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ANEK [815]
2 years ago
7

What local group of galaxies is the Milky Way part of?

Physics
1 answer:
Karo-lina-s [1.5K]2 years ago
5 0

The local group of galaxies which the Milky Way belong to is the Virgo Supercluster. C

<h3>What are Virgo Supercluster?</h3>

The Virgo Supercluster as known as the Local Supercluster is a mass concentration of galaxies that contains the Virgo Cluster and Local Group which in turn contains the Milky Way and Andromeda galaxies, as a host as others galaxies.

Therefore, the local group of galaxies which the Milky Way belong to is the Virgo Supercluster.

Learn more about Milky Way here:

brainly.com/question/1241789

#SPJ1

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The temperature of a gas is increased from 125 celsius inside a rigid container. The original pressure of a gas was 1.22atm, what will the pressure of a gas be after the temperature changes?

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A baseball is hit that just goes over a wall that is 45.4m high. If the baseball is traveling at 46.2 m/s at an angle of 32.7° b
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Answer:

54.9 m/s at 44.9 degrees

Explanation:

If the ball has a total velocity of 46.2 m/s, at an angle of -32.7 degrees, we can decompose its speed into its horizontal and vertical components.

Vx = V * cos(a) = 46.2 * cos(-32.7) = 38.9 m/s

Vy = V * sin(a) = 46.2 * sin(-32.7) = -25 m/s

SInce there is no force on the horizontal direction (omitting air drag), we can assume constant horizontal speed.

Since a ball thrown is at free fall, only affected by gravity (omitting air drag), we can say it is affected by constant acceleration, therefore we can use

Y(t) = Y0 + Vy0 *t + 1/2 * a * t^2

We consider t=0 as the moment when the ball was hit, so in this case Y0 = 1 m

If we take the first derivative of the equation of position, we get the equation for speed

V(t) = Vy0 + a * t

We know that being t2 the moment the ball goes over the wall

V(t2) = -25 m/s

Y(t2) = 45.4 m

So:

45.4 = 1 + Vy0 * t2 + 1/2 * a * t2^2

-25 = Vy0 + a * t2

Then:

Vy0 = -25 - a * t2

So:

45.4 = 1 + (-25 - a * t2) * t2 + 1/2 * a * t2^2

0 = -44.4 - 25 * t2 - 1/2 * a * t2^2

a = -9.81 m/s^2

0 = -44.4 - 25 * t2 + 4.9 * t2^2

Solving this quadratic equation we get:

t1 = -1.39 s

t2 = 6.5 s

Since we are looking for a positive value we disregard t1.

Now we can obtain Vy0:

Vy0 = -25 + 9.81 * 6.5 = 38.76 m/s

Since horizontal speed is constant Vx0 = 38.9 m/s

By Pythagoras theorem we obtain the value of the initial speed:

V0 = \sqrt{Vx0^2 + Vy0^2} = \sqrt{38.9^2 + 38.76^2} = 54.9 m/s

The angle is in the the first quadrant because both comonents ate positive, so: 0 < a < 90

a = atan(Vy0/Vx0) = 44.9 degrees

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