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gladu [14]
4 years ago
15

. A current-carrying gold wire has diameter 0.87 mm. The electric field in the wire is 0.54 V>m. What are (a) the current car

ried by the wire; (b) the potential difference between two points in the wire 7.0 m apart; (c) the resistance of a 7.0-m length of this wire?
Physics
1 answer:
mixas84 [53]4 years ago
6 0

Answer:

a) I = 13.156 A

b) V = 3.78 V

c) R = 0.2873 Ω

Explanation:

The diameter(d) of the wire is 0.87 mm = 0.87 × 10⁻³ m.

The electric field(E) = 0.54 V/m

The cross sectional area(A) = πd²/4 = π × (0.87 × 10⁻³)² /4 = 5.945 × 10⁺⁷ m²

Gold resistivity(ρ) = 2.44 × 10⁻⁸

(a) the current carried by the wire

E = Jρ

Where J is the current density = I/A and I is the current

E = Iρ/A

I = EA/ρ

Substituting values:

I = 0.54 × 5.945 × 10⁺⁷/  2.44 × 10⁻⁸ = 13.156 A

(b) the potential difference between two points in the wire 7.0 m apart

The potential difference(V) = EL

Where L is the length = 7 m

V = EL

Substituting values:

V = 0.54 × 7 = 3.78 V

(c) the resistance of a 7.0-m length of this wire

The resistance(R) = ρL/A

R =  ρL/A

Substituting values:

R = 2.44 × 10⁻⁸ × 7 /  5.945 × 10⁺⁷ = 0.2873 Ω

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When forces are balanced, what affect does that have on the motion of an object?
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Answer: Option (c) is the correct answer.

Explanation:

It is known that the expression for potential energy related to charges and distance between their separation is as follows.

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So, when both the charges are carry the same charge and r is small then the value of potential energy will be positive in nature.

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Answer:

The work required is -515,872.5 J

Explanation:

Work is defined in physics as the force that is applied to a body to move it from one point to another.

The total work W done on an object to move from one position A to another B is equal to the change in the kinetic energy of the object. That is, work is also defined as the change in the kinetic energy of an object.

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Ec=\frac{1}{2} *m*v^{2}

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The work (W) of this force is equal to the difference between the final value and the initial value of the kinetic energy of the particle:

W=\frac{1}{2} *m*v2^{2}-\frac{1}{2} *m*v1^{2}

W=\frac{1}{2} *m*(v2^{2}-v1^{2})

In this case:

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Replacing:

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W= -515,872.5 J

<u><em>The work required is -515,872.5 J</em></u>

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Answer:

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