Answer:
Explanation:
T = 2π√(L/g)
If you increase L to 2L, the period is increased by a factor of √2
T = 3.5√2 ≈ 4.9 s
Answer:
In a cardio kickboxing class, you will learn proper form for the famous jab, hooks,crosses, uppercuts, back kicks, front kicks.
Explanation:
Answer:
it evaporats
Explanation:
because the sun is so hot that the water will turn into gas hope i helped
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:
(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
