Na dame Dolla fo surrrreeeeee
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: 90 m/s
Explanation:
Given
mass of racecar 
velocity of racecar 
mass of still honeybadger 
after collision race car is traveling at a speed of 
conserving linear momentum
![Mu+m\times0=Mv_1+ mv_2\quad[v_2=\text{velocity of honeybadger after colllision}]](https://tex.z-dn.net/?f=Mu%2Bm%5Ctimes0%3DMv_1%2B%20mv_2%5Cquad%5Bv_2%3D%5Ctext%7Bvelocity%20of%20honeybadger%20after%20colllision%7D%5D)
15 min
Explanation:
take 0.25 and put it in for 1.00 and you will see its 0.25 but when you add it all 4 times it is 1.00 so then you would take that and do it to the hour ... how many times does four go into 60
