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vichka [17]
3 years ago
14

Select the properties that apply to superconductors.

Physics
2 answers:
makkiz [27]3 years ago
7 0
made from pure metals . . . no;
they've been made from all kinds of weird compounds and alloys.
 
conduct electricity with zero resistance . . . yes;
that's why they're called "superconductors".

produce a strong magnetic field . . . possible, but not because it's a superconductor;
just like any other conductor, the magnetic field depends on the current that's flowing in the conductor.

no loss of energy in the transfer of electricity . . .
there's no loss of energy in the current flowing in the superconductor;
but if you tried to transfer the current out of the superconductor into
something else, then there would be some loss.
Svetradugi [14.3K]3 years ago
4 0

The correct answers are:

conduct electricity with zero resistance

produce a strong magnetic field

no loss of energy in the transfer of electricity

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10. How far does a transverse pulse travel in 1.23 ms on a string with a density of 5.47 × 10−3 kg/m under tension of 47.8 ?????
KATRIN_1 [288]

Answer: Tension = 47.8N, Δx = 11.5×10^{-6} m.

              Tension = 95.6N, Δx = 15.4×10^{-5} m

Explanation: A speed of wave on a string under a tension force can be calculated as:

|v| = \sqrt{\frac{F_{T}}{\mu} }

F_{T} is tension force (N)

μ is linear density (kg/m)

Determining velocity:

|v| = \sqrt{\frac{47.8}{5.47.10^{-3}} }

|v| = \sqrt{0.00874 }

|v| = 0.0935 m/s

The displacement a pulse traveled in 1.23ms:

\Delta x = |v|.t

\Delta x = 9.35.10^{-2}*1.23.10^{-3}

Δx = 11.5×10^{-6}

With tension of 47.8N, a pulse will travel Δx = 11.5×10^{-6}  m.

Doubling Tension:

|v| = \sqrt{\frac{2*47.8}{5.47.10^{-3}} }

|v| = \sqrt{2.0.00874 }

|v| = \sqrt{0.01568}

|v| = 0.1252 m/s

Displacement for same time:

\Delta x = |v|.t

\Delta x = 12.52.10^{-2}*1.23.10^{-3}

\Delta x = 15.4×10^{-5}

With doubled tension, it travels \Delta x = 15.4×10^{-5} m

4 0
3 years ago
The electric potential in a region of space is \[V=350/\sqrt{x ^{2}+y ^{2}}\] where x and y are in meters. what is the strength
VARVARA [1.3K]
So the given value or the formula in getting the electric potential region of space is V=350/sqrt of x^2+y^2. So the given data is x and y is equals to 2.6 and 2.8. So in my calculation i came up with an answer of 91.6
8 0
3 years ago
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