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:
374 N
Explanation:
N = normal force acting on the skier
m = mass of the skier = 82.5
From the force diagram, force equation perpendicular to the slope is given as
N = mg Cos18.7
μ = Coefficient of friction = 0.150
frictional force is given as
f = μN
f = μmg Cos18.7
F = force applied by the rope
Force equation parallel to the slope is given as
F - f - mg Sin18.7 = 0
F - μmg Cos18.7 - mg Sin18.7 = 0
F = μmg Cos18.7 + mg Sin18.7
F = (0.150 x 82.5 x 9.8) Cos18.7 + (82.5 x 9.8) Sin18.7
F = 374 N
Answer:
16.2 s
Explanation:
Given:
Δx = 525 m
v₀ = 0 m/s
a = 4.00 m/s²
Find: t
Δx = v₀ t + ½ at²
525 m = (0 m/s) t + ½ (4.00 m/s²) t²
t = 16.2 s
Answer:
4/3
Explanation:
Shown in the picture attached
