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Finger [1]
2 years ago
15

Imagine that heike onnes cooled mercury down to absolute zero and there was still resistance. would he have discovered supercond

uctivity? yes, because he was working with mercury, which has superconductive properties yes, because he would have observed it at 4.2 k no, because superconductivity cannot occur in mercury since it is a liquid no, because superconductivity cannot occur if there is resistance
Physics
1 answer:
likoan [24]2 years ago
4 0

No, because superconductivity cannot occur if there is resistance

In addition to explaining electrical resistance, equilibrium distance theory also foretells the existence of superconductivity. According to its postulates, electrical resistivity decreases with distance from the equilibrium. There is only superconductivity at zero distance, with no resistance

<h3>What is Superconductivity ?</h3>

The ability of some materials to transmit electric current with virtually little resistance is known as superconductivity.

  • This ability has intriguing and maybe beneficial ramifications. Low temperatures are necessary for a material to exhibit superconductor behaviour. H. K. made the initial discovery of superconductivity in 1911.
  • Aluminum, magnesium diboride, niobium, copper oxide, yttrium barium, and iron pnictides are a few well-known examples of superconductors.

Learn more about Superconductivity here:

brainly.com/question/17166152

#SPJ4

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Complete Question: A charge q1 = 2.2 uC is at a distance d= 1.63m from a second charge q2= -5.67 uC. (b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as measured from q1.)

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d= 0.46 m

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The electric potential is defined as the work needed, per unit charge, to bring a positive test charge from infinity to the point of interest.

For a point charge, the electric potential, at a distance r from it, according to Coulomb´s Law and the definition of potential, can be expressed as follows:

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If we call x to the distance from q₁, the distance from q₂, will be the distance between both charges, minus x.

So, we can find the value of x, adding the potentials due to q₁ and q₂, in such a way that both add to zero:

V = \frac{k*q1}{x} +\frac{k*q2}{(1.63m-x)} = 0

⇒k*q1* (1.63m - x) = -k*q2*x:

Replacing by the values of q1, q2, and k, and solving for x, we get:

⇒ x = (2.22 μC* 1.63 m) / 7.89 μC = 0.46 m from q1.

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