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4 years ago

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How are galaxies created

1 answer:

marishachu [46]4 years ago

4 0

Hi There! :)

<span>How are galaxies created

</span>Instead of large gas clouds collapsing to form a galaxy<span> in which the gas breaks up into smaller clouds, it is proposed that matter started out in these “smaller” clumps (mass on the order of globular clusters), and then many of these clumps merged to form </span>galaxies<span>, which then were drawn by gravitation to form </span>galaxy<span> ...</span>

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Young _____ stars would most likely be found in a giant molecular cloud.

marysya [2.9K]

The correct answer is letter A. red. Young red stars would most likely be found in a giant molecular cloud. Red stars<span> are cooler than the sun, with surface temperatures of 3,500 K for a bright red star</span>

6 0

3 years ago

Two locomotives approach each other on parallel track. Each has speed of 95 km/h with respect to the ground.If they are initiall

irinina [24]

Answer:

It will take 2.68 minutes for them to reach each other.

Explanation:

We use the two following kinematic equations, making the final position the same (for the moment they meet each other):

locomotive 1 --> $x_f=v*t$

locomotive 2 --> $x_f=8.5\,-\,v*t$

we make the two xf equal, and solve for the time (t) using v = 95 km/h:

$95*t=8.5-95*t\\2*95*t=8.5\\t=8.5/(190)\\t = 0.0447\,\,hours$

converting the hours into minutes by multiplying this value times 60;

t = 2.68 minutes

3 0

3 years ago

In a Millikan oil-drop experiment, the condenser plates are spaced 2.00 cm apart, the potential across the plates is 4000 V, the

astraxan [27]

Answer:

Answer is in the following attachment

Explanation:

3 0

3 years ago

The purpose of this piece of equipment is to

frozen [14]

A. Because that is a digital balance

6 0

4 years ago

Tarzan is running with a horizontal velocity, along level ground. While running, he encounters a 2.21 m vine of negligible mass,

Lena [83]

Answer:

a) $a_c = 1.09m/s^2$

b) T = 720.85N

Explanation:

With a balance of energy from the lowest point to its maximum height:

$m*g*L(1-cos\theta)-1/2*m*V_o^2=0$

Solving for $V_o^2$ :

$V_o^2=2*g*L*(1-cos\theta)$

$V_o^2=2.408$

Centripetal acceleration is:

$a_c = V_o^2/L$

$a_c = 2.408/2.21$

$a_c = 1.09m/s^2$

To calculate the tension of the rope, we make a sum of forces:

$T - m*g = m*a_c$

Solving for T:

$T =m*(g+a_c)$

T = 720.85N

3 0

4 years ago