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Sagot :
A) An equation that relates F and s for the spring on the jumper is; F = ks
B) The number of weights necessary for the spring to stretch 7 centimeters is; 28 weights
Hook's Law
We are given;
Force applied; F = 8 weights
Stretch of spring; s = 2 cm = 0.02 m
- A) We are told that according to Hook's law, the amount that the spring is stretched, s, varies directly with the force applied to the spring by the weights. This can be represented as;
F ∝ s
Now, there is a constant of proportionality here that will make this a relatable equation and this is k. Thus, the equation that relates F and s for the spring on the jumper is; F = ks
- B) Initially, we are told that 8 of the weights produces a stretch of 0.02 m. Thus;
8 = 0.02k
k = 8/0.02
k = 400 weights/m
To now find the number of weights necessary for the spring to stretch 7 centimeters or 0.07 m, we will have;
F = 400 × 0.07
F = 28 weights
Read more about hooke's law at; https://brainly.com/question/25711925
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