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icang [17]
3 years ago
14

When distances were carefully measured from the Earth to globular clusters above and below the Milky Way plane (where our view o

f them is not obscured by interstellar dust and gas), their distribution was found to be
A) spherically symmetric about a point in the constellation Sagittarius and concentrated in that direction.
B) in a relatively flat disk almost perpendicular to the plane of the Galaxy, with a relatively higher density of clusters toward its center.
C) concentrated in the plane of the Milky Way and clustered around the Sun's position, indicating that the Sun is close to the Galaxy's center.
D) uniformly distributed throughout space, with no concentration in any area of the Milky Way.
Physics
1 answer:
uranmaximum [27]3 years ago
4 0

Answer:

A) Spherically symmetric about a point in the constellation Sagittarius and concentrated in that direction

Explanation:

The globular clusters are present mainly in the direction of Sagittarius with the center of the system of globular cluster being measured as a spherical cluster cloud such that the center of the Milky Way can be taken as being in the Sagittarius constellation

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A person in a kayak starts paddling, and it accelerates from 0 to 0.61 m/s in a distance of 0.39 m. If the combined mass of the
Iteru [2.4K]

Answer:

35.3 N

Explanation:

U = 0, V = 0.61 m/s, s = 0.39 m

Let a be the acceleration.

Use third equation of motion

V^2 = u^2 + 2 as

0.61 × 0.61 = 0 + 2 × a × 0.39

a = 0.477 m/s^2

Force = mass × acceleration

F = 74 × 0.477 = 35.3 N

6 0
3 years ago
When a burning stick of increase is moved fast in a circle a circle of red light is seen.​
anzhelika [568]

Answer:

The impression of the image on the retina lasts for about 1/16th of a second after the removal of the object. If a burning stick of incense is revolved at a rate of more than sixteen revolutions per second, we see a circle of red light due to persistence of vision.

Explanation:

7 0
3 years ago
a boy is standing 4 meter from a plane mirror how far and in what distance must te move so that he will be 4 meter from his imag
Alona [7]

Answer:

2 meters towards the mirror.

Explanation:

In a plane mirror the image distance is equal to the object distance. Therefore, by moving 2 meters towards the mirror, the boy reduces the distance between him and the mirror to two meters which is the object distance. The image distance is also 2 meters. add the two distances you will get four meters.

6 0
3 years ago
Two point charges, initially 2.0 cm apart, experience a 1.0 N force. If they are moved to a new separation of 0.25 cm, what is t
den301095 [7]

Explanation:

Th electric force between charges is inversely proportional to the square of distance between them. It means,

F\propto \dfrac{1}{r^2}

Initial distance, r₁ = 2 cm

Final distance, r₂ = 0.25 cm

Initial force, F₁ = 1 N    

We need to find the electric force between charges if the new separation of 0.25 cm. So,

\dfrac{F_1}{F_2}=(\dfrac{r_2}{r_1})^2\\\\F_2=\dfrac{F_1r_1^2}{r_2^2}\\\\F_2=\dfrac{1\times 2^2}{(0.25)^2}\\\\F_2=64\ N

So, the new force is 64 N if the separation between charges is 64 N.

7 0
3 years ago
We would like to place an object 45.0cm in front of a lens and have its image appear on a screen 90.0cm behind the lens. What mu
liubo4ka [24]

Answer:

the radii of curvature is 30 cm.

Explanation:

given,

object is place at = 45 cm

image appears at = 90 cm

focal length = ?

refractive index = 1.5

radii of curvature = ?

\dfrac{1}{f} = \dfrac{1}{u} +\dfrac{1}{v}

\dfrac{1}{f} = \dfrac{1}{45} +\dfrac{1}{90}

f = 30 cm

using lens formula

\dfrac{1}{f} = (n-1)(\dfrac{1}{R_1} -\dfrac{1}{R_2})

R_1 = R\ and\ R_2 = -R

\dfrac{1}{f} = (n-1)(\dfrac{1}{R} +\dfrac{1}{R})

R = (n -1)\ f

R = 2(1.5 -1)\ 30

R = 30 cm

hence, the radii of curvature is 30 cm.

3 0
3 years ago
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