1. Encoding Information
2. Storing Information
3. Retrieval Information
One day, as I was walking, I found some sandy soil beside the road.
2) Unbalanced. Mike will push the box with a force of 20 N. The forces would be balanced if the box responded with 30 N.
3) Balanced. Both boys are pulling with the same force. Neither is winning.
4) Unbalanced. The rope will move with 10 N to the west. The teachers are winning.
5) Unbalanced. The kids are pulling 220 N to the east. The kids are winning.
6) Balanced. You and the dog are pulling with the same force.
I would think A. Reduce Environmental Impacts.
If you build and populate more zoos, some of the endangered species might not be able to survive in captivity
If you introduce new predators they are going to kill all the endangered species and if you pass new laws about land use some people may kill the species
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
