All categories

3 years ago

What are the similarities and differences between novae and supernovae?

Physics

1 answer:

DochEvi [55]3 years ago

3 0

Answer:

Explanation:

A novae in astronomy means an explosion in the white dwarf star which had tapped enough gas from a companion star,hence it releases an incredible amount of energy which is Over a million times brighter than it normal stars.

A super novae on the other hand is a cosmic explosion that can be a billion times brighter than the normal.

From this one can see that a perculiar similarity between a novae and super novae is that both generate huge explosion and bright Ness, and a major difference is super novae release huge amount of brightness and energy more than the novae

You might be interested in

A tennis ball is thrown into

liraira [26]

Answer:

Velocity (v) is a vector quantity that measures displacement (or change in position, Δs) over the change in time (Δt), represented by the equation v = Δs/Δt. Speed (or rate, r) is a scalar quantity that measures the distance traveled (d) over the change in time (Δt), represented by the equation r = d/Δt.

Explanation:

7 0

2 years ago

Which of the following causes objects to orbit the Sun?

algol [13]

B. The suns gravity

hope this helps

7 0

3 years ago

Read 2 more answers

Define newton's 3rd law of motion

soldier1979 [14.2K]

Answer:

For every action, there is an equal and opposite reaction.

Explanation:

Let's say I threw a ball on the wall. The image shows the instance the ball comes into contact with the wall.

If you would look at the image above,

- F1 and F2 are acting in opposite directions.

- They both have the same magnitude too.

F1 is the force of the ball on the wall (action force) and F2 is the force of the wall on the ball (reaction force). This is called an action-reaction pair.

7 0

3 years ago

If it has enough kinetic energy, a molecule at the surface of the Earth can "escape the Earth's gravitation", in the sense that

user100 [1]

Answer:

$K_E=mgr_E$

Explanation:

By conservation of energy, the sum of the kinetic and gravitational potential energies at the surface of the Earth must be equal than their sum at infinity, so we have:

$K_E+U_E=K_\infty+U_\infty$

$\frac{mv_E^2}{2}-\frac{GM_Em}{r_E}=\frac{mv_\infty^2}{2}-\frac{GM_Em}{r_\infty}$

Where $G=6.67\times10^{-11}Nm^2/kg^2$ is the gravitational constant, $M_E=5.97\times10^{24}kg$ and $r_E=6371000m$ are the mass and radius of the Earth, <em>m </em>is the mass of the particle, $v_E$ its velocity at the surface of the Earth (which would be its escape velocity) and $v_\infty$ and $r_\infty$ are the velocities and distance at infinity, which would be null and infinity respectively, so the right hand side of our equation is 0J, which leaves us with:

$\frac{GM_Em}{r_E}=\frac{mv_E^2}{2}=K_E$

Also, since the force the molecule experiments is the force of gravity (disregarding drag), we can write its weight in terms of Newton's Law of Gravitation:

$F=mg=\frac{GM_Em}{r_E^2}$