PLEASE HELP ASAP BEST ANSWER WILL BE MARKED BRAINLIEST

You are making a 568b utp crossover cable that will be used to cascade two switches on an ethernet network. you have attached an

Elina [12.6K]

For a 568B crossover cable that already has wht-org and org on pins 1 and 2 of the connector at one end, the connector at the OTHER end should have wht-grn and grn on pins 1 and 2 respectively.

The wht-org and org at that end should drop to pins 3 and 6 respectively.

You're welcome, and good luck.

6 0

3 years ago

A cube of side 6.50 cm is placed in a uniform field E = 7.50 × 10^3 N/C with edges parallel to the field lines (field enters the

Svetllana [295]

Answer:

a) $\Phi_{net} = 0\,\frac{N\cdot m^{2}}{C}$ , b) $\Phi_{right} = -31.688\,\frac{N\cdot m^{2}}{C}$ , c) $\Phi_{left} = 31.688\,\frac{N\cdot m^{2}}{C}$

Explanation:

a) The net flux through the cube is:

$\Phi_{net}=-(7.50\cdot 10^{3}\,\frac{N}{C} )\cdot (0.065\,m)^{2}+(7.50\cdot 10^{3}\,\frac{N}{C} )\cdot (0.065\,m)^{2}$

$\Phi_{net} = 0\,\frac{N\cdot m^{2}}{C}$

b) The flux through the right face is:

$\Phi_{right}=-(7.50\cdot 10^{3}\,\frac{N}{C} )\cdot (0.065\,m)^{2}$

$\Phi_{right} = -31.688\,\frac{N\cdot m^{2}}{C}$

c) The flux through the left face is:

$\Phi_{left}=(7.50\cdot 10^{3}\,\frac{N}{C} )\cdot (0.065\,m)^{2}$

$\Phi_{left} = 31.688\,\frac{N\cdot m^{2}}{C}$

5 0

3 years ago

Read 2 more answers

Leslie incorrectly balances an equation as 2C4H10 + 12O2 → 8CO2 + 10H2O.

tatiyna

Answer:

13 behind o2

Explanation:

answer is in photo above

4 0

3 years ago

Two stationary point charges of 3.00 nC and 2.00 nC are separated by a distance of 50.0 cm. An electron is released from rest at

andrew-mc [135]

Answer:

1. the electric potential energy of the electron when it is at the midpoint is - 2.9 x $10^{-17}$ J

2. the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge is - 5.04 x $10^{-17}$ J

Explanation:

given information:

$q_{1}$ = 3 nC = 3 x $10^{-9}$ C

$q_{2}$ = 2 nC = 2 x $10^{-9}$ C

r = 50 cm = 0.5 m

the electric potential energy of the electron when it is at the midpoint

potential energy of the charge, F

F = k $\frac{q_{e}q}{r}$

where

k = constant (8.99 x $10^{9} Nm^{2} /C^{2}$ )

electron charge, $q_{e}$ = - 1.6 x $10^{-19}$ C

since it is measured at the midpoint,

r = $\frac{0.5}{2}$

= 0.25 m

thus,

F = $F_{1}+ F_{2}$

= k $\frac{q_{e} q_{1} }{r}$ + k $\frac{q_{e} q_{2} }{r}$

= $\frac{kq_{e} }{r}$ ( $q_{1} +q_{2}$ )

= (8.99 x $10^{9}$ )( - 1.6 x $10^{-19}$ )(3 x $10^{-9}$ +2 x $10^{-9}$ )/0.25

= - 2.9 x $10^{-17}$ J

the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge

$r_{1}$ = 10 cm = 0.1 m

$r_{2}$ = 0.5 - 0.1 = 0.4 m

F = k $\frac{q_{e} q_{1} }{r}$ + k $\frac{q_{e} q_{2} }{r}$

= $kq_{e}$ ( $\frac{q_{1} }{r_{1} }$ + $\frac{q_{2} }{r_{2} }$ )

= (8.99 x $10^{9}$ )( - 1.6 x $10^{-19}$ )(3 x $10^{-9}$ /0.1+2 x $10^{-9}$ /0.4)

= - 5.04 x $10^{-17}$ J

3 0

3 years ago

Select the correct arrow.<br> Which arrow correctly shows the flow of heat?<br> Reset<br> Next

julsineya [31]

Answer:

where is the image?

Explanation:

6 0

2 years ago

Read 2 more answers