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:
A.) 8 m/s
B.) 7.0 m
Explanation:
Given that a block is given an initial velocity of 8.0 m/s up a frictionless 28° inclined plane.
(a) What is its velocity when it reaches the top of the plane?
Since the plane is frictionless, the final velocity V will be the same as 8 m/s
The velocity will be 8 m/s as it reaches the top of the plane.
(b) How far horizontally does it land after it leaves the plane?
For frictionless plane,
a = gsinø
Acceleration a = 9.8sin28
Acceleration a = 4.6 m/s^2
Using the third equation of motion
V^2 = U^2 - 2as
Substitute the a and the U into the equation. Where V = 0
0 = 8^2 - 2 × 4.6 × S
9.2S = 64
S = 64/9.2
S = 6.956 m
S = 7.0 m
Explanation:
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Answer:
Theta1 = 12° and theta2 = 168°
The solution procedure can be found in the attachment below.
Explanation:
The Range is the horizontal distance traveled by a projectile. This diatance is given mathematically by Vo cos(theta) t. Where t is the total time of flight of the projectile in air. It is the time taken for the projectile to go from starting point to finish point. This solution assumes the projectile finishes uts motion on the same horizontal level as the starting point and as a result the vertical displacement is zero (no change in height).
In the solution as can be found below, the expression to calculate the range for any launch angle theta was first derived and then the required angles calculated from the equation by substituting the values of the the given quantities.