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lions [1.4K]
3 years ago
12

High voltage power is often carried in wire bundles made up of individual strands. In your initial post to the discussion, discu

ss the forces on the strands of wire due to the current flowing through them. What would happen if the force acted opposite of the known behavior
Physics
1 answer:
kicyunya [14]3 years ago
5 0

Answer:

<em>More current will be loss through the metal wire strands if the force on them was repulsive, and more stress will be induced on the wire strands due to internal and external flexing.</em>

<em />

Explanation:

A wire bundle is made up of wire strands bunched together to increase flexibility that is not always possible in a single solid metal wire conductor. In the strands of wire carrying a high voltage power, each strand carries a certain amount of current, and the current through the strands all travel in the same direction. <em>It is know that for two conductors or wire, separated by a certain distance, that carries current flowing through them in the same direction, an attractive force is produced on these wires, one on the other. This effect is due to the magnetic induction of a current carrying conductor.</em> The forces between these strands of the high voltage wire bundle, pulls the wire strands closer, creating more bond between these wire strands and reducing internal flex induced stresses.

If the case was the opposite, and the wires opposed themselves, the effect would be that a lot of cost will be expended in holding these wire strands together. Also, stress within the strands due to the repulsion, will couple with external stress from the flexing of the wire, resulting in the weakening of the material.

<em>The biggest problem will be that more current will be lost in the wire due to increased surface area caused by the repulsive forces opening spaces between the strand. This loss is a s a result of the 'skin effect' in wire transmission, in which current tends to flow close to the surface of the metal wire. The skin effect generates power loss as heat through the exposed surface area.</em>

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3 years ago
When is the electric flux on a section of a closed surface positive?.
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When the electric field pierces outward through the section.
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2 years ago
A sled is moving down a steep hill. The mass of the sled is 50 kg and the net force acting on it is 20 N. What must be done to f
amid [387]

You need to first measure the angle of descent, i.e. the angle the hill makes with the ground. Then identify the forces acting on the sled, split them up into horizontal and vertical components, or into components that are parallel and perpendicular to the hill, and use Newton's second law to determine the components of the sled's acceleration vector.

There are at least 2 forces acting on the sled:

• its weight, pointing downward with magnitude <em>W</em> = <em>m g</em>

• the normal force, pointing perpendicular to the hill and away from the ground with mag. <em>N</em>

The question doesn't specify, but there might also be friction to consider, indicated in the attachment by the vector <em>F</em> pointing parallel to the slope of the hill and opposing the direction of the sled's motion with mag. <em>F</em>.

Splitting up the forces into parallel/perpendicular components is less work. By Newton's second law, the net force (denoted with ∑ or "sigma" here) in a particular direction is equal to the mass of the sled times its acceleration in that direction:

∑ (//) = <em>W</em> (//) = <em>m</em> <em>a</em> (//)

∑ (⟂) = <em>W</em> (⟂) + <em>N</em> = <em>m </em><em>a</em> (⟂)

where, for instance, <em>W</em> (//) denotes the component of the sled's weight in the direction parallel to the hill, while <em>a</em> (⟂) denotes the component of the sled's acceleration perpendicular to the hill. If there is friction, you need to add -<em>F</em> to the first equation.

If the hill makes an angle of <em>θ</em> with flat ground, then <em>W</em> makes the same angle with the hill so that

<em>W</em> (//) = -<em>m g </em>sin(<em>θ</em>)

<em>W</em> (⟂) = -<em>m g</em> cos(<em>θ</em>)

So we have

<em>-m g </em>sin(<em>θ</em>) = <em>m</em> <em>a</em> (//)   →   <em>a</em> (//) = -<em>g </em>sin(<em>θ</em>)

<em>-m g</em> cos(<em>θ</em>) + <em>N</em> = <em>m </em><em>a</em> (⟂)   →   <em>a</em> (⟂) = 0

where the last equality follows from the fact that the normal force exactly opposes the perpendicular component of the weight. This is because the sled is moving along the slope of the hill, and not into the air or into the ground.

Then the acceleration vector is

<em>a</em> = <em>a</em> (//)

with magnitude

||<em>a</em>|| = <em>a</em> = <em>g </em>sin(<em>θ</em>).

6 0
3 years ago
How do you calculate mass using Newton’s 2nd Law?
deff fn [24]
The 2nd Law says F=ma, where F is the force in Newtons, m is mass and a is acceleration.  Earth's gravity is an acceleration, 9.8m/s^2.  So you can solve the equation for mass, m=F/a, or m=F/9.8 where you've measured the force (weight) in Newtons.
5 0
3 years ago
Read 2 more answers
Using a scale diagram, calculate the resultant force acting on a sailing boat when an easterly wind provides 2, point, 50, k, N,
EastWind [94]

Answer:

F = 3.6 kN, direction is 9.6º to the North - East

Explanation:

The force is a vector, so one method to find the solution is to work with the components of the vector as scalars and then construct the resulting vector.

Let's use trigonometry to find the component of the forces, let's use a reference frame where the x-axis coincides with the East and the y-axis coincides with the North.

Wind

X axis

          F₁ = 2.50 kN

Tide

         cos 30 = F₂ₓ / F₂

         sin 30 = F_{2y} / F₂

          F₂ₓ = F₂ cos 30

         F_{2y} = F₂ sin 30

         F₂ₓ = 1.20cos 30 = 1.039 kN

         F_{2y} = 1.20 sin 30 = 0.600 kN

the resultant force is

X axis

        Fₓ = F₁ₓ + F₂ₓ

        Fₓ = 2.50 +1.039

        Fₓ = 3,539 kN

        F_y = F_{2y}

        F_y = 0.600

to find the vector we use the Pythagorean theorem

         F = \sqrt{F_x^2 +F_y^2}

         F = \sqrt{ 3.539^2 + 0.600^2 }

         F = 3,589 kN

the address is

         tan θ = F_y / Fₓ

         θ = tan⁻¹ \frac{F_y}{F_x}

         θ = tan⁻¹  \frac{0.6}{3.539}0.6 / 3.539

         θ = 9.6º

the resultant force to two significant figures is

         F = 3.6 kN

the direction is 9.6º to the North - East

7 0
3 years ago
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