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3 years ago

Which of the following describes a major role of gravity in star formation?

Physics

2 answers:

artcher [175]3 years ago

8 0

Answer:

B. Gravity compresses gases at the center of a solar nebula until temperatures are high enough for nuclear fusion to occur.

Explanation:

Stars are born in stellar nurseries known as nebula. Sun, the star at the center of the solar system was born in the solar nebula.

Nebula is a molecular cloud of dust and gases. Due to gravity, this cloud collapses under its own weight and condenses to form the core. It gathers more mass till the temperature of the core rises enough to start the nuclear fusion reaction. A star is born then.

MrRissso [65]3 years ago

7 0

The ans is B) because gravity attracts and doesn't repel

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RUDIKE [14]

Answer:

I think deposition of C is the oldest deposit

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3 years ago

What is the compound oc2 called?

MAXImum [283]

The answer to your question is dioxygen carbide

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3 years ago

For a damped simple harmonic oscillator, the block has a mass of 1.2 kg and the spring constant is 9.8 N/m. The damping force is

ArbitrLikvidat [17]

Answer:

a) t=24s

b) number of oscillations= 11

Explanation:

In case of a damped simple harmonic oscillator the equation of motion is

m(d²x/dt²)+b(dx/dt)+kx=0

Therefore on solving the above differential equation we get,

x(t)=A₀ $e^{\frac{-bt}{2m}}cos(w't+\phi)=A(t)cos(w't+\phi)$

where A(t)=A₀ $e^{\frac{-bt}{2m}}$

A₀ is the amplitude at t=0 and

$w'$ is the angular frequency of damped SHM, which is given by,

$w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }$

Now coming to the problem,

Given: m=1.2 kg

k=9.8 N/m

b=210 g/s= 0.21 kg/s

A₀=13 cm

a) A(t)=A₀/8

⇒A₀ $e^{\frac{-bt}{2m}}$ =A₀/8

⇒ $e^{\frac{bt}{2m}}=8$

applying logarithm on both sides

⇒ $\frac{bt}{2m}=ln(8)$

⇒ $t=\frac{2m*ln(8)}{b}$

substituting the values

$t=\frac{2*1.2*ln(8)}{0.21}=24s(approx)$

b) $w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }$

$w'=\sqrt{\frac{9.8}{1.2}-\frac{0.21^{2}}{4*1.2^{2}}}=2.86s^{-1}$

$T'=\frac{2\pi}{w'}$ , where $T'$ is time period of damped SHM

⇒ $T'=\frac{2\pi}{2.86}=2.2s$

let $n$ be number of oscillations made

then, $nT'=t$

⇒ $n=\frac{24}{2.2}=11(approx)$

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3 years ago

If Anya decides to make the star twice as massive, and not change the length of any crossbar or the location of any object, what

charle [14.2K]

Answer:

She will make the mass of the smiley face twice as massive in order to keep the mobile in perfect balance.

Explanation:

mass of an object is directly proportional to the cube of its length. In this case the length is constant, the mass will also be constant for the smiley face, so that the mobile will be kept in perfect balance.

Therefore, If Anya decides to make the star twice as massive, and not change the length of any crossbar or the location of any object, she will make the mass of the smiley face twice as massive in order to keep the mobile in perfect balance.

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3 years ago

Satellites remain in orbit around earth because

svetoff [14.1K]

Earth's gravity and the satellite's velocity keeps it so that it stays in orbit. (there is a more complicated side, too...)

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3 years ago