Avg acceleration; under constant acceleration condition only; = Vf-Vi/time
a=Vf-Vi/time=(15m/s-6m/s)/(3s)=
(9m/s)/(3s)=3m/s^2
Hope this helps. Thanks.
The work done by the unstretched spring is 5 J.
<h3>What is the Hooke's law?</h3>
Hooke's law states that the extension of a given material is directly proportional to the force applied as log as the elastic limit is not exceeded. Thus we have to know that; F = Ke
F = force applied
K = force constant
e = extension
Using the graph;
K = F/e
F = 200N
e = 5 cm or 5 * 10^-2 m
K = 200N/ 5 * 10^-2 m
K = 4000 N/m
Now;
Work = 1/2Ke^2
Work = 0.5 * 4000 N/m * (5 * 10^-2 m)^2
Work = 5 J
Learn more about Hooke's law:brainly.com/question/13348278
#SPJ1
B. Starting with most recent
I think so not take my answer because I think the answer is b
Answer:
F=507.7N
Explanation:
According to Newton's second law:

in this case, the football players need to drag the coach at a constant velocity, thus means with no acceleration, so:

there are 20 degrees between the two ropes that means each player exerts a force 10 degrees from the zero reference.
