Answer:
A collision in which both total momentum and total kinetic energy are conserved
Explanation:
In classical physics, we have two types of collisions:
- Elastic collision: elastic collision is a collision in which both the total momentum of the objects involved and the total kinetic energy of the objects involved are conserved
- Inelastic collision: in an inelastic collision, the total momentum of the objects involved is conserved, while the total kinetic energy is not. In this type of collisions, part of the total kinetic energy is converted into heat or other forms of energy due to the presence of frictional forces. When the objects stick together after the collision, the collisions is called 'perfectly inelastic collision'
I think that the answer is friction
Answer electrons are negative. Protons are positive, and neutrons are neutral
Answer:
5.2791264*10¹³
Explanation:
Convert the 9 years to seconds and then multiple it by 186000
Answer:
Explanation:
(a) It is given that Joseph jogs on a straight road of 300m in a time interval of 2 minutes and 30 seconds, which is equal to 150seconds. Therefore, when Joseph jogs from point A to point B, he covers a distance of 300m in time of 150seconds. Hence, his average speed is 300m/150s=2ms^−1. Since it is a straight road and he jogs in a single direction in this case, his displacement is equal to 300m. Since it is a straight road and he jogs in a single direction in this case, his displacement is equal to 300m.
Hence, his average velocity is 300m/150s=2ms^−1
(b) Then it is given that he turns back and points B and jogs on the same road but in the opposite direction for a time interval for 1 minute and covers a distance of 100m.If we consider the whole motion of Joseph, i.e. from point A to point C, then he covers a total distance of 300m+100m=400m. And he covers this total distance in a time interval of 2.5min+1min=3.5min=210s.
Therefore, his average speed for this journey is 400m210s=1.9ms−1.
For the same journey is displacement is equal to the distance between the points A and C,i.e. 300m−100m=200m.
Hence, his average velocity for this case is 200m/210s=0.95ms^−1
