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uysha [10]
2 years ago
10

Which process is required to allow a gravitationally-collapsing gas cloud to continue to collapse?.

Physics
1 answer:
Annette [7]2 years ago
4 0

The gravitational pull must be greater than the heat pressure's outward push. A gas cloud that is falling builds up thermal energy, which creates thermal pressure that can stop the gravitational collapse.

<h3>What phenomenon prevents an interstellar gas cloud from collapsing gravitationally if it is massive enough to become a star?</h3>

Even when there is no fusion taking on in the star's core, degeneracy pressure can stop a star from contracting gravitationally.

<h3>What mechanism prevents the gravitational collapse of an interstellar gas cloud when it attains a star-like mass?</h3>

For there to be a large gravitational pull, there must be a lot of material present, and for there to be little gas pressure, there must be very low temperatures. Interstellar clouds of gas are typically.

To know more about gravitational visit:-

brainly.com/question/3009841

#SPJ4

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a bubble of air of volume 1cm^3 is released by a deep sea diver at a depth where the pressure is 4.0 atmospheres. assuming its t
lys-0071 [83]

Answer:

hope this helps!

Explanation:

Volume of the air bubble, V1=1.0cm3=1.0×10−6m3

Bubble rises to height, d=40m

Temperature at a depth of 40 m, T1=12oC=285K

Temperature at the surface of the lake, T2=35oC=308K

The pressure on the surface of the lake: P2=1atm=1×1.103×105Pa 

The pressure at the depth of 40 m: P1=1atm+dρg

Where,

ρ is the density of water =103kg/m3

g is the acceleration due to gravity =9.8m/s2

∴P1=1.103×105+40×103×9.8=493300Pa

We have T1P1V1=T2P2V2

Where, V2 is the volume of the air bubble when it reaches the surface.

V2=

8 0
3 years ago
The pilot directs the aircraft to fly due north at 600km/h. A side-wind blows at
loris [4]

Answer:

678.2 km/h and 80.54° north of east

Explanation:

From the question,

Using pythagoras theorem,

a² = b²+c²..................... Equation 1

Where a = resultant velocity

Given: b = 600 km/h, c = 100 km/h

Substitute these values into equation 1

R² = 600²+100²

R² = 360000+10000

R² = 460000

R  = √460000

R = 678.2 km/h.

And the direction is

tanθ = 600/100

tanθ = 6

tanθ = 6

θ = tan⁻¹(6)

θ = 80.54°.

Hence the resultant velocity of the aircraft is 678.2 km/h and 80.54° north of east

3 0
3 years ago
2 equal charges are kept at x=a and x=-a. there can be only 1 point where the electric field will be 0? how, can anyone explain?
svet-max [94.6K]

Answer:

At the midpoint of the line joining the two equal charges

Explanation:

The midpoint of the line joining the two equal charges has equal distance from each charge. Since the charges are at x = a and x = -a, at the midpoint of the line joining the two equal charges, the magnitude of the electric field experienced as a result of each charge is the same. But these two fields are in opposite direction, hence the net resulting electric field would be equal to zero.

6 0
3 years ago
Which of the following is an example in which kinetic energy is converted to potential energy?
Ivan

Answer:

C. A spring is stretched.

Explanation:

Kinetic energy is the energy of motion, potential energy is stored energy. The motion of stretching the spring is kinetic, the energy it has before it's released is potential.

3 0
4 years ago
Suppose you wish to fabricate a uniform wire from a mass m of a metal with density rhom and resistivity rho. If the wire is to h
slamgirl [31]

Answer:

Explanation:

Resistivity and resistance are proportional and depends of the length and the cross-sectional area of the wire:

R=\frac{L}{A}\rho

furthermore, the density is the mass divided by the volume, and the volume can be written as the area multiplyed by the length:

\rho_m=\frac{m}{A*L}

Now you have tw equations and two variables, so you can solve for each of them.

first, solve for A in both equations and replace them:

\frac{L}{R}\rho=\frac{m}{\rho_mL}

L^2=\frac{mR}{\rho_m \rho} \\L=

now replace this into any of the previous equiations:

R=\frac{\sqrt{\frac{mR}{\rho_m \rho} } \rho}{A} \\A=\sqrt{\frac{mR\rho^2}{\rho_m \rho R^2} }\\A=\sqrt{\frac{m\rho}{\rho_m R} }

If you assume the wire has circular cross-sectional area, then the area is:

A=\pi(\frac{d}{2} )^2

solving for d:

d=2\sqrt{\frac{A}{\pi} }

replacing A and simplifying:

d=2 \sqrt[4]{\frac{m\rho}{\rho_m R \pi^2} }

5 0
3 years ago
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