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skelet666 [1.2K]
3 years ago
15

¿Una corriente eléctrica sin resistencia puede fluir constantemente sin fuente alguna de potencia? Explique.

Physics
1 answer:
frosja888 [35]3 years ago
4 0

Answer:

No.

Explicación:

No, una corriente eléctrica sin resistencia no puede fluir constantemente sin ninguna fuente de energía porque la fuente de energía proporciona electrones para el flujo continuo de corriente en el circuito. Si fluye corriente por el cable, debe haber resistencia en el circuito. La fuente de energía proporciona corriente eléctrica continua al cable para que fluya y el cable tiene resistencia, por lo que parte de la corriente eléctrica se pierde durante su flujo debido a esta resistencia. Entonces podemos decir que la fuente de energía es necesaria para el flujo constante de corriente eléctrica.

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If you slosh the water back and forth in a bathtub at the correct frequency, the water rises first at one end and then at the ot
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Answer:

Velocity(v) = frequency(f) × wavelength

f = 0.3165

Wavelength = 2×length(L)

L = 157cm

Convert the length in centimetres to metre = 1.57m

v = 2×1.57 × 0.3165

v = 0.99m/s

Approx. 1m/s

Explanation:

The velocity of a wave is the product of its frequency and it's wavelength. The frequency is already known. The wavelength is the distance between two successive wave crests which is formed by sloshing water back and forth in the bath tub. Sloshing water to one end of the tub will produce a wave crest first at that end then the other completing a cycle. The wavelength will be twice the length of the bath tub as it is the distance that both crests are formed.

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3 years ago
How does the density of a medium affectthe speed of a wave traveling through it
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The denser the medium, the harder the sound struggles to travel through. The medium will determine how effectively the sound will travel, for example, large bodies of water has barely any sound for its density.
3 0
3 years ago
A distant galaxy is determined to be 150 million light years distant and moving away from us; using the Hubble law determine its
dlinn [17]

Your question kind of petered out there towards the end and you didn't specify
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The "Hubble Constant" hasn't yet been pinned down precisely, so let's pick a
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             <em>70 km per second per megaparsec</em>.

We'll also need to know that 1 parsec = about 3.262 light years.

So the speed of your receding galaxy is

         (Distance in LY) x (1 megaparsec / 3,262,000 LY) x (70 km/sec-mpsc) =

              (150 million) x  (1 / 3,262,000) x (70 km/sec) =

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4 0
3 years ago
The intensity of the radiation from the Sun measured on Earth is 1360 W/m2 and frequency is f = 60 MHz. The distance between the
Mama L [17]

Answer: (a) power output = 3.85×10²⁶W

(b). There is no relative change in power as it is independent from frequency

(c). 590 W/m²

Explanation:

given Radius between earth and sun to be = 1.50 × 10¹¹m

Intensity of the radiation from the sun measured on earth to be = 1360 W/m²

Frequency = 60 MHz

(a). surface area A of the sun on earth is = 4πR²

substituting value of R;

A = 4π(.50 × 10¹¹)² = 2.863 10²³×m²

A = 2.863 10²³×m²

now to get the power output of the sun we have;

<em>P </em>sun = <em>I </em><em>sun-earth </em><em>A </em><em>sun-earth</em>

where A = 2.863 10²³×m², and <em>I </em> is 1360 W/m²

<em>P </em>sun =  2.863 10²³ × 1360

<em>P </em>sun = 3.85×10²⁶W

(c). surface area A of the sun on mars is = 4πR²

now we substitute value of 2.28 ×10¹¹ for R sun-mars, we have

A sun-mars = 4π(2.28× 10¹¹)²

A sun-mars = 6.53 × 10²³m²

now to calculate the intensity of the sun;

<em>I </em><em>sun-mars = </em><em>P </em>sun / A sun-mars

where <em>P </em>sun = 3.85×10²⁶W and A sun-mars = 6.53 × 10²³m²

<em>I </em><em>sun-mars =  </em>3.85×10²⁶W / 6.53 × 10²³m²

<em>I </em><em>sun-mars = </em>589.6 ≈ 590 W/m²

<em>I </em><em>sun-mars = </em>590 W/m²

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Answer: because of air resistance. See explanation for further details.

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But in reality this is most likely not true because of air resistance and other fluid frictional effects in consideration.

If the experiment is performed in a vacuum, it will always be true that time is independent of masses of two falling objects.

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