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Sphinxa [80]
4 years ago
8

A very light ideal spring having a spring constant (force constant) of 8.2 N/cm is used to lift a 2.2-kg tool with an upward acc

eleration of 3.25 m/s2. If the spring has negligible length when it us not stretched, how long is it while it is pulling the tool upward?
Physics
1 answer:
vampirchik [111]4 years ago
6 0

Answer:

x = 3.5 cm

Explanation:

When spring is used as a lift tool

so the mass suspended on it will have spring force on it

So as per the force equation of mass we can say

F_{net} = ma

now net force on the mass is

F_{net} = kx - mg

F_{net} = kx - mg = ma

here we have

kx = mg + ma

now we have

x = \frac{mg + ma}{k}

x = \frac{2.2(9.8) + 2.2(3.25)}{8.2}

x = 3.5 cm

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Answer:

Option c

Explanation:

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Thus option C follows the definition of the soil tilth.

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4 years ago
Read 2 more answers
(1 point) A frictionless spring with a 6-kg mass can be held stretched 0.2 meters beyond its natural length by a force of 50 new
guapka [62]

Answer:

x(t) = 0.077cos(6.455t)

Explanation:

If the spring can be stretched 0.2 m by a force of 50 N, then the spring constant is:

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The equation of simple harmonic motion is as the following:

x(t) = Acos(\omega t - \phi)

where \omega = \sqrt{k/m} = \sqrt{250 / 6} = 6.455

We also know that the initial velocity is 0.5 m/s, which is also the maximum speed at the equilibrium:

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A = v_{max}/\omega = 0.5 / 6.455 = 0.077 m

\phi = 0 is the initial phase

Therefore, the position of the mass after t seconds is

x(t) = 0.077cos(6.455t)

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3 years ago
A shot-putter exerts an unbalanced force of 128 N on a shot giving it an acceleration of 19
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Answer:

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Explanation:

128

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3 years ago
Does the universe need explaining or can it just 'be' there?
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