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3 years ago

Red giants are smaller than main sequence stars, which are smaller than white dwarfs.

1 answer:

8 0

<span> False. A Red Giant evolves from a main sequence star after it commences Helium fusion and undergoes a great expansion; in the case of our Sun it would expand to encompass Earth's orbit. Thus, it would have approximately the same mass (a small amount being expelled in the helium flash), and vastly greater volume, equating to vastly lower density.

You're welcome :)
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Of all the planets, which is the most different from the others? Explain.

Alenkinab [10]

Answer: VENUS

Explanation:

Venus tiene una lenta rotación retrógrada, lo que significa que gira de este a oeste, en lugar de hacerlo de oeste a este como lo hacen la mayoría de los demás planetas mayores (Urano también tiene una rotación retrógrada, aunque el eje de rotación de Urano, inclinado 97.86°, prácticamente descansa sobre el plano.

5 0

3 years ago

Read 2 more answers

A cooking pot is sitting in a kitchen for several hours unused. Why does it feel cold when you touch it?

masha68 [24]

Answer:

I would go with 2

Explanation:

But i would also not go with my answer. Lol

8 0

3 years ago

Hello,help me with this out please i need it hurry but please ensure your answer is correct..I attach here with my question.

lilavasa [31]

Answer:

.409 N

Explanation:

For this to balance, the moments around the fulcrum must sum to zero.

On the left you have .21 ( is that down? I will assume it is)

Counterclockwise moments :

.21 * 40 + 1.0 * 20

Clockwise moments :

.5 * 20 + F * 45

these moments must equal each other

.21*40 + 1 *20 = .5 * 20 + F * 45

F = .409 N

7 0

2 years ago

A college dorm room measures 14 ft wide by 13 ft long by 6 ft high. What is the air in it under normal conditions?

kirza4 [7]

Complete question:

A college dormitory room measures 14 ft wide by 13 ft long by 6 ft high. Weight density of air is 0.07 lbs/ft3. What is the weight of air in it under normal conditions?

Answer:

the weight of the air is 76.44 lbs

Explanation:

Given;

dimension of the dormitory, = 14 ft by 13 ft by 6 ft

density of the air, = 0.07 lbs/ft³

The volume of the air in the dormitory room = 14 ft x 13 ft x 6 ft

= 1092 ft³

The weight of the air = density x volume

= 0.07 lbs/ft³ x 1092 ft³

= 76.44 lbs

Therefore, the weight of the air is 76.44 lbs

6 0

3 years ago

The charges and coordinates of two charged particles held fixed in the xy plane are: q1 = +3.3 µc, x1 = 3.5 cm, y1 = 0.50 cm, an

Readme [11.4K]

1) First of all, we need to find the distance between the two charges. Their distance on the xy plane is
$d= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$
substituting the coordinates of the two charges, we get
$d= \sqrt{(3.5+2)^2+(0.5-1.5)^2}=5.6~cm=0.056~m$

2) Then, we can calculate the electrostatic force between the two charges $q_1$ and $q_2$ , which is given by
$F=k_e \frac{q_1 q_2}{d^2}$
where $k_e=8.99\cdot10^{9} Nm^2C^{-2}$ is the Coulomb's constant.
Substituting numbers, we get
$F=8.99\cdot10^{9} Nm^2C^{-2} \frac{(3.3\cdot10^{-6}~C) (-4\cdot10^{-6}~C)}{(0.056~m)^2} =-37.8~N$
and the negative sign means the force between the two charges is attractive, because the two charges have opposite sign.

7 0

3 years ago