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Dmitriy789 [7]
3 years ago
13

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

rj-45 connector to one end of the cable and inserted the white with orange stripe wire into pin 1 of the connector. the orange wire has been inserted into pin 2 of the connector. which wire should be inserted into pin 1 of the rj-45 connector on the other end of the cable?
Physics
1 answer:
Elina [12.6K]3 years ago
6 0

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.

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If you walk 30 meters forwards, and then turn around and walk 25 meters backwards, what is the distance that you walked? What di
xeze [42]

Given :

Walk in forward direction is 30 m .

Walk in backward direction is 25 m .

To Find :

The distance and displacement .

Solution :

We know , distance is total distance covered and displacement is distance between final and initial position .

So , distance travelled is :

D = 30 + 25 m = 55 m .

Now , we first move 30 m in forward direction and then 25 m in backward direction .

So , displacement is :

D = 30 - 25 m = 5 m .

Therefore , distance and displacement covered is 55 m and 5 m respectively .

Hence , this is the required solution .

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Answer:

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Explanation:

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3 years ago
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A) Determine the x and y-components of the ball's velocity at t = 0.0s, 2.0, 3.0 secs.
malfutka [58]

The kinematic relationships we can find the position, acceleration and launch angle of the body on the planet Exidor.

a) the position are

      time (s)  x (m)   y(m)

        0            0          0

        2.0         3.6        1.2

        3.0         5.4        0.9

b) The aceleration is  g = 0.6 m / s²

c) The launch angle      θ = 33.7º

given parameters

  • the initial velocity of the body vₓ = 1.8m / s and v_y = 1.2 m / s
  • the movement times t = 1.0s, 2.0s and 3.0 s

to find

    a) position

    b) acceleration

    c) launch angle

Projectile launch is an application of kinematics to the movement of the body in two dimensions where there is no acceleration on the x axis and the y axis has the planet's gravity acceleration

b) To calculate the acceleration of the plant acting on the y-axis, we use that the vertical velocity of the body at the highest point is zero.

         v_y = v_{oy} - g t

where v and v({oy}  are the velocities of the body, g the acceleration of the planet's gravity and t the time

          0 = v_{oy} - gt

           g = v_{oy} / t

from the graph we observe that the highest point occurs for t = 2.0 s

           g = 1.2 / 2.0

           g = 0.6 m / s²

 

a) The position is requested for several times

X axis

in this axis there is no acceleration so we can use the uniform motion relationships

          vₓ = x / t

          x = vₓ t

where x is the position, vx is the velocity and t is the time

we calculate for the time

t = 0.0 s

          x₀ = 0

           

t = 2.0 s

          x₂ = 1.8 2

          x₂ = 3.6 m

t = 3.0 s

          x₃ = 1.8 3

          x₃ = 5.4 m

Y axis

In this axis there is the acceleration of the planet, let us use for the position the relation

          y = v_{oy} t - ½ g t²

t = 0.0 s

          y₀ = 0

          y₀ = 0 m

t = 2.0 s

         y₂ = 1.2 2 - ½ 0.6 2²

         y₂ = 1.2 m

t = 3.0 s

        y₃ = 1.2  3 - ½  0.6  3²

        y₃ = 0.9 m

c) the launch angle use the trigonometry relation

        tan θ = \frac{v_y}{v_x}

        θ = tan⁻¹ \frac{v_y}{v_x}

        θ = tan⁻¹ \frac{1.2}{1.8}

        θ = 33.7º

measured counterclockwise from the positive side of the x-axis

With the kinematic relationships we can find the position, acceleration and launch angle of the body on the planet Exidor.

a) the position are

      time (s)  x (m)   y(m)

        0            0          0

        2.0         3.6        1.2

        3.0         5.4        0.9

b) The aceleration is  g = 0.6 m / s²

c) The launch angle      θ = 33.7ºto)

learn more about projectile launch here:

brainly.com/question/10903823

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How much work is done lifting a 12-m chain that is initially coiled on the ground and has a density 2 kg/m so that its top end i
Misha Larkins [42]

Answer:

2587.2 J.

Explanation:

From potential energy,

The work done to lift the chain = potential energy of the chain.

W = mgh............... Equation 1

Where W = work done to lift the chain, m = mass of the chain, g = acceleration due to gravity of the chain, h = height of the chain.

But,

m = m'd............... Equation 2

Where m' = density of the chain, d = length of the chain.

Substitute equation 2 into equation 1

W = m'dgh................ Equation 3

Given: m' = 2 kg/m, d = 12 m, h = 11 m, g = 9.8 m/s²

Substitute into equation 3

W = 2(12)(11)(9.8)

W = 2587.2 J.

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3 years ago
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