Answer:
a and b
Explanation:
Hydro static equilibrium holds a star steady and balanced. Whenever a star stops burning hydrogen in its center, there must be evolutionary improvements to maintain equilibrium for the star Of example, if a star's internal pressure and temperature fall, gravity will take over and force the star to contract and heat up, restoring stability. By contrast, if a star's internal pressure and temperature rises, the extra pressure causes the star to widen and cool, restoring balance.
so, according to above explanation options a and b both are true
a) A small increase in the star's internal pressure and temperature causes the star's outer layers to expand and cool.
b) A small decrease in the star's internal pressure and temperature causes the star's outer layers to contract and heat up.
Let's take the analogy of the baseball pitcher a step farther. When a baseball is thrown in a straight line, we already said that the ball would fall to Earth because of gravity and atmospheric drag. Let's pretend again that there is no atmosphere, so there is no drag to slow the baseball down. Now, let's assume that the person throwing the ball throws it so fast that as the ball falls towards the Earth, it also travels so far, before falling even a little, that the Earth's surface curves away from the ball's path.
In other words, the baseball falls as it did before, but the ball is moving so fast that the curvature of the Earth becomes a factor and the Earth "falls away" from the ball. So, theoretically, if a pitcher on a 100 foot (30.48 m) high hill threw a ball straight and fast enough,the ball would circle the Earth at exactly 100 feet and hit the pitcher in the back of the head once it circled the globe! The bad news for the person throwing the ball is that the ball will be traveling at the same speed as when they threw it, which is about 8 km/s or several times faster than a rifle bullet. This would be very bad news if it came back and hit the pitcher, but we'll get to that in a minute.
To solve this problem we will apply the concepts related to the kinematic equations of linear motion. We will calculate the initial velocity of the object, and from it, we will calculate the final position. With the considerations made in the statement we will obtain the total height. Initial velocity of the acorn,
![u = 0m/s](https://tex.z-dn.net/?f=u%20%3D%200m%2Fs)
Also, it is given that the acorn takes 0.201s to pass the length of the meter stick.
![s = ut+\frac{1}{2} at^2](https://tex.z-dn.net/?f=s%20%3D%20ut%2B%5Cfrac%7B1%7D%7B2%7D%20at%5E2)
Replacing,
![1 = u(0.141)+ \frac{1}{2} (9.8)(0.141)^2](https://tex.z-dn.net/?f=1%20%3D%20u%280.141%29%2B%20%5Cfrac%7B1%7D%7B2%7D%20%289.8%29%280.141%29%5E2)
![u =6.4013m/s](https://tex.z-dn.net/?f=u%20%3D6.4013m%2Fs)
The height of the acorn above the meter stick can be calculated as,
![v^2 = u^2 +2gh](https://tex.z-dn.net/?f=v%5E2%20%3D%20u%5E2%20%2B2gh)
![h = \frac{v^2-u^2}{2g}](https://tex.z-dn.net/?f=h%20%3D%20%5Cfrac%7Bv%5E2-u%5E2%7D%7B2g%7D)
![h = \frac{6.4013^2-0^2}{2(9.8)}](https://tex.z-dn.net/?f=h%20%3D%20%5Cfrac%7B6.4013%5E2-0%5E2%7D%7B2%289.8%29%7D)
![h = 2.0906m](https://tex.z-dn.net/?f=h%20%3D%202.0906m)
Also the top of the meter stick is 1.87m above the ground hence the height of the acorn above the ground is
![h = 2.0906+1.87](https://tex.z-dn.net/?f=h%20%3D%202.0906%2B1.87)
![h = 3.9606m](https://tex.z-dn.net/?f=h%20%3D%203.9606m)
Answer:
The answer is simple machine!!