Answer:
Dynamic flexibility
Explanation:
Dynamic flexibility can be generally defined as the ability of the body muscles and joints to move in full range of motion. High flexibility in these joints and muscles leads to the decreasing pain and injury in different parts of the body.
Proper warm up exercises are needed to be carried out that involves both the combination of controlling movements and stretching of the body, and this directly enhances the dynamic flexibility of the body.
The athletes and sports persons possesses a good dynamic flexibility of their body as they carry our different types of body exercises.
Answer:

Explanation:
As we know that electric field due to infinite line charge distribution at some distance from it is given as

now we need to find the electric field at mid point of two wires
So here we need to add the field due to two wires as they are oppositely charged
Now we will have

now plug in all data



now we have



C is the diffraction angle.... step by step explanation= I think it’s that I might be wrong lol
The answer is A. Friction.
Friction is the resisting force between two objects that move against one another.
Hope this helps! :)