Question #133293
A spring 1.5m long is stretched through 0.0050m by a load of 0.75N the stiffness of the material is what
1
Expert's answer
2020-09-16T10:09:52-0400

F=kΔx→k=F/Δx=0.75/0.005=150(N/m)F=k\Delta x\to k=F/\Delta x=0.75/0.005=150(N/m)


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