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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:
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The cable of the 1800kg elevator cab in Fig. 8−56 snaps when the cab is at rest at the first floor, where the cab bottom is a di
drek231 [11]

a ) The speed of the cab just before it hits the spring = 7.4 m / s

b ) The maximum distance x that the spring is compressed = 0.9 m

c ) The distance that the cab will bounce back up the shaft = 2.8 m

a ) The speed of the cab just before it hits the spring,

Ki + Pi = K final + P final + W

1 / 2 m vo² + m g hi = 1 / 2 m v² + m g h + f d

0 + ( 1800 * 9.8 * 3.7 m ) = ( 1 / 2 * 1800 * v² ) + 0 + ( 4400 * 3.7 )

v = 7.4 m / s

b ) The maximum distance x that the spring is compressed,

Ki + Pi = K final + P final + W + Fs

1 / 2 m vo² + m g x = 1 / 2 m v² + m g h + f d + 1 /2 k x²

( 1/2 * 1800 * 7.4² ) + ( 1800 * 9.8 * x ) =0 + 0+ ( 4400 * x ) + ( 1/2 * 1800 * x² )

75000 x² + 13420 x - 50625 = 0

x = 0.9 m

c ) The distance that the cab will bounce back up the shaft,

Ki + Pi + Fs = K final + P final + W

1 / 2 m vo² + m g x + 1 /2 k x² = 1 / 2 m v² + m g h + f d

0 + 0 + ( 1 / 2 * 0.15 * 0.9² ) = 0 + ( 1800 * 9.8 * h ) + ( 4400 * h )

h = 2.8 m

Therefore,

a ) The speed of the cab just before it hits the spring = 7.4 m / s

b ) The maximum distance x that the spring is compressed = 0.9 m

c ) The distance that the cab will bounce back up the shaft = 2.8 m

To know more about law of conservation of energy

brainly.com/question/12050604

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3 0
1 year ago
The term pressure most nearly refers to which of the following?
Arte-miy333 [17]

The term pressure refers to the force per area created by the weight
of anything whose weight is distributed over an area, such as the
Earth's atmosphere, a lake, a gas inside a sealed jar, or a pointy
high heel.

5 0
4 years ago
Read 2 more answers
After creating a prototype, which of the following would not be and appropriate next step in the design process of a new technol
Dafna11 [192]

The answer would be to research the need.  This should have been done before the project began.

6 0
4 years ago
Read 2 more answers
If the room radius is 4.5 m, and the rotation frequency is 0.8 revolutions per second when the floor drops out, what is the mini
kondaur [170]
<span>The force of static friction F equals the coefficient of friction u times the normal force N the object exerts on the surface: F = uN. N is the centripetal force of the wall on the people; N = ma_N, where m is the mass of the people and a_N is the centripetal acceleration. The people will not slip down if F is greater than the force of gravitation: F = uma_N > mg, or u > g/a_N. a_N is the velocity v of the people squared divided by the radius of the room r: a_N = v^2/r. The circumference of the room is 2 pi r = 28.3 m. So v = 28.3 * 0.8 m/sec = 22.6 m/sec. So a_N = 114 m/sec^2. g = 9.81 m/sec^2, so u must be at least 9.81/114 = 0.086.</span>
3 0
3 years ago
What is the magnitude of electrical force of attraction between an copper nucleus (29 protons) and its innermost electron if the
Agata [3.3K]
The charge of the copper nucleus is 29 times the charge of one proton:
Q=29 q= 29 \cdot 1.6 \cdot 10^{-19}C=4.64 \cdot 10^{-18}C
the charge of the electron is
e=-1.6 \cdot 10^{-19}C
and their separation is
r=1.0 \cdot 10^{-12} m

The magnitude of the electrostatic force between them is given by:
F=k_e  \frac{Qe}{r^2}
where k_e is the Coulomb's constant. If we substitute the numbers, we find (we can ignore the negative sign of the electron charge, since we are interested only in the magnitude of the force)
F=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(4.64 \cdot 10^{-18}C)(1.6 \cdot 10^{-19}C)}{(1.0 \cdot 10^{-12} m)^2}=6.68 \cdot 10^{-3} N
3 0
3 years ago
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