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3 years ago

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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An oscillator consists of a block attached to a spring (k = 427 N/m). At some time t, the position (measured from the system's e

White raven [17]

Answer:

a) 4.49Hz

b) 0.536kg

c) 2.57s

Explanation:

This problem can be solved by using the equation for he position and velocity of an object in a mass-string system:

$x=Acos(\omega t)\\\\v=-\omega Asin(\omega t)\\\\a=-\omega^2Acos(\omega t)$

for some time t you have:

x=0.134m

v=-12.1m/s

a=-107m/s^2

If you divide the first equation and the third equation, you can calculate w:

$\frac{x}{a}=\frac{Acos(\omega t)}{-\omega^2 Acos(\omega t)}\\\\\omega=\sqrt{-\frac{a}{x}}=\sqrt{-\frac{-107m/s^2}{0.134m}}=28.25\frac{rad}{s}$

with this value you can compute the frequency:

$f=\frac{\omega}{2\pi}=\frac{28.25rad/s}{2\pi}=4.49Hz$

the mass of the block is given by the formula:

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\\\\m=\frac{k}{4\pi^2f^2}=\frac{427N/m}{(4\pi^2)(4.49Hz)^2}=0.536kg$

c) to find the amplitude of the motion you need to know the time t. This can computed by dividing the equation for v with the equation for x and taking the arctan:

$\frac{v}{x}=-\omega tan(\omega t)\\\\t=\frac{1}{\omega}arctan(-\frac{v}{x\omega })=\frac{1}{28.25rad/s}arctan(-\frac{-12.1m/s}{(0.134m)(28.25rad/s)})=2.57s$

Finally, the amplitude is:

$x=Acos(\omega t)\\\\A=\frac{0.134m}{cos(28.25rad/s*2.57s )}=0.45m$

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2 years ago

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Nataly_w [17]

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3 years ago

Squid use jet propulsion for rapid escapes. A squid pulls water into its body and then rapidly ejects the water backward to prop

Setler [38]

Answer:

a. FTh = 30 N

b. Fw = 30 N

c. a = 200 m/s2

Explanation:

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2 years ago

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A gas is compressed from an initial volume of 5.75 L to a final volume of 1.23 L by an external pressure of 1.00 atm. During the

nataly862011 [7]

Answer:

<em>The internal energy change is 330.01 J</em>

Explanation:

Given

$v_{1}$ the initial volume = 5.75 L

$v_{2}$ the final volume = 1.23 L

$P_{e}$ is the external pressure = 1.00 atm

q the heat energy removed = -128 J (since is removed from the system)

expansion against a constant external pressure is an example of an irreversible pathway, here pressure in is greater than pressure out and can be obtained thus;

W = - $P_{e}$ ΔV

W = -1.00 x(1.23 - 5.75)

W = -1.00 x -4.52

W = 4.52 L atm

converting to joules we have

W = 4.52 L atm x 101.33 J/ L atm = 458.01 J

The internal energy change during compression can be calculated thus;

ΔU = q + W

ΔU = -128 J + 458.01 J

ΔU = 330.01 J

Therefore the internal energy change is 330.01 J

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2 years ago