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3 years ago

I THINK I KNOW THE ANSWER BUT I NEED VERIFICATION!!!!!!!

2 answers:

4 0

Yes, the answer is in fact C. All of the above.

Variable stars are stars that change in brightness. The change could be due to the physical change in the star or it could also be caused when another star crosses another, or in other words eclipse.

Umnica [9.8K]3 years ago

4 0

The answer is All of the above. Variable Stars include White Dwarf Stars, Red Giant Stars & Binary Stars. A white dwarf is what stars like the Sun become after they have exhausted their nuclear fuel. A red giant star is a dying star in the last stages of stellar evolution. Binary stars are two stars orbiting a common center of mass. The brighter star is officially classified as the primary star, while the dimmer of the two is the secondary .

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1) research and provide a description of the "talk test" and "breath test" as it

insens350 [35]

In the test, it is observed that water that is not too deep If carbon dioxide levels are not low enough, blackout causes elevation.

<h3>What is research?</h3>

Unique and methodical activity performed to improve the reservoir of knowledge is known as the research. . A research project might be a continuation of previous work in the topic.

By the following way, the test is conducted;

1. The effect of the conversation test on exercise intensity prescription and monitoring in competitive athletes. The sound of your breath might assist you figure out what's wrong.

Athletes require oxygen to give energy and to eliminate carbon dioxide, which is a waste product produced when energy is produced.Because it is a reliable indicator of ventilatory efficiency in cardiac patients.

Ventilatory threshold performance is used to determine the lactate threshold. The maximal rate of oxygen consumption recorded during activity is 5.VO2 max. It is higher in trained people than in unskilled people.

Hence,in the test it is observed that water that is not too deep If carbon dioxide levels are not low enough, blackout causes elevation.

To learn more about the research, refer to the link;

brainly.com/question/18723483

#SPJ1

3 0

1 year ago

a train car of mass 444 kg moving at 5 m/s bounces into another car on the same tracks of mass 344 king. of the second car was m

Kamila [148]

Answer:

3.7 m/s

Explanation:

M = 444 kg

U = 5 m/s

m = 344 kg

u = - 5 m/s

Let the velocity of train is V and the car s v after the collision.

As the collision is elastic

By use of conservation of momentum

MU + mu = MV + mv

444 x 5 - 344 x 5 = 444 V + 344 v

500 = 444 V + 344 v

125 = 111 V + 86 v .... (1)

By using the formula of coefficient of restitution ( e = 1 for elastic collision)

$e = \frac{V-v}{u-U}$

-5 - 5 = V - v

V - v = - 10

v = V + 10

Substitute the value of v in equation (1)

125 = 111 V + 86 (V + 10)

125 = 197 V + 860

197 V = - 735

V = - 3.7 m/s

Thus, the speed of first car after collision is 3.7 m/s. negative sign shows that the direction is reverse as before the collision.

4 0

2 years ago

What is the magnitude of the electric force of attraction between two point charges + 2.8 mc and -1.2 mc if the distance

Anit [1.1K]

Answer:

Explanation:

it just a

8 0

2 years ago

A person starts at a position of 16 meters and finishes at a position of

DochEvi [55]

Answer:

38 is a good girl and a great place to work for u and I miss you y and I miss you much love your love and love to

7 0

2 years ago

You pull with a force of 295 N on a rope that is attached to a block of mass 22 kg, and the block slides across the floor at a c

Sergeeva-Olga [200]

Answer:

Fnet = 0

Explanation:

Since the block slides across the floor at constant speed, this means that it's not accelerated.
According Newton's 2nd Law, if the acceleration is zero, the net force on the sliding mass must be zero.
This means that there must be a friction force opposing to the horizontal component of the applied force, equal in magnitude to it:

$F_{appx} = F_{app} * cos \theta = 295 N * cos 35 = 242 N (1)$

In the vertical direction, the block is not accelerated either, so the sum of the normal force and the vertical component of the applied force, must be equal in magnitude to the force of gravity on the block:

$F_{appy} = F_{app} * cos \theta = 295 N * sin 35 = 169 N (2)$

⇒ 169 N + Fn = Fg = 216 N (3)

This means that there must be a normal force equal to the difference between Fappy and Fg, as follows:
Fn = 216 N - 169 N = 47 N (4)

6 0

2 years ago