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dexar [7]
3 years ago
12

If Jupiter were 10 times more massive, it would generate nuclear fusion in its core and be a star instead of a planet. If Jupite

r were 10 times more massive, it would generate nuclear fusion in its core and be a star instead of a planet.
A. True
B. False
Physics
2 answers:
Delicious77 [7]3 years ago
8 0

Answer:

See explanation

Explanation:

Jupiter’s diameter is in fact larger than that of the smallest star, at 140,000 kilometres against 121,000 km for the tiniest star.

However it is mass, not size, that counts. This determines the internal pressure that, if sufficiently high, can overcome the mutual repulsion of hydrogen nuclei and convert these to helium through nuclear fusion. This releases the huge amount of energy that makes stars shine.

If a large cloud of interstellar gas came Jupiter’s way, maybe the planet could gain enough extra mass to start fusion. Fusion would be short lived if it became a brown dwarf, an object midway between star and planet. If it accreted even more mass, just enough to become a true star, it would be a dim red dwarf. Its radiation would barely affect us and it wouldn’t look very different to now. A bigger worry would be Jupiter’s increased mass disrupting the solar system, not to mention the raised temperature of the sun, as a result of it capturing most of the gas cloud.

Exact figures are uncertain, but calculations suggest Jupiter would need to be 80 times as massive as it is to turn into a small red dwarf star. Another possibility, though, is a brown dwarf, which is a kind of half-star. This isn’t massive enough for ordinary hydrogen to fuse into helium as in most stars. Instead it uses the rarer hydrogen isotope deuterium.  It is estimated a brown dwarf needs to be about 13 times the mass of Jupiter.

OverLord2011 [107]3 years ago
5 0

Answer:

False

Explanation:

Let me put it in numbers. Jupiter has mass of 1.898x10^27Kg, make it 10 times as much and it becomes 1.898x10^28 Kg.

TRAPPIST-1, smallest star ever found has Mass of 1.77x10^29Kg, that is around 93 times larger than the mass of Jupiter.

Clearly 10 times as much is not enough to generate fusion reaction and become a star, in theory Jupiter must be at least 85 times larger than it's current mass to be able to generate fusion reaction and become a star.

So i think we can conclude that if Jupiter were to have 10 times of its own mass, it would not become a star, therefore the given statement is False.

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3 0
3 years ago
You are driving your car and the traffic light ahead turns red.You apply the breaks for 3s and the velocity of the car decreases
hammer [34]

Answer:

The car's displacement during this time is 25.65 meters.

Explanation:

Given that,

Final velocity of the car, v = 4.5 m/s

Deceleration of the car, a=-2.7\ m/s^2

Let u is the initial speed of the car. It is given by :

v=u+at

u=v-at

u=4.5-(-2.7)\times 3

u = 12.6 m/s

Let d is the car's displacement during this time. It can be calculated using second equation of motion as :

d=ut+\dfrac{1}{2}at^2

d=12.6\times 3+\dfrac{1}{2}\times (-2.7)\times 3^2

d = 25.65 meters

So, the car's displacement during this time is 25.65 meters. Hence, this is the required solution.                        

7 0
3 years ago
What is a factor that limits a technological design?
Ivanshal [37]
The answer is a constraint
5 0
3 years ago
Read 2 more answers
A uniform disk with radius 0.650 m
VashaNatasha [74]

Answer:

a = 13.758\,\frac{m}{s^{2}}

Explanation:

First, the instant associated to the angular displacement is:

(1.10\,\frac{rad}{s} )\cdot t + (6.30\,\frac{rad}{s^{3}} )\cdot t^{2} - 0.628\,rad = 0

Roots of the second-order polynomial are:

t_{1} \approx 0.240\,s, t_{2} \approx -0.415\,s

Only the first root is physically reasonable.

The angular velocity is obtained by deriving the angular displacement function:

\omega (0.240\,s) = 1.10\,\frac{rad}{s}+ (12.6\,\frac{rad}{s^{2}})\cdot (0.240\,s)

\omega (0.240\,s) = 4.124\,\frac{rad}{s}

The angular acceleration is obtained by deriving the previous function:

\alpha (0.240\,s) = 12.6\,\frac{rad}{s^{2}}

The resultant linear acceleration on the rim of the disk is:

a_{t} = (0.650\,m)\cdot (12.6\,\frac{rad}{s^{2}} )

a_{t} = 8.190\,\frac{m}{s^{2}}

a_{n} = (0.650\,m)\cdot (4.124\,\frac{rad}{s} )^{2}

a_{n} = 11.055\,\frac{m}{s^{2}}

a = \sqrt{a_{t}^{2}+a_{n}^{2}}

a = 13.758\,\frac{m}{s^{2}}

3 0
3 years ago
When two stars are bound together gravitationally and orbit a common mass, they're known as
garik1379 [7]
D. Binary Stars

Well, galaxies are not stars at all rather they are collection of stars, nebulae, globular clusters and stuff like this.
Meteoroids are something entirely different.
Constellations are the pattern of stars.
So, "Binary Stars" is the Right One.
3 0
3 years ago
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