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IRISSAK [1]
3 years ago
11

an electron is released from rest in a region of space with a nonzero electric field.1. As the electron moves, does the electric

potential energy of the system increase, decrease, or stay the same?a. increaseb. decreasec. stay the same2. Choose the best explanation from among the following:a. As the electron begins to move, its kinetic energy increases. The increase in kinetic energy is equal to the decrease in the electric potential energy of the system.b. Because the electron has a negative charge its electric potential energy doesn't decrease, as one might expect, but increases instead.c. The electron will move perpendicular to the electric field, and hence its electric potential energy will remain the same.
Physics
1 answer:
zvonat [6]3 years ago
5 0

Answer:

1. a. increase

2. Because the electron has a negative charge its electric potential energy does not decrease as one might expect, but increases instead.

Explanation:

Lets first consider the relation between the electric field and electric potential.

E = -ΔV/Δs

As this equation indicates that the electric field is due to the change in potential and change in the the position of charge. Electric field is directed towards the decreasing potential and the electron moves in the opposite direction of the electric field  where potential increases. Thats why the best explanation is that the electron has a negative charge it moves towards the positive region where the electric potential energy increases.

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A 6.41 $\mu C$ particle moves through a region of space where an electric field of magnitude 1270 N/C points in the positive $x$
Marat540 [252]

Answer:

The particle's velocity is 212.15 m/s.

Explanation:

Given that,

Charge of particle, q=6.41\ \mu C=6.41\times 10^{-6}\ C

The magnitude of electric field, E = 1270 N/C

The magnitude of magnetic field, B = 1.28 T

Net force, F=6.4\times 10^{-3}\ N

We need to find the magnitude of the particle's velocity. the net force acting on the particle is given by Lorentz force as :

F=qE+qvB\\\\v=\dfrac{F-qE}{qB}\\\\v=\dfrac{6.4\times 10^{-3}-6.41\times 10^{-6}\times 1270}{6.41\times 10^{-6}\times 1.28}\\\\v=-212.15\ m/s

So, the particle's velocity is 212.15 m/s.

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3 years ago
Determine if each phrase describes speed or velocity Sort each tile into the appropriate column
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Answer:

Speed

• distance per time

• a scalar quantity

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• displacement per time

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What evidence do we have that Earth has a magnetic field surrounding the planet?
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A spring is 6.0cm long when it is not stretched, and 10cm long when a 7.0N force is applied. What force is needed to make it 20c
Artist 52 [7]

Answer:

Approximately 25\; {\rm N} (assuming that this spring is ideal.)

Explanation:

The displacement of a spring is the new length of the spring relative to the original length.

For example:

  • When the 6.0\; {\rm cm}-spring in this question is stretched to 10\; {\rm cm}, the displacement is x = (10\; {\rm cm} - 6.0\; {\rm cm}).
  • Likewise, if this spring is stretched to 20\; {\rm cm}, the displacement would be (20\; {\rm cm} - 6\; {\rm cm}).

If this spring is ideal, the force on the spring would be proportional to the displacement of the spring. In other words, if a force of F_{\text{a}} displaces this spring by x_{\text{a}}, while a force of F_{\text{b}} displaces this spring by x_{\text{b}}, then:

\displaystyle \frac{F_{\text{a}}}{x_{\text{a}}} = \frac{F_{\text{b}}}{x_{\text{b}}}.

In this question, it is given that a force of F_{\text{a}} = 7.0 \; {\rm N} would stretch this spring by x_{\text{a}} = (10\; {\rm cm} - 6.0\; {\rm cm}). Thus, the force F_{\text{b}} required to stretch this spring by x_{\text{a}} = (20\; {\rm cm} - 6.0\; {\rm cm}) would satisfy:

\displaystyle \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}}= \frac{F_{\text{b}}}{20\; {\rm cm} - 6.0\; {\rm cm}}.

Rearrange and solve for F_{\text{b}}:

\begin{aligned} F_{\text{b}} &= \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}} \, (20\; {\rm cm} - 6.0\; {\rm cm}) \\ &\approx 25\; {\rm N}\end{aligned}.

7 0
2 years ago
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