Answer:
The angular acceleration is 
Explanation:
From the question we are told that
The moment of inertia is 
The net torque is 
Generally the net torque is mathematically represented as

Where
is the angular acceleration so

substituting values


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
1) In a circular motion, the angular displacement

is given by

where S is the arc length and r is the radius. The problem says that the truck drove for 2600 m, so this corresponds to the total arc length covered by the tire:

. Using the information about the radius,

, we find the total angular displacement:

2) If we put larger tires, with radius

, the angular displacement will be smaller. We can see this by using the same formula. In fact, this time we have:
Answer:
1,3,5
Explanation:
i think maybe dont come at me
Answer:
4m/s
Explanation:
May be different considering how long the pole is and how heavy the firefighter is.