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Alona [7]
4 years ago
13

The sparse area surrounding a spiral galaxy is called a

Physics
2 answers:
djyliett [7]4 years ago
3 0
<span>The sparse area surrounding a spiral galaxy is called a

Halo

</span>
Anit [1.1K]4 years ago
3 0

Answer: halo

Explanation: The sparse area surrounding a spiral galaxy is called a <em>halo</em>. The galactic halo extends beyond the main component. The galactic halo is composed by different elements of a galaxy. The stellar halo, the galactic corona and the dark matter halo.

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One clue we can observe that means a chemical reaction has occurred is the formation of light. What is one chemical reaction you
dimaraw [331]

Answer:

There are five signs of a chemical change:

Color Change.

Production of an odor.

Change of Temperature.

Evolution of a gas (formation of bubbles)

Precipitate (formation of a solid)

Explanation:

I just went ahead and gave you the five signs of chemical change hoped it helped

8 0
3 years ago
Which statement is incorrect when considering gases? 1. Gas molecules are in constant, random motion at room temperature 2. The
vova2212 [387]

Answer:

4

Explanation:

For gases :

1. The motion of gases molecule is in random manner at the room temperature.

2.The distance between the gas molecule is more and that is why gas can be compress.

3.The attraction force between the gas molecule is negligible or we can say that there is no any force between the gas molecules.that is why gas can be filled in the container.But the motion of the gas molecule does not stop they are still moving inside the container but the space for movement become less.When a gas container heated then the container start to vibrate because the movement of the gas molecule.

So the option 4 is incorrect.

6 0
3 years ago
Scientists use laser range-finding to measure the distance to the moon with great accuracy. A brief laser pulse is fired at the
Fofino [41]

Answer:

d = 2,042 10-3 m

Explanation:

The laser diffracts in the circular slit, so the process equation is

      d sin θ= m λ

The first diffraction minimum occurs for m = 1

We can use trigonometry in the mirror

        tan θ = Y / L

Where L is the distance from the Moon to Earth

Since the angle is extremely small

           tan θ = sin θ / cos θ

           Cos θ = 1

           tant θ = sin θ = y / L

We replace

           d y / L = λ

           d = λ L / y

Let's calculate

           d = 532 10⁻⁹ 3.84 10⁶/1 10³

           d = 2,042 10-3 m

5 0
3 years ago
A jet transport has a weight of 2.25 x 106 N and is at rest on the runway. The two rear wheels are 16.0 m behind the front wheel
Rudik [331]

Answer:

Explanation:

Given that,

Weight of jet

W = 2.25 × 10^6 N

It is at rest on the run way.

Two rear wheels are 16m behind the front wheel

Center of gravity of plane 10.6m behind the front wheel

A. Normal force entered on the ground by front wheel.

Taking moment about the the about the real wheel.

Check attachment for better understanding

So,

Clock wise moment = anti-clockwise moment

W × 5.4 = N × 16

2.25 × 10^6 × 5.4 = 16•N

N = 2.25 × 10^6 × 5.4 / 16

N = 7.594 × 10^5 N

B. Normal force on each of the rear two wheels.

Using the second principle of equilibrium body.

Let the rear wheel normal be Nr and note, the are two real wheels, then, there will be two normal forces

ΣFy = 0

Nr + Nr + N — W = 0

2•Nr = W—N

2•Nr = 2.25 × 10^6 — 7.594 × 10^5

2•Nr = 1.491 × 10^6

Nr = 1.491 × 10^6 / 2

Nr = 7.453 × 10^5 N

6 0
4 years ago
Consider a turnbuckle that has been tightened until the tension in wire AD is 350 N. Draw the FBD that is required to determine
Mars2501 [29]

Answer:

yes

Explanation:

yes

6 0
3 years ago
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