The area of the bar over r=4 is 0.468
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
Ernest Rutherford is the answer you are looking for my friend.
Answer:
It requires a measurement of altitude azimuth time.
Hope this helps, if it did, please give it a brainliest.
The box is accelerated from rest to 4 m/s in a matter of 2.5 s, so its acceleration <em>a</em> is such that
4 m/s = <em>a</em> (2.5 s) → <em>a</em> = (4 m/s) / (2.5 s) = 1.6 m/s²
Then the force applied to the box has a magnitude <em>F</em> such that
<em>F</em> = (10 kg) (1.6 m/s²) = 16 N
