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solmaris [256]
3 years ago
7

A parallel-plate capacitor is held at a potential difference of 250 V. A proton is fired toward a small hole in the negative pla

te with a speed of 3.0 x 105m/s. What is its speed when it emerges through the hole in the positive plate? (Hint: The electric potential outside of a parallel-plate capacitor is zero).Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.plate capacitor is held at a potential difference of 250 V. A proton is fired toward a small hole in the negative plate with a m/s. What is its speed when it emerges through the hole in the positive plate? (Hint: The electric potential outside plate capacitor is
Physics
1 answer:
MissTica3 years ago
5 0

Answer:

The speed of proton when it emerges through the hole in the positive plate is 2.05\times 10^5\ m/s.

Explanation:

Given that,

A parallel-plate capacitor is held at a potential difference of 250 V.

A A proton is fired toward a small hole in the negative plate with a speed of, u=3\times 10^5\ m/s

We need to find the speed when it emerges through the hole in the positive plate. It can be calculated using the conservation of energy as :

qV=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2\\\\1.6\times10^{-19}\times250=\dfrac{1}{2}mv^2-\frac{1}{2}\cdot1.67\times10^{-27}\cdot(3\times10^{5})^{2}\\\\\dfrac{1}{2}mv^2=3.515\cdot10^{-17}\\\\v=\sqrt{\dfrac{3.515\cdot10^{-17}\cdot2}{1.67\times10^{-27}}}\\\\v=2.05\times 10^5\ m/s

So, the speed of proton when it emerges through the hole in the positive plate is 2.05\times 10^5\ m/s.

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Explanation:

Assuming the wall is frictionless, there are four forces acting on the ladder.

Weight pulling down at the center of the ladder (mg).

Reaction force pushing to the left at the wall (Rw).

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∑τ = Iα

Rw (3.0 sin 60°) − mg (1.5 cos 60°) = 0

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Ff − Rw = 0

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3 years ago
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2.08 s

Explanation:

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Answer:

C. Heat and Pressure

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The arrow which is labeled A points from igneous rock to metamorphic rock.

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