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2 years ago

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How are 8-position, 8-contact (8P8C) modular connectors pinned to unshielded twisted-pair (UTP) cable, which is used for connect

ing computers and broadband modems to a local area network (LAN)?

Computers and Technology

1 answer:

AleksandrR [38]2 years ago

8 0

An unshielded cable are twisted pair is used to connect modem in telecommunication industry to connect computer through telephone cable, Ethernet cables it is called as UTP cables.

<u>Explanation:</u>

These UTP cables are twisted cables where conducted are used in form of single circuit.

These types of UTP cables used to connect modem to establish connection to other networks for internet or connecting to other side computer or desktop or laptops.

UTP cables uses rj-45 or rj-11 or rs232 or rs 499. Normally rs-45 with cat 5e cable is used to connect LAN. Rj-11 is used to connect to modem for dial purpose.

If we use ip networks RJ-45 will do both.

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2 years ago

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the volume of two similar solids are 1080cm and 1715cm .if the curved surface area of the smaller cone is 840cm .fond the curved

gavmur [86]

Answer:

$A_{big} = 1143.33cm^2$

Explanation:

The given parameters are:

$V_{small} = 1080$

$V_{big} = 1715$

$C_{small} = 840$

Required

Determine the curved surface area of the big cone

The volume of a cone is:

$V = \frac{1}{3}\pi r^2h$

For the big cone:

$V_{big} = \frac{1}{3}\pi R^2H$

Where

R = radius of the big cone and H = height of the big cone

For the small cone:

$V_{small} = \frac{1}{3}\pi r^2h$

Where

r = radius of the small cone and H = height of the small cone

Because both cones are similar, then:

$\frac{H}{h} = \frac{R}{r}$

and

$\frac{V_{big}}{V_{small}} = \frac{\frac{1}{3}\pi R^2H}{\frac{1}{3}\pi r^2h}$

$\frac{V_{big}}{V_{small}} = \frac{R^2H}{r^2h}$

Substitute values for Vbig and Vsmall

$\frac{1715}{1080} = \frac{R^2H}{r^2h}$

Recall that: $\frac{H}{h} = \frac{R}{r}$

So, we have:

$\frac{1715}{1080} = \frac{R^2*R}{r^2*r}$

$\frac{1715}{1080} = \frac{R^3}{r^3}$

Take cube roots of both sides

$\sqrt[3]{\frac{1715}{1080}} = \frac{R}{r}$

Factorize

$\sqrt[3]{\frac{343*5}{216*5}} = \frac{R}{r}$

$\sqrt[3]{\frac{343}{216}} = \frac{R}{r}$

$\frac{7}{6} = \frac{R}{r}$

The curved surface area is calculated as:

$Area = \pi rl$

Where

$l = slant\ height$

For the big cone:

$A_{big} = \pi RL$

For the small cone

$A_{small} = \pi rl$

Because both cones are similar, then:

$\frac{L}{l} = \frac{R}{r}$

and

$\frac{A_{big}}{A_{small}} = \frac{\pi RL}{\pi rl}$

$\frac{A_{big}}{A_{small}} = \frac{RL}{rl}$

This gives:

$\frac{A_{big}}{A_{small}} = \frac{R}{r} * \frac{L}{l}$

Recall that:

$\frac{L}{l} = \frac{R}{r}$

So, we have:

$\frac{A_{big}}{A_{small}} = \frac{R}{r} * \frac{R}{r}$

$\frac{A_{big}}{A_{small}} = (\frac{R}{r})^2$

Make $A_{big}$ the subject

$A_{big} = (\frac{R}{r})^2 * A_{small}$

Substitute values for $\frac{R}{r}$ and $A_{small}$

$A_{big} = (\frac{7}{6})^2 * 840$

$A_{big} = \frac{49}{36} * 840$

$A_{big} = \frac{49* 840}{36}$

$A_{big} = 1143.33cm^2$

<em>Hence, the curved surface area of the big cone is 1143.33cm^2</em>

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Answer:

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Explanation:

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How many bits can a memory chip with the below configuration support? For full credit, show how you got the answer.

tatiyna

Answer:

Total Memory= 4 KB = 4096 bytes = 32768 bits

Explanation:

<em><u>1. Data lines are 8 From D0 to D7</u></em>

so

Total memory at single address locations is 8 bits.

<em><u>2. Address lines are 12 (A0 to A11)</u></em>

There are 12 address lines but 3 out 12 are for selction of chip or memory bank.

so only 9 pins are there to address the locations on one chip.

Total No. of address locations on single chip = 2^9 = 512 locations

as 1 location is 1 byte so total memory of single chip is 512 bytes.

<u><em>3. Total Memory Bank </em></u>

There are total 3 selection pins for memory bank.

so

Total chips = 2^3 = 8.

<em><u>4. Total Memory </u></em>

Total size of 1 chip = 512 bytes

Total size of 8 chip = 8x512 bytes = 4096 bytes = 4096/1024 kb = 4 kb

<em>So total memory of system is 4 Kb = 4096 bytes = 32768 bits</em>

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2 years ago