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umka2103 [35]
3 years ago
8

A cylindrical solenoid (radius r=0.6 m, turns N=600, length l=0.5 m) carries 15 A of current. What is the inductance L of the co

il in Henries (Hint: use the infinite length approximation for Bsol)?
Physics
1 answer:
Anuta_ua [19.1K]3 years ago
8 0

Answer:

L = 1.023 H

Explanation:

given,

radius of the cylindrical solenoid = r = 0.6 m

Number of turns = N = 600

Length = l = 0.5 m

Current in the cylindrical solenoid = 15 A

Inductance in the coil = ?

using formula

L = \dfrac{\mu_0\ N^2\ A}{l}

L = \dfrac{4 \pi \times 10^{-7}\times 600^2\times \pi r^2}{0.5}

L = \dfrac{4 \pi \times 10^{-7}\times 600^2\times \pi 0.6^2}{0.5}

L = \dfrac{4 \pi \times 10^{-7}\times 600^2\times \pi 0.6^2}{0.5}

L = 1.023 H

the inductance L of the coil is = 1.023 H

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A train moving west with an initial velocity of 20 m/s accelerates at 4 m/s2 for 10 seconds. During this time, the train
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Answer:400m

Explanation: x=20⋅10+ 4.10^2/2                                                                                                        =400m

​

4 0
3 years ago
Which of the following is NOT one of Kirchhoff's laws?
Anon25 [30]

Answer:

C) A low-density, cool gas in isolation creates a continuous spectrum.

Explanation:

Kirchhoff’s laws established that:

  • A solid, liquid or dense incandescent gas emits a continuous spectrum.
  • A hot and diffuse gas produces bright spectral lines (emission lines).
  • A gas of lower temperature against a source of continuum spectrum, produces dark spectral lines (absorption lines) superposed in the continuum spectrum.

Stars are perfect examples for Kirchhoff’s laws. Since in the case of the stars, the photons that are received are not directly from the nucleus, but those that have traveled hundreds of thousands of years to reach the stellar atmosphere. Due to the stars are not at homogeneous temperature, density and pressure, but have gradients in different layers because of the nuclear reactions, superficial gravity or to its constant exchange of heat with its surroundings in an attempt to reach the thermodynamic equilibrium, the continuum observed in the stellar spectra comes from the inner layer of the photosphere, while absorption lines are formed in the outer layer of the photosphere and the stellar atmosphere. More accurately, a photon of the inner layer of the photosphere will be absorbed by an electron of an atom or ion that is in the outer layer, generating an electronic transition¹, the electron, upon returning to its base state will emit a photon or a series of photons that will not necessarily go in the same direction of the incident photon, creating an absorption line in the stellar spectrum.

On the other hand, in the case where the stars have surrounding material (diffuse gas), the atoms, molecules or ions in the medium are excited by the radiation that comes from the stellar atmosphere, thus producing an emission spectrum.

Key terms:

¹Electronic transition: When an electron passes from one energy level to another, either for the emission or absorption of a photon.

8 0
3 years ago
Which of the following statements is true? A. Both warming up and cooling down or important. B. It is more important warm up the
Alenkasestr [34]

Answer:

Both warming up and cooling down or not important

8 0
3 years ago
A piston–cylinder device initially contains steam at 3.5 MPa, superheated by 5°C. Now, steam loses heat to the surroundings and
velikii [3]

<u>Answer :</u>

(a) The Initial temperature of the steam is 247.557°C

(b) The enthalpy change per unit mass of the steam is -1771 kJ/kg

(c) The final pressure is 1555 kPa, and the liquid vapor mixture contains small mass of vapor.

<u>Explanation :</u>

From the given statements, we know

Initial steam pressure (P_1) = 3.5 MPa

Degree of Super heat = 5°C

Saturation Temperature (T_s_a_t) = 242.557°C [From Steam Table]

Now, the Initial temperature (T_1) of steam is given by-

T_1 = T_s_a_t + Degree of Superheat

 = 242.557 + 5 = 247.557°C

(b)To find: The enthalpy change per unit mass of the steam is  

We know enthalpy (h) change is given by

Δh = h_2 – h_1

From steam Table, properties of the gas at state 1 and state 2  

h_1 = 2821.1 kJ/kg

h_2 = 1049.7 kJ/kg

On substituting the values, we get

or, Δh = 1049.7 - 2821.1 = -1771 kJ/kg

(c) To Find: The final pressure of the liquid vapor mixture contains small mass of vapor.

We find that the gas at the final position, i.e., at position 3

Final Temperature (T_3) = 200°C [Given] ……………………………..(1)

Specific Volume (v_3) = Specific Volume (v_2) = 0.00123 m^3/kg ….....(2)

Using steam table for corresponding values of (1) and (2), we get

Final Pressure P_3 = 1555 kPa

Dryness fraction (x_3) = 0.0006.

6 0
3 years ago
IF U ANSWER I WILL PUT U AS BRAINIEST ANSWER AND GIVE U A THANK U
Bingel [31]
<span>
At the Earth's surface, warm air expands and rises, creating
what is known as an area of low pressure.

Cold air is dense and sinks to the surface to create what is
known as an area of high pressure.</span>


8 0
4 years ago
Read 2 more answers
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