Asked by Emzkidinn.tk
A spring of natural length 1.5m is extended 0.0050m by a force of 0.8N, what will its length be when the applied force is 3.2N?
Answers
Answered by
Henry
k = 0.8N/0.0050m = 160 N/m.
L = 1.5 + 3.5N*1m/160N. = 1.5219 m.
L = 1.5 + 3.5N*1m/160N. = 1.5219 m.
Answered by
praise
2*2356nm
Answered by
FUNKE
19.936
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Answered by
Lialian
I don't understand
Answered by
ace
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Answered by
Uche
Chukwu
Answered by
Emmanuel
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Answered by
AMINAT
THANKS
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