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JulsSmile [24]
3 years ago
12

Hello people ~

Physics
2 answers:
ASHA 777 [7]3 years ago
5 0

The charge would be in equilibrium so there would be no charge in the body of the conductor.


Answer:

(b) there cannot be any charge in the body of the conductor

Mamont248 [21]3 years ago
3 0

Answer:

B

Explanation:

Note the line that there is no charges anywhere else outside means the charge is placed in equilibrium position.

Hence there cannot be any charge in the body of conductor

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The object has a redshift of 7.6 and the JWST observes the object at a wavelength of 2 microme-
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b. The wavelength of light emitted by the object is 233 nm

c. The type of radiation originally emitted by the object is ultraviolet radiation.

To find the wavelength, we need to know what redshift is.

<h3>What is redshift?</h3>

Redshift is the increase in wavelength and the corresponding decrease of frequency and photon energy of electromagnetic radiation.

Redshift is given by z = λ'/λ - 1 where

  • λ' = observed wavelength and
  • λ = emitted wavelength.

Making λ subject of the formula, we have

λ = λ'/(1 + z)

Given that has a redshift of 7.6 and the JWST observes the object at a wavelength of 2 micrometres (mid-infrared light).

So,

  • z = 7.6 and
  • λ' = 2μm

<h3>(b) What is the wavelength of the light emitted by the object?</h3>

Substituting the values of the variables into the equation, we have

λ = λ'/(1 + z)

λ = 2μm/(1 + 7.6)

λ = 2μm/8.6

λ = 0.233 μm

λ = 233 nm

So, the wavelength of light emitted by the object is 233 nm

<h3>c. What type of radiation was originally emitted by the object?</h3>

Since the wavelength is 233 nm and the wavelength is in the range of ultraviolet radiation 200 nm - 315 nm.

So, the type of radiation originally emitted by the object is ultraviolet radiation.

Learn more about redshift here:

brainly.com/question/27915180

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