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Mashcka [7]
3 years ago
15

A new cat 6 cable run and key stone jack were installed for a workstation that was moved from the sales department to the shippi

ng department. since the move, the user has been complaining of slow transfer speeds on both internet and local network communications. the technician ran a test and is viewing the following output on the tester. which of the following is the most likely reason for the sluggish network speeds?a. the cable run exceeds the specifications for ethernet over twisted pairb. the nic in the workstation is statically set to 10 mbps/half-duplexc. the network switch port is statically set to 10 mbps/half-duplexd. the isp is experiencing issues with its service to this customer
Physics
1 answer:
Andrei [34K]3 years ago
8 0

Answer:

The cable run exceeds the specifications for Ethernet over twisted pair

Explanation:

The ethernet  network's router also serves as a bridge to the Internet. The router connects to the modem, which carries the Internet signal, sending and receiving data packet requests and routing them to the proper computers on the network.

Ethernet is a way of connecting computers together in a local area network or LAN. It has been the most widely used method of linking computers together in LAN s since the 1990 s.

The basic idea of its design is that multiple computers have access to it and can send data at any time.

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Read 2 more answers
A coil has 400 turns and self-inductance 7.50 mH. The current in the coil varies with time according to i = 1680 mA2 cos [πt/(0.
zhannawk [14.2K]

Answer:

(a) 1.58 V

(b) 0.0126 Wb

(c) 0.0493 V

Solution:

As per the question:

No. of turns in the coil, N = 400 turns

Self Inductance of the coil, L = 7.50 mH = 7.50\times 10^{- 3}\ H

Current in the coil, i = 1680cos[\frac{\pi t}{0.0250}] A

where

i_{max} = 1680\ mA = 1.680\ A

Now,

(a)  To calculate the maximum emf:

We know that maximum emf induced in the coil is given by:

e = \frac{Ldi}{dt}

e = L\frac{d}{dt}(1680)cos[\frac{\pi t}{0.0250}]

e = - 7.50\times 10^{- 3}\times \frac{\pi}{0.0250}\times \frac{d}{dt}(1680)sin[\frac{\pi t}{0.0250}]

For maximum emf, sin\theta should be maximum, i.e., 1

Now, the magnitude of the maximum emf is given by:

|e| = 7.50\times 10^{- 3}\times 1680\times 10^{- 3}\times \frac{\pi}{0.0250} = 1.58\ V

(b) To calculate the maximum average flux,we know that:

\phi_{m, avg} = L\times i_{max} = 7.50\times 10^{- 3}\times 1.680 = 0.0126\ Wb

(c) To calculate the magnitude of the induced emf at t = 0.0180 s:

e = e_{o}sin{\pi t}{0.0250}

e = 7.50\times 10^{- 3}\times sin{\pi \times 0.0180}{0.0250} = 2.96\times 10^{- 4} =0.0493\ V

7 0
3 years ago
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