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: 0.169 (3 s.f.)
Explanation:
Force = 76 N
Spring constant = 450 N/m
Extension/displacement = x
Hooke's law states that: F = kx
Therefore, 76 = 450 X x
76/450 = x
0.169 (3 s.f.) = x
To solve this problem we will use the kinematic equations of angular motion, starting from the definition of angular velocity in terms of frequency, to verify the angular displacement and its respective derivative, let's start:
The angular displacement is given as the form:
In the equlibrium we have to 
and in the given position we have to
Derived the expression we will have the equivalent to angular velocity
Replacing,
Finally
Therefore the maximum angular displacement is 9.848°
Answer:
The acceleration is 11.25 km/hr
Explanation:
Divide the velocity over time to find the acceleration. The bus is accelerating 11.25 km/hr per second in the range of 8 seconds
