Let's say that the teacher raised the lawn bowling ball to a height of 1.6 meters before they released it . THE LAWN BOWLING BA
1 answer:
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Answer:
0.130
Explanation:
From the given data, the coefficient of static friction for each trial are:
1. 0.053
2. 0.081
3. 0.118
4. 0.149
5. 0.180
6. 0.198
The sum of the coefficient of static friction = 0.053 + 0.081 + 0.118 + 0.149 + 0.180 + 0.198
= 0.779
So that;
the average coefficient of static friction =
=
= 0.12983
The average coefficient of static friction is 0.130
Answer:
the force exerted on the foot by the tibia would be 2975 N
Explanation:
Given the data in the question;
To maintain equilibrium between the foot and the ball vertically, the addition normal normal force (750 N) and the tension in the Achilles tendon
(2225 N) must be equal to the force exerted on the foot by the tibia;
so
| | + |
| = |
|
so force exerted on the foot by the tibia will be;
| | = |
| + |
|
so we substitute IN OUR VALUES
| | = 750 N + 2225 N
| | = 2975 N
Therefore, the force exerted on the foot by the tibia would be 2975 N