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Alexxx [7]
4 years ago
6

The magnetic field created by a large current passing through plasma (ionized gas) can force current-carrying particles together

. This pinch effect has been used in designing fusion reactors. It can be demonstrated by making an empty aluminum can carry a large current parallel to its axis. Let R represent the radius of the can and I the current, uniformly distributed over the can's curved wall. Determine the magnetic field just inside the wall.
Physics
1 answer:
RUDIKE [14]4 years ago
5 0

To develop this problem we must apply the concepts related to the Ampere Law. The line integral of B.ds around any closed path is equal to the current permeability constant, this current for the particular case passes through the 'internal' surface delimited by the closed path.

Ampere laws is defined as,

\oint Bds=mu_0 I

The radius of the circle is r, so if

r < R

And there is no current inside so ,

\oint Bds=mu_0 I

B (2\pi R) = 0

Therefore the magnetic field inside the wall is Zero.

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Weight is equal to the mass of the object times the acceleration due to gravity (which is 9.8m/s^{2} on earth)

So W=mg
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An organ pipe open at both ends is 1.5 m long. A second organ pipe that is closed at one end and open at the other is 0.75 m lon
telo118 [61]

Answer:

A. 110Hz,220Hz, 330 Hz

Explanation:

for organ open at open both ends;

the length of the organ for the fundamental frequency, L = A---->N + N----->A

A---->N  = λ /4 and N----->A = λ /4

L = λ /4 + λ /4 = λ /2

L = \frac{\lambda}{2} \\\\\lambda = 2L

λ  = 2 x 1.5m = 3.0 m

Wave equation is given by;

V = Fλ

Where;

V is the speed of sound

F is the frequency of the wave

F = V/ λ

F₀ = V / 2L

Where;

F₀  is the fundamental frequency

F₀ = 330 / 2(1.5)

F₀ = 330 / 3

F₀ = 110 Hz

the length of the organ for the first overtone, L = A---->N + N----->A + A----->N +  N----->A

L = 4λ /4

L = λ

λ = 1.5 m

F₁ = 330 / 1.5

F₁ = 220 Hz

Thus, F₁ = 2F₀

For open organ at one end

the length of the organ for the fundamental frequency, L = N------A

L = λ /4

λ = 4L

F₀ = V/4L

F₀ = 330 / (4 x 0.75)

F₀ = 110 Hz

the length of the organ for the first overtone, L = N-----N + N-----A

L = λ/2 + λ / 4

L = 3λ /4

F₁ = 3F₀

F₁ = 3 x 110

F₁ = 330 Hz

Thus the fundamental frequency for both organs is 110 Hz,

The first overtone for the organ open at both ends is 220 Hz

The first overtone for the organ open at one end is 330 Hz

The correct option is "A. 110Hz,220Hz, 330 Hz"

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3 years ago
What is the adaptation to the arctic fox
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How do we get this equation ??<br><br><br> H=V^2÷R
lesantik [10]

H=\frac{V^2}{R}\times t is the equation that represents the Joule's law of heating.

<h3>Explanation:</h3>

Joule's law of heating defines the heat generated by any current flowing conductor is directly proportional to  

1. Square of Current (I²),

2. Resistance of the conductor (R)

3. Time for which current is passed (t)

Hence, Heat generated = H = I^2 Rt .....................(1)

By Ohm's Law, the potential difference (V) across a conductor is directly proportional to the current(I) flowing through it. The constant of proportionality is termed as resistance of the conductor (R).

V\ \alpha\ I\\V=I\times R ...............................................(2)

From (2), Current (I) can be rewritten as

I = \frac{V}{R} ........................................................(3)

Substituting (3) in (1), we get

H = I^2\times R\times t \\=(\frac{V}{R} )^2\times R\times t\\\\=\frac{V^2}{R^2}\times R\times t = \frac{V^2}{R}\times t\\\\ H =\frac{V^2}{R}\times t

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