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?

Answer 1

k = 0.8N/0.0050m = 160 N/m.

L = 1.5 + 3.5N*1m/160N. = 1.5219 m.

Answer 2

2*2356nm

Answer 3

19.936

Answer 4

I the school.

Answer 5

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Answer 6

I don't understand

Answer 7

thanks

Answer 8

It's good but the works is not clear

Answer 9

Chukwu

Answer 10

Thanks for your help

Answer 11

THANKS