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dexar [7]
3 years ago
14

If you place a free electron and a free proton in the same electric field, how will the forces acting on them compare? 82. How w

ill the accelerations of the proton and the electron in the preceding problem compare? 83. How will the directions of travel compare for the electron and the proton in the preceding problem?
Physics
1 answer:
Paha777 [63]3 years ago
3 0

Answer:

the force experienced by both particles will be of same magnitude as the magnitude of charge is same for both particles but with different sign, the acceleration of both particles will be same in magnitude but opposite in direction, the direction of free electron will be opposite to the direction of electric field whereas the direction of proton will be in the direction of electric field,

Explanation:

the force experienced by both particles can be calculated by the formula

F=|q|E

the acceleration can be calculated by Newton's 2nd law i.e F=ma

the direction of electric field is always towards the positive charge so proton will be directed towards electric field and free electron will be directed away from the electric field.

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Express the kinetic energy K in terms of the potential energy U.<br><br><br> K=GMm/2R
max2010maxim [7]

Answer:

K = -½U

Explanation:

From Newton's law of gravitation, the formula for gravitational potential energy is;

U = -GMm/R

Where,

G is gravitational constant

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R is the distance between the two objects

Now, in the question, we are given that kinetic energy is;

K = GMm/2R

Re-rranging, we have;

K = ½(GMm/R)

Comparing the equation of kinetic energy to that of potential energy, we can derive that gravitational kinetic energy can be expressed in terms of potential energy as;

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5 0
2 years ago
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mr_godi [17]

Answer and Explanation:

The computation of the shortest wavelength in the series is shown below:-

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Where

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Now Substitute is

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now we will put the values into the above formula

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Now we will rewrite the answer in the term of \lambda

\lambda = \frac{1}{1218888.889} m\\\\ = 0.82\times 10^{-6} m

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