Which statement correctly describes variable stars?

Physics

1 answer:

Kryger [21]2 years ago

5 0

Answer:

Explanation:

The Period-Luminosity relationship tells us that luminosity increases with the period, and of course the more luminosity a star has the more far away they can be seen, so from this we know that:

A) False since lower luminosities can be observed when they are close.

B) False since longer periods means higher luminosities

C) False since lower luminosities can be observed when they are close.

D) True: Variable stars with shorter periods have lower luminosities, so they can only be observed when they are close.

You might be interested in

Which transfer of energy occurs mainly through the process of convection?

Olegator [25]

Heat due to the amount of density difference within a fluid

4 0

2 years ago

We measure the loudness of sound in decibels.<br> a. True<br> b. False

Tatiana [17]

Answer: The correct answer is True.

Explanation:

Loudness of sound is referred to how soft or loud a sound is for the listener.

This term is measured in a unit known as decibels referred to as dB.

This unit is used to measure the relative intensity of sounds on a scale from zero to 100 dB.

More the value of decibels, it will be uncomfortable for a person to hear that sound.

So Yes, the loudness of sound is measured in decibels.

7 0

2 years ago

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. choose the origin to be at the location where the bullet begin

lyudmila [28]

Part a) The work done by the gas on the bullet is the integral of the force in dx, where x is the distance covered by the bullet inside the barrel with respect to the origin:
$W= \int\limits^{0.540m}_{0} {F} \, dx = \int\limits^{0.540m}_{0} {(16000+10000x-26000x^2)} \, dx =$
$=16000x+10000 \frac{x^2}{2} - 26000 \frac{x^3}{3}$
By substituting the length of the barrel, L=0.540 m, we find the total work done by the gas on the bullet:
$W=16000(0.540m)+10000 \frac{(0.540m)^2}{2} - 26000 \frac{(0.540m)^3}{3} =$
$=8733 J=8.73 kJ$

part b) The resolution of the problem is the same, we just have to use the new length of the barrel (L=0.95 m) inside the final formula, and we find the new value of the work:
$W=16000(0.95m)+10000 \frac{(0.95m)^2}{2} - 26000 \frac{(0.95m)^3}{3} =$
$=12280 J=12.28 kJ$