Answer:
The work done in stretching the spring is 8 lb.ft
Step-by-step explanation:
Given;
Applied force, F = 192 lb
extension of the spring, x = 3 ft
Determine the spring constant from the applied force and extension;
When the spring is stretched 6 inches beyond its natural length, the work done is calculated as follows;
x = 6 inches = 0.5 ft
Therefore, the work done in stretching the spring is 8 lb.ft
Answer:
689 root 11
Step-by-step explanation:
simplify
16x4root11+5^4 root 11
64 root 11 +625 root 11
The only non-linear equation from your choices is the area of a circle as it is:
A=πr^2 and if you take the derivative of A you have:
dA/dr=2πr
So the rate of change changes as r changes, it is not constant thus the function has acceleration, so velocity changes.
This is in contrast to any linear equation which is of the form:
y=mx+b now taking the derivative you see that:
dy/dx=m, now m is a constant value, which means that there is no acceleration and the velocity remains constant.
3/4 = 6/8 = 9/12
4/5 = 8/10 = 12/15
