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3 years ago

What property of stars can be measured only by using binary-star systems?

Physics

1 answer:

drek231 [11]3 years ago

3 0

Answer:

mass

Explanation:

To expect an understanding of the nature and evolution of the stars, astronomers had to try to determine an important parameter: its mass. This is difficult to determine because neither the luminosity measurements nor the spectral analysis are of any help. The only solution is to resort to astrometry, the precise measurement of stellar positions, and apply it to what is called binary systems, that is, pairs of stars linked by their mutual gravitational attraction and in orbit around each other.

In the solar system there is a law, called the third law of Kepler, that connects the size and period of each planetary orbit and that makes the Sun's mass intervene. This law can be generalized in all bodies in orbit, in particular the members of a binary system. Instead of the mass of the Sun, it is the total mass of the couple that counts. Thus, if it was possible to measure the period and size of a binary system by observation, it would be enough to apply this law in order to calculate the total mass of the pair.

To determine the mass of each star, not just that of the couple, the astronomer had to study in more detail the relative movement of the two members. This allowed him to determine the proportion of each star in the total of the couple and finally obtain the mass of each body.

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An proton-antiproton pair is produced by a 2.20 × 10 3 MeV photon. What is the kinetic energy of the antiproton if the kinetic e

timama [110]

Answer:

K = 80.75 MeV

Explanation:

To calculate the kinetic energy of the antiproton we need to use conservation of energy:

$E_{ph} = E_{p} + E_{ap} = E_{0p} + K_{p} + E_{0ap} + K_{ap} = m_{0p}c^{2} + K_{p} + m_{0ap}c^{2} + K_{ap}$

<em>where $E_{ph}$ : is the photon energy, $E_{0p} and E_{0ap}$ : are the rest energies of the proton and the antiproton, respectively, equals to m₀c², $K_{p} and K_{ap}$ : are the kinetic energies of the proton and the antiproton, respectively, c: speed of light, and m₀: rest mass.</em>

Therefore the kinetic energy of the antiproton is:

$K_{ap} = E_{ph} - m_{0p}c^{2} - K_{p} - m_{0ap}c^{2}$

<u>The proton mass is equal to the antiproton mass, so</u>:

$K_{ap} = E_{ph} - 2m_{0p}c^{2} - K_{p}$

$K_{ap} = 2.20 \cdot 10^{3}MeV - 2(1.67 \cdot 10^{-27}kg)(3\cdot 10^{8} \frac {m}{s})^{2} - 242.85MeV$

$K_{ap} = 2.20 \cdot 10^{3}MeV - 2(1.67 \cdot 10^{-27}kg)(3\cdot 10^{8} \frac {m}{s})^{2}(\frac{1eV}{1.602 \cdot 10^{-19}J})(\frac{1 MeV}{10^{6}eV}) - 242.85MeV$

$K_{ap} = 80.75 MeV$

Hence, the kinetic energy of the antiproton is 80.75 MeV.

I hope it helps you!

3 0

3 years ago

Phenotype describes

vivado [14]

Answer is C...........

5 0

3 years ago

Please someone help me with this!!!

exis [7]

Answer: Gravitational potential energy changes.

Explanation: This is because depending on the amount of mass in an object that’s the amount of gravity pulling you down to the center of the earth

5 0

2 years ago

Which takes place as water cycles from the bottom of the pot toward the top?

Elenna [48]

The option that takes place as water cycles from the bottom of the pot toward the top is that A. thermal energy is transferred.
As the pot gets warmer and warmer, the heat flows everywhere inside the pot, ultimately reaching the top, and heating the water at the top as well. There is no chemical energy here, and molecules don't gain thermal energy, it is just transferred to the top of the pot.

8 0

3 years ago

Read 2 more answers

An asteroid orbiting the Sun has a mass of 4.00×1016 kg. At a particular instant, it experiences a gravitational force of 3.14×1

Ksivusya [100]

<h2>The asteroid is 4.11 x 10¹¹ m far from Sun</h2>

Explanation:

We have gravitational force

$F=\frac{GMm}{r^2}$

Where G = 6.67 x 10⁻¹¹ N m²/kg²

M = Mass of body 1

M = Mass of body 2

r = Distance between them

Here we have

M = Mass of Sun = 1.99×10³⁰ kg

m = Mass of asteroid = 4.00×10¹⁶ kg

F = 3.14×10¹³ N

Substituting

$F=\frac{GMm}{r^2}\\\\3.14\times 10^{13}=\frac{6.67\times 10^{-11}\times 1.99\times 10^{30}\times 4\times 10^{16}}{r^2}\\\\r=4.11\times 10^{11}m$

The asteroid is 4.11 x 10¹¹ m far from Sun

3 0

3 years ago