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3 years ago

14

When the core of a star like the sun uses up its supply of hydrogen for fusion, the core begins to ________?

2 answers:

pav-90 [236]3 years ago

4 0

When the core of a star like the sun uses up its supply of hydrogen for fusion, the core begins to contract in size. The gravitational forces becomes strong and cause the star to reduce in size. as the star contracts, the temperature of the star also rises since the heat is distributed inside smaller Volume now.

Marrrta [24]3 years ago

3 0

Its core contracts, but its outer layers expand and the star becomes bigger and brighter.

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Focusing on irelevant information can negatively impact the ability to reason and problem solve effectively.

My name is Ann [436]

Answer:

true

Explanation:

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3 years ago

Assume we’re able to travel to your planet and decide to take some fireworks with us to celebrate our journey.

julia-pushkina [17]

Answer:

The horizontal distance covered by the firework will be $\frac{1876.8}{g}$

Explanation:

Let acceleration due to gravity on the planet be g, initial velocity of the firework be u and angle made with the horizontal be ∅.

writing equation of motion in vertical direction:

$v_{y}=u_{y}+(-g) t$

$u_{y}= u\sin \phi$

and $v_{y}=0$

therefore $\frac{u\sin \phi }{g} =t$

writing equation of motion in horizontal direction:

$s_{x}=u_{x}t$

$u_{x} = u\cos \phi$

therefore the equation becomes $s_{x}=\frac{u^{2} \sin \phi \cos \phi}{g}$

therefore horizontal distance traveled = $\frac{u^{2}\sin 2\alpha \phi }{2g}=\frac{1876.8}{g}\frac{m}{s}$

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3 years ago

A stone is dropped into a river from a bridge 41.7 m above the water. Another stone is thrown vertically down 1.80 s after the f

hram777 [196]

Answer:

31.75 m/s

Explanation:

h = 41.7 m

Let the initial velocity of the second stone is u

Let the time taken to reach to the bottom by the first stone is t then the time taken by the second stone to reach the ground is t - 1.8.

For first stone:

Use second equation of motion

$h=ut+\frac{1}{2}gt^2$

Here, u = 0, g = 9.8 m/s^2 and t be the time and h = 41.7

So, 41.7= 0 + 0.5 x 9.8 x t^2

41.7 = 4.9 t^2

t = 2.92 s ..... (1)

For second stone:

Use second equation of motion

$h=ut+\frac{1}{2}gt^2$

Here, g = 9.8 m/s^2 and time taken is t - 1.8 = 2.92 - 1.8 = 1.12 s, h = 41.7 m and u be the initial velocity

$h=u\left ( t-1.8 \right )+4.9\left ( t-1.8 \right )^2$ .... (2)

By equation the equation (1) and (2), we get

$41.7=1.12 u +4.9 \times 1.12^{2}$

u = 31.75 m/s

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3 years ago

Which of these are part of our solar system? select all that apply

Len [333]

A)planets
b)the sun
c)moons
e)comets
f)asteroids

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Reil [10]

Answer:

just trace a picture of it.

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3 years ago