Ask question

Ask question

All categories

3 years ago

7

What type of galaxy has mostly old stars, without a lot of star formation?

2 answers:

erma4kov [3.2K]3 years ago

6 0

Elliptical galaxies, such as M87 (left), have very little gas and dust. Because gas and dust are found in the clouds that are the birthplaces of stars, we should expect to see very few young stars in elliptical galaxies. In fact, elliptical galaxies contain primarily old, red stars (also known as Population II stars).

Elliptical galaxies vary widely in size. Both the largest and the smallest known galaxies are elliptical. Very large elliptical galaxies can reach 300 million light years in diameter. Dwarf ellipticals, which are very common, may contain only 1/100,000th as many stars as the Milky Way!

Anettt [7]3 years ago

4 0

Spiral Galaxies, Irregular Galaxies and Elliptical Galaxies?

You might be interested in

An object of mass 3.00 kg, moving with an initial velocity of 5.05 m/s, collides with and sticks to an object of mass 2.76 kg wi

Lynna [10]

Answer:

0.752 m/s

Explanation:

m1 = 3.00kg

u1 = 5.05m/s

m2 = 2.76kg

u2 = -3.66m/s

According to the law of conservation of momentum,

m1u1 + m2u2 = (m1+m2)v

3(5.05) + 2.76(-3.66) = (5.05+2.76)v

15.15 - 9.2736 = 7.81v

5.8764 = 7.81v

v = 5.8764/7.81

v = 0.752m/s

6 0

2 years ago

75 kg + 1352 g = ______________g?

JulsSmile [24]

Answer is 76,352 just look it up

8 0

2 years ago

Read 2 more answers

Help I need a answer to this

natta225 [31]

Answer:

1700 Joules

Explanation:

Work=force x distance

Force = 170 kg

Distance= 10 Meters

170 x 10 = 1700 Joules of work

3 0

2 years ago

Can someone help asap with this

kondaur [170]

Answer:

B

Explanation:

i had a test on this and got it correct

7 0

2 years ago

Normal atmospheric pressure is 1.013 105 Pa. The approach of a storm causes the height of a mercury barometer to drop by 27.1 mm

Burka [1]

Answer:

The atmospheric pressure is $0.97622\times10^{5}\ Pa$ .

Explanation:

Given that,

Atmospheric pressure $P_{atm}= 1.013\times10^{5}\ Pa$

drop height h'= 27.1 mm

Density of mercury $\rho= 13.59 g/cm^3$

We need to calculate the height

Using formula of pressure

$p = \rho g h$

Put the value into the formula

$1.013\times10^{5}=13.59\times10^{3}\times9.8\times h$

$h =\dfrac{1.013\times10^{5}}{13.59\times10^{3}\times9.8}$

$h=0.76\ m$

We need to calculate the new height

$h''=h - h'$

$h''=0.76-27.1\times10^{-3}$

$h''=0.76-0.027$

$h''=0.733\ m$

We need to calculate the atmospheric pressure

Using formula of atmospheric pressure

$P=\rho g h$

Put the value into the formula

$P= 13.59\times10^{3}\times9.8\times0.733$

$P=0.97622\times10^{5}\ Pa$

Hence, The atmospheric pressure is $0.97622\times10^{5}\ Pa$ .

7 0

2 years ago