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alexandr402 [8]
3 years ago
15

A star begins as a ____, a large cloud of gas and dust.

Physics
2 answers:
Vlada [557]3 years ago
6 0

A. Nebula . Is The Answer !


serious [3.7K]3 years ago
3 0

Answer: a. nebula

A nebula is a large cloud that is formed by dust and gas present in space. Some of the nebulae may also form by the explosion of a dying star called as supernova. Also, some nebulae may prove as a precursor of new star formation. The gases which are responsible for the formation of nebulae are helium and hydrogen.  The dust and gases which forms the nebula are far but they are brought together as a single mass by the action of gravity. The clumps of gases and dust get bigger and the gravitational force acting on them also increases.

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If different groups of scientist have access to the same data, how can they draw different conclusions?
vazorg [7]

If a group of scientists have access to one data, from the data they can draw conclusions either through mathematics or just thought experiments.

Those thought experiments is different for any scientist, no one thinks the same especially when the topic is difficult.

For example when talking about parallel universes, scientists have come up with the weirdest examples of a multiverse. Some thinking of a brane universe, while others say that its a landscape universe, quilted universe. All of their 'evidence' seems correct but they have opposite meanings.

A weird analogy is 'religion'. All the religions seem to have 'evidences' (hardly) that attract people towards it, they all make sense but that doesn't mean that their evidence is right.

----

Now if they're trying to break down the data using maths, there could be a great uncertainty and measurement error that if done enough could change the whole idea behind the data.

Interesting question, I can babble for days for this but lets keep it as that

6 0
3 years ago
Read 2 more answers
In an extrasolar planetary system containing a single planet, the parent star is measured to move about its center of mass every
Mademuasel [1]

Answer:

Orbital Time Period is 24 years

Explanation:

This can be explained by the definition of time period.

Time period can be defined as the time taken by an object to complete one cycle, here, time taken to complete one revolution.

Also, we know that an extra solar planet which is also called as an exo planet is that planet which is outside our solar system and orbits any star other than our sun. The system in consideration is extra solar system with a single planet.

Therefore, the time taken by the parent star to move about its mass center is the orbital time period that is 24 years.

4 0
3 years ago
Which of the following is NOT useful when trying to overcome an eating disorder? Continuing the same eating patterns Asking for
Mariana [72]

Answer:

continuing the same eating patterns.

Explanation:

since it will only make the suituition go on for longer.

7 0
3 years ago
Read 2 more answers
whats grater than god,more evil than the devil,the poor have it the rich need it and if you eat it you die.what am I
dolphi86 [110]

Answer:

Nothing

Explanation:

The rich need nothing they already have a lot of stuff, the poor have nothing they are poor and do not have much. Nothing is greater than god and nothing is more evil than the devil.

6 0
3 years ago
We can reasonably model a 90-W incandescent lightbulb as a sphere 5.7 cm in diameter. Typically, only about 5% of the energy goe
o-na [289]

Answer:

See answer

Explanation:

Given quantities:

\eta = 0.05\\ W=90[W]\\r=0.0285[m]

where \eta is the efficiency of the lightbulb (visible light is 5% of the total power), W is the total power of the lightbulb, r is the radius of the lightbulb in meters.

Intensity is power divided by area:

I =\frac{P}{A}

a) Now the effective power is \eta*W, therefore:

I =\frac{\eta*W}{\pi r^2}=\frac{0.05*90}{4\pi (0.0285)^2}=440.87[W/m^2]

b) Now the intensity is the average poynting vector is related to the magnitudes of the  maximum electric field and magnetic field amplitudes, following:

S_{average}= \frac{EB}{2\mu_{0}}[W/m]

now E and B are related:

E=cB\\ B=\frac{E}{c} and c=\frac{1}{\sqrt{\epsilon_{0} \mu_{0}}}

replace in S_{average}

S_{average}=I= \frac{c \epsilon_{0}E^2}{2}[W/m]

we replace the values and we get:

E= \sqrt{\frac{2I}{\epsilon_{0}c}}

E = \sqrt{\frac{2(440.8)}{8.85*10^{-12}3*10^8}}=576.24[V/m]

therefore

B=\frac{E}{c}=\frac{576.24}{3*10^{8}}=1.92*10^{-6}[T]

8 0
3 years ago
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