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Pepsi [2]
1 year ago
12

niobium metal becomes a superconductor when cooled below 9 k. its superconductivity is destroyed when the surface magnetic field

exceeds 0.100 t. in the absence of any external magnetic field, determine the maximum current a 2.60-mm-diameter niobium wire can carry and remain superconducting
Physics
1 answer:
Furkat [3]1 year ago
5 0

Niobium wire with a 2.60 mm diameter has a maximum current capacity of 500 A while still remaining superconducting.

<h3>Describe the present.</h3>

Current is the rate at which charge passes from one point on a circuit to another. In a circuit, a significant current flows when several coulombs or charge pass over the cross section of a wire. When the charge carriers are firmly packed inside the wire, high currents can be generated at low speeds.

<h3>What do current and electron actually mean?</h3>

Electron movement is referred to as electron current. The positive terminal receives electrons that are released by the negative terminal. Traditional current, usually referred to as just current, exhibits behavior consistent with positive charge carriers being the source of current flow. Regular current is received at the positive end and then flows to a negative terminal.

To know more about current visit:
brainly.com/question/15141911

#SPJ4

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Which layer of earth is a solid rock?
Sholpan [36]

Answer:

Crust

Explanation:

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In this wave, there is a vertical line which represents what part of the wave?
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Answer:Height

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Read 2 more answers
A horizontal force FFaa���⃗ of magnitude 20.0 N is applied to a 3.00 kg psychology book as the book slides a distance d = 0.500
alina1380 [7]

Answer:

(a) W = 8.66 J

(b) Velocity = 2.40 m/s

Explanation:

(a) Work done is given as the product of force and displacement. That is:

W = F * d * cos(A)

Where F = force applied

d = distance moved

A = angle of ramp

Therefore, work done is:

W = 20 * 0.5 * cos30

W = 8.66 J

(b) Work done is equal to change in Kinetic energy. Since the initial kinetic energy is zero:

W = KE(final)

W = ½ * m * v²

Where v = final velocity

=> 8.66 = ½ * 3 * v²

v² = 5.773

v² = 2.40 m/s

7 0
3 years ago
You need to design a banked curve at the new circular Super 100 Raceway. The radius of the track is 800 m and cars typically tra
uysha [10]

Answer:

33.1^{\circ}

Explanation:

Let's start by writing the equations of the forces along the two directions:

- Vertical:

N cos \theta = mg

where

N is the normal reaction

\theta is the angle between the road and the horizontal

(mg) is the weight of the car, with m being its mass and g the acceleration of gravity

- Horizontal:

N sin\theta = m \frac{v^2}{r}

where

v is the speed of the car

r is the radius of the turn

Dividing the 2nd equation by the 1st one, we get:

tan \theta = \frac{v^2}{rg}

In this problem:

r = 800 m (radius of the turn)

v=160 mi/h = 71.5 m/s is the speed

g=9.8 m/s^2

Substituting, we find:

\theta= tan^{-1} (\frac{v^2}{rg})=tan^{-1}(\frac{(71.5)^2}{(800)(9.8)})=33.1^{\circ}

8 0
4 years ago
Two simple pendulums are in two different places. The length of the second pendulum is 0.4 times the length of the first pendulu
faltersainse [42]

Answer:

\sqrt{\frac{4}{9}}

Explanation:

The frequency of a simple pendulum is given by:

f=\frac{1}{2\pi}\sqrt{\frac{g}{L}}

where

g is the acceleration of gravity

L is the length of the pendulum

Calling L_1 the length of the first pendulum and g_1 the acceleration of gravity at the location of the first pendulum, the frequency of the first pendulum is

f_1=\frac{1}{2\pi}\sqrt{\frac{g_1}{L_1}}

The length of the second pendulum is 0.4 times the length of the first pendulum, so

L_2 = 0.4 L_1

while the acceleration of gravity experienced by the second pendulum is 0.9 times the acceleration of gravity experienced by the first pendulum, so

g_2 = 0.9 g_1

So the frequency of the second pendulum is

f_2=\frac{1}{2\pi}\sqrt{\frac{g_2}{L_2}}=\frac{1}{2\pi} \sqrt{\frac{0.9 g_1}{0.4 L_1}}

Therefore the ratio between the two frequencies is

\frac{f_1}{f_2}=\frac{\frac{1}{2\pi}\sqrt{\frac{g_1}{L_1}}}{\frac{1}{2\pi} \sqrt{\frac{0.9 g_1}{0.4 L_1}}}=\sqrt{\frac{0.4}{0.9}}=\sqrt{\frac{4}{9}}

8 0
3 years ago
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