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3 years ago

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The Electric Potential of the Earth The Earth has a vertical electric field with a magnitude of approximately 100 V/m near its s

urface.
What is the magnitude of the potential difference between a point on the ground and a point on the same level as the top of the Washington Monument (555 ft high)?

1 answer:

maksim [4K]3 years ago

4 0

Answer:

16916.4 V

Explanation:

Electric potential: This is the work done in bringing a unit positive charge from infinity to that point in against the action of the field. The S.I unit of Electric potential is V.

mathematically, Electric potential can be expressed as

P = E×d ....................................... Equation 1.

Where P = Electric potential, E = Electric Field, d = distance/height of the level at the top of the Washington Monument.

Given: E = 100 V/m, d = 555 ft = 555×0.3048 m = 169.164 m.

Substitute into equation 1

P = 100×169.164

P = 16916.4 V.

Thus the potential difference = 16916.4 V.

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An object with mass 3.5 kg is attached to a spring with spring stiffness constant k = 270 N/m and is executing simple harmonic m

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Answer:

Part a)

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Part b)

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As we know that angular frequency of spring block system is given as

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here we know

m = 3.5 kg

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$\omega = \sqrt{\frac{270}{3.5}}$

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Part a)

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Answer:

The value is $E_i = 1.5596 *10^{-18} \ J$

Explanation:

From the question we are told that

The wavelength is $\lambda = 48.2 nm = 48.2 *10^{- 9 }\ m$

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Generally the energy of the incident light is mathematically represented as

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Here c is the speed of light with value $c = 3.0 *10^{8} \ m/s$

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So

$E = \frac{6.62607015 * 10^{-34 }* 3.0 *10^{8}}{48.2 *10^{- 9 }}$

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Generally the kinetic energy is mathematically represented as

$E_k = \frac{1}{2} * m_e * v^2$

=> $E_k = \frac{1}{2} * 9.109*10^{-31} * (2.371*10^6 )^2$

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Answer:

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Given:

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