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Leokris [45]
3 years ago
5

Which portion of the electromagnetic spectrum is used to identify fluorescent minerals?

Physics
1 answer:
S_A_V [24]3 years ago
6 0
<h2>Answer: Ultraviolet Light </h2>

Ultraviolet light, whose wavelength is approximately between 100 nm and 380 nm; is a type of electromagnetic radiation that is not visible to the human eye.

This light is used for many purposes, among which is the identification of fluerescent minerals.

In this sense, fluorescence is a property that certain materials have in which they absorb energy in the form of short wavelength not visible electromagnetic radiation (the ultraviolet, for example) and then emit some of that energy in the form of longer wavelength electromagnetic radiation (in the visible spectrum). This is also called luminiscence.

Hence, the correct option is a.

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In solving a physics problem you end up with m in the numerator and m/s in the denominator. The units for your answer are :_____
Vladimir [108]

Answer:

s

Explanation:

It is given that, a physics problem you end up with m in the numerator and m/s in the denominator. It would mean that,

\dfrac{\text{numerator}}{\text{denominator}}=\dfrac{m}{m/s}

m is metre and s is second

or

=\dfrac{m}{\dfrac{m}{s}}\\\\=\dfrac{s}{1}

so, the unit of the answer is <u>s</u>.

7 0
4 years ago
What is the frequency of a photon with an energy of 4. 56 x 10^-19 j
Sauron [17]

The frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.

<h3>What is a frequency?</h3>

The number of waves that travel through a particular point in a given length of time is described by frequency. So, if a wave takes half a second to pass, the frequency is 2 per second.

Given that the energy of the photon is 4.56 x 10⁻¹⁹ J. Therefore, the frequency of the photon can be written as,

\rm \gamma = \dfrac{E}{h} = \dfrac{4.56x10^{-19} J}{6.626 \times 10^{-34}\ Jsec^{-1}}\\\\\\\gamma  = 6.88 \times 10^{14}\ s^{-1}

Hence, the frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.

Learn more about Frequency:

brainly.com/question/5102661

#SPJ4

5 0
2 years ago
Read 2 more answers
Which of the following statements would NOT result in an eclipse? a. The Earth is positioned directly between the Sun and the Mo
vovangra [49]

c. The Moon is positioned directly between the earth and the sun is the statement that does not result in an eclipse.

Explanation:

  • The Sun is completely blocked in a solar eclipse because the Moon passes between Earth and the Sun.
  • it is just the right distance away from Earth, the Moon can fully blocks the Sun's light from Earth's perspective
3 0
4 years ago
Molecules can be made of atoms of the same element only. different elements only. the same or different elements.
Ira Lisetskai [31]

Both, there are two different types of molecules to distinguish that

8 0
4 years ago
Read 2 more answers
In a nuclear experiment a proton with kinetic energy 3.0 MeV moves in a circular path in a uniformmagnetic field. What energy mu
dlinn [17]

Answer:

a)     K = 3 MeV   b)   K=  1.5 MeV

Explanation:

We can solve this experiment using the equation of the magnetic force with Newton's second law, where the acceleration is centripetal.

        F = q v x B

We can also write this equation based on the modules of the vectors

        F = qv B sin θ

With Newton's second law

       F = ma

       F = m v² / r

       q v B = m v² / r

       v = q B r / m

The kinetic energy is

       K = ½ m v²

Substituting

       K = ½ m (q B r/ m)²

       K = ½ B² r²  q² / m

       K = (½ B² R²)  q²/m

The amount in brackets does not change during the experiment

      K = A  q² / m

For the proton

     K = 3.0 10⁶eV (1.6 10⁻¹⁹ J / 1eV) = 4.8 10⁻¹³ J

With this data we can find the amount we call A

    A = K  m/q²

    A = 4.8 10⁻¹³ 1.67 10⁻²⁷ /(1.6 10⁻¹⁹)²

    A = 3.13 10⁻²

With this value we can write the equation

    K = 3.13 10⁻²  q² / m

Alpha particle

    m = 4 uma = 4 1.66 10⁻²⁷ kg

   K = 3.13 10⁻² (2 1.6 10⁻¹⁹)² / 4.0 1.66 10⁻²⁷

   K = 4.82 10⁻¹³ J ((1 eV / 1.6 10⁻¹⁹ J) = 3 10⁶ eV

   K = 3 MeV

Deuteron

   K = 3.13 10⁻² (1.6 10⁻¹⁹)²/2 1.66 10⁻²⁷

   K = 2.4 10⁻¹³ J (1eV / 1.6 10⁻¹⁹J)

   K = 1.5 10⁶ eV

   K=  1.5 MeV

6 0
3 years ago
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