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2 years ago

The range of diameters and luminosities displayed by elliptical galaxies is ____.

1 answer:

8 0

The range of diameters and luminosities displayed by elliptical galaxies is 3000 to 700,000 light years and Mv-23 down to Mv- -16 mag respectively

<h3>What are elliptical galaxies?</h3>

Elliptical galaxies are type of galaxies with a smooth ellipsoidal shape.

They are the most abundant galaxies in the universe.

Their diameter ranges from 3, 000 to 700, 000 light years.

Their luminosity ranges from Mv-23 down to Mv- -16 mag

Therefore, the range of diameters and luminosities displayed by elliptical galaxies is 3000 to 700,000 light years and Mv-23 down to Mv- -16 mag respectively.

Learn more about elliptical galaxies here:

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A person can jump 1.5m on the earth. How high could the person jump on a planet having the twice the mass of the earth and twice

MrMuchimi

F=mg=Gm1m2/r^2
g=Gm2/r^2
g=2Gm2/(2r)^2=2Gm2/4r^2=Gm2/2r^2
So since there is half times the gravity on this unknown planet that has twice earth's mass and twice it's radius, then the person can jump twice as high. 1.5*2= 3m high

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4 years ago

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In 1965, a group of students wore black arm bands to school in protest of American policies in Vietnam. Administrators banned th

Ede4ka [16]

Answer:

protected under students first amendment rights

Explanation:

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3 years ago

A mobile starts with a speed of 250m / s and begins to decelerate at a rate of 3m / s². How fast is it after 45s?

Korvikt [17]

$\large{ \underline{ \underline{ \bf{ \purple{Given}}}}}$

Speed of the mobile = 250 m/s
It starts decelerating at a rate of 3 m/s²
Time travelled = 45s

$\large{ \underline{ \underline{ \bf{ \green{To \: find}}}}}$

Velocity of mobile after 45 seconds

$\large{ \underline{ \underline{ \red{ \bf{Now, \: What \: to \: do?}}}}}$

We can solve the above question using the three equations of motion which are:-

v = u + at
s = ut + 1/2 at²
v² = u² + 2as

So, Here a is acceleration of the body, u is the initial velocity, v is the final velocity, t is the time taken and s is the displacement of the body.

$\large{ \bf{ \underline{ \underline{ \orange{Solution:}}}}}$

We are provided with,

u = 250 m/s
a = -3 m/s²
t = 45 s

By using 1st equation of motion,

⇛ v = u + at

⇛ v = 250 + (-3)45

⇛ v = 250 - 135 m/s

⇛ v = 115 m/s

✤ Final velocity of mobile = 115 m/s

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3 years ago

A block weighing 10 newtons is resting on a plane inclined 30° to the horizontal. What is the magnitude of the normal force acti

aleksley [76]

Magnitude of normal force acting on the block is 7 N

Explanation:

10N = 1.02kg

Mass of the block = m = 1.02 kg

Angle of incline Θ = 30°

Normal force acting on the block = N

From the free body diagram,

N = mgCos Θ

N = (1.02)(9.81)Cos(30)

N = 8.66 N

Rounding off to nearest whole number,

N = 7 N

Magnitude of normal force acting on the block = 7 N

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3 years ago

Given the function f(x) = 8x3 − 3x2 − 5x 8, what part of the function indicates that the left and right ends point in opposite d

ale4655 [162]

Answer: The degree of the first term.

Explanation:

The function:

$f(x) = 8x^3-3x^2-5x^8$

The left and right ends would be indicated when x is changed to -x. When this is substituted, the change is indicated by the first term because only the degree of first term is odd.

Let the left hand side be donated by -x.

Then,

$f(-x) =8(-x)^3-3(-x)^2-5(-x)^8\\ \Rightarrow f(-x)=-8x^3-3x^2-5x^8$

Hence, the correct option is the degree of the first term indicates the left and right end points of the function.

8 0

4 years ago

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