What is the best approximate value for the elastic potential energy (epe) of the spring elongated by 3.0 meters?

4.5 j
6.8 j
7.5 j
14 j

1/0.14nm = 10^10/1.4 a lot

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answer: a. same unit

explanation:

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm