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larisa [96]
4 years ago
7

Diffinition of nebula

Physics
1 answer:
GrogVix [38]4 years ago
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 <span> [hide] </span><span><span>1Observational history</span><span>2Formation</span><span><span>3Types of nebulae</span><span><span>3.1Classical types</span><span>3.2Diffuse nebulae</span><span><span>3.3Planetary nebulae</span><span>3.3.1Protoplanetary nebula</span></span><span>3.4Supernova remnants</span></span></span><span><span>4Notable named nebulae</span><span>4.1Nebula catalogs</span></span><span>5See also</span><span>6References</span><span>7<span>External links

</span></span></span>

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A meteorologist plans to release a weather balloon from ground level, to be used for high-altitude atmospheric measurements. The
Slav-nsk [51]

Answer:

563.86 N

Explanation:

We know the buoyant force F = weight of air displaced by the balloon.

F = ρgV where ρ = density of air = 1.29 kg/m³, g = acceleration due to gravity = 9.8 m/s² and V = volume of balloon = 4πr/3 (since it is a sphere) where r = radius of balloon = 2.20 m

So, F = ρgV = ρg4πr³/3

substituting the values of the variables into the equation, we have

F =  1.29 kg/m³ × 9.8 m/s² × 4π × (2.20 m)³/3

= 1691.58 N/3

= 563.86 N

8 0
3 years ago
Two 3 m long conductors are separated from each other by 5 mm and carry a current of 10 A dc. Calculate the force that one condu
makkiz [27]

Answer:

0.012 N

Explanation:

The formula to apply is that adopted from the Ampere law which is;

F= μ* I₁*I₂*l /2πd   where

F is force that one conductor exerts on the other.

μ = magnetic permeability of free space = 4π×10⁻⁷ T. m/A

I₁ = current in conductor one=10 A

I₂ = current in conductor two= 10 A

l= length of conductor= 3 m

d= distance between the conductors = 5 mm = 0.005 m

Applying the values in the equation

F= 4π×10⁻⁷ *10*10*3 / 2π*0.005

F= 6 * 10⁻⁵ / 0.005

F=0.012 N

7 0
3 years ago
Which statement describes a property of a proton?
lesantik [10]

Answer:

They have no charge and are present in the nucleus of an atom. They have a negative charge and travel around the nucleus of an atom. They have a positive charge and travel around the nucleus of an atom

Explanation:

7 0
3 years ago
A policeman in a stationary car measures the speed of approaching cars by means of an ultrasonic device that emits a sound with
Ymorist [56]

Answer:

the frequency of the beats is 8.7687 kHz

Explanation:  

Given the data in the question;

The frequency for stationary source and moving observer is;

f' = f( 1 +( v_observer/v_sound))

we know that, speed of sound in dry air =  343 m/s

so we substitute

f' = 41.2 kHz( 1 + (33.0 m/s / 343 m/s) ) = 41.2kHz( 1 + 0.0962) = 41.2kHz(1.0962)

f' = 45.1634 kHz

Now the frequency for stationary observer and moving source with frequency f' will be

f" = f'( 1 / (1 - ( v_observer/v_sound)))

45.1634 kHz( 343 / 343 - 33)

we substitute

f" = 45.1634 kHz( 1 / (1 - (33.0 m/s / 343 m/s)))

f" = 45.1634 kHz( 1 / (1 - 0.0962))

f" = 45.1634 kHz( 1 / 0.9038 )

f" = 45.1634 kHz( 1.1064 )

f" = 49.9687 kHz

Now the beat frequency will be;

f_beat = f' - f

we substitute

f_beat = 49.9687 kHz - 41.2 kHz

f_beat = 8.7687 kHz

Therefore, the frequency of the beats is 8.7687 kHz

7 0
3 years ago
In a real system of levers, wheels, or pulleys, the AMA is less than the IMA because _____.
grandymaker [24]
In a real system of levers, wheel or pulleys, the AMA (actual mechanical advantage) is less than the IMA (ideal mechanical advantage) because of the presence of friction.

In fact, the IMA and the AMA of a machine are defined as the ratio between the output force (the load) and the input force (the effort):
IMA= \frac{F_{out}}{F_{in}}
however, the difference is that the IMA does not take into account the presence of frictions, while the AMA does. As a result, the output force in the AMA is less than the output force in the IMA (because some energy is dissipated due to friction), and the AMA is less than the IMA.
5 0
3 years ago
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