Answer:
the resulting angular acceleration is 15.65 rad/s²
Explanation:
Given the data in the question;
force generated in the patellar tendon F = 400 N
patellar tendon attaches to the tibia at a 20° angle 3 cm( 0.03 m ) from the axis of rotation at the knee.
so Torque produced by the knee will be;
T = F × d⊥
T = 400 N × 0.03 m × sin( 20° )
T = 400 N × 0.03 m × 0.342
T = 4.104 N.m
Now, we determine the moment of inertia of the knee
I = mk²
given that; the lower leg and foot have a combined mass of 4.2kg and a given radius of gyration of 25 cm ( 0.25 m )
we substitute
I = 4.2 kg × ( 0.25 m )²
I = 4.2 kg × 0.0626 m²
I = 0.2625 kg.m²
So from the relation of Moment of inertia, Torque and angular acceleration;
T = I∝
we make angular acceleration ∝, subject of the formula
∝ = T / I
we substitute
∝ = 4.104 / 0.2625
∝ = 15.65 rad/s²
Therefore, the resulting angular acceleration is 15.65 rad/s²
Answer: i can see properly
Explanation:
D. Equal to zero.
Because the forces balance each other.
Answer:
The focal length of a lens is refers to the distance from the center of the lens to the principal foci.
With constant angular acceleration 
, the disk achieves an angular velocity 
at time 
according to
and angular displacement 
according to
a. So after 1.00 s, having rotated 21.0 rad, it must have undergone an acceleration of
b. Under constant acceleration, the average angular velocity is equivalent to
where 
and 
are the final and initial angular velocities, respectively. Then
c. After 1.00 s, the disk has instantaneous angular velocity
d. During the next 1.00 s, the disk will start moving with the angular velocity 
equal to the one found in part (c). Ignoring the 21.0 rad it had rotated in the first 1.00 s interval, the disk will rotate by angle according to
which would be equal to
