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statuscvo [17]
4 years ago
7

Higher mass protostars enter the main sequence: at the same rate, but at a higher luminosity and temperature. slower and at a lo

wer luminosity and temperature. faster and at a higher luminosity and temperature. slower and at a higher luminosity and temperature. faster and at a lower luminosity and temperature.
Physics
1 answer:
Morgarella [4.7K]4 years ago
8 0

Answer:

<em>faster and at a higher luminosity and temperature.</em>

Explanation:

A protostar looks like a star but its core is not yet hot enough for fusion to take place. The luminosity comes exclusively from the heating of the protostar as it contracts. Protostars are usually surrounded by dust, which blocks the light that they emit, so they are difficult to observe in the visible spectrum.

A protostar becomes a main sequence star when its core temperature exceeds 10 million K. This is the temperature needed for hydrogen fusion to operate efficiently.

Stars above about 200 solar masses (Higher mass) generate power so furiously that gravity cannot contain their internal pressure. These stars blow themselves apart and do not exist for long if at all. A protostar with less than 0.08 solar masses never reaches the 10 million K temperature needed for efficient hydrogen fusion. These result in “failed stars” called brown dwarfs which radiate mainly in the infrared and look deep red in color. They are very dim and difficult to detect, but there might be many of them, and in fact they might outnumber other stars in the universe.

That is why higher mass protostars enter the main sequence at a <em>faster and at a higher luminosity and temperature.</em>

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Answer:

a

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b

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c

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Explanation:

From the question we are told that the

       The magnitude of electric field is E = 5000N/C

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The total charge can be mathematically represented as

            q = N_p * e

Where e is the charge on one proton which has a value of =1.6*10^{-19}C

      Substituting values

           q = 1 *10^{15} * 1.6*10^{-19} = 1.6 *10^{-4}C

The change in potential energy is mathematically represented ads

         \Delta U = -(qE)d

where the negative sign shows that the work done by the electric force is against the electric field

                Substituting values

         \Delta U = - 1.6*10^{-4} * 5000 * 0.05

                = -4*10^{-2}J

 

   The mass of the bead is given as 0.05kg

  The change in potential due to gravity is mathematically given as

                    \Delta U = -mgh

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And g =9.8m/s^2

                  Making h the subject

        h = \frac{\Delta U}{mg}

Substituting values

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4 years ago
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1 m/s^2

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Answer:

The magnitude or strength of an electric field in the space surrounding a source charge is related directly to the quantity of charge on the source charge and inversely to the distance from the source charge. The direction of the electric field is always directed in the direction that a positive test charge would be pushed or pulled if placed in the space surrounding the source charge.

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