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djverab [1.8K]
3 years ago
9

A positive charge is placed at rest at the center of a region of space in which there is a uniform electric field. (A uniform fi

eld is one whose strength and direction are the same at all points within the region.) What happens to the electric potential energy of the positive charge, after the charge is released from rest in the uniform electric field?A. It will remain constant because the electric field is uniform.B.It will decrease because the charge will move in the opposite direction of the electric field.C.It will remain constant because the charge remains at rest.D.It will decrease because the charge will move in the direction of the electric field.E.It will increase because the charge will move in the direction of the electric field.
Physics
1 answer:
Harlamova29_29 [7]3 years ago
7 0

Answer:

.D.It will decrease because the charge will move in the direction of the electric field

Explanation:

<em>which there is a uniform electric field. (A uniform field is one whose strength and direction are the same at all points within the region.) What happens to the electric potential energy of the positive charge, after the charge is released from rest in the uniform electric field?A. It will remain constant because the electric field is uniform.B.It will decrease because the charge will move in the opposite direction of the electric field.C.It will remain constant because the charge remains at rest.D.It will decrease because the charge will move in the direction of the electric field.E.It will increase because the charge will move in the direction of the electric field.</em>

solution

The potential energy decreases, converted to kinetic energy

The charge will feel a force in the direction of the electric field (F=Eq) and thus it will accelerate with

a constant acceleration. (Just like releasing an object above the earth's surface - constant acceleration,

at least until it hits something.)

tus te answer will be

.D.It will decrease because the charge will move in the direction of the electric field

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Explanation:

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2 years ago
N which order did the events forming our solar system occur?
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Explanation:

hope this helps

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What is the equivalent resistance between the points A and B of the network?​
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Explanation:

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3 0
2 years ago
A horizontal spring with spring constant 130 N/m is compressed 17 cm and used to launch a 2.8 kg box across a frictionless, hori
olasank [31]

Explanation:

The given data is as follows.

        k = 130 N/m,       \Delta x = 17 cm = 0.17 m   (as 1 m = 100 cm)

     mass (m) = 2.8 kg

When the spring is compressed then energy stored in it is as follows.

             Energy = \frac{1}{2}kx^{2}

Now, spring energy gets converted into kinetic energy when the box is launched.

So,    \frac{1}{2}kx^{2} = \frac{1}{2}mv^{2}

   \frac{1}{2} \times 130 \times (0.17)^{2} = \frac{1}{2} \times 2.8 \times v^{2}

          v^{2} = \frac{3.757}{2.8}

                     = 1.34

                v = 1.15 m/sec

Now,

           Frictional force = \mu \times mg

                                    = 0.15 \times 2.8 \times 9.8

                                    = 4.116 N

Also,  Kinetic energy = work done by friction

           \frac{1}{2}mv^{2} = F_{f} \times d

           \frac{1}{2} \times 2.8 \times (1.15)^{2} = 4.116 \times d

             1.8515 = 4.116 \times d

                 d = 0.449 m

Thus, we can conclude that the box slides 0.449 m across the rough surface before stopping.

8 0
3 years ago
A horizontal 810-N merry-go-round of radius 1.60 m is started from rest by a constant horizontal force of 55 N applied tangentia
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Answer:

576 joules

Explanation:

From the question we are given the following:

weight = 810 N

radius (r) = 1.6 m

horizontal force (F) = 55 N

time (t) = 4 s

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where MI is the moment of inertia and ω is the angular velocity

MI = 0.5 x m x r^2

mass = weight ÷ g = 810 ÷ 9.8 = 82.65 kg

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therefore

angular velocity (ω) = a x t = 0.83 x 4 = 3.33 rad/s

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K.E = 0.5 x 105.8 x 3.33^{2} = 576 joules

6 0
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