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:
Price=$7×10⁷
Explanation:
Step 1: Estimate the volume of the pile,
Step 2: Multiply it by the density to get its mass
Step 2: Then multiply the mass by the price per gram to get the total price
So
The average pile dimensions are=45.7×45×172.7

Price=m×$10
Price=(7×10⁶)×$10
Price=$7×10⁷
Answer:
5600N
Explanation:
Given parameters:
Mass of car = 700kg
Initial velocity = 10m/s
Final velocity = 30m/s
Displacement = 50m
Unknown:
Net force acting on the car = ?
Solution:
To find the force acting on a body, it is pertinent we know the mass and acceleration.
Force = mass x acceleration
Now;
Let us find the acceleration from the kinematics equations:
v² = u² + 2aS
v is the final velocity
u is the initial velocity
a is the acceleration
S is the distance
30² = 10² + (2 x a x 50)
900 = 100 + 100a
100a = 800
a = 8m/s²
Therefore;
Force = 700 x 8 = 5600N
Answer:
Explanation:
If i'm not wrong and late it might be F