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algol13
3 years ago
9

A middle-age star that burns hydrogen is a A.neutron star B.white dwarf c.main sequence star D.pusar

Physics
1 answer:
Usimov [2.4K]3 years ago
4 0
Hello!

The correct answer is C.

As we should know, a neutron star is a small cluster of neutrons. It is neither middle aged or able to burn hydrogen.
A white dwarf is a “mock planet” and has no impact on hydrogen.
A pulsar is a form of a neutron star, making C the only possible answer.
Main sequence stars are both middle aged and able to use hydrogen to form other substances.

I hope I helped!
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vivado [14]

Answer:

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Explanation:

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A recent study found that electrons that have energies between 3.45 eV and 19.9 eV can cause breaks in a DNA molecule even thoug
vlada-n [284]

Answer:

The Minimum wavelength is  \lambda_{min}= 382.2nm

The Maximum wavelength is \lambda_{max}= 624.2nm

Explanation:

From the question we are told that  

              The energy range is  E_r = 3.25eV \ and  \ 19.9eV

   Considering E = 19.9eV

When a single photon is transferred to to an electron the energy obtained can be calculated as follows

              E = 19.9eV = 19.9 *1.6 *10^{-19}J

This energy is mathematically represented as

                    E = \frac{hc}{\lambda_{max}}

Here h is the Planck's constant with value of  h= 6.625*10^{-34}J\cdot s

        c is the speed of light with value of  c = 3*10^8 m/s

Substituting values and making \lambda the subject of the formula

                       \lambda_{max} = \frac{hc}{E}

                         = \frac{6.625*10^{-34} * 3.0*10^{8}}{19.9*1.6*10^{-19}}

                         \lambda_{max}= 624.2nm

  Considering E = 3.25eV

When a single photon is transferred to to an electron the energy obtained can be calculated as follows

              E = 19.9eV = 3.25 *1.6 *10^{-19}J

This energy is mathematically represented as

                    E = \frac{hc}{\lambda_{min}}

Substituting values and making \lambda the subject of the formula

                       \lambda_{min} = \frac{hc}{E}

                           = \frac{6.625*10^{-34} * 3.0*10^{8}}{3.25*1.6*10^{-19}}

                           \lambda_{min}= 382.2nm

3 0
3 years ago
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5 0
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What is the x-component of the force on the charge located at x = 8 cm given that q = 1.1 μC in N?
jonny [76]

Answer:

Answer is = 12.3 x 10^6 N

Explanation:

We need to calculate force on the charge which is in Newtons:

Formula is = K q1q2/r^2

k is constant depeneds on the permitivity of the medium and its value is 9x10^9 Nm^2/C^2

q1 and q2 are charges and r is the distance between these two charges so just by putting values we get F= 9x10^9 x 1.1x10^-6/(8x10^-2)^2 (we get the above answer in newton)

6 0
4 years ago
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