The relative velocity of the athlete relative to the ground is 5.2 m/s
The given parameters;
constant velocity of the athlete, V = 5.2 m/s
let the velocity of the ground = Vg = 0
The relative velocity concept helps us to determine the velocity of a moving object relative to a stationary observer.
The athlete is the moving object in this question while the ground is stationary.
The relative velocity of the athlete relative to the ground is calculated as follows;
Thus, the relative velocity of the athlete relative to the ground is 5.2 m/s
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The ideal spring equation is
Stretch = K times Force .
This says that the stretch is directly proportional to the force.
In simple English, that means that if you double the force, then
you double the stretch, and if you multiply the force by π or
any other number, you multiply the stretch by the same number.
So you can always write a proportion for a spring:
Stretch₁ / Force₁ = Stretch₂ / Force₂ .
Part A:
In Part-A of this question, the force is increased to (2.5 / 2.0) = 1.25 times .
So the stretch is also increased to 1.25 times .
(1.25) x (6.1 cm) = 7.625 cm .
Answer:
0.1143m
Explanation:
W=f×s
8=70s
make s the subject of the formula
s=8/70
=0.1143m
Answer:
Acceleration is a vector quantity that is defined as the rate at which an object changes its velocity. An object is accelerating if it is changing its velocity.
may be it helped you
