Answer:
the knee extensors must exert 15.87 N
Explanation:
Given the data in the question;
mass m = 4.5 kg
radius of gyration k = 23 cm = 0.23 m
angle ∅ = 30°
∝ = 1 rad/s²
distance of 3 cm from the axis of rotation at the knee r = 3 cm = 0.03 m
using the expression;
ζ = I∝
ζ = mk²∝
we substitute
ζ = 4.5 × (0.23)² × 1
ζ = 0.23805 N-m
so
from; ζ = rFsin∅
F = ζ / rsin∅
we substitute
F = 0.23805 / (0.03 × sin( 30 ° )
F = 0.23805 / (0.03 × 0.5)
F F = 0.23805 / 0.015
F = 15.87 N
Therefore, the knee extensors must exert 15.87 N
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Answer:
F = 100 N
Explanation:
The torque is given by the expression
τ = F x r
where bold letters indicate vectors, the magnitude of this expression is
τ = F r sin θ
In general, when tightening a nut, the force is applied perpendicular to the arm, therefore θ = 90 and sin 90 = 1
τ = F r
F = τ / r
calculate
F = 30 / 0.30
F = 100 N
Answer:
The equation used to calculate the work done is: work done = force × distance. W = F × d. This is when: work done (W) is measured in joules (J)
