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Sedaia [141]
2 years ago
10

When a 2.40-kg object is hung vertically on a certain light spring described by Hooke's law, the spring stretches 2.92 cm.(a) Wh

at is the force constant of the spring
Physics
1 answer:
Anastasy [175]2 years ago
3 0

Answer:

805.48N/m

Explanation:

According to Hookes law

F = Ke

F is the force = mg

F = 2.4×9.8 = 23.52N

e is the extension = 2.92cm = 0.0292m

Force constant K = F/e

K = 23.52/0.0292

K = 805.48N/m

Hence the force constant of the spring is 805.48N/m

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Answer:

T=51.64^\circ F

t=180.10s

Explanation:

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Applying (1) for t=1 min=60s:

T=10^\circ F+(60^\circ F)e^{-0.0061*60}\\T=10^\circ F+(60^\circ F)0.694\\T=10^\circ F+41.64^\circ F\\T=51.64^\circ F

Applying (1) for T=30^\circ F:

30^\circ F=10^\circ F+(60^\circ F)e^{-0.0061t}\\30^\circ F-10^\circ F=(60^\circ F)e^{-0.0061t}\\-0.0061t=ln(\frac{20}{60})\\t=\frac{ln(\frac{20}{60})}{-0.0061}=180.10s

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