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dedylja [7]
3 years ago
15

Earlier in this lesson, you spent some time thinking about events that involve matter and energy. Consider balls of different sh

apes and materials that reach different heights as they bounce off the same floor. What aspect of this event relates to matter? What aspect relates to energy? Font Sizes
Physics
1 answer:
Marianna [84]3 years ago
5 0

Answer:

See the answer below

Explanation:

T<em>he aspect of the event that relates to matter is </em><em>the materials and different shapes of the balls</em><em> while the aspect that relates to energy is t</em><em>he heights reached by the balls as they bounce off the same floor.</em>

Matter is generally defined as any substance that has weight and is able to occupy space. <u>Hence, the balls being made up of different materials and shapes means that they are made up of matter. </u>

Energy, however, is generally defined as the capacity to do work. The more the energy of a system, the more the work that can be done by the system. <u>The height reached by the balls is thus an indication of the amount of energy possessed by the balls.</u>

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

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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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