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Kryger [21]
3 years ago
8

Calcular la resistencia de una varilla de grafito de 170 cm de longitud y 60 mm2. Resistividad grafito 3,5 10-5 Ωm

Physics
1 answer:
ozzi3 years ago
3 0

Answer:

R = 0.992 Ω

Explanation:

En esta pregunta, dada la información que contiene, debemos calcular la resistencia de la varilla de grafito.

Matemáticamente,

Resistencia = (resistividad * longitud) / Área De la pregunta;

Resistividad = 3,5 * 10 ^ -5 Ωm

longitud = 170 cm = 1,7 m

Área = 60 mm ^ 2 = 60/1000000 = 6 * 10 ^ -5 m ^ 2

Conectando estos valores a la ecuación anterior, tenemos;

Resistencia = (3.5 * 10 ^ -5 * 1.7) / (6 * 10 ^ -5) =

(3.5 * 1.7) / 6 = 0.992 Ω

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What is the potential energy of an object 20 m in the air with a<br> mass of 600 kg?
Lana71 [14]

Answer:

Ep = 117600 J

Explanation:

Data:

  • Mass (m) = 600 kg
  • Height (h) = 20 m
  • Gravity (g) = 9.8 m/s²
  • Potential Energy (Ep) = ?

Use formula:

  • Ep = m * g * h

Replace:

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Multiply operations, and units:

  • Ep = 117600 J

What is the potential energy?

The potential energy is <u>117600 Joules.</u>

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You have given a power supply, copper wire, and an iron nail. What should you do to decrease the strength of the electro magnet?
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By calculating its wavelength (in nm), show that the second line in the Lyman series is UV radiation.
Rashid [163]

Answer:

 λ = 102.78  nm

This radiation is in the UV range,

Explanation:

Bohr's atomic model for the hydrogen atom states that the energy is

           E = - 13.606 / n²

where 13.606 eV   is the ground state energy and n is an integer

an atom transition is the jump of an electron from an initial state to a final state of lesser emergy

            ΔE = 13.606 (1 / n_{f}^{2} - 1 / n_{i}^{2})

the so-called Lyman series occurs when the final state nf = 1, so the second line occurs when ni = 3, let's calculate the energy of the emitted photon

            DE = 13.606 (1/1 - 1/3²)

            DE = 12.094 eV

let's reduce the energy to the SI system

            DE = 12.094 eV (1.6 10⁻¹⁹ J / 1 ev) = 10.35 10⁻¹⁹ J

let's find the wavelength is this energy, let's use Planck's equation to find the frequency

            E = h f

             f = E / h

            f = 19.35 10⁻¹⁹ / 6.63 10⁻³⁴

            f = 2.9186 10¹⁵ Hz

now we can look up the wavelength

           c = λ f

           λ = c / f

           λ = 3 10⁸ / 2.9186 10¹⁵

           λ = 1.0278  10⁻⁷ m

let's reduce to nm

            λ = 102.78  nm

This radiation is in the UV range, which occurs for wavelengths less than 400 nm.

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