Given that:
k = 500 n/m,
work (W) = 704 J
spring extension (x) = ?
we know that,
Work = (1/2) k x²
704 = (1/2) × 500 × x²
x = 1.67 m
A spring stretched for 1.67 m distance.
Momentum = mass x velocity
Thus Option A is the correct answer
Momentum (dog) = 10 kg x (0.447 x 30) m/s
= 134.1 Kg m/s
Momentum ( bullet) = 0.02 kg x (0.447 x 800) m/s
= 7.152 Kg m/s
Momentum ( truck) = 0, as v = 0
tightrope has both low mass and low speed, thus its momentum will be low
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
