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Fudgin [204]
3 years ago
15

Will the sun explode in future?​

Physics
1 answer:
anzhelika [568]3 years ago
6 0

Answer:

Explanation:

The Sun won't explode. ... Eventually (in five or seven billion years time), the Sun's life will come to an end. Our star will swell up, becoming something called a “Red Giant” star. It might even get so big that it swallows the Earth whole.

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What is the magnitude of the magnetic dipole moment of the bar magnet
Annette [7]

The magnitude of the magnetic dipole moment of the bar magnet is 1.2 Am²

<h3> Magnetic dipole moment of the bar magnet</h3>

The magnitude of the magnetic dipole moment of the bar magnet at distance from its axis is calculated as follows;

B = \frac{2\mu_0m}{4\pi r^3} \\\\m = \frac{4\pi r^3 B}{2\mu_0}

where;

  • B is magnetic field
  • m is dipole moment
  • μ is permeability of free space

m = (4π x 0.1³ x 2.4 x 10⁻⁴)/(2 x 4π x 10⁻⁷)

m = 1.2 Am²

The complete question is below:

What is the magnitude of the magnetic dipole moment of the bar magnet from 0.1 m of its axis and magnetic field strength of 2.4 x 10⁻⁴ T.

Learn more about dipole moment here: brainly.com/question/27590192

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2 years ago
Deprecion como la ejerzo
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3 years ago
A battery with an emf of 12.0 V shows a terminal voltage of 11.7 V when operating in a circuit with two lightbulbs, each rated a
wariber [46]
<h2>Answer:</h2>

0.46Ω

<h2>Explanation:</h2>

The electromotive force (E) in the circuit is related to the terminal voltage(V), of the circuit and the internal resistance (r) of the battery as follows;

E = V + Ir                      --------------------(a)

Where;

I = current flowing through the circuit

But;

V = I x Rₓ                    ---------------------(b)

Where;

Rₓ = effective or total resistance in the circuit.

<em>First, let's calculate the effective resistance in the circuit:</em>

The effective resistance (Rₓ) in the circuit is the one due to the resistances in the two lightbulbs.

Let;

R₁ = resistance in the first bulb

R₂ = resistance in the second bulb

Since the two bulbs are both rated at 4.0W ( at 12.0V), their resistance values (R₁ and R₂) are the same and will be given by the power formula;

P = \frac{V^{2} }{R}

=> R = \frac{V^{2} }{P}             -------------------(ii)

Where;

P = Power of the bulb

V = voltage across the bulb

R = resistance of the bulb

To get R₁, equation (ii) can be written as;

R₁ = \frac{V^{2} }{P}    --------------------------------(iii)

Where;

V = 12.0V

P = 4.0W

Substitute these values into equation (iii) as follows;

R₁ = \frac{12.0^{2} }{4}

R₁ = \frac{144}{4}

R₁ = 36Ω

Following the same approach, to get R₂, equation (ii) can be written as;

R₂ = \frac{V^{2} }{P}    --------------------------------(iv)

Where;

V = 12.0V

P = 4.0W

Substitute these values into equation (iv) as follows;

R₂ = \frac{12.0^{2} }{4}

R₂ = \frac{144}{4}

R₂ = 36Ω

Now, since the bulbs are connected in parallel, the effective resistance (Rₓ) is given by;

\frac{1}{R_{X} } = \frac{1}{R_1} + \frac{1}{R_2}       -----------------(v)

Substitute the values of R₁ and R₂ into equation (v) as follows;

\frac{1}{R_X} = \frac{1}{36} + \frac{1}{36}

\frac{1}{R_X} = \frac{2}{36}

Rₓ = \frac{36}{2}

Rₓ = 18Ω

The effective resistance (Rₓ) is therefore, 18Ω

<em>Now calculate the current I, flowing in the circuit:</em>

Substitute the values of V = 11.7V and Rₓ = 18Ω into equation (b) as follows;

11.7 = I x 18

I = \frac{11.7}{18}

I = 0.65A

<em>Now calculate the battery's internal resistance:</em>

Substitute the values of E = 12.0, V = 11.7V and I = 0.65A  into equation (a) as follows;

12.0 = 11.7 + 0.65r

0.65r = 12.0 - 11.7

0.65r = 0.3

r = \frac{0.3}{0.65}

r = 0.46Ω

Therefore, the internal resistance of the battery is 0.46Ω

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