Delis cactus is the correct way to write it.
I don't know about it your answer will give another people
Answer:
Explanation:
The heaviside function is defined as:
so we see that the Heaviside function "switches on" when
, and remains switched on when 
If we want our heaviside function to switch on when 
, we need the argument to the heaviside function to be 0 when
Thus we define a function f:
The 
term inside the heaviside function makes sure to displace the function 5 units to the right.
Now we just need to add a scale up factor of 240 V, because thats the voltage applied after the heaviside function switches on. (
when 
, so it becomes just a 1, which we can safely ignore.)
Therefore our final result is:
I have made a sketch for you, and added it as attachment.
The elastic potential energy stored in the stretched spring is 1 J.
<h3>What is Hooke's law?</h3>
Hooke's law states that; provided the elastic limit is not exceeded, the extension of the spring is directly proportional to the force on the spring.
Given that;
Force on the spring = 350 Newton
Distance stretched = 7 centimeters or 0.07 m
Hence;
F = ke
k = F/e = 350 Newton/0.07 m = 5000 N/m
Work done in stretching a spring = 1/2ke^2
= 0.5 × 5000 × (2 × 10^-2)^2 =1 J
Learn more about elastic potential energy: brainly.com/question/156316
