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sleet_krkn [62]

3 years ago

14

When a 0.350-kg package is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The package is now di

splaced from its equilibrium position and undergoes simple harmonic oscillations when released. What is the period of the oscillations?

1 answer:

AleksandrR [38]3 years ago

4 0

To solve the problem it is necessary to use the concepts related to the calculation of periods by means of a spring constant.

We know that by Hooke's law

$F=kx$

Where,

k = Spring constant

x = Displacement

Re-arrange to find k,

$k= \frac{F}{x}$

$k= \frac{mg}{x}$

$k= \frac{(0.35)(9.8)}{12*10^{-2}}$

$k = 28.58N/m$

Perioricity in an elastic body is defined by

$T = 2\pi \sqrt{\frac{m}{k}}$

Where,

m = Mass

k = Spring constant

$T = 2\pi \sqrt{\frac{0.35}{28.58}}$

$T = 0.685s$

Therefore the period of the oscillations is 0.685s

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Read 2 more answers

A car battery with a 12 V emf and an internal resistance of 0.11 Ω is being charged with a current of 56 A. What are (a) the pot

denpristay [2]

Answer:

Part a)

V = 18.16 V

Part b)

$P_r = 345 Watt$

Part c)

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Part d)

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Part a)

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here we have

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Part b)

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Part c)

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$P_{emf} = 672 Watt$

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Part d)

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Part e)

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