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
Torque = Force applied x lever arm. It twists the object or changes the rotational motion of things
Hope it helps
Pivots Points are price levels chartists can use to determine intraday support and resistance levels
Complete Question
A proton is located at <3 x 10^{-10}, -5*10^{-10} , -5*10^{-10}> m. What is r, the vector from the origin to the location of the proton
Answer:
The vector position is 
Explanation:
From the question we are told that
The position of the proton is 
Generally the vector location of the proton is mathematically represented as
So substituting values
Answer:
(d) ATP molecules are produced in the cytosol as glucose is converted into pyruvate.
