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ivanzaharov [21]
3 years ago
6

What forces shape the stars in galaxies? (Astronomy question, but didn't fit into any categories) PLEASE 

Physics
1 answer:
MaRussiya [10]3 years ago
4 0

Answer:

Force of gravity

Explanation:

when the force of gravity pulls large gas clouds and dust together, the concentrated gas clouds and dust collapse under the force of gravity forming stars.

There are many galaxies out there in the universe, each galaxy has its own solar systems, stars, and collection of gas and dust. We (earth) belong to the Milky Way galaxy, our galaxy got this name from the Romans. They called in 'via lactea', which directly translates to 'road of milk' because of the milky patch they saw at night.

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Vera_Pavlovna [14]

Answer:

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8 0
3 years ago
Three point charges lie in a straight line along the y-axis. a charge of q1 = -9.10 µc is at y = 6.30 m, and a charge of q2 = -7
inysia [295]

Answer:

 Electric field E = kQ/r^2  

Distance between charges = 6.30 - (-4.40) = 10.70m  

Say the neutral point, P, is a distance d from q1. This means it is a distance (10.70 - d) from q2.  

Field from q1 at P = k(-9.50x^10^-6) / d^2  

Field from q2 at P = k(-8.40x^10^-6) / (10.70-d)^2  

These fields are in opposite directions and are equal magnitudes if the resultant field = 0  

k(-9.50x^10^-6) / d^2 = k(-8.40x^10^-6) / (10.70-d)^2  

9.50 / d^2 =8.40 / (10.70-d)^2  

d^2 / (10.70-d)^2 = 9.50/8.40 = 1.131  

d/(10.70-d) = sqrt(1.1331) = 1.063  

d = 1.063 ((10.70-d)  

= 10.63 - 1.063d  

2.063d = 10.63  

d = 5.15m  

The y coordinate where field is zero is 6.30 - 5.15 = 1.15m

Explanation:

4 0
3 years ago
El peso de un cuerpo en la luna es?
8_murik_8 [283]

Answer:

1/6 del peso en la tierra.

6 0
3 years ago
A student designs an electromagnet, as shown in the picture. The electromagnet is only able to pick up 1 paper clip. List 2 modi
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-- Change the battery to one with higher voltage.

-- Wrap more turns of wire around the spike.

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6 0
3 years ago
if an object 18mm high is placed 12mm from a diverging lens and the image is formed 4mm in front of the lens what is the height
brilliants [131]

To solve this problem we use an amplification formula for divergent lenses

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Where:

i: distance of the image to the lens

o: Distance from the object to the lens

h = height of the object

h '= height of the image

M = \frac{4mm}{12mm} = \frac{h'}{18mm}

h'= 18mm\frac{4mm}{12mm}

h '= 6 mm

The height is 6 mm

5 0
3 years ago
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