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3 years ago

Which of the following is the most likely end for a star that formed from an extremely large nebula?

2 answers:

7 0

The choice that is most likely end for a star that formed from an extremely large nebula would be a black hole. Gravity pulls so much in a black hole that no light exists.

dezoksy [38]3 years ago

6 0

The correct answer of this question is : Black hole

EXPLANATION :

As per the question, the star is formed from the large nebula. The star which has formed from large nebula due to the gravitational pull might have masses greater than 3.5 solar masses.

At the end of the star i.e when it is run out of its fuel, it will be converted into black hole as per Chandrasekhar limit which depicts that a dying star will be a black hole if its mass is greater than 3.5 solar masses.

Hence, the correct answer is black hole.

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One particle has mass m and a second particle has mass 2m. The second particle is moving with speed v and the first with speed 2

choli [55]

Answer:

Explanation:

The formula for kinetic energy to be used here is 1/2mv².

If the first particle is "particle a" and the second particle is "particle n"; there kinetic energies (K.E) will be

K.Eₐ = 1/2.m2v² = mv²

K.Eₙ = 1/2.2mv² = mv²

From the above, <u>it can be said that there kinetic energies are the same</u>.

NOTE that the m and v used in the question means mass and velocity respectively.

3 0

3 years ago

3)(20pts) Un barco invasor se encuentra a 560m de un puerto de defensa, el puerto tiene cañones que

Eduardwww [97]

a) Las balas de cañon disparadas desde el puerto deben tener una ángulo de 27.39° para que puedan impactar al barco, con una velocidad inicial de 82 m/s.

b) El tiempo de vuelo de las balas de cañon para alcanzar al barco que está a 560 m de distancia es de 7.69 s.

c) Sabiendo que el ángulo calculado en el inciso a) es para una distancia de 560 m, el barco debe estar a una distancia mayor para que las balas no lo alcancen.

Podemos usar las ecuaciones de tiro parabólico para encontrar el ángulo que permita derribar al barco invasor.

$x=\frac{v_{i}^{2}sin(2\alpha)}{g}$ (1)

Donde:

v(i) es la velocidad inicial del cañon (82 m/s)
α es el ángulo de tiro
g es la gravedad (9.81 m/s²)
x es el desplazamiento total (560 m)

Lo que debemos hacer es depejar α de la ecuación 1

$sin(2\alpha)=\frac{xg}{v_{i}^{2}}$

$sin(2\alpha)=\frac{560*9.81}{82^{2}}$

$sin(2\alpha)=0.82$

$\alpha=\frac{sin^{-1}(0.82)}{2}$

$\alpha=27.39^{\circ}$

Por lo tanto, el ángulo para que el cañón impacte en el barco es de 27.39 °.

Sabemos que la componente de la velocidad en el eje x es constante, así que podemo usar la siguiente ecuación.

$v_{x}=\frac{x}{t}$

La componente x de la velocidad es V(x) = V(i)cos(α) y sabiendo la distancia total de 560 m, el tiempo será:

$v_{i}cos(\alpha)=\frac{x}{t}$

$t=\frac{x}{v_{i}cos(\alpha)}$

$t=\frac{560}{82cos(27.39)}$

$t=7.69\: s$

El tiempo total de vuelo de las balas de cañon es de 7.69 s.

Sabemos que el ángulo calculado en el inciso a) es de 27.39 °, y ese valor fue calcualdo para una distancia de 560 m, por lo tanto el barco debe estar a una distancia mayor que esa para que las balas no lo alcancen.

Puedes encontrar más información sobre tiro parabólico aquí:

https://brainly.lat/tarea/3605927

Espero te haya sido de ayuda!

3 0

3 years ago

I WILL MARK BRAINLIEST TO EXPLAINED AND CORRECT ANSWER

Ket [755]

The answer would be: Winter
The Southern hemisphere is tilted to the left, few degrees away from the sun. This condition will result in more(to maximum) exposure to the sunlight, lead to the summer. The condition is the opposite of the northern hemisphere where it has less area that exposed to sunlight, lead to winter.

6 0

3 years ago

When charged objects are brought close (not touching) to uncharged objects, what occurs?

seraphim [82]

They will become equally negative charges, I believe.

6 0

3 years ago

When magma flows on the surface on the surface, it is already called lava<br><br>TRUE OR FALSE

castortr0y [4]

Answer:

True

Explanation:

I guess you made a mistake on question.

but I understood what you wanted to say.

Hope this helps... :)

6 0

3 years ago