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2 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]2 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.

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A negative charge of 20 x 10-6C and another charge of 15 x 10-6C are separated by as distance of 0.7 m.

denpristay [2]

Answer:

Approximately $5.5\; \rm N$ , assuming that the volume of these two charged objects is negligible.

Explanation:

Assume that the dimensions of these two charged objects is much smaller than the distance between them. Hence, Coulomb's Law would give a good estimate of the electrostatic force between these two objects regardless of their exact shapes.

Let $q_1$ and $q_2$ denote the magnitude of two point charges (where the volume of both charged object is negligible.) In this question, $q_1 = 20 \times 10^{-6}\; \rm C$ and $q_2 = 15 \times 10^{-6}\; \rm C$ .

Let $r$ denote the distance between these two point charges. In this question, $r = 0.7\; \rm m$ .

Let $k$ denote the Coulomb constant. In standard units, $k \approx 8.98755\times 10^{9}\; \rm kg \cdot m^{3}\cdot s^{-2}\cdot C^{-2}$ .

By Coulomb's Law, the magnitude of electrostatic force (electric force) between these two point charges would be:

$\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^{2}}\end{aligned}$ .

Substitute in the values and evaluate:

$\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^{2}}\\ &\approx 8.98755 \times 10^{9}\; \rm kg \cdot m^{3}\cdot s^{-2}\cdot C^{-2} \\ &\quad \times 20\times 10^{-6}\; \rm C\\ &\quad \times 15\times 10^{-6}\; \rm C \\ &\quad \times \frac{1}{{(0.7\; \rm m)}^{2}}\\ &\approx 5.5\; \rm N \end{aligned}$ .

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