It depends on the size of the star. If it's size was normal then it cools down into White dwarf, then a black dwarf. If a really huge star dies, then we can see a "Supernova" from that.
Hope this helps!!
Answer:
Option B:
A child sitting on a swing.
Explanation:
When we hear the word oscillator, a good example is the pendulum bob of a grandfather clock. We can picture the motion to get a perfect understanding of its path of motion and relate it to other systems of motion in our everyday life.
An oscillator is a system that moves in such a way that it reverses its direction after a period of time. It can be seen as a "to-and-fro" motion.
From the options, a child sitting on a swing is the perfect example of an oscillating system because the child will be moving forwards and backwards, alternately reversing the direction of motion with time.
Answer:
D) The ball exerts a force on the wall and the wall exerts a force back.
Explanation:
Newton's third law of motion states that:
"When an object A exerts a force on another object B, then object B exerts an equal and opposite force on object A"
In this problem, we can identify (for instance) object A with tha ball and object B with the wall. Therefore, if we apply Newton's third law, we get:
The ball (object A) exerts a force on the wall (object B), therefore the wall (object B) exerts an equal and opposite force on the ball (object A). So, option D is the correct one.
Answer:
a = 0m/s²
Explanation:
Average acceleration = (change in velocity)/(time it takes). Since the car's change in velocity is zero, its acceleration is zero.
Answer:
The ball would have landed 3.31m farther if the downward angle were 6.0° instead.
Explanation:
In order to solve this problem we must first start by doing a drawing that will represent the situation. (See picture attached).
We can see in the picture that the least the angle the farther the ball will go. So we need to find the A and B position to determine how farther the second shot would go. Let's start with point A.
So, first we need to determine the components of the velocity of the ball, like this:
we pick the positive one, so it takes 0.317s for the ball to hit on point A.
so now we can find the distance from the net to point A with this time. We can find it like this:
Once we found the distance between the net and point A, we can similarly find the distance between the net and point B:
t= -0.9159s or t=0.468s
we pick the positive one, so it takes 0.468s for the ball to hit on point B.
so now we can find the distance from the net to point B with this time. We can find it like this:
So once we got the two distances we can now find the difference between them:
so the ball would have landed 3.31m farther if the downward angle were 6.0° instead.
