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aivan3 [116]
3 years ago
14

1. Which type of galaxy is the Milky Way? (Points : 1). elliptical. . spiral. . irregular. . satellite. . . 2. In which part of

the Milky Way galaxy is our sun located? (Points : 1). halo. . bulge. . center. . spiral. . . 3. From which of the following do spiral galaxies form? (Points : 1). rapidly cooling protogalactic clouds. . high-density protogalactic clouds. . rapidly spinning protogalactic clouds. . colliding/merging protogalactic clouds
Physics
2 answers:
docker41 [41]3 years ago
8 0

Here are the answers to the following questions:

1. spiral

2. spiral

3. rapidly spinning protogalactic clouds

I am hoping that these answers have satisfied your queries and it will be able to help you in your endeavors, and if you would like, feel free to ask another question.

Musya8 [376]3 years ago
7 0

Answer:

1. Which type of galaxy is the Milky Way?

spiral

2. In which part of the Milky Way galaxy is our sun located?

spiral

3. From which of the following do spiral galaxies form?

rapidly spinning protogalactic clouds

Explanation:

The Milky Way has a spiral shape formed by two arms. In one of them, called the Orion Arm is the solar system that is a planetary system composed of eight planets that revolve around the sun, the only star of the system.

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Answer:

none

Explanation:

5 0
2 years ago
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A toy car on a string is pulled across a table horizontally. The string is at an angle
ddd [48]
Which way is it being pulled?
6 0
3 years ago
A coin with a diameter 3.00 cm rolls up a 30.0 inclined plane. The coin starts out with an initial angular speed of 60.0 rad/s
sweet [91]

This question is in complete.The question is

A coin with a diameter 3.00 cm rolls up a 30.0° inclined plane. The coin starts out with an initial angular speed of 60.0 rad/s and rolls in a straight line without slipping. If the moment of inertia of the coin is(1/2) MR² , how far will the coin roll up the inclined plane (length along the ramp)? Hint: Conservation of mechanical energy.

Answer:

distance=0.124 m

Explanation:

mgh=mglSin\alpha =(1/2)Iw_{i}^{2}+(1/2)mv^{2}\\   v=wR\\Solve for L\\L=((1/2)(1/2)0.015^{2}*60^{2}+(1/2)(60*0.015^{2} ))/9.8Sin30\\   L=0.124m

6 0
3 years ago
1) A record is spinning at the rate of 25 rpm. If a ladybug is sitting 10 cm from the
Sladkaya [172]

<h2>distance = 523 cm</h2>

Explanation:

( a ) The rotational speed of the ladybug = 25 r.p.m = 25/60 r.p.s

= 5/12 rev/sec

( b ) The definition of frequency is the number of rotations per second .

Here the number of rotations per second is 5/12 . Thus frequency = 5/12 Hz

( c ) The tangential speed is v = angular velocity x radius of rotation

The angular velocity ω = 2π x n , where n is the number of rotations per second

Thus angular velocity = 2π x 5/12   = 5π/6 rad/sec

The linear velocity = angular velocity x distance from center of record

Thus tangential speed = 5π/6 x 10 = 25π/3 cm/sec

Angular displacement in 20 sec = ω x t = 5π/6 x 20 = 50π/3 rad

Linear displacement = angular displacement x distance from center of record

= 50π/3 x 10 = 500π/3 = 523 cm

8 0
2 years ago
After flying for 15 min in a wind blowing 42 km/h at an angle of 19° south of east, an airplane pilot is over a town that is 48
masha68 [24]

Answer:

The speed of the airplane relative to the air is 209.47km/hr

Explanation:

Whenever we are solving a physics problem, it's really useful to start by drawing a diagram of the problem (See picture attached). It will help us visualize the problem better.

Now, we know that the plane flew for an amount of time of 15 minutes. For our dimensions to be the same, we need to turn those 15min to hours, like this:

15min*\frac{1hr}{60min}=0.25hr

Once our time is rewritten as hours, we can now calculate the velocity towards north of the plane.

V=\frac{distance}{time}

the plane traveled a distance to the north of 48km so the velocity is:

V=\frac{48km}{0.25hr}

so

V=192km/hr j

Now, we can calculate the x and y-components of the velocity of the wind. The problem states that the wind is blowing at 42km/hr at an angle of 19° south of east, so the x and y-components of the velocity of the wind are:

V_{x}=42km/hr*cos(-19^{o} )=39.71 i

and

V_{y}=42km/hr*sin(-19^{o} )=-13.67 j

So the velocity of the wind can be expressed as a vector as:

V_{wind}=(39.71i - 13.67j)km/hr

Once we know this, we can find the velocity of the plane with respect of the wind on x and on y:

V_{plane x}=V_{plane/wind x}+V_{wind x}

V_{plane/wind x}=V_{plane x}-V_{wind x}

V_{plane/wind x}=(0-39.71 i)km/hr

V_{plane/wind x}= -39.71 i km/hr

and

V_{plane y}=V_{plane/wind y}+V_{wind y}

V_{plane/wind y}=V_{plane y}-V_{wind y}

V_{plane/wind y}=192km/hr j - (- 13.67j)km/hr

V_{plane/wind x}= 205.67 j km/hr

So the velocity of the plane with respect to the wind can be rewritten as:

V_{plane/wind x}= (-39.71i + 205.67 j) km/hr

Since the problem asks us to find the speed of the plane with respect to the wind, this means that we need to find the magnitude of the velocity, since the speed is a scalar defined to be the magnitude of the velocity.

so:

speed=\sqrt{(-39.71)^{2}+(205.67)^{2}  }

speed= 209.47 km/hr

Therefore, the speed of the airplane relative to the air is 209.47km/hr

6 0
3 years ago
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