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fiasKO [112]
1 year ago
13

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

stars and stars. What condition would allow a protostar to become a stable star?.
Physics
1 answer:
Natasha_Volkova [10]1 year ago
3 0

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:

brainly.com/question/12534975

#SPJ4

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A manufacturer of CD-ROM drives claims that the player can spin the disc as frequently as 1200 revolutions per minute.
Leya [2.2K]

a. 7.0 m/s

First of all, we need to convert the angular speed (1200 rpm) from rpm to rad/s:

\omega = 1200 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=125.6 rad/s

Now we know that the row is located 5.6 cm from the centre of the disc:

r = 5.6 cm = 0.056 m

So we can find the tangential speed of the row as the product between the angular speed and the distance of the row from the centre of the circle:

v=\omega r = (125.6 rad/s)(0.056 m)=7.0 m/s

b.  875 m/s^2

The acceleration of the row of data (centripetal acceleration) is given by

a=\frac{v^2}{r}

where we have

v = 7.0 m/s is the tangential speed

r = 0.056 m is the distance of the row from the centre of the trajectory

Substituting numbers into the formula, we find

a=\frac{(7.0 m/s)^2}{0.056 m}=875 m/s^2

3 0
3 years ago
A point charge of 4.0 µC is placed at a distance of 0.10 m from a hard rubber rod with an electric field of 1.0 × 103 . What is
yuradex [85]
Since

Electric potential energy = qV

Where V = Ed

Hence
Electric potential energy = q(Ed) --- (1)

Since E = 1.0 * 10^3 N/C
d = 0.10 m
q = 4 * 10^-6 C

Plug in the values in (1)
(1) => Electric potential energy =  4 * 10^-6(1.0 * 10^3 * 0.10)
Electric potential energy = 400 μJ
3 0
4 years ago
A mass m = 75 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius R = 19.2 m and finally
dalvyx [7]

Answer:

The velocity is 19.39 m/s

Solution:

As per the question:

Mass, m = 75 kg

Radius, R = 19.2 m

Now,

When the mass is at the top position in the loop, then the necessary centrifugal force is to keep the mass on the path is provided by the gravitational force acting downwards.

F_{C} = F_{G}

\frac{mv^{2}}{R} = mg

where

v = velocity

g = acceleration due to gravity

v = \sqrt{2gR} = \sqrt{2\times 9.8\times 19.2} = 19.39\ m/s

4 0
3 years ago
An expensive vacuum system can achieve a pressure as low as 1.00×10^{-7} N/m^2 at 20ºC . How many atoms are there in a cubic cen
marta [7]

Answer:

24.70818432141\times 10^7\ atoms

Explanation:

P = Pressure = 1\times 10^{-7}\ N/m^2

V = Volume = 1 cm³

n = Amount of substance

N = Number of atoms

N_A = Avogadro's constant = 6.022\times 10^{23}\ /mol

R = Gas constant = 8.314 J/k mol

T = Temperature = 273.15+20 = 293.15 K

From the ideal gas law

PV=nRT\\\Rightarrow n=\frac{PV}{RT}

n=\frac{N}{N_A}

\frac{N}{N_A}=\frac{PV}{RT}\\\Rightarrow N=N_A\times \frac{PV}{RT}\\\Rightarrow N=\frac{1\times 10^{-7}\times 1\times 10^{-6}}{8.314\times 293.15}\times 6.022\times 10^{23}\\\Rightarrow N=24708184.32141\ atoms=24.70818432141\times 10^7\ atoms

The number of atoms is 24.70818432141\times 10^7\ atoms

8 0
4 years ago
Which of the following criteria can help Linda classify gases as greenhouse gases?
liubo4ka [24]

gas molecules having at least one oxygen atom

7 0
3 years ago
Read 2 more answers
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