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GalinKa [24]
2 years ago
7

Does the transmission of electric energy require electrical conductors between the source and receiver? Cite an example to defen

d your answer.
Physics
1 answer:
kolezko [41]2 years ago
3 0

No, electrical conductors are not necessary for the electrical transmission between the recipient and the source. Through a vacuum, sunlight carries electromagnetic energy to Earth.

What is meant by the transmission of electric energy?

  • Electrical transmission refers to the process of transferring produced power to the transmission network situated in populous regions, typically across large distances. Transformers, which raise voltage levels to enable long-distance transmission, is a crucial component of this process.
  • Electric power can be transported primarily via one of two systems: the electrical transmission method for high voltage DC electrical transmission method for high AC.

Therefore it is concluded that theelectrical transmission does not need any conductor.

Learn more about electrical transmission here:

brainly.com/question/949576

#SPJ4

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A 2.0-mm-diameter copper ball is charged to 40 nC . What fraction of its electrons have been removed? The density of copper is 8
murzikaleks [220]

Answer:

0.02442 × 10⁻⁹

Explanation:

Given:

Diameter of copper ball = 2.00 mm = 0.002 m

Charge on ball = 40 nC = 40 × 10⁻⁹ C

Density of copper = 8900 Kg/m³

Now,

The number of electrons removed, n = \frac{\textup{Charge on ball}}{\textup{Charge of an electron}}

also, charge on electron = 1.6 × 10⁻¹⁹ C

Thus,

n = \frac{40\times10^{-9}}{1.6\times10^{-19}}

or

n = 25 × 10¹⁰ Electrons

Now,

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Or

Mass of copper ball =  \frac{4}{3}\pi(\frac{d}{2})^3  × 8900

or

Mass of copper ball =  \frac{4}{3}\pi(\frac{0.002}{2})^3  × 8900

or

Mass of copper ball = 0.03726 grams

Also,

molar mass of copper = 63.546 g/mol

Therefore,

Number of mol of copper in  0.03726 grams = \frac{ 0.03726}{63.546}

or

Number of mol of copper in  0.03726 grams = 5.86 × 10⁻⁴ mol

and,

1 mol of a substance contains = 6.022 × 10²³ atoms

Therefore,

5.86 × 10⁻⁴ mol of copper contains = 5.86 × 10⁻⁴ × 6.022 × 10²³ atoms.

or

5.86 × 10⁻⁴ mol of copper contains = 35.88 × 10¹⁹ atoms

Now,

A neutral copper atom has 29 electrons.  

Therefore,

Number of electrons in ball = 29 × 35.88 × 10¹⁹ = 1023.37 × 10¹⁹ electrons.

Hence,

The fraction of electrons removed = \frac{25\times10^{10}}{1023.37\times10^{19}}

or

The fraction of electrons removed = 0.02442 × 10⁻⁹

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