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igor_vitrenko [27]
3 years ago
13

An electron is released form rest in a region of space where a uniform electric field is present. Joanna claims that its kinetic

and potential energies both increase as it moves from its initial position to its final position. Sonya claims that they both decrease. Which one, if either, is correct?
Joanna, because the electron moves opposite to the direction of the field
Sonya, because the electron moves opposite to the direction of the field.
Joanna, because the electron moves in the direction of the field.
Sonya, because the electron moves in the direction of the field.
Neither, because the kinetic energy increases while the electron moves to a point at a higher potential.
Physics
1 answer:
djyliett [7]3 years ago
4 0

Answer:

Neither.

Explanation:

When an electron is released from rest, in an uniform electric field, it will accelerate moving in a direction opposite to the field (as the field has the direction that it would take a positive test charge, and the electron carries a negative charge).

It will move towards a point  with a higher potential, so its kinetic energy will increase, while its potential energy will decrease:

⇒ ΔK + ΔU = 0 ⇒ ΔK = -ΔU = - (-e*ΔV)

As ΔV>0, we conclude that the electric potential energy decreases while the kinetic energy increases in the same proportion, in order to energy be conserved, in absence of non-conservative forces.

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Newton's Second Law is applied because of the acceleration caused by natural forces as the egg is plummeting to the earth. And the amount of acceleration the egg has will be largely affected by the amount of force Mr. Baker uses to hurl the egg to the ground.

Explanation:

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A crow is flying horizontally with a constant speed of 2.70m/s when it releases a claim from its beak. The clan lands on the roc
jenyasd209 [6]

Given:

Speed = 2.70 m

Time, t = 2.10 seconds

Let's solve for the following:

• (a) The horizontal component of the velocity.

To find the horizontal component, apply the formula:

V_{ox}=V_o\cos \theta

Where:

Vo is the initial speed = 2.70 m

θ = 0 degrees

Hence, we have:

\begin{gathered} V_{ox}=2.70\cos 0 \\  \\ V_{ox}=2.7\text{ m/s} \end{gathered}

The horizontal component of the velocity just before it lands is 2.70 m/s.

• (b) The vertical component of the velocity.

To find the vertical component, apply the formula:

V_{oy}=V_{0y}-gt=\text{V}_{oy}\text{ sin}\Theta-gt

Where:

g is the acceleration due to gravity = 9.8 m/s²

t is the time = 2.10 s

Hence, we have:

\begin{gathered} V_{oy}=V_{oy}\sin \theta-gt \\  \\ V_{oy}=2.70\sin 0-9.8(2.10) \\  \\ V_{oy}=0-20.58 \\  \\ V_{oy}=-20.58\text{ m/s} \end{gathered}

The vertical component of the velocity just before it lands is -20.58 m/s.

(c) Here, the initial speed is equal to the constant horizontal speed.

Therefore, in part (a) the horizontal component will increase in the x-direction if the speed of the crow is increased.

The initial vertical velocity is 0 m/s in both cases.

Therefore, in part (b) the vertical component will remain constant.

ANSWER:

(a) 2.70 m/s

(b) -20.58

(c) In part (a) the horizontal component will increase, while in part (b) the vertical component will remain constant.

6 0
1 year ago
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