All categories

3 years ago

Diffinition of nebula

1 answer:

7 0

nebulae, nebulæ, or nebulas) is an interstellar cloud of dust, hydrogen, helium and other ionized gases. Originally, nebula was a name for any diffuse astronomical object, including galaxies beyond the Milky Way. The Andromeda Galaxy, for instance, was once referred to as the Andromeda Nebula (and spiral galaxies in general as "spiral nebulae") before the true nature of galaxies was confirmed in the early 20th century by Vesto Slipher, Edwin Hubble and others.

Most nebulae are of vast size, even hundreds of light years in diameter.[3] Although denser than the space surrounding them, most nebulae are far less dense than any vacuum created on Earth – a nebular cloud the size of the Earth would have a total mass of only a few kilograms. Many nebulae are visible due to their fluorescence caused by the embedded hot stars, while others are so diffuse they can only be detected with long exposures and special filters. Some nebulae, are variably illuminated by T Tauri variable stars. Nebulae are often star-forming regions, such as in the "Pillars of Creation" in the Eagle Nebula. In these regions the formations of gas, dust, and other materials "clump" together to form denser regions, which attract further matter, and eventually will become dense enough to form stars. The remaining material is then believed to form planets and other planetary system objects.

The range of objects called nebula are very diverse, have diverse origins, and final ends.

Contents [hide] 1Observational history2Formation3Types of nebulae3.1Classical types3.2Diffuse nebulae3.3Planetary nebulae3.3.1Protoplanetary nebula3.4Supernova remnants4Notable named nebulae4.1Nebula catalogs5See also6References7External links

You might be interested in

An electric current of 0.25 A passes through a circuit that has a resistance of

Katen [24]

Answer:

The voltage is V = 37.5 [V]

Explanation:

To solve this problem we must use ohm's law which tells us that the voltage is equal to the product of the current by the resistance.

V = I*R

where:

V = voltage [Volt]

I = current = 0.25[amp]

R = resistance = 150 [ohm]

V = 0.25*150 = 37.5 [V]

8 0

3 years ago

Read 2 more answers

Which resource would be the best choice to learn more information about studying martial arts?

zhuklara [117]

Answer:

Dance studio

Explanation:

Martial art use to defend ourself from any dangerous. Dance is a way to learn it

3 0

3 years ago

A published hypothesis:

kow [346]

Answer:

should be tested by the scientific community

4 0

3 years ago

Continuous and aligned fiber-reinforced composite with cross-sectional area of 340 mm2 (0.53 in.2) is subjected to a longitudina

Alecsey [184]

(a) 23.4

The fiber-to-matrix load ratio is given by

$\frac{F_f}{F_m}=\frac{E_f V_f}{E_m V_m}$

where

$E_f = 131 GPa$ is the fiber elasticity module

$E_m = 2.4 GPa$ is the matrix elasticity module

$V_f=0.3$ is the fraction of volume of the fiber

$V_m=0.7$ is the fraction of volume of the matrix

Substituting,

$\frac{F_f}{F_m}=\frac{(131 GPa)(0.3)}{(2.4 GPa)(0.7)}=23.4$ (1)

(b) 44,594 N

The longitudinal load is

F = 46500 N

And it is sum of the loads carried by the fiber phase and the matrix phase:

$F=F_f + F_m$ (2)

We can rewrite (1) as

$F_m = \frac{F_f}{23.4}$

And inserting this into (2):

$F=F_f + \frac{F_f}{23.4}$

Solving the equation, we find the actual load carried by the fiber phase:

$F=F_f (1+\frac{1}{23.4})\\F_f = \frac{F}{1+\frac{1}{23.4}}=\frac{46500 N}{1+\frac{1}{23.4}}=44,594 N$

(c) 1,906 N

Since we know that the longitudinal load is the sum of the loads carried by the fiber phase and the matrix phase:

$F=F_f + F_m$ (2)

Using

F = 46500 N

$F_f = 44594 N$

We can immediately find the actual load carried by the matrix phase:

$F_m = F-F_f = 46,500 N - 44,594 N=1,906 N$

(d) 437 MPa

The cross-sectional area of the fiber phase is

$A_f = A V_f$

where

$A=340 mm^2=340\cdot 10^{-6}m^2$ is the total cross-sectional area

Substituting $V_f=0.3$ , we have

$A_f = (340\cdot 10^{-6} m^2)(0.3)=102\cdot 10^{-6} m^2$

And the magnitude of the stress on the fiber phase is

$\sigma_f = \frac{F_f}{A_f}=\frac{44594 N}{102\cdot 10^{-6} m^2}=4.37\cdot 10^8 Pa = 437 MPa$

(e) 8.0 MPa

The cross-sectional area of the matrix phase is

$A_m = A V_m$

where

$A=340 mm^2=340\cdot 10^{-6}m^2$ is the total cross-sectional area

Substituting $V_m=0.7$ , we have

$A_m = (340\cdot 10^{-6} m^2)(0.7)=238\cdot 10^{-6} m^2$

And the magnitude of the stress on the matrix phase is

$\sigma_m = \frac{F_m}{A_m}=\frac{1906 N}{238\cdot 10^{-6} m^2}=8.0\cdot 10^6 Pa = 8.0 MPa$

(f) $3.34\cdot 10^{-3}$

The longitudinal modulus of elasticity is

$E = E_f V_f + E_m V_m = (131 GPa)(0.3)+(2.4 GPa)(0.7)=41.0 Gpa$

While the total stress experienced by the composite is

$\sigma = \frac{F}{A}=\frac{46500 N}{340\cdot 10^{-6}m^2}=1.37\cdot 10^8 Pa = 0.137 GPa$

So, the strain experienced by the composite is

$\epsilon=\frac{\sigma}{E}=\frac{0.137 GPa}{41.0 GPa}=3.34\cdot 10^{-3}$

3 0

3 years ago

Can tectonic plates switch places with other tectonic plates, Im generally curious

Bingel [31]

Answer:

Earth's crust, called the lithosphere, consists of 15 to 20 moving tectonic plates. The heat from radioactive processes within the planet's interior causes the plates to move, sometimes toward and sometimes away from each other. This movement is called plate motion, or tectonic shift.

beainliest?

4 0

3 years ago

Read 2 more answers