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3 years ago

What type of galaxy is M82 based on its appearance in the visible-light view?

Physics

1 answer:

SSSSS [86.1K]3 years ago

4 0

<h2>Answer: irregular</h2>

According to Hubble galaxies are classified into elliptical, spiral and irregular.

It should be noted this classification is based only on the visual appearance of the galaxy, and does not take into account other aspects, such as the rate of star formation or the activity of the galactic nucleus.

The classification is as follows:

1. Elliptical galaxies: Their main characteristic is that the concentration of stars decreases from the nucleus, which is small and very bright, towards its edges. In addition, they contain a large population of old stars, usually little gas and dust, and some newly formed stars.

2. Spiral galaxies: They have the shape of flattened disks containing some old stars and also a large population of young stars, enough gas and dust, and molecular clouds that are the birthplace of the stars.

3. Irregular Galaxies: Galaxies that do not have well-defined structure and symmetry.

In this context, galaxy M82 does not match with the first two types of galaxies, because it has not a defined shape.

Therefore, M82 is an irregular galaxy.

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1) A force of 157 N is applied to a box 25o above the horizontal. If it is applied over a distance of 14.5 m how much work is do

vfiekz [6]

Work done is given by

$W = F.d cos\theta$

here we know that

F = 157 N

d = 14.5

$\theta = 25^0$

now from the above formula

$W = (157)(14.5)cos25 = 2063.2 J$

By energy conservation

Initial total potential energy of egg = final total kinetic energy of egg

$mgh = \frac{1}{2}mv^2$

$v = \sqrt{2gh}$

$v = \sqrt{2(9.8)(8.2)}$

$v = 12.7 m/s$

By momentum conservation we know that

$m_1v_1 = (m_1 + m_2)v$

$215(9.5) = (215 + 19.7) v$

$v = \frac{215 (9.5)}{215 + 19.7}$

$v = 8.7 m/s$

by momentum conservation we know that

$m_1v_1 + m_2v_2 = (m_1 + m_2) v$

let direction of velocity of Jay Z car motion is positive direction

now we will say

$1100(-20) + 1475v = (1100 + 1475)(7)$

$1475 v = 2575(7) + 1100(20)$

$v = 27.14 m/s$

Yes Jay Z is overspeeding as his speed is more than speed limit

Tension force in the string will be centripetal force

so here we can say

$T = \frac{mv^2}{R}$

here we know that

m = 0.63 kg

v = 7.8 m/s

R = 0.23 m

$T = \frac{0.63 (7.8)^2}{0.23}$

$T = 166.6 N$

6 0

3 years ago

A proton moves with a speed of 1.00 x 106 m/s perpendicular to a magnetic field, B. As a result, the proton moves in circle of r

Softa [21]

Answer:

0.0109 m ≈ 10.9 mm

Explanation:

proton speed = 1 * 10^6 m/s

radius in which the proton moves = 20 m

<u>determine the radius of the circle in which an electron would move </u>

we will apply the formula for calculating the centripetal force for both proton and electron ( Lorentz force formula)

For proton :

Mp*V^2 / rp = qp *VB ∴ rp = Mp*V / qP*B ---------- ( 1 )

For electron:

re = Me*V/ qE * B -------- ( 2 )

Next: take the ratio of equations 1 and 2

re / rp = Me / Mp ( note: qE = qP = 1.6 * 10^-19 C )

∴ re ( radius of the electron orbit )

= ( Me / Mp ) rp

= ( 9.1 * 10^-31 / 1.67 * 10^-27 ) 20

= ( 5.45 * 10^-4 ) * 20

= 0.0109 m ≈ 10.9 mm

5 0

3 years ago

28. Ken and Musa shared a cake such that Ken got twice the size

ratelena [41]

Answer:

$Musa = \frac{1}{3}$

$Ken = \frac{2}{3}$

Explanation:

Given

$Ken = 2 * Musa$ --- Ken's share

Required

The fraction each got

Since they both shared a cake, we have:

$Ken + Musa = 1$

Substitute: $Ken = 2 * Musa$

$2 * Musa+ Musa = 1$

Factorize

$Musa(2+ 1)= 1$

$Musa(3)= 1$

Divide both sides by 3

$Musa = \frac{1}{3}$

Recall that: $Ken = 2 * Musa$

$Ken = 2 * \frac{1}{3}$

$Ken = \frac{2}{3}$

3 0

3 years ago

In 2000, NASA placed a satellite in orbit around an asteroid. Consider a spherical asteroid with a mass of 5.00�1015kg and a rad

irakobra [83]

Answer:

a). $v_{1}=13.49 \frac{m}{s}$

b). $v_{2}=17.54\frac{m}{s}$

Explanation:

Using the conservation of energy and the kinetic energy of the satellite around the asteroid can model the motion in

$\frac{1}{2}*m*v^2=\frac{G*M*m}{a_{o}}$

$v^2=\frac{2*G*M*m}{a_{o}}$

$G=6.673x10^{-11}\frac{m^3}{kg*s^2}$

$M=1.20x10^{16}kg$

a).

$a_{o}=8.8km*\frac{1000m}{1km}=8800m$

$v=\sqrt{\frac{2*6.673x10^{-11}\frac{m^3}{kg*s^2}*1.2x10^{10}kg}{8800m}}$

$v_{1}=13.49 \frac{m}{s}$

b).

$a_{f}=5.2km*\frac{1000m}{1km}=5200m$

$v=\sqrt{\frac{2*6.673x10^{-11}\frac{m^3}{kg*s^2}*1.2x10^{10}kg}{5200m}}$

$v_{2}=17.54\frac{m}{s}$

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3 years ago

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Another element is silvery-white with a shiny luster, is very brittle, and forms ions with a −2 charge. give one possible identi

Lesechka [4]

The element is magnesium ( Mg )

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3 years ago

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