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3 years ago

120J 5A 3s xΩ x=? Pls help

Physics

1 answer:

Mrrafil [7]3 years ago

3 0

Answer:

can I plz show the the photo of this question

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3 years ago

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Explanation:

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4 years ago

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The stiffness of a particular spring is 42 N/m. One end of the spring is attached to a wall. A force of 2 N is required to hold

Liula [17]

Answer:

$L_{o}=0.1224m$

Explanation:

Given data

Force F=2 N

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Spring Constant k=42 N/m

To find

Relaxed length of the spring

Solution

From Hooke's Law we know that

$F_{spring}=k_{s}|s|\\F_{spring}=k_{s}(L-L_{o})\\ 2N=(42N/m)(0.17m-L_{o})\\2=7.14-42L_{o}\\-42L_{o}=2-7.14\\42L_{o}=5.14\\L_{o}=(5.14/42)\\L_{o}=0.1224m$

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4 years ago

What happens at a convergent boundary?

LenaWriter [7]

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3 years ago

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