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: Option D: 5.5×10²Joules
Explanation:
Work done is the product of applied force and displacement of the object in the direction of force.
W = F.s = F s cosθ
It is given that the force applied is, F = 55 N
The displacement in the direction of force, s = 10 m
The angle between force and displacement, θ = 0°
Thus, work done on the object:
W = 55 N × 10 m × cos 0° = 550 J = 5.5 × 10² J
Hence, the correct option is D.
It would be C. the color of the pot. its pretty obvious that i would not effect the project.
Force = mass * acceleration = 1500kg * 8m/s²