In what state of matter are particles vibrating and moving at high speeds?

Stellar nurseries, such as the orion nebula, contain hundreds or more fragmenting and contracting regions, as well as many proto

Natasha_Volkova [10]

The hydrogen fusion process will begin after the protostar reaches a temperature of 10 million degrees kelvin, and it will then turn into a stable star.

<h3>How does a protostar become a stable star?</h3>

The interstellar medium can sometimes be gathered into a large nebula, which is a cloud of gas and dust. A nebula can span a number of light years. These nebulae are where gas and dust can combine to produce stars. Until a star can combine hydrogen into helium, it cannot be considered a star. They are referred to as protostars before then. As gravity starts to gather the gases into a ball, a protostar is created. Accrution is the term for this procedure.

Gravitational energy starts to heat the gasses as gravity draws them into the ball's core, which causes the gasses to radiate radiation. Radiation initially just dissipates into space. However, much of the radiation is retained inside the protostar as it draws in stuff and becomes denser, which causes the protostar to heat up even more quickly.

The hydrogen fusion process will begin after the protostar reaches a temperature of 10 million degrees kelvin, and it will then turn into a star.

Learn more about a protostar here:

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3 0

11 months ago

Initially, a 2.00-kg mass is whirling at the end of a string (in a circular path of radius 0.750 m) on a horizontal frictionless

drek231 [11]

Answer:

$v_f = 15 \frac{m}{s}$

Explanation:

We can solve this problem using conservation of angular momentum.

The angular momentum $\vec{L}$ is

$\vec{L} = \vec{r} \times \vec{p}$

where $\vec{r}$ is the position and $\vec{p}$ the linear momentum.

We also know that the torque is

$\vec{\tau} = \frac{d\vec{L}}{dt} = \frac{d}{dt} ( \vec{r} \times \vec{p} )$

$\vec{\tau} = \frac{d}{dt} \vec{r} \times \vec{p} + \vec{r} \times \frac{d}{dt} \vec{p}$

$\vec{\tau} = \vec{v} \times \vec{p} + \vec{r} \times \vec{F}$

but, as the linear momentum is $\vec{p} = m \vec{v}$ this means that is parallel to the velocity, and the first term must equal zero

$\vec{v} \times \vec{p}=0$

so

$\vec{\tau} = \vec{r} \times \vec{F}$

But, as the only horizontal force is the tension of the string, the force must be parallel to the vector position measured from the vertical rod, so

$\vec{\tau}_{rod} = 0$

this means, for the angular momentum measure from the rod:

$\frac{d\vec{L}_{rod}}{dt} = 0$

that means :

$\vec{L}_{rod} = constant$

So, the magnitude of initial angular momentum is :

$| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i| cos(\theta)$

but the angle is 90°, so:

$| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i|$

$| \vec{L}_{rod_i} | = r_i * m * v_i$

We know that the distance to the rod is 0.750 m, the mass 2.00 kg and the speed 5 m/s, so:

$| \vec{L}_{rod_i} | = 0.750 \ m \ 2.00 \ kg \ 5 \ \frac{m}{s}$

$| \vec{L}_{rod_i} | = 7.5 \frac{kg m^2}{s}$

For our final angular momentum we have:

$| \vec{L}_{rod_f} | = r_f * m * v_f$

and the radius is 0.250 m and the mass is 2.00 kg

$| \vec{L}_{rod_f} | = 0.250 m * 2.00 kg * v_f$

but, as the angular momentum is constant, this must be equal to the initial angular momentum

$7.5 \frac{kg m^2}{s} = 0.250 m * 2.00 kg * v_f$

$v_f = \frac{7.5 \frac{kg m^2}{s}}{ 0.250 m * 2.00 kg}$

$v_f = 15 \frac{m}{s}$

8 0

2 years ago

Plate Tectonic Theory

Vitek1552 [10]

Answer:

its d

Explanation:

6 0

3 years ago

It is known that heat is added to a gas in sealed container. The container is fitted with a moveable piston.

jasenka [17]

Answer:

d. Not enough information is given to answer this question.

Explanation:

From first law of thermodynamics

Q= W + ΔU

Q=Heat ,W= Work , ΔU=Change in internal energy

If work done by the gas :

It means that W and Q both are positive

Q- W = ΔU

Ii Q > W ,then temperature of the gas will increase.

If Q< W ,Then temperature of the gas will decreases.

If work done on the gas:

Q positive but W will be negative

Q- W = ΔU

Q= W or Q>W or Q< W ,then temperature of the gas will increase.

There are three cases because they did not give any information about the work.That is why option d is correct.

3 0

3 years ago

Find the west component of 45 m 19º S of W

BaLLatris [955]

Answer:

The west component of the given vector is - 42.548 meters.

Explanation:

We need to translate the sentence into a vectoral expression in rectangular form, which is defined as:

$(x, y) = (r_{x}, r_{y})$

Where:

$r_{x}$ - Horizontal component of vector distance, measured in meters.

$r_{y}$ - Vertical component of vector distance, measured in meters.

Let suppose that east and north have positive signs, then we get the following expression:

$(x, y) = (-45\cdot \cos 19^{\circ}, -45\cdot \sin 19^{\circ})\,[m]$

$(x, y) = (-42.548,-14.651)\,[m]$

The west component corresponds to the first component of the ordered pair. That is to say:

$x = -42.548\,m$

The west component of the given vector is - 42.548 meters.

8 0

3 years ago