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3 years ago

Can someone please help me out with this ?

Physics

1 answer:

storchak [24]3 years ago

7 0

Answer:

The atomic number is the number over the symbol of the element in the periodic table and the atomic mass the number under it. An isotope is an atom with more or less neutrons than its regular form. a neon atom with mass number of 22 will have 12 neutrons in the nucleus and 10 protons, draw it with its 10 electrons. Two electrons go in the first level of energy and eight in the second. Draw a nucleus with 10 protons and 12 neutrons and an arc with 2 electrons around it and another arc with 8 electrons over the first arc.

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Vector A⃗ points in the positive y direction and has a magnitude of 12 m. Vector B⃗ has a magnitude of 33 m and points in the ne

gavmur [86]

vector A has magnitude 12 m and direction +y

so we can say

$\vec A = 12 \hat j$

vector B has magnitude 33 m and direction - x

$\vec B = -33 \hat i$

Now the resultant of vector A and B is given as

$\vec A + \vec B = 12 \hat j - 33 \hat i$

now for direction of the two vectors resultant will be given as

$\theta = tan^{-1}\frac{12}{-33}$

$\theta = 160 degree$

so it is inclined at 160 degree counterclockwise from + x axis

magnitude of A and B will be

$R = \sqrt{A^2 + B^2}$

$R = \sqrt{12^2 + 33^2} = 35.11 m$

so magnitude will be 35.11 m

6 0

3 years ago

Consider two insulating balls with evenly distributed equal and opposite charges on their surfaces, held with a certain distance

siniylev [52]

Answer:

interest point:

1) Point on the left side

2) Point within the radius r₁ of the first sphere

3) Point between the two spheres

4) point within the radius r₂ of the second sphere

5) Right side point

Explanation:

In this case, the total electric field is the vector sum of the electric fields of each sphere, to simplify the calculation on the line that joins the two spheres

We will call the sphere on the left 1 and it has a positive charge Q with radius r1, the sphere on the right is called 2 with charge -Q with radius r2. The total field is

E_ {total} = E₁ + E₂

E_{ total} = $k \frac{Q}{x_1^2} + k \frac{Q}{x_2^2}$

the bold indicate vectors, where x₁ and x₂ are the distances from the center of each sphere. If the distance that separates the two spheres is d

x₂ = x₁ -d

E total = $k \frac{Q}{x_1^2} - k \frac{Q}{(x_1 - d)^2}$

Let's analyze the field for various points of interest.

1) Point on the left side

in this case

E_ {total} = $k Q \ ( \frac{1}{x_1^2} - \frac{1}{(x_1 +d)2} )$

E_ {total} = $k \frac{Q}{x_1^2}$ $( 1 - \frac{1}{(1 + \frac{d}{x_1} )^2 } )$

We have several interesting possibilities:

* We can see that as the point is further away the field is more similar to the field created by two point charges

* there is a point where the field is zero

E_ {total} = 0

x₁² = (x₁ + d)²

2) Point within the radius r₁ of the first sphere.

In this case, according to Gauus' law, the charge is on the surface of the sphere at the point, there is no charge inside so this sphere has no electric field on its inner point

E_ {total} = $-k \frac{Q}{x_2^2} = -k \frac{Q}{((d-x_1)^2}$