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11111nata11111 [884]

3 years ago

12

A 2.0 kg hanging mass stretches a coiled spring by 0.15 m. The spring constant, k, is: (A) 0.075 N/m, (B) 2.9 N/m (C) 131 N/m, (

D) 1,742 N/m, (E) none of the above.

1 answer:

Alla [95]3 years ago

3 0

Answer:

C

Explanation:

Givens

m = 2 kg

F = 2 * 9.81

F = 19.62 N

x = 0.15 m

Formula

F = k*x

Solution

19.62 = k*0.15

k = 19.62/0.15

k = 130.8 which rounded to the nearest given answer is C

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Answer:

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Explanation:

Solution:

As we know that:

F = Kx

Here,

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First, they are displacing it with 6 cm from its rest position for which Time period of the oscillation is T = 2 seconds.

But next, they want to know the effect on the time period of the oscillation if the displacement x is doubled from 6cm to 12 cm.

First of all, let us see the equation of the time period of the oscillation.

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Answer:

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