The apparent brightness of a star depends on how bright it really IS, and also on its distance from Earth. Just like matches, flashlights, and fireflies, stars that are closer to us look brighter to us. (Think about the Sun.)
Answer:
+5.7 m/s
Explanation:
According to the law of conservation of momentum is that the momentum before the collision is equal to the momentum after the collision. In an equation form it would look like this:
M₁V₁+M₂V₂ = M₁V₁'+M₂V₂'
Where:
M₁ = mass of object 1 (kg)
V₁ = velocity of object 1 before the collision (m/s)
V₁' = Final velocity of object 1 after the collision (m/s)
M₂ = mass of object 2 (kg)
V₂ = velocity of object 2 before the collision (m/s)
V₂' = Final velocity of object 2 after the collision (m/s)
According to your problem you have the following given:
M₁ = 5 g = 0.005kg
V₁ = 3 m/s
V₁' = -5m/s (It bounced off so it is going the other direction)
M₂ = 6g = 0.006kg
V₂ = -1 m/s (It is coming from the opposite direction of the 3-ball)
V₂' = ?
So we plug in what we know and solve for what we don't know.
Answer:
+5.7 m/s
Explanation:
Just did the assignment :)
(b) How far away from the dart board is the dart released?
The basketball do not bounce back to the same height after each drop due to the fact of loss of kinetic energy.
Explanation:
When the basketball or any ball is dropped off from a certain height it falls down under the influence of gravity with the kinetic energy which has been transferred to the ball with hand by dribble.
When the basketball hits the ground with the kinetic energy, it loses some of the energy to the ground and then with the action reaction pair with the ground, is bounced back but with lower energy to a lower height.
Hi!
So if you drop a basketball from a higher height, it has more time to gather more energy, which it then forces against the ground and causes the ball to bounce.
Hope this helps! Sorry I didn't really use any technical terms!