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3 years ago

What is the purpose of the ism/wlan indicator light on the backplane of the router? what does it refer to?

2 answers:

4 0

SM/LAN which would demonstrate that there is a Cisco Internal Services Module or a remote LAN card. LAN interfaces the PC equipment in a confined territory, for example, an office or home. Normally, LANs utilize wired associations with connect the PCs to each other and to an assortment of fringe gadgets, for example, printers. LAN clients can speak with each other by talk or email.

sweet-ann [11.9K]3 years ago

4 0

The purpose of the ISM/WLAN indicator light on the back plane of the router is The Cisco 1941 router can support a Cisco Internal Services Module that can enhance the intelligence and abilities of the router to perform activities like intrusion prevention scanning. The Cisco 1941 router can be also equipped with a WLAN Card or Wireless LAN card for supporting wireless local area networks.

You might be interested in

If a true heading of 135° results in a ground track of 130° and a true airspeed of 135 knots results in a groundspeed of 140 kno

vladimir2022 [97]

Answer:

245.1° and 13 knots

Explanation:

The given parameters are;

The true heading = 135°

The resultant ground track = 130°

The true airspeed = 135 knots

The ground speed = 140 knots

Given that the true airspeed the ground speed and the wind direction and magnitude form a triangle, we have;

From cosine rule, we have;

a² = b² + c² - 2×b×c×cos(A)

Where

a = The magnitude of the wind speed in knot

b = The true airspeed = 135 knots

c = The ground speed = 140 knots

A = The angle in between the true heading and the resultant ground track heading = 5°

Which gives;

a² = 135² + 140² - 2×135×140×cos(5 degrees) = 168.84 knots²

a = √168.84 = 12.9934 ≈ 13 knots

We have;

135 × sin(135 degrees) - 140× sin(130 degrees) = -11.7868

135 × cos(135 degrees) - 140× cos(130 degrees) = -5.469

Tan(θ) = -11.8/-5.5 = 2.155

θ = tan⁻¹(2.155) = 65.108°

Given that the wind is moving in opposite direction (slowing down the airplane, we add 180°, to get

Therefore, the angle direction = 180 + 65.108 = 245.1

Therefore, we have;

245.1° and 13 knots

3 0

3 years ago

a crane does 9,500 J of work to lift a crate straight up using a force of 125 N. how high does the crane lift the crate?

Lina20 [59]

Answer:

The crane lifts the crate up to 76 m high

Explanation:

Work Done by a Force

The work done by a force of magnitude F that displaces an object by a distance y is given by

$W=F.y$

We know a crane does a work of 9,500 Joule to lift a crate and is using a force of 125 Nw.

The above equation can be solved to know the value of y in terms of the work and the force:

$\displaystyle y=\frac{W}{F}$

Plugging in the given values

$\displaystyle y=\frac{9,500}{125}$

$y=76\ m$

The crane lifts the crate up to 76 m high

7 0

2 years ago

Is direction the length of the route between two points ?

zepelin [54]

Answer:

No distance is the length between two routes

Explanation:

Distance is the length of the route between two points. ... Direction is just as important as distance in describing motion. A vector is a quantity that has both size and direction. It can be used to represent the distance and direction of motion.

6 0

2 years ago

A 800 kg safe is 2.1 m above a heavy-duty spring when the rope holding the safe breaks. The safe hits the spring and compresses

stellarik [79]

Answer:

k = 17043.5 N/m = 17.04 KN/m

Explanation:

First we need to find the force applied by safe pn the spring:

F = Weight of Safe

F = mg

where,

F = Force Applied by the safe on the spring = ?

m = mass of safe = 800 kg

g = 9.8 m/s²

Therefore,

F = (800 kg)(9.8 m/s²)

F = 7840 N

Now, using Hooke's Law:

F = kΔx

where,

K = Spring Constant = ?

Δx = compression = 46 cm = 0.46 m

Therefore,

7840 N = k (0.46 m)

k = 7840 N/0.46 m

k = 17043.5 N/m = 17.04 KN/m

5 0

3 years ago

For a home sound system, two small speakers are located so that one is 52 cm closer to the listener than the other. What is the

galina1969 [7]

Answer:

The first frequency of audible sound in the speed sound is

f = 662 Hz

Explanation:

vs = 344 m/s

x = 52 cm * 1 / 100m = 0.52m

The wave length is the distance between the peak and peak so

d = 2x

d = 2*0.52 m

d = 1.04 m

So the frequency in the speed velocity is

f = 1 / T

f = vs / x = 344 m/s / 0.52m

f ≅ 662 Hz

7 0

3 years ago