All categories

2 years ago

Explain how astronomers might use spectroscopy to determine the composition and temperature of a star.

1 answer:

4 0

Everything starts from spectroscopy. Astronomers only have concentrated information at wavelengths that are emitted from the stars. What they do with this information is to obtain the frequency range of the stars and through spectroscopes they are responsible for dividing the radiation beams and determining the coincidence with the emission of those same waves, of chemical elements. From these observation techniques it is possible to obtain the composition and according to the color, obtaining characteristics such as temperature. The spectrum of stars consists of dark and bright lines called Fraunhofer lines. This spectrum is compared to the spectrum of different elements to find the composition of the stars. This is possible because the elements emit or absorb only specific wavelengths.

You might be interested in

In an experiment to estimate the acceleration due to gravity, a student drops a ball at a distance of 1 m above the floor. His l

anyanavicka [17]

Answer:

$9.45179\ m/s^2$

$s=4.725895t^2$

Explanation:

t = Time taken = 0.46

u = Initial velocity

v = Final velocity

s = Displacement = 1 m

a = Acceleration

$s=ut+\frac{1}{2}at^2\\\Rightarrow 1=0\times 0.46+\frac{1}{2}\times a\times 0.46^2\\\Rightarrow a=\frac{1\times 2}{0.46^2}\\\Rightarrow a=9.45179\ m/s^2$

The acceleration due to gravity is 9.45179 m/s²

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=\frac{1}{2}at^2\\\Rightarrow s=\dfrac{1}{2}9.45179t^2\\\Rightarrow s=4.725895t^2$

The function is $s=4.725895t^2$

3 0

2 years ago

What is the current in a circuit with a 4.5 V battery and a 9 Ω resistor?

horsena [70]

Answer:

V=I×R

4.5 = I×9

 I. = 4.5/9

 I. = 0.5 A

current is 0.5 A

3 0

2 years ago

Read 2 more answers

The maximum distance from the Earth to the Sun (at aphelion) is 1.521 1011 m, and the distance of closest approach (at perihelio

LUCKY_DIMON [66]

Answer:

29274.93096 m/s

$2.73966\times 10^{33}\ J$

$-5.39323\times 10^{33}\ J$

$2.56249\times 10^{33}\ J$

$-5.21594\times 10^{33}$

Explanation:

$r_p$ = Distance at perihelion = $1.471\times 10^{11}\ m$

$r_a$ = Distance at aphelion = $1.521\times 10^{11}\ m$

$v_p$ = Velocity at perihelion = $3.027\times 10^{4}\ m/s$

$v_a$ = Velocity at aphelion

m = Mass of the Earth = 5.98 × 10²⁴ kg

M = Mass of Sun = $1.9889\times 10^{30}\ kg$

Here, the angular momentum is conserved

$L_p=L_a\\\Rightarrow r_pv_p=r_av_a\\\Rightarrow v_a=\frac{r_pv_p}{r_a}\\\Rightarrow v_a=\frac{1.471\times 10^{11}\times 3.027\times 10^{4}}{1.521\times 10^{11}}\\\Rightarrow v_a=29274.93096\ m/s$

Earth's orbital speed at aphelion is 29274.93096 m/s

Kinetic energy is given by

$K=\frac{1}{2}mv_p^2\\\Rightarrow K=\frac{1}{2}\times 5.98\times 10^{24}(3.027\times 10^{4})^2\\\Rightarrow K=2.73966\times 10^{33}\ J$

Kinetic energy at perihelion is $2.73966\times 10^{33}\ J$

Potential energy is given by

$P=-\frac{GMm}{r_p}\\\Rightarrow P=-\frac{6.67\times 10^{-11}\times 1.989\times 10^{30}\times 5.98\times 10^{24}}{1.471\times 10^{11}}\\\Rightarrow P=-5.39323\times 10^{33}$

Potential energy at perihelion is $-5.39323\times 10^{33}\ J$

$K=\frac{1}{2}mv_a^2\\\Rightarrow K=\frac{1}{2}\times 5.98\times 10^{24}(29274.93096)^2\\\Rightarrow K=2.56249\times 10^{33}\ J$

Kinetic energy at aphelion is $2.56249\times 10^{33}\ J$

Potential energy is given by

$P=-\frac{GMm}{r_a}\\\Rightarrow P=-\frac{6.67\times 10^{-11}\times 1.989\times 10^{30}\times 5.98\times 10^{24}}{1.521\times 10^{11}}\\\Rightarrow P=-5.21594\times 10^{33}$