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insens350 [35]
2 years ago
12

If the two particles that make up the dipole are 2.5 mm apart, what is the magnitude of the charge on each particle

Physics
1 answer:
ad-work [718]2 years ago
7 0

This question is incomplete, the complete question is;

The electric force due to a uniform external electric field causes a torque of magnitude 20.0 × 10⁻⁹ N⋅m on an electric dipole oriented at 30° from the direction of the external field. The dipole moment of the dipole is 7.5 × 10⁻¹² C⋅m.

What is the magnitude of the external electric field?

If the two particles that make up the dipole are 2.5 mm apart, what is the magnitude of the charge on each particle?

Answer:

- the magnitude of the external electric field is 5333.3 N/C

- the magnitude of the charge on each particle is 3.0 × 10⁻¹² C  ≈ 3 nC

Explanation:

Given that;

Torque = 20.0 × 10⁻⁹ N⋅m

dipole moment = 7.5 × 10⁻¹²

∅ = 30°

The moment T of restoring couple is;

T = PEsin∅

E = T/Psin∅

we substitute

E = 20.0 × 10⁻⁹ N⋅m / (7.5 × 10⁻¹²) sin(30°)

E = 20.0 × 10⁻⁹ / 3.75 × 10⁻¹²

E =  5333.3 N/C

Therefore, the magnitude of the external electric field is 5333.3 N/C

The dipole moment is given by the expression;

p = ql

q = p / l

given that l = 2.5 mm = 0.0025 m

we substitute

q = 7.5 × 10⁻¹² / 0.0025

q = 3.0 × 10⁻¹² C ≈ 3 nC

Therefore, the magnitude of the charge on each particle is 3.0 × 10⁻¹² C ≈ 3 nC

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The heat capacity of the silver spoon at the given temperature difference is 817.65 J.

<h3>Heat capacity of the silver spoon</h3>

The heat capacity of theb silver spoon is the quantity of heat absorbed by the silver spoon. The heat capacity of the silver spoon at the given temperature difference is calculated as follows;

Q = mcΔθ

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Q = 50 x 0.237 x (89 - 20)

Q = 817.65 J

Thus, the heat capacity of the silver spoon at the given temperature difference is 817.65 J.

Learn more about heat capacity here: brainly.com/question/16559442

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2 years ago
Jack has two boxes: one is 148g and one is 78g. if jack pushes both boxes with the same amount of force which will accelerate fa
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1) A The 78g
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5 0
3 years ago
A constant voltage of 12.00 V has been observed over a certain time interval across a 1.20 H inductor. The current through the i
kiruha [24]

Answer:

The time is 0.5 sec.

Explanation:

Given that,

Voltage V= 12.00 V

Inductance L= 1.20 H

Current = 3.00 A

Increases rate = 8.00 A

We need to calculate change in current

\Delta A = 8.00-3.00= 5.00\ A

We need to calculate the time interval

Using formula of inductor

V=L\dfrac{\Delta A}{\Delta t}

\Delta t =\dfrac{L\Delta A}{V}

Where, \Delta A = change in current

V = voltage

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Put the value into the formula

\Delta t=\dfrac{1.20\times5.00}{12.00}

\Delta t=0.5\ sec

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3 years ago
A quarterback back pedals 3.3 meters southward and then runs 5.7 meters northward. For this motion, what is the distance moved?
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Answer:

The distance moved is 9 meters

The magnitude of the displacement is 2.4 meters

The direction of the displacement is northward

Explanation:

- <em><u>Distance</u></em> is the length of the actual path between the initial and the

  final position. Distance is a scalar quantity

- <u><em>Displacement</em></u> is the change in position, measuring from its starting

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The quarterback pedals 3.3 meters southward

That means it moves down 3.3 meters

Then runs 5.7 meters northward

That means it runs up 5.7 meters

The distance = 3.3 + 5.7 = 9 meters

<em></em>

<em>The distance moved is 9 meters</em>

It moves southward (down) for 3.3 meters and then moves northward

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It moves from zero to 3.3 down and then moves up to 5.7

The displacement = 5.7 - 3.3 = 2.4 meters northward

<em />

<em>The magnitude of the displacement is 2.4 meters</em>

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8 0
2 years ago
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Answer:

0.027m

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the bolt loses contact with the piston only when acceleration due to gravity equals acceleration of piston

ω² * A = g where ω is angular velocity, A amplitude, g acceleration due to gravity

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i used 10m/s² in this answer

5 0
3 years ago
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