Ask question

Ask question

All categories

Elena-2011 [213]

3 years ago

7

or Ethernet, if an adapter determines that a frame it has just received is addressed to a different adapter it discards the fram

e without sending an error message to the network layer it sends a NACK (not acknowledged frame) to the sending host it delivers the frame to the network layer, and lets the network layer decide what to do it discards the frame and sends an error message to the network layer

1 answer:

Artist 52 [7]3 years ago

5 0

For Ethernet, if an adapter determines that a frame it has just received is addressed to a different adapter

a. it discards the frame without sending an error message to the network layer

b. it sends a NACK (not acknowledged frame) to the sending host

c. it delivers the frame to the network layer, and lets the network layer decide what to do

d. it discards the frame and sends an error message to the network layer

Answer:

Option A

Explanation:

The nodal address has to match the signal message address for it to function well but if the it doesn't match the nodal receiver address, it disregards it.

You might be interested in

A train at a constant 60.0 km/h moves east for 40.0 min, then in a direction 50.0 degrees east of due north for 20.0 min, and th

son4ous [18]

Answer:

Part a)

$v = 7.57 km/h$

Part b)

$\theta = 67.5 degree$ North of East

Explanation:

Speed of train towards East = 60 km/h

displacement towards East is given as

$d_1 = 40 km$

now it turns towards 50 degree East of North

so its distance is given as

$d_2 = 20 km(sin50 \hat i + cos50\hat j)$

$d_2 = 15.3 \hat i + 12.8 \hat j$

then finally it moves towards west for 50 min

$d_3 = -50 \hat i$

Now the total displacement of the train is given as

$d = d_1 + d_2 + d_3$

$d = (40 + 15.3 - 50)\hat i + 12.8 \hat j$

$d = 5.3\hat i + 12.8 \hat j$

now total time duration of the motion is given as

$T = 40 min + 20 min + 50 min$

$T = 1.83 h$

now average velocity is given as

$v_{avg} = \frac{5.3\hat i + 12.8\hat j}{1.83}$

$v_{avg} = 2.89\hat i + 6.99\hat j$

Part a)

magnitude of the average velocity is given as

$v = \sqrt{v_x^2 + v_y^2}$

$v = \sqrt{2.89^2 + 6.99^2}$

$v = 7.57 km/h$

Part b)

Direction of the velocity is given as

$tan\theta = \frac{v_y}{v_x}$

$tan\theta = \frac{6.99}{2.89}$

$\theta = 67.5 degree$ North of East

6 0

4 years ago

As an intern with an engineering firm, you are asked to measure the moment of inertia of a large wheel, for rotation about an ax

AysviL [449]

Answer:

$I=2.766\ kg.m^2$

Explanation:

We have:

diameter of the wheel, $d=0.88\ m$

weight of the wheel, $w_w=280\ N$

mass of hanging object to the wheel, $m_o=6.32\ kg$

speed of the hanging mass after the descend, $v_o=4\ m.s^{-1}$

height of descend, $h=2.5\ m$

(a)

moment of inertia of wheel about its central axis:

$I=\frac{1}{2} m.r^2$

$I=\frac{1}{2} \frac{w_w}{g}.r^2$

$I=\frac{1}{2} \times \frac{280}{9.8}\times 0.44^2$

$I=2.766\ kg.m^2$

3 0

3 years ago

Does the tide that the moon raises on the earth different?

alukav5142 [94]

Answer:

No the gravity of the moon pulls the water making high tide

Explanation:

7 0

3 years ago

A race car accelerates on a straight road from rest to speed of 180 km/hr in 25s . Assuming uniform acceleration of the car thro

adelina 88 [10]

X-x0=0.5(v0+v)t ergo x=0.5(50)25=625m

3 0

4 years ago

Please help I'm confused on where to start and with the graph.

tankabanditka [31]

I don't think anyone knows the answer

3 0

3 years ago