Asked by Rosie
Kati observes a force of 15 Newtons stretch a spring a distance of 0.450m. a) what is the elasticity coefficient of the spring and b) how much force must be done to stretch the spring a distance of 0.750m?
a) F=Kx
15=K(0.450)
15/0.450=k
33N=k
It seems off to me know sure if done this correct
b) F=15*0.750
F=11N
please help
a) F=Kx
15=K(0.450)
15/0.450=k
33N=k
It seems off to me know sure if done this correct
b) F=15*0.750
F=11N
please help
Answers
Answered by
Rosie
Not sure if done correct!
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