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2 years ago

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Two very large parallel metal plates, separated by 0.20 m, are connected across a 12-V source of potential. An electron is relea

sed from rest at a location 0.10 m from the negative plate. When the electron arrives at a distance 0.050 m from the positive plate, how much kinetic energy (J) has the electron gained

1 answer:

Semmy [17]2 years ago

8 0

Answer:

${\rm K} = 2.4\times 10^{-19}~J$

Explanation:

The electric field inside a parallel plate capacitor is

$E = \frac{Q}{2\epsilon_0 A}$

where A is the area of one of the plates, and Q is the charge on the capacitor.

The electric force on the electron is

$F = qE = \frac{qQ}{2\epsilon_0 A}$

where q is the charge of the electron.

By definition the capacitance of the capacitor is given by

$C = \epsilon_0\frac{A}{d} = \frac{Q}{V}\\\frac{Q}{\epsilon_0 A} = \frac{V}{d} = \frac{12}{0.20} = 60$

Plugging this identity into the force equation above gives

$F = \frac{qQ}{2\epsilon_0 A} = \frac{q}{2}(\frac{Q}{\epsilon_0 A}) = \frac{q}{2}60 = 30q$

The work done by this force is equal to change in kinetic energy.

W = Fx = (30q)(0.05) = 1.5q = K

The charge of the electron is $1.6 \times 10^{-19}$

Therefore, the kinetic energy is $2.4\times 10^{-19}$

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Darya [45]

Answer:

$y = -8.37 cm$

Explanation:

As we know that the equation of SHM is given as

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1 year ago

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antiseptic1488 [7]

Answer:

Rod 1 has greater initial angular acceleration; The initial angular acceleration for rod 1 is greater than for rod 2.

Explanation:

For the rod 1 the angular acceleration is

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Similarly, for rod 2

$\alpha_2 = \dfrac{\tau_2}{I_2}.$

Now, the moment of inertia for rod 1 is

$I_1 = \dfrac{1}{3}ML^2$ ,

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$\boxed{\alpha_1 = \dfrac{3g}{2L} }$

Now, for rod 2 the moment of inertia is

$I_2 = \dfrac{1}{3}(2M)(2L)^2$

$I_2 = \dfrac{8}{3} ML^2,$

and the torque acting is (about the center of mass)

$\tau _2 = (2M)g \dfrac{(2L)}{2}$

$\tau _2 = 2MgL;$

therefore, the angular acceleration $\alpha_2$ is

$\alpha_2 = \dfrac{2MgL;}{\dfrac{8}{3} ML^2,}.$

$\boxed{\alpha_2 = \dfrac{3g}{4L}}$

We see here that

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therefore

$\boxed{\alpha_1 > \alpha_2.}$

In other words , the initial angular acceleration for rod 1 is greater than for rod 2.

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If you'd like to know more about the different boundaries, read on:

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Lastly, divergent boundaries occur when two plates move apart. The separation creates a way for magma to come up. New crust is formed when the magma that seeps out is cooled by its cooler surroundings. This is observed in the mid oceanic ridge.

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