Question
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
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.
praise
2*2356nm
FUNKE
19.936
AMARACHI
I the school.
AMARACHI
I like this school. because it is good school.
Lialian
I don't understand
ace
thanks
Cent
It's good but the works is not clear
Uche
Chukwu
Emmanuel
Thanks for your help
AMINAT
THANKS