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Bogdan [553]
2 years ago
10

Plz answer asap I need all of the answer

Physics
1 answer:
Ugo [173]2 years ago
3 0

Answer:

Turned up side down

Explanation:

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A force F=0.12N is aplied on spring and spring elongates by 3cm . specific constant of spring ​
PilotLPTM [1.2K]

The spring constant is 4 N/m

Explanation:

When a spring is stretched/compressed by the application of a force, the relationship between the magnitude of the force applied and the elongation of the spring is given by Hooke's law:

F=kx

where

F is the magnitude of the spring applied

k is the spring constant

x is the elongation of the spring, relative to its equilibrium position

For the spring in this problem, we have:

F = 0.12 N (force applied)

x = 3 cm = 0.03 m (elongation of the spring)

Therefore, we can solve the formula for k to find the spring constant:

k=\frac{F}{x}=\frac{0.12}{0.03}=4 N/m

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#LearnwithBrainly

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3 years ago
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