Answer:
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Explanation:
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Answer:
Maximum force, F = 1809.55 N
Explanation:
Given that,
Diameter of the anterior cruciate ligament, d = 4.8 mm
Radius, r = 2.4 mm
The tensile strength of the anterior cruciate ligament, 
We need to find the maximum force that could be applied to anterior cruciate ligament. We know that the unit of tensile strength is Pa. It must be a type of pressure. So,

So, the maximum force that could be applied to anterior cruciate ligament is 1809.55 N
Answer:
An object moves with constant velocity .
Explanation:
•2nd law