31]

solution

In This question we have given ,

Charge on each ball= $q=6.7\times 10^{-9} C$

force of repulsion between balls is $F=8.2\times 10^{-7}N$

Let distance between ball be x

We know by Coulombs law,

$F=\frac{k\times q_{1}q_{2}}{x^2}$ ..............(1)

here, $k = 9.0\times 10^9 Nm^2C^{-2}$

Put values of k and charges in equation 1

$8.2\times 10^{-7}N=\frac{9.0\times 10^9 Nm^2C^{-2}\times 6.7^2\times 10^{-18} C^2}{x^2}$

$8.2\times 10^{-7}N\times x^2=9.0\times 10^9 Nm^2C^{-2}\times 6.7^2\times 10^{-18} C^2}$

$x^2=\frac{9.0\times 10^9 Nm^2C^{-2}\times 6.7^2\times 10^{-18} C^2}{8.2\times 10^{-7}N}$

$x=\frac{3\times 6.7}{\sqrt{820} }m$

$x=.701m$

therefore distance between two given charges is =.701m

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3 years ago

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