Certain materials can become superconductors under what conditions?
1 answer:
Option (b) is correct. The certain materials can act as superconductors at a very low temperature.
Explanation:
The Superconductivity of a material is the property of the particular substance to provide no resistance to the path of the current through it. The current does not face any hindrance due to the motion of the other particles inside the conductor.
The superconductivity of a substance leads to the flow of current through the substance or the material without any hindrance. The resistance of a substance is due to the motion of the particles inside it. If all the motion occurring inside the substance are seized, there will be no resistance offered by the conductor.
Therefore, the superconductivity can be defined as the state where all the motions inside the a material are seized. This occurs at a very low temperature. The absolute zero temperature is regarded as the temperature where there is no motion in any particle.
Thus, Option (b) is correct. The certain materials can act as superconductors at a very low temperature.
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Answer Details:
Grade: High School
Subject: Physics
Chapter: Conductivity
Keywords:
material, superconductor, motion, pressure, cold, temperature, substance, material, conditions, superconductivity.
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Answer:
13.7m
Explanation:
Since there's no external force acting on the astronaut or the satellite, the momentum must be conserved before and after the push. Since both are at rest before, momentum is 0.
After the push
Where is the mass of the astronaut,
is the mass of the satellite,
is the speed of the satellite. We can calculate the speed
of the astronaut:
So the astronaut has a opposite direction with the satellite motion, which is further away from the shuttle. Since it takes 7.5 s for the astronaut to make contact with the shuttle, the distance would be
d = vt = 1.83 * 7.5 = 13.7 m
Answer:
a) ΔV = 25.59 V, b) ΔV = 25.59 V, c) v = 7 10⁴ m / s, v/c= 2.33 10⁻⁴ ,
v/c% = 2.33 10⁻²
Explanation:
a) The speed they ask for electrons is much lower than the speed of light, so we don't need relativistic corrections, let's use the concepts of energy
starting point. Where the electrons come out
Em₀ = U = e DV
final point. Where they hit the target
Em_f = K = ½ m v2
energy is conserved
Em₀ = Em_f
e ΔV = ½ m v²
ΔV = mv²/e (1)
If the speed of light is c and this is 100% then 1% is
v = 1% c = c / 100
v = 3 10⁸/100 = 3 10⁶6 m/ s
let's calculate
ΔV =
ΔV = 25.59 V
b) Ask for the potential difference for protons with the same kinetic energy as electrons
K_p = ½ m v_e²
K_p = 9.1 10⁻³¹ (3 10⁶)²
K_p = 40.95 10⁻¹⁹ J
we substitute in equation 1
ΔV = Kp / M
ΔV = 40.95 10⁻¹⁹ / 1.6 10⁻¹⁹
ΔV = 25.59 V
notice that these protons go much slower than electrons because their mass is greater
c) The speed of the protons is
e ΔV = ½ M v²
v² = 2 e ΔV / M
v² =
v² = 49,035 10⁸
v = 7 10⁴ m / s
Relation
v/c =
v/c= 2.33 10⁻⁴