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Ber [7]
3 years ago
6

The Andromeda Galaxy is faintly visible to the naked eye in the constellation Andromeda. Suppose instead it were located in the

same direction in space as the center of the Milky Way Galaxy (but still at its current distance). How would it appear to the eye in that case?
a) We could not see it at all.
b) It would look about the same, but would be in the constellation Sagittarius instead of Andromeda.
c) It would be much brighter, because it would be illuminated by the many stars in the center of our galaxy.
d) It would look about the same, but it would be harder to pick out because its cloud-like appearance would make it blend in with the cloud-like appearance of the Milky Way in our sky.
Physics
1 answer:
Kaylis [27]3 years ago
8 0

Answer:

a) We could not see it at all.

Explanation:

The most distant object that can be seen is the andromeda galaxy, which we may have a slight view of. The andromeda galaxy is a large galaxy that along with the previous two is also part of the local group. Spiral-type galaxy that is approximately 250,000 light years in diameter (more than twice the diameter of the Milky Way!) And is about 2.9 million light years away from our galaxy. Because of its distance, we have difficulty visualizing this galaxy, we would have this difficulty even if the andromeda galaxy was in the center of the Milky Way, but maintaining its current distance. That is, even if the andromeda galaxy were located in the same direction in space as the center of the Milky Way (but still at its current distance), we could not see it at all.

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stich3 [128]

Answer:

500 Newton’s 3 meters high

Explanation:

3 0
2 years ago
In this problem, you will answer several questions that will help you better understand the moment of inertia, its properties, a
scoundrel [369]

Answer:

a)  Total mass form, density and axis of rotation location are  True

b)   I = m r²

Explanation:

a) The moment of inertia is the inertia of the rotational movement is defined as

       I = ∫ r² dm

Where r is the distance from the pivot point and m the difference in body mass

In general, mass is expressed through density

        ρ = m / V

        dm = ρ dV

From these two equations we can see that the moment of inertia depends on mass, density and distance

Let's examine the statements, the moment of inertia depends on

- Linear speed       False

- Acceleration angular False

-  Total mass form True

-  density True

- axis of rotation location   True

b) we calculate the moment of inertia of a particle

For a particle the mass is at a point whereby the integral is immediate, where the moment of inertia is

          I = m r²

4 0
3 years ago
A plane comes in for a landing at a velocity of 80 meters per second west. As it touches down, it decelerates at a constant rate
azamat

Answer:

Answer D : about 1067 meters

Explanation:

There are two steps to this problem:

1) First find the time it takes the plane to stop using the equation for the acceleration:

a=\frac{Vf-Vi}{t}

Where Vf is the final velocity of the plane (in our case: zero )

Vi is the initial velocity of the plane (in our case: 80 m/s)

a is the acceleration (in our case -3 m/s^2 - notice negative value because the velocity is decreasing)

a=\frac{Vf-Vi}{t}\\-3=\frac{0-80}{t}\\t=\frac{-80}{-3} = \frac{80}{3}

with units corresponding to seconds given the quantities involved in the calculation.

2) Second knowing the time it took the plane to stop, now use that time in the equation for the distance traveled under accelerated motion:

Xf-Xi=Vi*t+\frac{1}{2} a t^{2} \\Xf-Xi= 80 (\frac{80}{3}) +\frac{1}{2} (-3) (\frac{80}{3}) ^{2}=1066.666666...

Where the answer results in units of meters given the quantities used in the calculation.

We round this to 1067 meters

7 0
3 years ago
The masses of the Moon and Earth are 7.35 x 1022 kg and 5.97 x 1024 kg, respectively. The strength of the gravitational force be
bixtya [17]

Answer:

Distance between centre of Earth and centre of Moon is 3.85 x 10⁸ m

Explanation:

The attractive force experienced by two mass objects is known as Gravitational force.

The gravitational force is determine by the relation:

F=\frac{Gm_{1} m_{2} }{d^{2} }      ....(1)

According to the problem,

Mass of Moon, m₁ = 7.35 x 10²² kg

Mass of Earth, m₂ = 5.97 x 10²⁴ kg

Gravitational force experienced by them, F = 1.98 x 10²⁰ N

Universal gravitational constant, G = 6.67 x 10⁻¹¹ Nm²kg⁻²

Substitute these values in equation (1).

1.98\times10^{20} =\frac{6.67\times10^{-11}\times7.35\times10^{22}\times5.97\times10^{24} }{d^{2} }

d^{2}=\frac{2.93\times10^{37}}{1.98\times10^{20}}

d=\sqrt{1.48\times10^{17}}

d = 3.85 x 10⁸ m

3 0
3 years ago
At which temperature does the motion of atoms and molecules stop?
Scorpion4ik [409]

Answer:

0 Kelvin

Explanation:

Atoms in absolute temperature get approximatelly motionless since 0 Kelvin is -273 degrees Celcius. The kinetic energy of atoms/particles in matter has the possible lowest value ( almost zero), so that there is nothing colder than 0 Kelvin.

4 0
3 years ago
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