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Montano1993 [528]
3 years ago
6

What is the first step in the formation of a protostar?

Physics
1 answer:
Fittoniya [83]3 years ago
4 0

Star formation begins in relatively small molecular clouds called dense cores.[7] Each dense core is initially in balance between self-gravity, which tends to compress the object, and both gas pressure and magnetic pressure, which tend to inflate it. As the dense core accrues mass from its larger, surrounding cloud, self-gravity begins to overwhelm pressure, and collapse begins. Theoretical modeling of an idealized spherical cloud initially supported only by gas pressure indicates that the collapse process spreads from the inside toward the outside.[8] Spectroscopic observations of dense cores that do not yet contain stars indicate that contraction indeed occurs. So far, however, the predicted outward spread of the collapse region has not been observed.[9]

The gas that collapses toward the center of the dense core first builds up a low-mass protostar, and then a protoplanetary disk orbiting the object. As the collapse continues, an increasing amount of gas impacts the disk rather than the star, a consequence of angular momentum conservation. Exactly how material in the disk spirals inward onto the protostar is not yet understood, despite a great deal of theoretical effort. This problem is illustrative of the larger issue of accretion disk theory, which plays a role in much of astrophysics.

Regardless of the details, the outer surface of a protostar consists at least partially of shocked gas that has fallen from the inner edge of the disk. The surface is thus very different from the relatively quiescent photosphere of a pre-main sequence or main-sequence star. Within its deep interior, the protostar has lower temperature than an ordinary star. At its center, hydrogen is not yet undergoing nuclear fusion. Theory predicts, however, that the hydrogen isotope deuterium is undergoing fusion, creating helium-3. The heat from this fusion reaction tends to inflate the protostar, and thereby helps determine the size of the youngest observed pre-main-sequence stars.[11]

The energy generated from ordinary stars comes from the nuclear fusion occurring at their centers. Protostars also generate energy, but it comes from the radiation liberated at the shocks on its surface and on the surface of its surrounding disk. The radiation thus created most traverse the interstellar dust in the surrounding dense core. The dust absorbs all impinging photons and reradiates them at longer wavelengths. Consequently, a protostar is not detectable at optical wavelengths, and cannot be placed in the Hertzsprung-Russell diagram, unlike the more evolved pre-main-sequence stars.

The actual radiation emanating from a protostar is predicted to be in the infrared and millimeter regimes. Point-like sources of such long-wavelength radiation are commonly seen in regions that are obscured by molecular clouds. It is commonly believed that those conventionally labeled as Class 0 or Class I sources are protostars.[12][13] However, there is still no definitive evidence for this identification.

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On a balanced seesaw, a boy three times as heavy as his partner sits
slega [8]

Answer:

1/3 the distance from the fulcrum

Explanation:

On a balanced seesaw, the torques around the fulcrum calculated on one side and on another side must be equal. This means that:

W_1 d_1 = W_2 d_2

where

W1 is the weight of the boy

d1 is its distance from the fulcrum

W2 is the weight of his partner

d2 is the distance of the partner from the fulcrum

In this problem, we know that the boy is three times as heavy as his partner, so

W_1 = 3 W_2

If we substitute this into the equation, we find:

(3 W_2) d_1 = W_2 d_2

and by simplifying:

3 d_1 = d_2\\d_1 = \frac{1}{3}d_2

which means that the boy sits at 1/3 the distance from the fulcrum.

8 0
3 years ago
For a given wave IF frequency doubles the wavelength
motikmotik

Answer:

C is halved

Explanation:

The frequency and the wavelength of a wave are related by the equation:

v=f\lambda

where

v is the speed of the wave

f is the frequency

\lambda is the wavelength

From the equation above, we see that for a given wave, if the wave is travelling in the same medium (and so, its speed is not changing), then the frequency and the wavelength are inversely proportional to each other.

Therefore, if the frequency doubles, the wavelength will halve in order to keep the speed constant:

\lambda = \frac{v}{f}\\\lambda' = \frac{v}{f'}=\frac{v}{2f}=\frac{1}{2}(\frac{v}{f})=\frac{\lambda}{2}

7 0
3 years ago
Plz help!!!!! Due tomorrow Analyze the weather map in 2 to 3 complete sentences. Describe what the weather will be in Olympia, W
lana66690 [7]

Answer:

Very sorry if this is late

Explanation:

Currently, in Olympia, Washington, it is raining because a cold front with high pressure has met an area with low pressure. This occurs due to the fact that storms are caused by sudden differences in air pressure.

8 0
2 years ago
A rock is projected upward from the surface of the moon, at time t = 0.0 s, w a velocity of 30 m/s. The acceleration due to grav
Vinvika [58]
<h2>Answer: 277.777 m</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told that the rock was<u> projected upward from the surface</u>, we will only use the equations related to the Y axis.

In this sense, the movement equations in the Y axis are:

y-y_{o}=V_{o}.t+\frac{1}{2}g.t^{2}    (1)

V=V_{o}-g.t    (2)

Where:

y  is the rock's final position

y_{o}=0  is the rock's initial position

V_{o}=30\frac{m}{s} is the rock's initial velocity

V is the final velocity

t is the time the parabolic movement lasts

g=1.62\frac{m}{s^{2}}  is the acceleration due to gravity at the surface of the moon

As we know y_{o}=0 , equation (2) is rewritten as:

y=V_{o}.t+\frac{1}{2}g.t^{2}    (3)

On the other hand, the maximum height  is accomplished when V=0:

V=V_{o}-g.t=0    (4)

V_{o}-g.t=0    

V_{o}=g.t    (5)

Finding t:

t=\frac{V_{o}}{g}    (6)

Substituting (6) in (3):

y=V_{o}(\frac{V_{o}}{g})+\frac{1}{2}g(\frac{V_{o}}{g})^{2}    (7)

y_{max}=\frac{{V_{o}}^{2}}{2g}    (8)  Now we can calculate the maximum height of the rock

y_{max}=\frac{{(30m/s)}^{2}}{(2)(1.62m/s^{2})}   (9)

Finally:

y_{max}=277.777m  

4 0
3 years ago
A garage door opener has a power rating of 350 watts. If the door is in operation for 30 seconds, how many joules of energy are
Ksju [112]
Because of the hint we can conclude what equation we need to solve this problem. We have power and duration that means that we need to express energy:

1 joule = 1watt * 1 second
or
E (energy) = P (power) * t (time duration)
E = 350 * 30 = 10500 joules.
7 0
2 years ago
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