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dusya [7]
4 years ago
13

A mass on a spring vibrates in simple harmonic motion at an amplitude of 8.0 cm. If the mass of the object is 0.20 kg and the sp

ring constant is 130 N/m, what is the frequency?
Physics
1 answer:
Dmitry [639]4 years ago
5 0

Answer:

Frequency, f = 3.73Hz

Explanation:

The frequency of a simple harmonic 6is given by:

f = w/2pi

But w= Sqrt( k/m)

Where k is the spring constant

And m is the mass

Given:

Mass=0.20kg

Spring constant, k=130N/m

w= Sqrt(130/0.20)

w= Sqrt(650)

w= 25.50 m

Frequency, f = w/2pi

f = 25.50/(2×3.142)

f = 3.73Hz

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Answer:

The voltage drop through which the proton moves is 39.1 V.

Explanation:

Given that,

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Time t=1.10\times10^{-6}\ s

We need to calculate the acceleration

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s = ut+\dfrac{1}{2}at^2

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Put the value in the equation

4.76\times10^{-2}=\dfrac{1}{2}\times a \times(1.10\times10^{-6})^2

a=\dfrac{2\times 4.76\times10^{-2}}{(1.10\times10^{-6})^2}

a=7.87\times10^{10}\ m/s^2

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Using newton's second law

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Put the value of F in equation (I) from equation (II)

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Put the value into the formula

V=\dfrac{1.67\times10^{-27}\times7.87\times10^{10}\times4.76\times10^{-2}}{1.6\times10^{-19}}

V=39.1\ V

Hence, The voltage drop through which the proton moves is 39.1 V.

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