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harkovskaia [24]
3 years ago
7

A star having a luminosity 10,000 times that of the sun and a surface temperature only two-thirds of that of the sun, or 4,000k,

will fit into the red ____ group of stars on the hertzsprung-russell diagram.
Physics
1 answer:
ELEN [110]3 years ago
7 0
The hertzsprung-russel diagram is a diagram that correlates the different factors such as effective temperature in Kelvin,absolute mangnitude, relative luminosity of the star compared to the sun and the spectral class of the star. The temperature of the star moves from the left to right that is from blue to red star. To determine the group of stars, we just have to trace the factors in the diagram. We can tell from the diagram that the area surrounding the point are supergiants including Betelguese and Antares. The spectral class is between KO and MO. The class where the star belongs is the supergiants.
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When do we experience conservation of energy
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Estimate the kinetic energy of the earth with respect to the sun as the sum of two terms.
nekit [7.7K]

The definition of kinetic energy allows to find the result for the relationship between the energy of the sun and the Earth is:

  • The kinetic energy ratio is   \frac{K_{Sum} }{K_{Earth}} = 5.3 \ 10^2
<h3 /><h3 /><h3> Kinetic enrgy.</h3>

Kinetic energy is the energy due to the movement of bodies, it is given by the relation

          K = ½ m v²

where K is the kinetic energy, m the mass of the body and v the velocity of the body.

In a compound motion it is common to separate energy into parts to simplify calculations.

  • Translational kinetic energy. Due to the linear movement of the body

            K_{tras} =\frac{1}{2} m v^2

  • Rotational kinetic energy. Due to the rotational movement of the body.

            K_{rot} = \frac{1}{2} I w^2

Where I is the inrtia momentum and w the angular velocity.

They indicate that we compare the kinetic energy of the sun and the Earth.

The Earth has two movements, one of rotation about its axis with a period of T = 24 h and one of translation with respect to the Sun with a period of T= 365 days, therefore the kinetic energy of the Earth.

           K_{earth} = K_{tras} + K_{rot}

Linear and rotational speed are related.

           v = w r

The Earth is an almost spherical body therefore the moment of inertia of a solid sphere.

           I = \frac{2}{5 }  m r^2  

Let's  subatitute.

         

          K_{earth} = \frac{1}{2} \  m r^2_{tras} w^2_{tras} + \frac{1}{2} ( \frac{2}{5} m r^2_{earth}) w^2_{rot}  

The movement of the Earth around the sun is almost circular, therefore we can use the relations of the uniform circular movement, where the angle for one revolution is 2π radians and the time is called the period.

       w = \frac{2 \pi}{T}  

Let's substitute.

        K_{earth} = \frac{1}{2} m ( \frac{2\pi r^2_{tras}}{T_{tras}})^2  \ + \frac{1}{5} m (\frac{2\pi r^2_{earth} }{T^2_{rot}})^2  

        K_{earth} = 4 \pi^2 \ m \ ( \frac{1}{2} [ \frac{r_{tras}}{T_{tras}y} ]^2 + \frac{1}{5} [ \frac{r_{rot}}{T_{rot}}]^2)  

Data for Earth are tabulated:

  • Mass m = 5.98 1024 kg
  • Radius r = 6.37 10⁶ m
  • Radius orbits tras = 1.496 10¹¹ m
  • Rotation period T_{rot} = 24 h (\frac{3600s}{1h}) = 8.64 10⁴s
  • Translation period  T_{tras} = 365 d (\frac{24h}{1 d}) (\frac{3600s}{1h}) = 3.15 10⁷ s

Let's calculate.

        K_{earth} = 4 \pi^2 5.98 \ 10^{24}  ( \frac{1}{2} ( \frac{1.496 \ 10^{11}}{3.15 \ 10^7 } )^2  \ +  \frac{1}{5}( \frac{6.37 \ 10^6 }{8.64 \ 10^4})^2 )

        K_{earth} = 2.36 \ 10^{26 } \ (1.128 \ 10^7 + 1.087 \ 10^3)

        K_{earth}= 2.66 \ 10^{33} J

Let's analyze the kinetic energy for the Sun, this is inside the solar system therefore it has no translation movement and is approximately a sphere with a rotation period of T_{Sum} = 27 days.

The kinetic energy of the sun is;

          K_{sum} = K_{rot} =  \frac{1}{2} I w^2  

          K_{sum} = \frac{1}{2} (\frac{2}{5} M R^2) (\frac{2\pi}{T_{sum}})^2  

          K_{sum} = \frac{4\pi^2 }{5} M (\frac{R}{T_{rot}})^2  

The tabulated data for the sun are:

  • Mass m = 1,991 1030 kg.
  • Radius R = 6.96 10⁸ m
  • Period T = 27 d (\frac{24h}{1 d} ) (\frac{3600s}{1h}) = 2.33 10⁶ s

         

Let's calculate.

           

          K_{sum} = 1.40 \ 10^{36} J

The relationship of the kinetic energy of the sun and the Earth is:

        \frac{K_{sum}}{K_{earth}} = \frac{1.40 \ 10^{36}}{2.66 \ 10^{33}}  

       \frac{K_{sum}}{K_{earth}} =  5.3 \ 10^2  

In conclusion using the definition of kinetic energy we can shorten the result for the relationship between the energy of the sun and the Earth is:

  • The kinetic energy ratio is:  \frac{K_{Sum}}{K_{Earth}} = 5 \ 10^2

Learn more about kinetic energy here: brainly.com/question/25959744

5 0
3 years ago
6. Four people are trying to help push and lift a refrigerator up a series of stairs. Jim pushes
professor190 [17]

Answer:

Net force due to all on the refrigerator is 400 N

Explanation:

As we know that two people are pushing the refrigerator while other two people are lifting the refrigerator

So net push force on the refrigerator is given as

F = F_1 + F_2

F = 60 + 210

F_1 = 270 N

Similarly net lift force on the refrigerator is given as

F = F_1' + F_2'

F = 200 + 95

F_2 = 295 N

Now net force on the refrigerator is given as

F_{net} =\sqrt{F_1^2 + F_2^2}

F_{net} = \sqrt{270^2 + 295^2}

F_{net} = 400 N

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ludmilkaskok [199]
Factor out 8 and then facotr and u get

8/9(9x+1)(9x-1
7 0
3 years ago
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