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3 years ago

1) a cathode-ray tube is a type of ______ tube.

1 answer:

galina1969 [7]3 years ago

4 0

1) Vacuum tube
2) silicon
3) integrated, I.C.s
4) this one is worded weird. Transistors are the main part of voltage regulators but they use (require) diodes and capacitors, too. And they are all put onto a microchip in the power distribution section. sorry for the long explanation.
5) away from
6) directly, inversely

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3 years ago

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3 years ago

A moving particle encounters an external electric field that decreases its kinetic energy from 9970 eV to 6340 eV as the particl

marusya05 [52]

Answer:

q = -1.61x10⁻¹⁷ C

Explanation:

The charge of the particle can be found using the definition of the work done by electric force:

$W = -q\Delta V$ (1)

Where:

q: is the charge

ΔV: is the difference in electric potential

The work is also equal to:

$W = E_{p_{A}} - E_{p_{B}}$ (2)

Where:

$E_{p_{A}}$ and $E_{p_{B}}$ are the electric potential energy of the points A and B, respectively.

Now, by conservation of energy we have:

$K_{A} + E_{p_{A}} = K_{B} + E_{p_{B}}$ (3)

Where:

$K_{A}$ and $K_{B}$ are the kinetic energy of the points A and B, respectively.

Rearranging equation (3):

$K_{B} - K_{A} = E_{p_{A}} - E_{p_{B}}$

$K_{B} - K_{A} = W$

$K_{B} - K_{A} = -q\Delta V$

Solving the above equation for q:

$q = -\frac{K_{B} - K_{A}}{V_{B} - V_{A}} = -\frac{6340 eV - 9970 eV}{19.0 V - 55.0 V} = -100.83 e \cdot \frac{1.6 \cdot 10^{-19} C}{1 e} = -1.61 \cdot 10^{-17} C$

Therefore, the charge of the particle is -1.61x10⁻¹⁷ C.

I hope it helps you!

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