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denis-greek [22]
3 years ago
13

What happens when a star blows up and it is next another star will it blow up too?

Physics
2 answers:
babunello [35]3 years ago
8 0

Answer:

no most of the time no it will just cause a shift in the position of the star that was next to it

Explanation:

stars have a gravitational pull and it would probably absorb the force and take in some of the gasses and bit and parts of what was left of the star that was blown up and burn them to fuel itself

FinnZ [79.3K]3 years ago
4 0

Answer:

Sometimes

Explanation:

Sometimes meaning in occasions. If it is a "vampire star" not likely because they also cause they also make novas happen by sucking the gas from another star.

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The answer to your question is 33
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SI is considered a consistent system because it what
NemiM [27]

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because it is a worldwide system....

Explanation:

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3 years ago
A stone with a mass of 0.100kg rests on a frictionless, horizontal surface. A bullet of mass 2.50g traveling horizontally at 500
jolli1 [7]

Answer:

Explanation:

Given that:

mass of stone (M) = 0.100 kg

mass of bullet (m) = 2.50 g = 2.5  ×10 ⁻³ kg

initial velocity of stone (u_{stone}) = 0 m/s

Initial velocity of bullet (u_{bullet}) = (500 m/s)i

Speed of the bullet after collision (v_{bullet}) = (300 m/s) j

Suppose we represent (v_{stone}) to be the velocity of the stone after the truck, then:

From linear momentum, the law of conservation can be applied which is expressed as:

m*u_{bullet} + M*{u_{stone}}= mv_{bullet}+Mv_{stone}

(2.50*10^{-3} \ kg) (500)i+0 = (2.50*10^{-3} \ kg)(300 \ m/s)j + (0.100 \ kg)v_{stone}

(2.50*10^{-3} \ kg) (500)i- (2.50*10^{-3} \ kg)(300 \ m/s)j=  (0.100 \ kg)v_{stone}

v_{stone}= (1.25\  kg.m/s)i-(0.75\ kg m/s)j

v_{stone}= (12.5\  m/s)i-(7.5\ m/s)j

∴

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\mathbf{v_{stone}= 14.6 \ m/s}

Using the tangent of an angle to determine the direction of the velocity after the struck;

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\theta = tan^{-1} (\dfrac{-7.5}{12.5})

\mathbf{\theta = 30.96^0 \ below \ the \ horizontal\ level}

5 0
3 years ago
A closely wound circular coil has a radius of 6.00 cmand carries a current of 2.65 A. How many turns must it have if the magneti
ad-work [718]

Answer:

Given:

radius of the coil, R = 6 cm = 0.06 m

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Magnetic field at the center, B = 6.31\times 10^{4} T

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To find the number of turns, N, we use the given formula:

B = \frac{\mu_{o}NI}{2R}

Therefore,

N = \frac{2BR}{\mu_{o}I}

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8 0
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