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2 years ago

The potential energy at x = 8 m is -2000 V and at x = 2 m is 400 V. What is the magnitude and direction of the electric field?

1 answer:

3 0

The potential energy at x = 8 m is -2000 V and at x = 2 m is 400 V. The magnitude and direction of the electric field will be 400 V/m directed parallel to the +x-axis

Electric field, an electric property associated with each point in space when charge is present in any form. The magnitude and direction of the electric field are expressed by the value of E, called electric field strength or electric field intensity or simply the electric field.

The electric potential energy of any given charge or system of changes is termed as the total work done by an external agent in bringing the charge or the system of charges from infinity to the present configuration without undergoing any acceleration.

The relation between electric field and electric potential can be generally expressed as – “Electric field is the negative space derivative of electric potential.”

Electric field = - d V / dx

-(-2000-400) = $\int\ {E} \, dx$

2400 = E (8-2)

2400 V = E (6)

E = 400 V/m

To learn more about electric potential energy here

brainly.com/question/16890427

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A solid sphere rolling without slipping on a horizontal surface. If the translational speed of the sphere is 2.00 m/s, what is i

koban [17]

Answer:

The total kinetic energy is 2.8m J. (NOTE: m is mass of the sphere)

Explanation:

The total kinetic energy of a sphere is given by the sum of the rotational kinetic energy and the translational kinetic energy. That is,

$K_{Total} = K_{R} + K_{T}$

The rotational kinetic energy $K_{R}$ is given by

$K_{R} = \frac{1}{2}I\omega^{2}$

Where $I$ is the moment of inertia

and $\omega$ is the angular velocity

The translational kinetic energy $K_{T}$ is given by

$K_{T} = \frac{1}{2}mv^{2}$

Where $m$ is the mass

and $v$ is the translational speed (velocity)

∴ $K_{Total} = \frac{1}{2}I\omega^{2} + \frac{1}{2}mv^{2}$

But, the moment of inertia $I$ of a sphere is given by

$I = \frac{2}{5}mr^{2}$

Where $m$ is mass

and $r$ is radius

∴ $K_{Total} = \frac{1}{2}\times \frac{2}{5}mr^{2} \omega^{2} + \frac{1}{2}mv^{2}$

$K_{Total} = \frac{1}{5}mr^{2} \omega^{2} + \frac{1}{2}mv^{2}$

Also, $\omega = \frac{v}{r}$

∴ $\omega^{2} = \frac{v^{2} }{r^{2} }$

Then,

$K_{Total} = \frac{1}{5}mr^{2} \times \frac{v^{2} }{r^{2} } + \frac{1}{2}mv^{2}$

$K_{Total} = \frac{1}{5}mv^{2} + \frac{1}{2}mv^{2}$

∴ $K_{Total} = \frac{7}{10}mv^{2}$

From the question, $v = 2.00 m/s$

Then,

$K_{Total} = \frac{7}{10}m(2.00)^{2}$

$K_{Total} = \frac{7}{10}m\times 4.00$

$K_{Total} = 2.8m J$

Hence, the total kinetic energy is 2.8m J. (NOTE: m is mass of the sphere)

8 0

3 years ago

How much force would you need to accelerate a 528 kg motorcycle at 3 m/s2

mart [117]

Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 528 × 3

We have the final answer as

Hope this helps you

4 0

3 years ago

A car with a speed of 50ms is stopped when its brakes are pressed. The displacement passed before stopping is 150 m. How much de

monitta

I dont think it can be solved as time is not mentioned

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3 years ago

Find the value of 02?

TiliK225 [7]

My educated guess : 21.2 deg

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3 years ago

Two charged particles are a distance of 1.62 m from each other. One of the particles has a charge of 7.10 nc, and the other has

k0ka [10]

Answer:

A. F=107.6nN

B. Repulsive

Explanation:

According to coulombs law, the force between two charges is express as

F=(Kq1q2) /r^2

If the charges are of similar charge the force will be repulsive and if they are dislike charges, force will be attractive.

Note the constant K has a value 9*10^9

Hence for a charge q1=7.10nC=7.10*10^-9, q2=4.42*10^-9 and the distance r=1.62m

If we substitute values we have

F=[(9×10^9) ×(7.10×10^-9) ×(4.42×10^-9)] /(1.62^2)

F=(282.4×10^-9)/2.6244

F=107.6×10^-9N

F=107.6nN

B. Since the charges are both positive, the force is repulsive

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3 years ago