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DaniilM [7]
3 years ago
12

A proton is released from rest at the origin in a uniform electric field that is directed in the positive x direction with magni

tude 950 \text{ N/C}950 N/C. What is the change in the electric potential energy of the proton-field system when the proton travels to x = 2.5 \text{ m}x=2.5 m
Physics
1 answer:
Vika [28.1K]3 years ago
4 0

Explanation:

It is given that,

Electric field, E=950\ N/C

We need to find the change in the electric potential energy of the proton-field system when the proton travels to x = 2.5 m

From the conservation of energy, the loss in potential energy is equal to the gain in kinetic energy and kinetic energy is is equal to the work done.

W=F\times x

W=q\times E\times x

W=1.6\times 10^{-19}\times 950\times 2.5

W=3.8\times 10^{-16}\ J

So, the change in electric potential energy of the proton field system is 3.8\times 10^{-16}\ J. Hence, this is the required solution.

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Lapatulllka [165]

Answer:

Wavelength, \lambda=0.011\ m

Explanation:

Given that,

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\lambda=\dfrac{v}{f}

\lambda=\dfrac{0.00925\ m/s}{0.79\ Hz}

\lambda=0.011\ m

So, the wavelength of the longitudinal pulse is 0.011 meters. Hence, this is the required solution.

8 0
2 years ago
A ball, which has a mass of 1.25 kg, is thrown straight up from the top of a building 225 meters tall with a velocity of 52.0 m/
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First we will find the speed of the ball just before it will hit the floor

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\frac{1}{2}(1.25)(52)^2 + 1.25(9.8)(225) = \frac{1}{2}(1.25)v_f^2

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3 0
3 years ago
Find the electric energy density between the plates of a 225-μF parallel-plate capacitor. The potential difference between the p
PSYCHO15rus [73]

Answer:

Energy density will be 14.73 J/m^3

Explanation:

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Potential difference between the plates = 365 V

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We know that there is relation between electric field and potential

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Energy density is given by E=\frac{1}{2}\varepsilon _0E^2=\frac{1}{2}\times 8.85\times 10^{-12}\times (1.825\times 10^6)^2=14.73J/m^3

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Answer:

A

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