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DerKrebs [107]
3 years ago
15

Points A, B, and C lie along a line from left to right, respectively. Point B is at a lower electric potential than point A. Poi

nt C is at a lower electric potential than point B. Which one of the following statements best describes the subsequent motion, if any, of a positively-charged particle released from rest at point B?
A) The particle will accelerate in the direction of point C.
B) The particle will move at constant velocity in the direction of point C.
C) The particle will remain at rest.
D) The particle will move at constant velocity in the direction of point A.
E) The particle will accelerate in the direction of point A.
Physics
1 answer:
djyliett [7]3 years ago
6 0

Answer:

A) The particle will accelerate in the direction of point C.

Explanation:

As we know that

potential at points A, B,C and D as V_A, V_B, V_C, V_D  and it is clear from the question that

V_A>V_B>V_C

And we know that flow is always from higher to lower potential (for positive charge due to positive potential energy).

So the charge will accelerate from B toward C.

Hence, the correct option is A.

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Ksivusya [100]

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5 0
2 years ago
An electron is accelrated by a unifor electric field (1000v/m) pointing vertically upward. Use energy methods to get the magnitu
ExtremeBDS [4]

Explanation:

In the given situation two forces are working. These are:

1) Electric force (acting in the downward direction) = qE

2) weight (acting in the downward direction) = mg

Therefore, work done by all the forces = change in kinetic energy

Hence, qE \times S + mg \times S = 0.5 \times mv^{2}

     1.6 \times 10^{-19} \times 1000 + 9.1 \times 10^{-31} \times 9.8 \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2}

It is known that the weight of electron is far less compared to electric force. Therefore, we can neglect the weight  and the above equation will be as follows.

   (1.6 \times 10^{-19} \times 1000) \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2
}

         v = sqrt{\frac{1.6 \times 10^{-19}}{(0.5 \times 9.1 \times 10^{-31})}

           = 592999 m/s

Since, the electron is travelling downwards it means that it looses the potential energy.

8 0
3 years ago
I need it now aaaasssssaaaappp!!!!
Vera_Pavlovna [14]

A) Pitch depends on the frequency of a sound wave. The higher the pitch, the higher the frequency and the lower the pitch, the lower the frequency. Sound waves have different pitches due to varying frequency levels.

B) Sound is created as the pen vibrates. This vibrations also interacts with the air atoms and molecules, causing them to vibrate too, therefore, creating sound waves. The reason the brain grows annoyed at this continual sound is because the brain will focus on this sound only, causing the brain to go into overdrive, creating annoyance.

C) Sound waves are travelling vibrations of particles. Space does not have any atoms or particles for the sound vibrations to interact with, therefore creating zero sound waves to travel.

Hope this helps!! :))

5 0
2 years ago
What is the potential energy of a 3kg ball that is on the ground?
ELEN [110]

This is where we have to admit that gravitational potential energy is
one of those things that depends on the "frame of reference", or
'relative to what?'.

         Potential energy = (mass) x (gravity) x (<em>height</em>).

So you have to specify <em><u>height above what</u></em> .

-- With respect to the ground, the ball has zero potential energy.
(If you let go of it, it will gain zero kinetic energy as it falls to
the ground.)

-- With respect to the floor in your basement, the potential energy is

                 (3) x (9.8) x (3 meters) = 88.2 joules.

(If you let go of it, it will gain 88.2 joules of kinetic energy as it falls
to the floor of your basement.)

-- With respect to the top of that 10-meter hill over there, the potential
energy is
                    (3) x (9.8) x (-10) = -294 joules

(Its potential energy is negative. After you let go of it, you have to give it
294 joules of energy that it doesn't have now, in order to lift it to the top of
the hill <em>where it will have zero</em> potential energy.)


5 0
3 years ago
In a shipping company distribution center, an open cart of mass 50.0 kg is rolling to the left at a speed of 5.00 m/s. Ignore fr
spin [16.1K]

Answer:

a) v_p=9.35m/s

Explanation:

From the question we are told that:

Open cart of mass   M_o=50.0 kg

Speed of cart   V=5.00m/s

Mass of package   M_p=15.0kg

Speed of package at end of chute V_c=3.00m/s

Angle of inclination   \angle =37

Distance of chute from bottom of cart   d_x=4.00m

a)

Generally the equation for work energy theory is mathematically given by

  \frac{1}{2}mu^2+mgh=\frac{1}{2}mv_p^2

Therefore

  \frac{1}{2}u^2+gh=\frac{1}{2}v_p^2

  v_p=\sqrt{2(\frac{1}{2}u^2+gh)}

  v_p=\sqrt{2(\frac{1}{2}v_c^2+gd_x)}

  v_p=\sqrt{2(\frac{1}{2}(3)^2+(9.8)(4))}

  v_p=9.35m/s

4 0
2 years ago
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