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a certain spring stretches 7.0 cm when a load of 36 n is suspended from it. how much will the spring stretch if 48 n is suspended from it (and it doesn't reach its elastic limit)?

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Final answer:

Using Hooke's Law and the proportionality of force to displacement, a spring that stretches 7 cm under 36 N will stretch 9.3 cm when 48 N is applied, given that the elastic limit is not exceeded.

Step-by-step explanation:

The subject question involves understanding Hooke's Law in physics, which states that the force F needed to extend or compress a spring by some distance X is proportional to that distance. That is, F = kX, where k is the force constant of the spring. In the given problem, we're told that a spring stretches 7.0 cm when a 36 N force is applied. Assuming the spring doesn't reach its elastic limit, we can find out how much it will stretch under a different force using the formula:

X2 = (F2 / F1) × X1

where:

  • X2 is the new extension of the spring (what we want to find).
  • F2 is the new force applied (48 N).
  • F1 is the original force applied (36 N).
  • X1 is the original extension (7.0 cm).

Plugging in the values, we get:

X2 = (48 N / 36 N) × 7 cm = (4/3) × 7 cm = 9.3 cm

Therefore, if a 48 N force is applied to the spring, it will stretch 9.3 cm, assuming it does not surpass its elastic limit.

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