Which particle is commonly used to initiate a fission chain reaction

two point charges of magnitude 4.0 μc and -4.0 μc are situated along the x-axis at x1 = 2.0 m and x2 = -2.0 m, respectively. wha

user100 [1]

The electric potential at the origin of the xy coordinate system is negative infinity

<h3>What is the electric field due to the 4.0 μC charge?</h3>

The electric field due to the 4.0 μC charge is E = kq/r² where

k = electric constant = 9.0 × 10 Nm²/C²,
q = 4.0 μC = 4.0 × 10 C and
r = distance of charge from origin = x₁ - 0 = 2.0 m - 0 m = 2.0 m

<h3>What is the electric field due to the -4.0 μC charge?</h3>

The electric field due to the -4.0 μC charge is E = kq'/r² where

k = electric constant = 9.0 × 10 Nm²/C²,
q' = -4.0 μC = -4.0 × 10 C and
r = distance of charge from origin = 0 - x₂ = 0 - (-2.0 m) = 0 m + 2.0 m = 2.0 m

Since both electric fields are equal in magnitude and directed along the negative x-axis, the net electric field at the origin is

E" = E + E'

= -2E

= -2kq/r²

<h3>What is the electric potential at the origin?</h3>

So, the electric potential at the origin is V = -∫₂⁰E".dr

= -∫₂⁰-2kq/r².dr

Since E and dr = dx are parallel and r = x, we have

= -∫₂⁰-2kqdxcos0/x²

= 2kq∫₂⁰dx/x²

= 2kq[-1/x]₂⁰

= -2kq[1/x]₂⁰

= -2kq[1/0 - 1/2]

= -2kq[∞ - 1/2]

= -2kq[∞]

= -∞

So, the electric potential at the origin of the xy coordinate system is negative infinity

Learn more about electric potential here:

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3 0

2 years ago

A 5.22×104 kg railroad car moves on frictionless horizontal rails until it hits a horizontal spring stopper with a force constan

In-s [12.5K]

To solve this problem we will apply the principles of conservation of energy, for which we have to preserve the initial kinetic energy as elastic potential energy at the end of the movement. If said equality is maintained then we can affirm that,

$\text{Initial Energy}=\text{Final Energy}$

$\frac{1}{2} mv^2=\frac{1}{2} kx^2$

Here,

m = mass

k = Spring constant

x = Displacement

v = Velocity

Rearranging to find the velocity,

$mv^2 = kx^2$

$v^2 = \frac{kx^2}{m}$

$v = \sqrt{\frac{kx^2}{m}}$

Our values are,

$m = 5.22*10^4kg$

$k = 4.58*10^5N/m$

$x = 32cm = 0.32m$

Replacing our values we have,

$v = \sqrt{\frac{(4.58*10^5)(5.22*10^4)}{0.32}}$

$v = 2.733*10^5m/s$

Therefore the velocity is $2.733*10^5m/s$

8 0

3 years ago

Object height 10cm is placed in front of plane mirror.The height of the image will be_______?

coldgirl [10]

Object height 10cm is placed in front of plane mirror. The height of the image will also be 10cm as the height of image is same as height of object in the case of plane mirror

7 0

3 years ago

A box has a mass of 5.8kg. The box is lifted from the garage floor and placed on a shelf 2.5m off the ground. How much gravitati

Damm [24]

Gravitational potential energy =

(mass) x (gravity) x (height)

= (5.8 kg) x (9.8 m/s²) x (2.5 m)

= 142.1 Joules (C)

5 0

3 years ago

A concave mirror always forms a virtual image of a real object. O True O False

kifflom [539]

Answer:

The given statement is false.

Explanation:

The spherical mirrors are the mirror that are a part of a sphere. Concave and convex mirrors are two types of spherical mirrors.

A concave mirror always forms real and inverted image. A convex mirror forms real and virtual images.

For concave mirror, the value of magnification is less that 1. Also, the focal length is negative for concave mirrors.

So, the given statement is false as a concave mirror always forms a real and inverted image. Hence, this is the required solution.

4 0

3 years ago