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Rudiy27
2 years ago
13

An electron moves 4.5 m in the direction of an electric field of strength 325 N/C. What is the change in electrical potential en

ergy?
Physics
1 answer:
Vinvika [58]2 years ago
6 0
Calculate q* E * d 
<span>Put q = 1.6 x 10^-19 </span>
<span>E = 325 </span>
<span>d = 4.5 


I hope this helps!</span>
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Using the right-hand rule from your lessons, determine the directions of the electrical current and magnetic field of the electr
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Answer:

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Explanation:

This is an accurate representation of something you are working on!

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2 years ago
Steep safety ramps are built beside mountain highways to enable vehicles with defective brakes to stop safely. A truck enters a
Furkat [3]

Answer:

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Explanation:

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2 years ago
An 1876 N crate is being pushed across a level force at a constant speed by a force of 747 N. What is the coefficient of kinetic
nekit [7.7K]

The crate only moves horizontally, so its net vertical force is 0. The only forces acting in the vertical direction are the crate's weight (pointing downward) and the normal force of the surface on the crate (pointing upward). By Newton's second law, we have

∑ <em>F</em> (vertical) = <em>n</em> - <em>mg</em> = 0   →   <em>n</em> = <em>mg</em> = 1876 N

where <em>n</em> is the magnitude of the normal force.

In the horizontal direction, the crate is moving at a constant speed and thus with no acceleration, so it's completely in equilibrium and the net horizontal force is also 0. The only forces acting on it in this direction are the 747 N push (pointing in the direction of the crate's motion) and the kinetic friction opposing it (pointing in the opposite direction). By Newton's second law,

∑ <em>F</em> (horizontal) = 747 N - <em>f</em> = 0   →   <em>f</em> = 747 N

The frictional force is proportional to the normal force by a factor of the coefficient of kinetic friction, <em>µ</em>, such that

<em>f</em> = <em>µn</em>   →   <em>µ</em> = <em>f</em> / <em>n</em> = (747 N) / (1876 N) ≈ 0.398188 ≈ 0.40

8 0
2 years ago
A long coaxial cable (Fig. 2.26) carries a uniform volume charge density rho on the inner cylinder (radius a), and a uniform sur
Yuki888 [10]

Answer:

a) E = ρ / e0

b) E = ρ*a / (e0 * r)

c) E = 0

Explanation:

Because of the geometry, the electric field lines will all have a radial direction.

Using Gauss law

Q/e0 = \int \int E * dA

Using a Gaussian surface that is cylinder concentric to the cable, the side walls will have a flux of zero, because the electric field lines will be perpendicular. The round wall of the cylinder will have the electric field lines normal to it.

We can make this cylinder of different radii to evaluate the electric field at different points.

Then:

A = 2*π*r (area of cylinder per unit of length)

Q/e0 = 2*π*r*E

E = Q / (2*π*e0*r)

Where Q is the charge contained inside the cylinder.

Inside the cable core:

There is a uniform charge density ρ

Q(r) = ρ * 2*π*r

Then

E = ρ * 2*π*r / (2*π*e0*r)

E = ρ / e0 (electric field is constant inside the charged cylinder.

Between ther inner cilinder and the tube:

Q = ρ * 2*π*a

E = ρ * 2*π*a / (2*π*e0*r)

E = ρ*a / (e0 * r)

Outside the tube, the charges of the core cancel each other.

E=0

4 0
2 years ago
Suppose that now you want to make a scale model of the solar system using the same ball bearing to represent the sun. How far fr
Dahasolnce [82]

Answer:

d = 0.645 m <em>(assuming a radius of the ball bearing of 3 mm)</em>

Explanation:

<u>The given information is:</u>

  • <em>The distance from the center of the sun to the center of the earth is 1.496x10¹¹m = d_{e}</em>
  • <em>The radius of the sun is 6.96x10⁸m = r_{s}</em>

<u>We need to assume a radius for the ball bearing, so suppose that the radius is 3 mm = r_{b}</u>.  

First, we need to find how many times the radius of the sun is bigger respect to the radius of the ball bearing, which is given by the following equation:

\frac{r_{s}}{r_{b}} = \frac{6.96\cdot 10^{8}m}{3\cdot 10^{-3}m} = 2.32\cdot 10^{11}

Now, we can calculate the distance from the center of the sun to the center of the sphere representing the earth, d_{s}:  

[tex] d_{s} = \frac{d_{e}}{r_{s}/r_{b}} = \frac{1.496 \cdot 10^{11} m}{2.32\cdot 10^{11}} = 0.645 m

I hope it helps you!

7 0
2 years ago
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