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
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The coefficient of linear expansion, given that the length of the pipe increased by 1.5 cm is 1.67×10¯⁵ /°F
<h3>How to determine the coefficient of linear expansion</h3>
From the question given above, the following data were obtained
- Original diameter (L₁) = 10 m
- Change in length (∆L) = 1.5 cm = 1.5 / 100 = 0.015 m
- Change in temperature (∆T) = 90 °F
- Coefficient of linear expansion (α) =?
The coefficient of linear expansion can be obtained as illustrated below:
α = ∆L / L₁∆T
α = 0.015 / (10 × 90)
α = 0.015 / 900
α = 1.67×10¯⁵ /°F
Thus, we can conclude that the coefficient of linear expansion is 1.67×10¯⁵ /°F
Learn more about coefficient of linear expansion:
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