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Alona [7]
3 years ago
13

The sparse area surrounding a spiral galaxy is called a

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

Halo

</span>
Anit [1.1K]3 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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What's the difference between a direct relationship and a positive one?
borishaifa [10]
A positive or direct relationship is one in which the two variables (we will generally call them x and y) move together, that is, they either increase or decrease together. In a negative or indirect relationship, the two variables move in opposite directions, that is, as one increases, the other descremases
5 0
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What did copernicus say about the motion of the sun?.
eimsori [14]

the sun was stationary in the center of the universe and the earth revolved around it....

8 0
2 years ago
What is perfect machine?Why is this machine not possible in the real life?​
andrezito [222]

Explanation:

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5 0
3 years ago
The specification limits for a product are 8 cm and 10 cm. A process that produces the product has a mean of 9.5 cm and a standa
My name is Ann [436]

Answer:

The value of Cpk is 0.83.

Explanation:

Given that,

Upper specification limits = 10 cm

lower specification limits = 8 cm

Mean = 9.5

Standard deviation = 0.2 cm

We need to calculate the process capability

Using formula of Cpk

Cpk=min(\dfrac{USL-mean}{3\times SD}, \dfrac{mean-LSL}{3\times SD})

Put the value into the formula

Cpk=min(\dfrac{10-9.5}{3\times0.2}, \dfrac{9.5-8}{3\times0.2})

Cpk=min(0.83,2.5)

Cpk=0.83

Hence, The value of Cpk is 0.83.

4 0
3 years ago
Jack drops a stone from rest off of the top of a bridge that is 24.4 m above the ground. After the stone falls 6.6 m, Jill throw
alukav5142 [94]
-17.555m/s

first I found the time it took for jacks stone to reach the bottom, using the formula vf = vi + at, vf and vi are final and initial velocities.

then i found the velocity at 6.6m using vf^2 = vi^2 + 2ad
and I found the time it took to get to 6.6m, so that I knew how long Jill waited to throw her stone, I used the formula d = t(vi+vf)/2, then i done total time - the time she waited, to get the time it took for there stones to hit the ground at the same time.

then to find the initial velocity of her throw I used the formula d = vit + (at^2)/2
4 0
3 years ago
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