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
I think it is a, b, and e. Hope it helps! :)
Answer:
B AND D
HOPE ITS HELF FUL .........
Na + Cl
An ionic bond is made between a metal and a non-metal
Answer:
t = 5.89 s
Explanation:
To calculate the time, we need the radius of the pulley and the radius of the sphere which was not given in the question.
Let us assume that the radius of the pulley (
) = 0.4 m
Let the radius of the sphere (r) = 0.5 m
w = angular speed = 150 rev/min = (150 × 2π / 60) rad/s = 15.708 rad/s
Tension (T) = 20 N
mass (m) = 3 kg each
Substituting values:
