Tag: evolution and end stages of stars

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Multiple choice physics constellations and galaxies light year evolution and end stages of stars in the world of stars

The average distance between Earth and the Sun is $1.496\times {10}^{8}\ km$ and the speed of light coming from the Sun is $3\times {10}^{8}\ m/s$. How much time will it take for Sun's rays to reach Earth?

  1. $3\ min$
  2. $498.66\ s$
  3. $8\ min$ $30\ s$
  4. $554\ s$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Distance between Earth and Sun$=1.496\times {10}^{8}\ km=1.496\times {10}^{11}\ m$
Speed of light $=3\times {10}^{8}\ m/s$
By using the formula:
$speed=\cfrac{Distance}{Time}$
or $3\times {10}^{8}\ m/s=\cfrac{1.496\times {10}^{11}m}{Time}$
So,  $Time=\cfrac{1.496\times {10}^{11}\ m}{3\times {10}^{8}\ m/s}$ = $\cfrac{1496}{3}s$ $=498.66\ s$ 

Multiple choice physics constellations and galaxies light year evolution and end stages of stars in the world of stars

If light travelling from the Sun at the speed of $3\times {10}^{8}\ m/s$, reach a planet $A$ in $25\ min\  30\ sec$. Then what is the distance between the Sun and the planet? 

(1 light year $=9.461\times {10}^{12}\ km$)

  1. $3$ light minutes
  2. $0.48\times {10}^{-4}$ light year
  3. $1.96\times {10}^{4} $light year
  4. $2.5$ light years
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Speed of light $=3\times {10}^{8}\ m/s$
Time taken $=25\ min\ 30\ sec = 1530\ sec$
By using the formula, 
$Speed=\cfrac{Distance}{Time}$
or 

$Distance=Speed \times Time$ $=3\times {10}^{8}\times 1530$ $=4590\times {10}^{8}\ m$ $=4590\times {10}^{5}\ km$
Distance (in light year) $=\cfrac{4590\times {10}^{5}}{9.461\times {10}^{12}}$ $=0.48\times {10}^{-4}$ light year.

Multiple choice physics constellations and galaxies light year evolution and end stages of stars in the world of stars

If the light from star $A$ takes $15$ min to reach star $B$ and the speed of light is $3\times {10}^{8}m/s$, then what is the distance between the stars?

  1. $2.7\times {10}^{8}km$
  2. $2,9\times {10}^{11}km$
  3. $2.7\times {10}^{8}m$
  4. $36\times {10}^{9}km$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given $t=15$min
Speed of light $=3\times {10}^{8}m/s$
$\therefore$ $t=15min=15\times 60=900 sec$
By using the formula
$Speed=\cfrac{Distance}{Time}$
$Distance=Speed \times time$ $=3\times {10}^{8}m/s\times 900 sec$ $=2700\times {10}^{8}m$ $=2.7\times {10}^{8}km$

Multiple choice physics constellations and galaxies light year evolution and end stages of stars in the world of stars

How does astronomer calculate distance of star?

  1. Comparing apparent brightness of star to true brightness

  2. Comparing true brightness of star to apparent brightness

  3. Comparing true brightness of star to true brightness

  4. None
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The brightness of the stars changes over time. The difference in the brightness (difference in the apparent brightness to the true brightness) over the time allows calculating the distance to the star. This is the cepheid variable stars method that is used to measure the distance to stars beyond 100 light years.

Comparing the apparent brightness of the star to true brightness

Multiple choice physics life cycle of stars evolution and end stages of stars the stars stars

Match the columns

A B
(1) Speed of light (a) 4.3 light years away from earth
(2) Light year                (b) 300,000 km/s
(3) Sun (c) Nearest star
(4) Alpha Centauri (d) distance travelled by light in one year
(e) 18 light minutes away from the Earth

  1. 1-d 2-d, 3-a, 4-e

  2. 1-a, 2-b, 3-c, 4-e

  3. 1-d, 2-b, 3-e, 4-c

  4. 1-b, 2-d, 3-c, 4-a
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The speed of light is 300,000 km/s.
Light year is the distance travelled by light in one year.
Sun is the nearest star from Earth.
Alpha Centauri is 4.3 light years away from Earth.

Multiple choice physics life cycle of stars evolution and end stages of stars the stars stars

The observed wavelength of light coming from a distant galaxy is found to be increased by $0.5\%$ as compared with that coming from a terrestrial source. The galaxy is.

  1. Stationary with respect t to the earth

  2. Approching the earth with velocity of light

  3. Receding from the earth with velocity of light
  4. Receding from the earth with a velocity equal to $1.5\times 10^{6}m/s$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\displaystyle\frac{\Delta \lambda}{\lambda}=\frac{v}{c}$
Now, $\Delta\lambda=\displaystyle\frac{0.5}{100} \lambda$
$\Rightarrow \displaystyle\frac{\Delta\lambda}{\lambda}=\frac{0.5}{100}$
$\therefore v=\displaystyle\frac{0.5}{100}\times c=\frac{0.5}{100}\times 3\times 10^8$
$=1.5\times 10^6m/s$
increase in $\lambda$ indicates that the star is receding.

Multiple choice physics life cycle of stars evolution and end stages of stars the stars stars

$1$ parsec equals to:

  1. $3.26$ light years
  2. $3.1 \times 10^{13}\ km$
  3. $1.92 \times 10^{13}\ miles$

  4. all of the above
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$1$ parsec = $3.26$ light years

$= 3.6 \times 9.46 \times 10^{12}\ km$
$\approx 3.1 \times 10^{13}\ km$
$=\dfrac{3.1 \times 10^{13}}{1.6}\ miles$

             
 $\approx 1.92 \times 10^{13}\ miles$

Multiple choice physics life cycle of stars evolution and end stages of stars the stars stars

The evidence for the rotation of stars comes mainly from the.

  1. Stark effect

  2. Photoelectric effect

  3. Doppler effect

  4. Zeeman effect
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The Doppler effect causes shifts in the spectral lines of a rotating star. As one side of the star rotates toward the observer and the other away, the light is blueshifted and redshifted respectively, causing spectral line broadening.