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aleksklad [387]
3 years ago
8

Several 4-conductor cables are installed in a ladder cable tray without a cover and without exceeding the cable tray fill limit.

Each 4-conductor cable comprises four 1/0 AWG THHN copper conductors, each of which is considered current-carrying. What is the allowable ampacity of each 1/0 AWG THHN copper conductor?
Physics
1 answer:
8090 [49]3 years ago
6 0

Answer: The allowable ampacity is 136 amperes

Explanation:According to the National Electrical Code the ampacity for insulated conductors of type 1/0 AWG THHN is 170 amperes(from tables) for cables no exceeding 3

But we are presented with 4 cables in a tray,

Going by the National Electrical Code where conductors exceeds 3, the allowable ampacity shall be reduced to 80% the of the ampacity gotten from table.

Hence the allowable ampacity for 4 copper cables of type 1/0 AWG THHN is (80/100) *170=136 Amperes

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A ball is kicked off of a roof at 23 m/s [R 25° U]. What is the height of
Ymorist [56]

Answer:

Explanation:

Considering the fact that we ave been given an angle of inclination here, we best use it! That means that the velocity of 23 m/s is actually NOT the velocity we need; I tell my students that it is a "blanket" velocity but is not accurate in either the x or the y dimension of parabolic motion. In order to find the actual velocity in the dimension in which we are working, which is the y-dimension, we use the formula:

v_{0y}=v_0sin\theta and filling in:

v_{0y}=23sin(25) which gives us an upwards velocity of 9.7 m/s. So here's what we have to work with in its entirety:

v_{0y}=9.7m/s

a = -9.8 m/s/s

t = 2.8 seconds

Δx = ?? m

The one-dimensional motion equation that utilizes all of these variables is

Δx = v_0t+\frac{1}{2}at^2 and filling in:

Δx = 9.7(2.8)+\frac{1}{2}(-9.8)(2.8)^2 I am going to do the math according to the correct rules of significant digits, so to the left of the + sign and to 2 sig fig, we have

Δx = 27 + \frac{1}{2}(-9.8)(2.8)^2 and then to the right of the + sign and to 2 significant digits we have

Δx = 27 - 38 so

Δx = -11 meters. Now, we all know that distance is not a negative value, but what this negative number tells us is that the ball fell 11 meters BELOW the point from which it was kicked, which is the same thing as being kicked from a building that is 11 meters high.

6 0
3 years ago
Why are the largest craters we find on the Moon and Mercury so much larger than the largest craters we find on the Earth
Veronika [31]

Answer:

Because Moon and Mars has no atmosphere.

Explanation:

Moon and Mars has no atmosphere, so there is no friction on the falling object due to the atmosphere. The speed of the falling object is more at Moon and Mars.

When a small object impact on the surface of moon or Mars with high speed, the size of crater is large than the earth as out earth has atmosphere.

3 0
3 years ago
Tarzan, who weighs 825 N, swings from a cliff at the end of a 19.7 m vine that hangs from a high tree limb and initially makes a
kodGreya [7K]

Answer:

a) T = (281.47 i ^ + 714.56 j ^) N , b) F_net = (281.47 i ^ - 110.44 j ^) N ,

c)  F = 281.70 N, d)    θ = 338.58º , e)  a = 3,588 m / s² , f)  θ = 201.45º

Explanation:

For this exercise we will use Newton's second law on each axis

X axis

         -Tₓ = m aₓ

Y Axisy

          T_{y} –W = m a_{y}

Let's use trigonometry to find the components of force

          sin 21.5 = Tₓ / T

          cos 21.5 = T_{y} / T

          Tₓ = T sin 21.5

          T_{y} = T cos 21.5

          Tₓ = 768 sin 21.5 = 281.47 N

          T_{y} = 768 cos 21.5 = 714.56 N

a) the force of the rope on Tarzan is

          T = (281.47 i ^ + 714.56 j ^) N

b) The net force is the subtraction of the tension minus the weight of Tarzan

Y  Axis   F_net = 714.56 - 825 = -110.44 N

              F_net = (281.47 i ^ - 110.44 j ^) N

c) Let's use Pythagoras' theorem

      F = √ (Fₓ² + T_{y}²)

      F = √ (281.47² + 110.44²)

      F = 281.70 N

d) Let's use trigonometry

     tan θ = F_{y} / Fₓ

      θ = tan⁻¹ F_{y} / Fₓ

      θ = tan⁻¹ (-110.44 / 281.47)

       θ = -21.42º

This angle is average clockwise, for counterclockwise measurement

       θ = 360 - 21.42

       θ = 338.58º

Acceleration

X axis

             Tₓ = m aₓ

             aₓ = Tₓ / m

The mass of Tarzan is

             m = W / g

             m = 825 / 9.8 = 84.18 kg

             

             aₓ = 281.47 / 84.18

             aₓ = -3.34 m / s2

Y Axis

            T_{y}-W = m a_{y}

            a_{y} = (T_{y} -W) / m

            a_{y} = (714.56-825) / 84.18

            a_{y} = - 1,312 m / s²

Acceleration Module

             a = √ aₓ² + a_{y}²

             a = √ (3.34² +1.312²)

             a = 3,588 m / s²

The angle

          θ = tan⁻¹ a_{y} / aₓ

          θ = tan⁻¹ (-1312 / -3.34)

          θ = 21.45º

Notice that the two components of the acceleration are negative, so the angle is in the third quadrant, to measure from the x-axis

          θ = 180 + 21.45

          θ = 201.45º

3 0
3 years ago
Suppose a 48-N sled is resting on packed snow. The coefficient of kinetic friction is 0.10. If a person weighing 660 N sits on t
Annette [7]

Assume the snow is uniform, and horizontal.

Given:

coefficient of kinetic friction = 0.10 = muK

weight of sled = 48 N

weight of rider = 660 N

normal force on of sled with rider = 48+660 N = 708 N = N

Force required to maintain a uniform speed

= coefficient of kinetic friction * normal force

= muK * N

= 0.10 * 708 N

=70.8 N


Note: it takes more than 70.8 N to start the sled in motion, because static friction is in general greater than kinetic friction.


8 0
3 years ago
What is a secondary color of light?
jasenka [17]
It is Yellow, Cyan, Magenta . The answer is D.
5 0
3 years ago
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