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n200080 [17]
3 years ago
7

. Common household appliances are rated at 110 V, but power companies deliver voltage in the kilovolt range and then step the vo

ltage down using transformers to 110 V to be used in homes. You will learn in later chapters that transformers consist of many turns of wire, which warm up as current flows through them, wasting some of the energy that is given off as heat. This sounds inefficient. Why do the power companies transport electric power using this method
Physics
1 answer:
natulia [17]3 years ago
8 0

Answer: High voltage transmission minimizes energy losses during electricity transmission.

Explanation:

When electricity is to be transmitted over a long distance, high voltage transmission is preferred to minimize energy losses due to heat.

The higher the transmission voltage, the lower the current and the lesser the resitance in the wire and the lesser the energy lost due to heat during transmission.

High voltage transmission in kilowatts enables light weight cables to be used for long distance electricity transmission.

This explains why, even though only 110 V may be required at home by some appliances, but electricity is transmitted in kilovolts and often require a stepdown transformer

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Answer:

salt dissolving in water

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Select the correct answer.
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Doubly ionized lithium Li2+ (Z = 3) and triply ionized beryllium Be3+ (Z = 4) each emit a line spectrum. For a certain series of
EleoNora [17]

Answer:

tex]\lambda_{Be}[/tex] = 22.78 nm

Explanation:

Bohr's model for the hydrogen atom has been used by other atoms with a single electric charge by changing the number of charges by the charge of the new atom (atomic number)

      E_{n}= k e² / 2a₀ (1 /n²)

      ao = h'² / k m e²               h' = h/2πi

For another atom with a single electron in the last layer

      a₀ ’= h’² / k m (Ze)²  

      a₀ ’= a₀ / Z²

Therefore, when replacing in the equation

      E_{n} = - Z²  Eo/n²

     E₀ = 13,606 eV

The transition occurs when the electron stops from one level to another

         E_{n} -  E_{m} = Z² E₀ (1 / n² - 1 / m²) = Z² ΔE

Let's relate this expression to the wavelength

       c = λ f

      E = h f

      E = h c /λ

      h c / λ = Z² ΔE

     λ = 1 / Z² (hc / ΔE)

     λ = 1 / Z² λ_hydrogen

Let's apply this last equation to our case

Lithium Z = 3

     E_{n} = - 9 Eo / n²

     

      40.5 10-9 = 1/9 λ_hydrogen

Beryllium Z = 4

      λ = 1/16 λ_hydrogen

Let's write our two equations is and solve

     40.5 10-9 = 1/9 λ_hydrogen

    tex]\lambda_{Be}[/tex] = 1/ 16 λ_hydrogen

      40.5 10⁻⁹ = 1/9 (16 \lambda_{Be} )

    tex]\lambda_{Be}[/tex] = 40.5 9/16

  tex]\lambda_{Be}[/tex] = 22.78 nm

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Round your answers to one decimal place.this parallel circuit has two resistors at 15 and 40 ohms. what is the total resistance?
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\frac{1}{R_{eq}}= \frac{1}{R_1}+ \frac{1}{R_2}
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2) If we have a battery of 12 V connected to the circuit, the current in the circuit will be given by Ohm's law, therefore:
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