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mario62 [17]
3 years ago
15

An engineer is considering time of convergence in a new Layer 3 environment design. Which two attributes must be considered? (Ch

oose two)
A.Addition of a valid forwarding path
B.Loss of a valid forwarding path
C.SPT timers update
D.OSPF database updates
E.Forwarding table updates
Engineering
1 answer:
mestny [16]3 years ago
7 0

Answer:

Options B and E

Explanation:

To give sustainable environmental design when considering time of convergence in layer 3, an engineer must consider the loss of a valid forwarding path and the table updates since these will determine whether the design becomes fit or not.

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If you are in a tornado situation, which of the following actions would put you in danger?
anzhelika [568]
Answer: Going outside to film it, not taking cover, trying to run away from the tornado, driving away, etc.

Please mark as brainliest
4 0
3 years ago
A long rod of 60-mm diameter and thermophysical properties rho=8000 kg/m^3, c=500J/kgK, and k=50 W/mK is initally at a uniform t
Monica [59]

Answer:

Tc = 424.85 K

Explanation:

Given that,

D = 60 mm = 0.06 m

\rho = 8000 kg/m^3

k = 50 w/m . kc = 500 j/kg.k

h_{\infty} = 1000 w/m^2t_{\infity} = 750 kt_w = 500 K

surface area = As = \pi dL \\\frac{As}{L} = \pi D = \pi \timeS 0.06

HEAT FLOW Q  is

Q = h_{\infty} As (T_[\infty} - Tw)  = 1000 \pi\times 0.06 (750-500)

 = 47123.88 w per unit length of rod

volumetric heat rate

q = \frac{Q}{LAs}

= \frac{47123.88}{\frac{\pi}{4} D^2 \times 1}

q = 1.66\times 10^{7} w/m^3

Tc = \frac{- qR^2}{4K} + Tw

= \frac{ - 1.67\times 10^7 \times (\frac{0.06}{2})^2}{4\times 50} +  500

  = 424.85 K

3 0
2 years ago
What is the relative % change in P if we double the absolute temperature of an ideal gas keeping mass and volume constant?
Contact [7]

Answer:  100% (double)

Explanation:

The question tells us two important things:

  1. Mass remains constant
  2. Volume remains constant

(We can think in a gas enclosed in a  closed bottle, which is heated, for instance)

In this case we know that, as always the gas can be considered as ideal, we can apply the general equation for ideal gases, as follows:

  1. State 1 (P1, V1, n1, T1)  ⇒ P1*V1 = n1*R*T1
  2. State 2 (P2, V2, n2, T2) ⇒ P2*V2 = n2*R*T2

But we know that V1=V2 and that n1=n2, som dividing both sides, we get:

P1/P2 = T1/T2, i.e, if T2=2 T1, in order to keep both sides equal, we need that P2= 2 P1.

This result is just reasonable, because as temperature measures the kinetic energy of the gas molecules, if temperature increases, the kinetic energy will also increase, and consequently, the frequency of collisions of the molecules (which is the pressure) will also increase in the same proportion.

6 0
3 years ago
Technician A says when the brakes are applied in a vacuum booster, the vacuum control valve is closed. Technician B says the vac
Wewaii [24]

Answer:

Technician A only

Explanation:

The application of the breaks by stepping on the break pedal moves the pedal pushrod and plunger forward within the diaphragm plate, bringing about the contact between the vacuum port seal and the vacuum valve that closes the vacuum port and the passage that connects the left and right chambers such that the pressure in one chamber and te vacuum in the other chamber are held steady.

4 0
2 years ago
Please answer question #2
natka813 [3]
Your answer is correct the last procedure of a vehicle starting is selecting the appropriate gear for the right situation. (D)
5 0
3 years ago
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