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
In pounds? Cuz if so 2.2 x 4.3 = 9.46
Answer:
8.2 m/s²
Explanation:
m = mass of the block
μ = Coefficient of kinetic friction = 0.17
= Normal force on the block by the ramp
= kinetic frictional force
Force equation perpendicular to ramp surface is given as
Kinetic frictional force is given as
Force equation parallel to ramp surface is given as
m/s²
Answer:
Twice the initial value
Explanation:
Let the current be = I
the width of the conducting strips be = a
We know that magnetic field between two plates is given by
If the direction of this magnetic field is same between the two plates, then
= 0
And when the currents runs opposite at each plate, then
Hence the magnetic field will be twice the initial value of the magnetic field that runs between the plates.
