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katen-ka-za [31]
3 years ago
5

The Hertzsprung-Russel diagram is used to show the relationship between which two characteristics of stars?

Physics
2 answers:
yuradex [85]3 years ago
8 0

Answer:

Absolute Magnitude (Luminosity) and Spectral class (Temperature)

Explanation:

Hertzsprung-Russell diagram, popularly known as HR Diagram, shows the relation between the absolute magnitude (or Luminosity) and temperature of stars (spectral class). HR diagram helps in understanding the stellar evolution. One can predict the life-cycle of a star with the help of this plot. One can see, with the help of HR Diagram, that stars spend 90% of their life time in main sequence stage where the fusion of Hydrogen takes place in the core.

Ejnar Hertzsprung and Henry Norris Russell created this plot in 1910.

andre [41]3 years ago
3 0
The Hertzsprung- Russel diagram is used to how the relationship between the absolute magnitudes of stars and their effective temperatures. 
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Please select the word from the list that best fits the definition site of sea-floor spreading
serious [3.7K]
<h3><u>Answer;</u></h3>

Mid-ocean ridges

<h3><u>Explanation</u>;</h3>
  • Sea-floor spreading is the process by which molten material adds new oceanic crust to the ocean floor.
  • Mid-ocean ridge is an undersea mountain chain where new ocean floor is produced at a divergent plate boundary.
  • Mid ocean ridge occurs when convection currents rise in the mantle beneath the oceanic crust and create magma where two tectonic plates meet at a divergent boundary.

8 0
3 years ago
The starships of the Solar Federation are marked with the symbol of the Federation, a circle, whereas starships of the Denebian
Llana [10]

Complete question

The complete question is shown on the first uploaded image  

Answer:

The velocity is  v = c* \sqrt{1 -  \frac{1}{n^2} }

Explanation:

From the question we are told that

           a = nb

The length of the minor axis  of  the symbol of the Federation, a circle, seen by the observer at velocity v must be equal to the minor axis(b) of the  Empire's symbol, (an ellipse)

Now this length seen by the observer can be mathematically represented as

        h = t \sqrt{1 - \frac{v^2}{c^2} }

Here t  is the actual length of the major axis of of the  Empire's symbol, (an ellipse)

So t = a = nb

and  b is the length of the minor axis of the symbol of the Federation, (a circle) when seen by an observer at velocity v which from the question must be the length of the minor axis of the of the  Empire's symbol, (an ellipse)

 i.e    h = b

So

    b  =  nb  [\sqrt{1 - \frac{v^2}{c^2} } ]  

     [\frac{1}{n} ]^2 =  1 -  \frac{v^2}{c^2}

      v^2 =c^2 [1- \frac{1}{n^2} ]

       v^2 =c^2 [\frac{n^2 -1}{n^2} ]

        v = c* \sqrt{1 -  \frac{1}{n^2} }

     

     

6 0
3 years ago
A mass of 148 g stretches a spring 13 cm. The mass is set in motion from its equlibrium position with a downward velocity of 10
Charra [1.4K]

Answer:

u(t)=1.15 \sin (8.68t)cm

0.3619sec

Explanation:

Given that

Mass,m=148 g

Length,L=13 cm

Velocity,u'(0)=10 cm/s

We have to find the position u of the mass at any time t

We know that

\omega_0=\sqrt{\frac{g}{L}}\\\\=\sqrt{\frac{980}{13}}\\\\=8.68 rad/s

Where g=980 cm/s^2

u(t)=Acos8.68 t+Bsin 8.68t

u(0)=0

Substitute the value

A=0\\u'(t)=-8.68Asin8.68t+8.68 Bcos8.86 t

Substitute u'(0)=10

8.68B=10

B=\frac{10}{8.68}=1.15

Substitute the values

u(t)=1.15 \sin (8.68t)cm

Period =T = 2π/8.68

After half period

π/8.68 it returns to equilibruim

π/8.68 = 0.3619sec

6 0
3 years ago
A rope of negligible mass passes over a uniform cylindrical pulley of 1.50kg massand 0.090m radius. The bearings of the pulley h
boyakko [2]

Answer:

Explanation:

mass of pulley, m3 = 1.5 kg

Radius of pulley, R = 0.09 m

mass of monkey, m2 = 4.5 kg

mass of banana bunch, m1 = 3 kg

Let a is teh acceleration ans T1 and T2 be the tension in the rope.

The moment of inertia of the pulley

I = 0.5 x m3 x R² = 0.5 x 1.5 x 0.09 x 0.09 = 0.006075 kgm²

According to Newton's second law

T1 - m1 g = m1 x a .... (1)

m2 g - T2 = m2 x a ..... (2)

(T2 - T1 ) x R = I x α    

where, α is the angular acceleration

α = a / R

(T2 - T1)R = 0.5 x m3 x R² x a / R

T2 - T1 = 0.5 x m3 x a ..... (3)  

from (1), (2) and (3)

a = \frac{m_{2}-m_{1}}{m_{1}+m_{2}+\frac{m_{3}}{2}}\times g

a = \frac{4.5-3}{3+4.5+0.75}\times 9.8

a = 1.78 m/s²

from equation (1)

T1 = m1 ( g + a) = 3 ( 9.8 + 1.78) = 34.77 N

from equation (2)

T2 = m2 (g - a) = 4.5 (9.8 - 1.78) = 36.13 N

4 0
3 years ago
Can someone help me asap this is due at 2 50
Rom4ik [11]
The data is inadequate
8 0
3 years ago
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